---
title: "Au B Sc Ii Chemistry Iok"
book: "test"
category: "MA"
publisher: "Ratan Prakashan Mandir Pvt. Ltd."
type: "Educational Material"
---
According to Latest Syllabus
For Dr. Bhimrao Ambedkar University (A.U.) Examination
# VIKAS
## GUIDE BOOK
B.Sc. II
### CHEMISTRY-I
Km. Pankaj
Published by
Ratan Prakashan Mandir Pvt. Ltd.
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B. Sc. II Chemistry I (Inorganic Chemistry)
Syllabus
Unit I : Chemistry of Elements of First Transition Series.
Characteristic properties of d-block elements.
Properties of the elements of the first transition series, their binary compounds and complexes illustrating relative stability of their oxidation states, coordination number and geometry.
Unit II : Chemistry of Elements of Second and Third Transition Series.
General Characteristics, comparative treatment with their 3d analogous in respect of ionic radii, oxidation states, magnetic, behaviour spectral properties and stereochemistry.
Oxidation and Reduction.
Use of redox potential data-analysis of redox cycle, redox stability in water-Frost, Latimer and Pourbaix diagrams, Principles involved in the extraction of the elements.
Unit III : Coordination Compounds.
Werner’s coordination theory and its experimental verification, effective atomic number concept, chelates, nomenclature of coordination compounds, isomerism in coordination compounds, valence bond theory of transition metal complexes.
Unit IV : Chemistry of Lanthanide Elements.
Electronic structure, oxidation states and ionic radii and lanthanide contraction, complex formation, occurrence and isolation, lanthanide compounds.
Chemistry of Actinides.
General features and chemistry of actinides, chemistry of separation of Np, Pu and Am from U, similarities between the later actinides and the later lanthanides.
Unit V : Acids and Bases.
Arrhenius, Bronsted-Lowry, the Lux-Flood, solvent system and Lewis concepts of acids and bases.
Non-aqueous Solvents.
Physical properties of a solvent, types of solvents and their general characteristics, reactions in non-aqueous solvents with reference to liquid NH3 and liquid SO2.
Bond and Band theories of metals.
Question Index
S. No. P. No.
Unit - I
Section ‘A’
Objective Type Questions (1 to 54)
Section ‘B’
11-16
Short Answer Type Questions
1\. Transition elements possess catalytic properties. Explain why.
16
2\. Give four important characteristics of d-block elements.
17
3\. Explain why TiO2 is white, while TiCl3 is coloured ?
17
- 4\. Explain why in Mn +2 while in Fe +3 oxidation state is more stable ?
- 5\. Explain why the atomic radii of d-block elements decrease upto VIII
17
and then increase ?
17
6\. Explain why the compounds of Ce(IV) are more stable ?
Section ‘C’
Long Answer Type Questions
1\. Discuss complex formation tendency of transition metals with
18
suitable examples.
- 2\. Why are transition elements called d-block elements? Explain briefly
with reasons the following properties of transition elements.
- (a) Paramagnetic properties
- (b) Colour
18
(c) Oxidation states.
Unit-II
20
Section ‘A’
Objective Type Questions (1 to 85)
Section ‘B’
Short Answer Type Questions
1\. Whether zinc will displace copper from its salt solution or not ?
22-31
(E°Zn2+/Zn = ‒ 0∙76 volt; E°Cu2+/Cu = + 0∙34 Volt)
31
2\. Describe van Arkel method for the purification of metals.
31
3\. What is slag ?
31
4\. Explain the term roasting and smelting.
31
B. Sc. II, Chemistry I
5
5\. Whether Ag metal will reduce Sn2+ or not ?
(E°Ag+/Ag = + 0∙80 Volt; E°Sn2+/Sn = ‒ 0∙14 Volt)
32
6\. Write a note on froth floatation process.
32
7\. The EMF diagram for Fe is given as :
\+ 2.20 + 0.77 ‒ 0.445
EA° (volts) FeO42---> Fe3+--> Fe2+--> Fe°
\+ 1.08
Determine the value of E° .
FeO42 / Fe2+
32
8\. Explain what is meant by flux.
33
9\. Whether Ag will displace H|2 gas from HCl or not ?
(E°Ag+/Ag = 0∙80 volt; E° H+/H2 = 0∙0 volt)
Section ‘C’
33
Long Answer Type Questions
- 1\. Compare the following properties of elements of second and third transition series with those of first series.
- (a) Oxidation States
- (b) Ionic Radii
- (c) Magnetic Properties
(d) Stereochemistry
33
2\. The reduction potential diagram (Latimer diagram) for Cu in acid solution is :
+0∙15 Volt +0∙50 Volt
Cu2+---------> Cu+---------> Cu
E° = x Volt
Calculate x. Does Cu+ ion disproportionate in solution ?
34
3\. Discuss in brief the various steps involved in the extraction of
metals from their ores.
35
4\. Discuss the stability of various oxidation states of an element in
aqueous solution.
38
Unit III
Section ‘A’
Objective Type Questions (1 to 150) 41-58
Section ‘B’
Short Answer Type Questions
- 1\. What is effective atomic number ? Calculate the effective atomic number of the central metal atom in the following complex compounds :
- (i) \[Cr(en)3\]Cl3
- (ii) K3\[Co(C2O4)2Cl2\]
- (iii) Ni(CO)4
- [(iv) K3\[Cu(CN)4\]58](#bookmark18)
- [2. Explain why \[Co(NH3)6\]3+ is diamagnetic and octahedral ?59](#bookmark19)
- 3\. Indicate ‘primary’ and auxillary valencies of central metal ion in the following complexes.
- [(a) K4\[Fe(CN)6\] (b) \[Co(NH3)6\]Cl359](#bookmark20)
- 4\. Write the IUPAC name of the following complex compounds :
- (a) K2\[PtCl6\]
- (b) Na3\[Ag(S2O3)2\]
- (c) K3\[Co(C2O4)2Cl2\]
- (d) \[Ag(NH3)2\]Cl
- (e) K4\[Fe(CN)6\]
- (f) \[Cr(H2O)4 Cl2\]NO2
- (g) \[Cr(en)3\]Cl3
- (h) \[Co(NH3)5Cl\]Cl2
- (i) \[Cu(NH3)4\]SO4
- (j) Na2\[CrOF4\]
- (k) \[Co(NH3)6\] \[CuCl5\]
- (l) \[Cr(C6H6)2\]
NH2
- (m) \[(NH3)5Co Co(NH3)5\] (NO3)5
- (n) NH4\[Cr(NH3)2(NCS)4\]
- (o) K3\[Co(NO2)6\].
- (p) Fe(CO)5
- (q) LiAlH4 60
- 5\. Write the possible isomers of the following complex compounds and explain in brief the isomerism involved :
- (i) \[Co(NH3)6\] \[Cr(CN)6\]
- (ii) \[Cr(H2O)6\]Cl3
- (iii) \[Pt(NH3)2Cl2\]
- (iv) \[Pt(NH3) (Py)Cl Br\]
- (v) \[Co(NH3)5Br\]SO4 61
- 6\. Name the type of isomerism involved in the following complex compounds, giving reason :
- (i) \[Co(NH3)5Br\]SO4 and \[Co(NH3)5SO4\]Br
OH
- (ii) \[(NH3)4Co Co(NH3)2Cl2\]SO4
OH
OH
and \[Cl(NH3)3Co Co(NH3)3Cl\]SO4
OH
- (iii) \[Pd(dipy)(SCN)2\] and \[Pd(dipy)(NCS)2\]
- (iv) \[Co(NH3)6\]\[Cr(CN)6\] and \[Cr(NH3)6\]\[Co(CN)6\]
NO2
[(v) H3N2](#bookmark21)
[(v)](#bookmark22)
NO2
H3NNH
[andCo /](#bookmark23)
H3N NH3
NO2
H3N NH3
NH3
_files/AU20B.20Sc.20II20Chemistry20I(OK)-1.png)
_files/AU20B.20Sc.20II20Chemistry20I(OK)-2.png)
- (vii) \[Cr(H2O)5Cl\]Cl2.H2O and \[Cr(H2O)4Cl2\]Cl2H2O 61
- 7\. Write a note on chelation and its application. 62
- 8\. Write the formulae of the following complex compounds :
- (a) Potassium hexacyanoferrate (III)
- (b) Ammonium trioxalatocobaltate (III)
- (c) Bis (acetyl) acetanatooxovanadium (IV)
- (d) Potassium dichlorodioxalatocobaltate (III)
- (e) Ammonium diamminetetrathiocyanato-S chromate (III)
- (f) Potassium tetracyanonickelate (o)
- (g) Triamminechlorocyanonitrocobalt (III)
- (h) Tris (ethylenediamine) chromium (III) chloride
- (i) Tetrammineplatinum (II) tetrachloroplatinate (II)
- (j) Dichlorobis (urea) copper (II)
- (k) Potassium tetracyanonickelate (II)
- (l) Sodium bis(thiosulphato) argentate (I)
- (m) Dichlorobis(ethylenediamine) cobalt (III) chloride
- (n) Tetraaquodichlorochromium (III) nitrite
- (o) µ-amido-µ-hydroxo bis \[tetraammine cobalt (III)\] bromide
- [(p) µ-amido bis (pentaammine cobalt (III)) nitrate63](#bookmark24)
- [9. Explain hydrate isomerism with example.64](#bookmark25)
- [10. Explain why \[Ni(CN)4\]2‒ ion is diamagnetic and square planar ?65](#bookmark26)
11\. Explain why Cr3+ ions always form paramagnetic complexes.
65
12\. Explain which one of the following has maximum magnetic moment ? \[Co(H2O)6\]3+, \[CoF6\]3‒ and \[Co(CN)6\]3‒
66
13\. Explain ionization isomerism with example.
66
14\. Explain why K3\[Fe(CN)6\] is paramagnetic ?
66
15\. Ni(CO)4 is diamagnetic and tetrahedral. Explain why ?
67
Section ‘C’
Long Answer Type Questions
1\. Explain the geometry of the following on the basis of valence bond theory :
(i) \[Fe(CN)6\]3‒ (ii) \[Co(NH3)6\]3+
(iii) \[Ni(CN)4\]2‒
68
- 2\. Explain whether the following complexes are low spin or high spin :
- (i) \[Co(NH3)6\]Cl3
- (ii) K3\[Fe(CN)6\]
- (iii) \[CoF6\]3‒
70
3\. Give a detailed account of Werner’s coordination theory.
72
4\. Discuss the various types of isomerism shown by complex compounds with suitable examples.
73
5\. Discuss valence bond theory of metal-ligand bonding with suitable examples. What are the limitation of this theory ?
77
Unit IV
Section ‘A’
Objective Type Questions (1 to 35) 81-85
Section ‘B’
Short Answer Type Questions
1\. Write a notes on lanthanide contraction.
86
2\. Discuss in brief the magnetic and spectroscopic properties of actinides.
86
3\. Discuss the various oxidation states of lanthanides.
86
4\. Discuss complex ion formation in actinides.
87
Section ‘C’
Long Answer Type Questions
1\. Discuss the various methods that have been used for the separation of lanthanides.
87
2\. Discuss the chemistry of the separation of Np, Pu and Am from U.
89
3\.
4\.
5\.
Compare the oxidation states of actinides and lanthanides. 90
What are actinides ? Discuss their electronic configuration. 90
What are lanthanides and why are they also known as ‘rare earths’.
Discuss the following properties of these elements:
- (a) Electronic configuration
- (b) Colour
- (c) Atomic and Ionic radii
- (d) Magnetic and spectral properties. 92
Unit V
Section ‘A’
Objective Type Questions (1 to 103)
Section ‘B’
Short Answer Type Questions
1\.
Give equations for the auto-ionization of liquid ammonia and liquid sulphur dioxide. 105
2\.
Describe what will happen if CH3COOH is dissolved in liquid ammonia. 105
3\.
Liquid ammonia is a better solvent for organic compounds than water. 105
4\.
5\.
Discuss in brief Lux and Flood concept of acids and bases. 106
Classify with reasons the following as Lewis acids and bases : ROH, BF3, NH3, AlCl3, SnCl4, H3O+, Ag+, HCl, H2O, M and L in a complex compound \[ML6\] where M is a metal ion and L is any ligand, CO and Fe in Fe(CO)5. 106
6\.
7\.
Write a note on classification of solvents. 107
Discuss Arrhenius concept of acids and bases. What are the main limitations of this concept. 107
8\.
9\.
Solutions of alkali metals in liquid ammonia are coloured. 108
Discuss Lewis concept of acids and bases. Give merits and demertis of this concept. 108
10\.
Arrange the members in the following groups of acids or bases in order of decreasing acidity or basicity.
- (i) HCl, HClO, HClO2, HClO3
- (ii) NH3, PH3, AsH3
- (iii) HF, HCl, HBr, HI
- (iv) H2O, H3O+, OH‒
- (v) NH3, NH2‒, NH4+
- (vi) HCl, HOBr, HOI 109
- 11\. Write the conjugate base of the following acids :
CH4, C2H5OH, NH4+, H2O, HCl, H3O+, HCO3‒, \[Al(H2O)6\]3+ 109
- 12\. What are non-aqueons solvents ? Mention the advantages og liquid ammonia over water as a solvent. 109
- 13\. Discuss in brief the solvent system concept of acids and bases. 110
Section ‘C’
Long Answer Type Questions
- 1\. Discuss Bronsted-Lowry and Lewis concepts of acids and bases.
Give merits and demerits of this concept. 110
- 2\. Discuss with examples the behaviour of liquid ammonica as nonaqueous solvent with reference to the following :
- (i) Precipitation reactions
- (ii) Ammonolysis
- (iii) Redox reactions
- (iv) Complex formaiton reactions
- [(v) Neutralization reactions111](#bookmark29)
- [3. Discuss with examples the various types of reactions in liquid sulphur dioxide.114](#bookmark30)
[Very Short Answer Type Questions (1-54)](#bookmark31)
□
Section ‘A’
Objective Type Questions
- 1\. Which of the following ions exhibit highest magnetic moment ?
- (a) Cu2+ (b) Ti3+
- (c) Ni2+ (d) Mn2+.
- 2\. Which one of the following pairs of ions are colourless ?
- (a) Ti3+, Cu2+ (b) Sc3+, Zn2+
- (c) Co2+, Fe3+ (d) Ni2+, V3+.
- 4\. Transition metals exhibit enthalpies of atomization which are :
- (a) low (b) zero
- (c) high (d) abnormal.
- 4\. For the first transition series, the magnetic moment, µ is related to the number of unpaired electrons, n by the expression :
A A
- (a) — = nn (n + 1) (b) — = nn (n + 2)
' ' Ab 7 7 Ab
A A
- (c) — = n (n + 1) (d) — = n (n + 2)
Ab 7 7 Ab
- 5\. The calculated magnetic moment (in Bohr Magnetons) of Cu2+ ion is :
- (a) 1∙73 (b) zero
- (c) 2∙6 (d) 3∙4.
- 6\. The number of electron in the outermost shell of the 3d-transition elements generally remains :
- (a) (n‒1) dn (b) ndn
- (c) ns2 (d) (n‒1) s2.
- 7\. Which of these ions has the greatest tendency to form complex ions ?
- (a) Hg2+ (b) Zn2+
- (c) Co2+ (d) Na+
- 8\. The basic character of the transition metal monoxides follows the order :
- (a) TiO > VO > CrO > FeO (b) VO > CrO > TiO > FeO
- (c) CrO > VO > FeO > TiO (d) TiO > FeO > VO > CrO
(Atomic Nos. Ti = 22, V = 23, Cr = 24, Fe = 26)
- 9\. Transition elements are less reactive because of :
- (a) higher ionization energies and higher reduction electrode potential
- (b) low ionization energies and low melting points
- (c) high electronegativities
- (d) higher ionization energies and single oxidation states.
10\.
11\.
12\.
13\.
14\.
15\.
16\.
17\.
18\.
19\.
Of the ions Zn2+, Ni2+ and Cr3+ :
\[At. Nos. Zn = 30, Ni =28, Cr = 24\]
- (a) only Zn2+ is colourless and Ni2+ and Cr3+ are coloured
- (b) all three are coloured
- (c) all three are colourless
- (d) only Ni2+ is coloured and Zn2+ and Cr3+ are colourless.
The aqueous solution containing which one of the following ions will be
colourless ?
- (a) Ti3+ (b) Mn2+
- (c) Sc3+ (d) Fe2+
(Atomic Numbers : Sc = 21, Fe = 26, Ti = 22, Mn = 25)
General electronic configuration of the transition elements is given by:
- (a) ns2nd1‒10 (b) ns2, np6 nd1‒10
- (c) (n‒1)d1‒10 np6 (d) (n‒1)d1‒10 ns0 ‒ 2
.
Transition elements exhibit variable oxidation states because they release electrons from the following orbit :
- (a) ns and np orbits (b) (n‒1)d and ns orbits
- (c) (n‒1)d orbit (d) ns orbit.
Which one of the following ions exhibits colour in aqueous solution ?
- (a) Sc3+ (b) Ni2+
- (c) Ti4+ (d) Zn2+.
The tendency towards formation of coloured ions is the maximum in :
(a) s-block elements
(b) d-block elements
(c) p-block elements
Transition metals :
(d) f-block elements.
(a) exhibit diamagnetism (b) undergo inert pair effect
(c) do not form alloys (d) show variable oxidation states.
Which one of the following ionic species will impart colour to an aqueous solution ?
- (a) Ti4+ (b) Cu+
- (c) Zn2+ (d) Cr3+.
The tendency of 3d-metal ions to form stable complexes is due to their :
- (a) variable oxidation state
- (b) strong electronegative nature
- (c) high charge/size ratio and vacant d orbitals
- (d) very low ionization energies.
The transition elements are more metallic than the representative elements because they have :
- (a) the electron in d orbitals
- (b) electron pairs in d orbitals
- (c) availability of d orbitals for bonding
- (d) unpaired electrons in metallic orbitals.
- 20. Which of the following has the maximum number of unpaired electrons ?
- (a) Mg2+ (b) Ti3+
- (c) V3+ (d) Fe2+.
- 21. Which of the following shows highest magnetic moment ?
- (a) V3+ (b) Cr3+
- (c) Mn2+ (d) Fe2+
- 22. What is the magnetic moment for Mn2+ ion in low spin state ?
- (a) 5.90 BM (b) 3.90 BM
- (c) 2.8 BM (d) 1.73 BM.
- 23. The following is not a coloured ion :
- (a) Cr3+ (1s2 2s2 2p6 3s2 3p6 3d3)
- (b) Cr2+ (1s2 2s2 2p6 3s2 3p6 3d4)
- (c) Cu+ (1s2 2s2 2p6 3s2 3p6 3d10)
- (d) Cu2+ (1s2 2s2 2p6 3s2 3p6 3d9)
- 24. The 3d-metal ions are paramagnetic in nature because :
- (a) they are reducing agents
- (b) they form coloured salts
- (c) they have one or more paired s electrons
- (d) they have one or more unpaired d electrons.
- 25. Characteristic of transition element is that............ is incomplete :
- (a) ‘d’ orbital (b) ‘p’ orbital
- (c) ‘f’ orbital (d) ‘s’ orbital.
- 26. Which one of the following properties is not shown by a transitional element ?
- (a) Paramagnetism (b) Constant oxidation states
- (c) Complex salt formation (d) Formation of alloys
- 27. Which of the following is expected to form colourless complex ?
- (a) Ni2+ (b) Cu+
- (c) Ti3+ (d) Fe3+
- 28. Which one of the transition metal ions is coloured ?
- (a) Cu+ (b) Zn2+
- (c) Sc3+ (d) V4+.
- 29. Which of the following statement is incorrect ? The d-block elements :
- (a) have atomic radii larger than s-and p-block elements
- (b) have high melting points, boiling points and tensile strength
- (c) have variable oxidation states
- (d) exhibit catalytic properties.
30\.
Which one of the following is not a transition metal ?
- (a) Zinc (b) Tungsten
- (c) Copper (d) Vanadium.
31\.
Transition elements show :
(a) variable valency (b) paramagnetism
(c) complex formation (d) all of these.
32\.
Out of the following which is a d-block element :
(a) Sodium (b) Calcium
(c) Copper (d) Argon.
33\.
Of the following outer electronic configurations of atoms, the highest oxidation state is achieved by which one of them ?
(a) (n ‒ 1)d8 ns2 (b) (n ‒ 1)d5 ns1
(c) (n ‒ 1)d3 ns2 (d) (n ‒ 1)d5 ns2
34\.
Which of the following statements is not correct ?
- (a) Compounds of transition metals are generally coloured
- (b) Transition metals form a number of interstitial compounds
- (c) Most of the transition metals exhibit large number of oxidation states
- (d) Compounds of transition metals are generally diamagnetic in behaviour.
35\.
Of the following transition metals, the maximum numbers of oxidations states are exhibited by :
(a) chromium (Z = 24) (b) manganese (Z = 25)
(c) iron (Z = 26) (d) titanium (Z = 22)
36\.
Elements which generally exhibit multiple oxidation states and whose ions are usually coloured are :
- (a) metalloids (b) transitional elements
- (c) non-metals (d) gases.
37\.
Which of the following is not a transition metal ?
(a) Gold (b) Mercury
(c) Scandium (d) Indium.
38\.
Which one of the following characteristics of the transition metals is associated with their catalytic activity ?
- (a) Variable oxidation states (b) High enthalpy of atomization
- (c) Paramagnetic behaviour (d) Colour of hydrate ions.
39\.
Which of the following is not true for transition metals ?
- (a) They are malleable and ductile
- (b) They have high boiling and melting points
- (c) They crystallize with body-centered cubic and hexagonal close-packed structures only
- (d) They show variable oxidation states, although not always.
40\.
The d-block elements consist mostly of :
- (a) monovalent metals
- (b) all non-metals
- (c) elements which generally form stoichiometric metal oxide
- (d) many metals with catalytic properties.
41\.
Which one of the given transition metal ions is diamagnetic ?
(a) Co2+ (b) Ni2+
(c) Cu2+ (d) Zn2+.
42\.
The 3d-elements show variable oxidation states. What is the maximum oxdiation state shown by the element Mn ?
- (a) + 4 (b) + 5
- (c) + 6 (d) + 7.
43\.
Which of the following transition metal cations has the minimum number of unpaired d electrons ?
(At. Nos. of Mn, Fe, Co = 25, 26, 27 respectively)
- (a) Mn2+ (b) Co2+
- (c) Co3+ (d) Fe3+.
44\.
Maximum number of oxidation states of transition metal is derived from the following configuration :
- (a) ns electron (b) (n‒1)d electron
- (c) (n+1)d electron (d) ns + (n‒1)d electron.
45\.
The colourless species is :
(a) VCl3 (b) VOSO4
(c) Na3VO4 (d) \[V(H2O)6\]SO4∙H2O
46\.
In which of the following pairs are both the ions coloured in aqueous solution ?
- (a) Sc3+, Ti3+ (b) Sc3+, Co2+
- (c) Ni2+, Cu+ (d) Ni2+, Ti3+
(At. Nos. : Sc = 21, Ti = 22, Ni = 28, Cu = 29, Co = 27)
47\.
The ‘spin only’ magnetic moment in units of Bohr magnetons (µB) of
Ni2+ in aqueous solution would be (At. No. Ni = 28) :
- (a) 0 (b) 1.73
- (b) 2.84 (c) 4.90
48\.
Which one of the following is diamagnetic ?
(a) Co2+ (b) Cu2+
(c) Mn2+ (d) Sc3+.
49\.
Which of the following pair of ions have same paramagnetic moment ?
(a) Cu2+, Ti3+ (b) Na2+, Cu2+
(c) Ti4+, Cu2+ (d) Ti3+, Ni2+.
- 50. Which one is not coloured ?
- (a) Cu+ion (b) Cu2+ ion
- (c) Fe3+ ion (d) Ni2+ ion.
- 51. Which of the ions in solution will be coloured ?
- (a) Cu+ (b) Mn2+
- (c) Ti4+ (d) Na+.
- 52. Which one of the following is an example of non-typical transition elements ?
- (a) Li, K, Na (b) Be, Al, Pb
- (c) Zn, Cd, Hg (d) Ba, Ca, Sr.
- 53. Which of the following has the tendency towards complex formation ?
- (a) d-block elements (b) p-block elements
- (c) s-block elements (d) elements of zero group.
- 54. Choose the correct answer for transition elements :
- (a) Transition elements have low melting points
- (b) Transition elements do not show catalytic activity
- (c) Transition elements exhibit variable oxidation states
- (d) Transition elements exhibit inert pair effect.
Answers
1\. (d)
2\. (b)
3\. (c)
4\. (b)
5\. (a)
6\. (c)
7\. (c)
8\. (a)
9\. (a)
10\. (a)
11\. (c)
12\. (d)
13\. (b)
14\. (b)
15\. (b)
16\. (d)
17\. (d)
18\. (c)
19\. (c)
20\. (d)
21\. (c)
22\. (d)
23\. (c)
24\. (b)
25\. (a)
26\. (b)
27\. (b)
28\. (d)
29\. (a)
30\. (a)
31\. (d)
32\. (c)
33\. (b)
34\. (d)
35\. (b)
36\. (b)
37\. (d)
38\. (a)
39\. (b)
40\. (d)
41\. (d)
42\. (d)
43\. (b)
44\. (d)
45\. (c)
46\. (d)
47\. (b)
48\. (d)
49\. (a)
50\. (a)
51\. (b)
52\. (c)
53\. (a)
54\. (c)
Section ‘B’
Short Answer Type Questions
Q. 1. Transition elements possess catalytic properties. Explain why.
Ans. Transition metals, their alloys and compounds possess marked catalytic activity e.g., iron, nickel, platinum, vanadium pentaoxide etc. In some cases, the transition metals with their variable valency may form unstable
intermediate compounds. In other cases, the transition metal provides a suitable reaction surface.
Q. 2. Give four important characteristics of d-block elements.
Ans. (i) They generally form coloured compounds. This is due to the presence of incomplete d-sub-shell. Thus for example monovalent silver ion is colourless since Ag+= 2, 8, 18, 18; monovalent cuprous ion is colourless since Cu+ = 2, 8, 18 while cupric ion is coloured (blue) since Cu2+ = 2, 8, 17 and nickel ion is also coloured (green) since Ni2+ = 2, 8, 16.
- (ii) These metals, their alloys and compounds possess marked catalytic activity, e. g., iron, nickel, platinum, vanadium pentaoxide etc. In some cases of the transition elements with their variable valency may form unstable intermediate compounds. In other cases the transition metal provides a suitable reaction surface.
- (iii) They form complex compounds. This is due to the presence of highly charged ions with vacant orbitals ready to accept lone pair of electrons donated by ligands.
- (iv) They are generally paramagnetic i.e., attract magnetic lines of forces. This property is due to the presence of unpaired electrons in the d-sub-shell . On account of the absence of unpaired electrons in d-sub-shell Zn2+etc. don’t exhibit paramagnetism. In the case of Fe, Co and Ni the unpaired electron spins are more pronounced resulting in the reinforcement of paramagnetism. These elements are, therefore, much more paramagnetic than rest of the elements and are said to be ferromagnetic.
Q. 3. Explain why TiO2 is white while TiCl3 is coloured ?
Ans. In TiO2 the oxidation state of Ti is +4 while in TiCl3 it is +3. The electronic configuration of Ti4+ is 2, 8, 8 while of Ti3+ is 2, 8, 9. Since in Ti4+ d-sub-shell is absent, no d-d electronic transition is possible. Hence, TiO2 will be white. On the other hand in Ti3+ there is one electron in d-sub-shell i.e., the d-sub-shell is incomplete. Hence d-d electronic transitions are possible and the energy absorbed in these transitions will correspond to a wavelength in the visible region. Therefore, TiCl3 will be coloured.
Q. 4. Explain why in Mn +2 while in Fe +3 oxidation state is more stable ?
Ans. The electronic configuration of Mn in +2 oxidation state will be 3d5 and of Fe in +3 oxidation state will also be 3d5 Since in both 3d is half-filled it has extra stability. Therefore, the + 2 state in Mn and the + 3 state in Fe is more stable.
Q. 5. Explain why the atomic radii of d-block elements decrease upto VIII and then increase ?
Ans. On going from Ni2+ with the configuration t2q6 eq2 to Cu2+ and Zn2+ electrons are added to eg orbitals where their screening power is abnormally high and the radii cease to decrease and actually show small increase.
Q. 6. Explain why the compounds of Ce(IV) are more stable ?
Ans. The electronic configuration of Ce is \[Xe\] 4f1 5d1 6s2. Hence the configuration of Ce (IV) will be that of noble gas xenon. Since this configuration is most stable, Ce (IV) compounds will be more stable.
Section ‘C’
Long Answer Type Questions
Q. 1. Discuss complex formation tendency of transition metals with suitable examples.
Ans. The transition metals possess remarkable ability to form complex compounds. This is attributed to the presence of highly charged ions with vacant orbitals ready to accept lone pair of electrons donated by other groups called ligands. The complex formation is not limited to inorganic compounds but many organic chelating ligands have got privilege to join with transition metals e.g., ferrocene (C5H5∙Fe∙C5H5). The molecules or ions which are added to the cation of transition metal are called ligands. These ligands always possess lone pairs of electrons which they can donate easily and complete the vacant orbitals of the transition metals through the formation of coordinate covalent bonds. The metal ions Cr3+, Co3+, Pt4+ and Pt2+ form thousands of complex ions. The structure usually found in such complex ions is linear, square planar, tetrahedral or octahedral.
The complex formation tendency of transition metals is attributed to two factors :
- 1\. Their small size. It is responsible for their high positive charge density which facilitates the acceptance of lone pairs of electrons from ligands.
- 2\. Existence of vacant orbitals. These possess right type of energy to accept lone pairs of electrons to complete their vacant orbitals.
A few examples to illustrate the formation of complex ions have been briefly discussed below.
Complex of Ag+ with ammonia, \[Ag(NH3)2\]+. The nitrogen atom in the ammonia molecule has a lone pair of electrons. The configuration of silver atom is 2, 8, 18, 4s2p6d10 5s1. The configuration of the outer orbitals, 4d10 5s1 has been shown in figure (a) and that of silver ion Ag+ 4d10 5s0 in figure (b) i.e., the 5s orbital is vacant. Hybridization takes place between the 5s orbital and one of the three 5p orbital whereby one lone pair of electrons from each molecule of ammonia is transferred to each of the hybridized orbital resulting in the formation of the complex ion \[Ag(NH3)2\]+ as shown in figure (c).
_files/AU20B.20Sc.20II20Chemistry20I(OK)-3.png)
5s
_files/AU20B.20Sc.20II20Chemistry20I(OK)-4.png)5s
5p
_files/AU20B.20Sc.20II20Chemistry20I(OK)-5.png)
5p
_files/AU20B.20Sc.20II20Chemistry20I(OK)-6.png)
sp hybridization.
The electrons from the silver ion are represented by arrows while those from the ligands, i.e., ammonia molecules are represented by dots. This is a case of sp hybridization. Since the angle between the sp hybridized bonds is equal to 180º, the ion is linear.
Complex of cupric ion with ammonia. The outer electronic configuration of copper atom is 3d10 4s1 as shown in fig. (a). Cupric ion is formed by the loss of one 3d electron and the 4s electron. The outer configuration of cupric ion, Cu2+ is 3d9 4s0 as shown in fig. (b). The structure of the hybridized complex ion is square planar which is characteristic of dsp2 hybridization.
3d
(a)
_files/AU20B.20Sc.20II20Chemistry20I(OK)-7.png)
4s
_files/AU20B.20Sc.20II20Chemistry20I(OK)-8.png)4s
4p
3d
(b)
_files/AU20B.20Sc.20II20Chemistry20I(OK)-9.png)
4p
3d 4s 4p
(c)
_files/AU20B.20Sc.20II20Chemistry20I(OK)-10.png)ds p2 hybridization.
Therefore, it appears that hybridization of one 4s, two 4p and one 3d orbitals takes place and the lone electron of the 3d orbital shifts to the remaining 4p orbital as shown in fig. (c).
2+
_files/AU20B.20Sc.20II20Chemistry20I(OK)-11.png)The square planar structure of \[Cu(NH3)4\]2+ complex ion is shown in the figure.
Q. 2. Why are transition elements called d-block elements? Explain briefly with reasons the following properties of transition elements.
- (a) Paramagnetic properties (b) Colour
- (c) Oxidation states.
Ans. Transition metals are called ‘d’ block elements because in them there is filling of ‘d’ sub-shell of the penultimate shell as is evident from the electronic configuration of the elements of first transition series.
Sc 21—1s2 2s2 2p6 3s2 3p6 3d1 4s2
Ti 22—1s2 2s2 2p6 3s2 3p6 3d2 4s2
V 23—1s2 2s2 2p6 3s2 3p6 3d3 4s2
Cr 24—1s2 2s2 2p6 3s2 3p6 3d5 4s1
Mn 25—1s2 2s2 2p6 3s2 3p6 3d5 4s2
Fe 26—1s2 2s2 2p6 3s2 3p6 3d6 4s2
Co 27—1s2 2s2 2p6 3s2 3p6 3d7 4s2
Ni 28—1s2 2s2 2p6 3s2 3p6 3d8 4s2
Cu 29—1s2 2s2 2p6 3s2 3p6 3d10 4s1
Zn 30—1s2 2s2 2p6 3s2 3p6 3d10 4s2
The discrepancies of electron filling in Cr and Cu are explained by the fact that the stability of half-filled or completely filled sub-shells is relatively more.
- (a) Paramagnetic properties
Most of the transition metals are paramagnetic, i.e., attract magnetic lines of force. The property of paramagnetism is associated with the existence of unpaired electrons in the d-sub shell. The magnitude of paramagnetism depends upon the number of unpaired electrons. Iron, cobalt and nickel are so strongly paramagnetic that they are called ferromagnetic.
- (b) Colour
The transition metals usually form coloured metal ions and compounds. If an ion is coloured it means that it absorbs light in the visible spectrum. In general only those atoms, ions or molecules would be coloured which have incomplete inner orbits and wherein the transition of electrons is possible by the absorption of energy in the visible regions. Thus for example, monovalent silver ions is colourless since Ag+ = 2, 8, 18, 18, monovalent cuprous ion is colourless since Cu+ = 2, 8, 18 while cupric ion is blue since Cu2+ = 2, 8, 17 and nickel ion is green since Ni2+ = 2, 8, 16. Almost all inorganic coloured substances are compounds of transition elements.
- (c) Oxidation states
The transition metals show a large number of variable oxidation states which are related to their electronic structures. Thus, titanium has a oxidation state of +2, +3 and +4. In the +2 state only the 4s2 electrons are used while in the +3 state one 3d electron and in +4 state both 3d electrons are used. Generally, the maximum oxidation number is equal to the total number of electrons available for bond formation and this, in turn, is equal to the number of the group to which the elements belong. The most common oxidation states are +2 and +3. The +1 oxidation state corresponding to the loss of a single s electron, is rate except for Cu, Ag and Au—the group IB elements. Only Sc, Y and La in group IIIA, Ti and Hf in group IVA, Ag in group IB, Zn and Cd in group IIB exhibit single oxidation states of +3, +4, +1 and +2 respectively. In the lower oxidation states, compounds are ionic. In the higher oxidation states, where a greater number of electrons are used in bonding, compounds are covalent and many transition elements form anion e.g., chromate, CrO42‒, dichromate, Cr2O72‒, manganate, MnO42‒, permanganate, MnO4‒, ferrocyanide, Fe(CN)64‒ and ferricyanide, Fe(CN)63‒ where the elements are in positive oxidation states. The various oxidation states of the elements of the first transition series are shown in the table.
Element
Outer Electronic Configuration
Oxidation States
Sc
3d1, 4s2
\+ 3
Ti
3d2, 4s2
\+ 2 + 3 + 4
V
3d3, 4s2
\+ 2 + 3 + 4 + 5
Cr
3d5, 4s1
\+ 2 + 3 + 4 + 6
Mn
3d5, 4s2
\+ 2 + 3 + 4 + 6 + 7
Fe
3d6, 4s2
\+ 2 + 3 + 4 + 6
Co
3d7, 4s2
\+ 2 + 3 + 4
Ni
3d8, 4s2
\+ 2 + 3
Cu
3d10, 4s1
\+ 1 + 2
Zn
3d10, 4s2
\+ 2
22
Vikas, 2009 (A. U.)
Unit-II
Section ‘A’
Objective Type Questions
- 1\. Van Arkel method of purification of metals involves converting the metal
to :
- (a) volatile stable compound
- (b) volatile unstable compound
- (c) non-volatile stable compound
- (d) none of the above.
- 2\. Which one of the following ores is best concentrated by froth floataion
process ?
- (a) Magnetite (b) Cassiterite
- (c) Galena (d) Malachite.
- 3\. Which of the following statements is correct regarding the slag obtained
during the extraction of a metal like copper or iron ?
- (a) The slag is lighter and lower melting than the metal
- (b) The slag is heavier and lower melting than the metal
- (c) The slag is lighter and higher melting than metal
- (d) The slag is heavier and higher melting than the metal.
- 4\. An ore like zinc blende is concentrated by :
- (a) froth floatation (b) magnetic separation
- (c) leaching (d) washing with water.
- 5\. Which metal is extracted by electrolytic reduction ?
- (a) Iron (b) Copper
- (c) Silver (d) Aluminium.
- 6\. Which of the following metals is obtained by using a solution of sodium cyanide and then precipitating the metal by addition of zinc dust ?
- (a) Platinum (b) Titanium
- (c) Vanadium (d) Gold.
- 7\. The substance which is added in ores for removal of impurities is known
as :
- (a) slag (b) gangue
- (c) flux (d) catalyst.
- 8\. Which one of the following is used as an acidic flux in metallurgy ?
- (a) CaO (b) SiO2
- (c) Na2CO3 (d) SO2.
- 9\. The cheap and having high melting point compound used in furnace is :
- (a) PbO (b) CaO
- (c) HgO (d) ZnO.
10\.
Gold is extracted by hydrometallurgical process, based on its property :
- (a) to form complexes which are water soluble
- (b) to form salts which are water soluble
- (c) of being electropositive
- (d) of being less reactive.
11\.
Which of the following metals in obtained by leaching out process using a solution of sodium cyanide and then precipitating the metal by addition of zinc dust ?
- (a) Copper (b) Silver
- (c) Nickel (d) Iron.
12\.
By the process of the poling, the following metal is purified :
- (a) Iron (b) Magnesium
- (c) Copper (d) Aluminium.
13\.
Which one of the following metals is extracted by a reduction process ?
- (a) Copper (b) Iron
- (c) Aluminium (d) Magnesium.
14\.
In blast furnace, the highest temperature is in :
- (a) reduction zone (b) slag zone
- (c) fusion zone (d) combusion zone.
15\.
Flux is added to the ore to :
- (a) reduce the ore (b) remove the moisture from ore
- (c) make the ore porous (d) remove impurities from the ore.
16\.
Thermal decomposition of limestone into lime and carbon dioxide is an example of :
- (a) calcination (b) leaching
- (c) levigation (d) smelting.
17\.
When a metal is to be extracted from its ore, if the gangue associated with the ore is silica :
- (a) an acidic flux is needed (b) a basic flux is needed
- (c) either of the two can be used (d) neither of them is needed.
18\.
While extracting an element from its ore the ore is ground and leached with dil. potassium cyanide soln. to form the soluble product potassium argentocyanide. The element is :
- (a) Lead (b) Chromium
- (c) Manganese (d) Silver.
19\.
Which one of the following reactions is an example of calcination process ?
- (a) 2Ag + 2HCl + \[O\] → 2AgCl + H2O
- (b) 2Zn + O2 → 2ZnO
- (c) 2ZnS + 3O2 → 2ZnO + 2SO2
- (d) MgCO3 → MgO + CO2.
- 20\. Froth floatation process for the concentration of ores is an illustration of the practical application of :
- (a) adsorption (b) absorption
- (c) coagulation (d) sedimentation.
- 21\. NaCN is used in the extraction of :
- (a) Iron (b) Copper
- (c) Magnesium (d) Gold.
- 22\. A substance that is added to the furnance charge during smelting operation to remove gangue is known as :
- (a) slag (b) flux
- (c) frother (d) concentrate.
- 23\. The ores that are concentrated by froth floatation method are :
- (a) carbonates (b) sulphides
- (c) oxides (d) phosphates.
- 24\. Heating pyrites in air to remove sulphur is known as :
- (a) roasting (b) calcination
- (c) smelting (d) fluxing.
- 25\. The extraction of metals from the sulphide ores is usually done by :
- (a) electrolysis (b) smelting process
- (c) metal displacement method (d) froth floatation process.
- 26\. The process of zone refining is used in the purification of :
- (a) Al (b) Si
- (c) Cu (d) Ag.
- 27\. Which of the following statement is correct ?
- (a) A mineral cannot be an ore (b) All ores cannot be mineral
- (c) All ores are mineral (d) All minerals are ores.
- 28\. In the reaction
CaO + SiO2 → CaSiO3
the product is known as :
- (a) ore (b) slag
- (c) flux (d) gangue.
- 29\. Chromium is obtained by reducing purified chromite ore with :
- (a) red hot coke (b) gaseous hydrogen
- (c) aluminium powder (d) carbon monoxide.
- 30\. The impurities of zinc and lead from gold are removed by :
- (a) zone refining (b) Van Arkel method
- (c) distillation (d) cupellation.
- 31\. Roasting is generally done in case of the following :
- (a) oxide ores (b) silicate ores
- (c) sulphide ores (d) carbonate ores.
- 32\. Hydrometallurgy is useful in the extraction of :
- (a) Sn (b) Al
- (c) Hg (d) Ag.
- 33\. Which of the following metals is refined by distillation ?
- (a) Copper (b) Tin
- (c) Silver (d) Mercury.
- 34\. Which of the following metals cannot be extracted by carbon process :
- (a) Pb (b) Al
- (c) Hg (d) Zn.
- 35\. Calcination is used in metallurgy for removal of :
- (a) water and sulphide (b) water and CO2
- (c) CO2 and H2S (d) H2O and H2S.
- 36\. The role of calcination in metallurgical operation is :
- (a) to remove moisture (b) to decompose carbonate
- (c) to drive off organic matter (d) to achieve all the above.
- 37\. Refractory materials are used in furnaces because they :
- (a) possess great structural strength
- (b) can withstand high temperature
- (c) are chemically inert
- (d) do not require replacement.
- 38\. One of the following metals forms a volatile carbonyl compound and this property is taken advantage of for its extraction. The metal is : (a) iron (b) nickel
- (c) cobalt (d) titanium.
- 39\. The method used for enrichment of sulphide ores of copper is :
- (a) magnetic separation (b) froth floatation process
- (c) electrorefining (d) smelting.
- 40\. The substance which reacts with the gangue to form a fusible mass is called :
- (a) flux (b) catalyst
- (c) ore (d) slag
- 41\. Chemical reduction is not suitable for converting :
- (a) bauxite into aluminium (b) cuprite into copper
- (c) haematite into iron. (d) zinc oxide into zinc.
- 42\. The commonest method for extraction of metals from oxide ores involves : (a) reduction with carbon (b) reduction with aluminium
- (c) reduction with hydrogen (d) electrolytic method.
- 43. In the reverberatory furnace :
- (a) the flames don’t come in contact with the charge
- (b) the flames come in contact with the charge
- (c) only hot gases come in contact with the charge
- (d) the flames are not there at all.
- 44. Amalgamation method of extracting metals is used for :
- (a) sodium, potassium, etc., which form heavy amalgams
- (b) lead, tin, etc., which form heavy amalgams
- (c) aluminium, magnesium, etc., which form light amalgams
- (d) silver, gold, etc., which occur free in nature.
- 45. The process of calcination and roasting is carried out in :
- (a) blast furnace (b) open furnace
- (c) reverberatory furnace (d) muffle furnace.
- 46\. The process of heating the ore strongly in excess of air so that the volatile impurities are removed and the ore changes to oxide is known as :
- (a) calcination (b) roasting
- (c) froth floatation (d) leaching.
- 47. Cupellation process is used in the metallurgy of :
- (a) copper (b) silver
- (c) aluminium (d) iron.
- 48. ‘Smelting’ is the process in which :
- (a) ore is heated in absence of air
- (b) ore is cooled
- (c) ore is heated in presence of air
- (d) ore is melted.
- 49. In metallurgy ‘flux’ is a substance used to convert :
- (a) infusible impurities to fusible
- (b) soluble particles to insoluble impurities
- (c) fusible impurities to infusible material
- (d) mineral into sillicates.
- 50. The metallurgical operation in which metal is obtained in fused state, is called :
- (a) froth floatation process (b) calcination
- (c) roasting (d) smelting.
- 51. In metallurgical processes the flux used for removing for acidic impurities is :
- (a) silica (b) sodium chloride
- (c) limestone (d) sodium carbonate.
- 52\. The electrolytic method of reduction is employed for the preparation of metals that :
- (a) are weakly electropositive
- (b) are moderately electropositive
- (c) are strongly electropositive
- (d) form oxides.
- 53\. When a metal is to be extracted from its ore, if the gangue associated with the ore is silica, then :
- (a) an acidic flux is needed
- (b) a basic flux is needed
- (c) both acidic and basic flux are needed
- (d) neither of them is needed.
- 54\. In electrorefining of metal the impure metal is made the anode and a strip of pure metal the cathode during the electrolysis of an aqueous solution of a complex metal salt. This method cannot be used for refining of :
- (a) silver (b) copper
- (c) aluminium (d) zinc.
- 55\. Flux is used to :
- (a) remove all impurities from ores
- (b) reduce metal oxide
- (c) remove silica
- (d) remove silica and other undesirable metal oxide.
- 56\. Mac Arthur Forrest cyanide process is used for the extraction of :
- (a) Vanadium (b) Titanium
- (c) Chromium (d) Gold.
- 57\. Suggest a purification method for obtaining highly pure silicon used as semiconductor material from the following :
- (a) Zone refining method (b) Oxidation method
- (c) Electrochemical refining method (d) Reduction method.
- 58\. Concentration of the ore is done :
- (a) to make the ore porous.
- (b) to make the ore fit for electrolysis
- (c) to drive off the volatile impurities
- (d) to free it from as many as impurities as possible.
- 59\. The purpose of smelting an ore is :
- (a) to oxidize it (b) to reduce it
- (c) to separate volatile impurities (d) to obtain an alloy.
- 60\. If the impurity in a metal has a greater affinity for oxygen and is more easily oxidized than the metal, then the purification of metal may be carried out by :
- (a) poling (b) zone refining
- (c) electrolytic refining (d) cupellation.
- 61\. Which ones of the following are prepared by electrolytic method :
- (1) Mg (2) Sn
- (3) Sulphur (4) F2.
- (a) 1 and 2 (b) 2 and 3
- (c) 3 and 4 (d) 1 and 4.
- 62. The impurities associated with minerals used in metallurgy are called collectively :
- (a) slag (b) flux
- (c) gangue (d) ore.
- 63. A naturally occuring substance by which an element can be extracted profitably is called :
- (a) mineral (b) compound
- (c) ore (d) salt.
- 64. Reverberatory furnace is employed in the metallurgical process mainly for :
- (a) reduction of oxide ores
- (b) smelting of sulphide ores
- (c) conversion of chloride to sulphate
- (d) getting magnetic materials.
- 65. Metallurgy is the process of :
- (a) concentrating the ore
- (b) roasting the ore
- (c) extracting the metal from the ore
- (d) adding carbon to the ore in blast furnace.
- 66. The method of zone refining of metals is based on the principle of :
- (a) greater solability of the impurity in the molten state than in the solid
- (b) greater mobility of the pure metal than that of the impurity
- (c) higher melting point of the impurity than that of the pure metal
- (d) greater noble character of the solid metal than that of the impurity.
- 67. Which metal cannot be obtained from electrolysis ?
- (a) Cu (b) Mg
- (c) Cr (d) Ni.
- 68. In metallurgy ‘flux’ means :
- (a) the gangue material in the ore
- (b) the substance added to the ore to remove the gangue material
- (c) the form in which the gangue material is eliminated
- (d) the metal obtained after the gangue is eliminated.
- 69. The inner lining of a blast furnace is made of :
- (a) graphite bricks (b) silica bricks
- (c) fireclay bricks (d) basic bricks.
- 70. The oil used in the floatation method for the purification of ores is :
- (a) coconut oil (b) olive oil
- (c) pine oil (d) none of these.
- 71. Electrometallurgical process (electrolysis of fused salt) is employed to extract :
- (a) Iron (b) Lead
- (c) Sodium (d) Silver
- 72. The main function of roasting is :
- (a) to remove the volatile matter (b) oxidation
- (c) reduction (d) to make slag
- 73. A mineral is known as an ore of a metal if the metal :
- (a) cannot be produced from it
- (b) can be produced from it
- (c) is very costly
- (d) can be produced from it profitably.
- 74. In which of the following furnaces, the highest temperature can be achieved ?
- (a) Reverberatory furnace (b) Electric furnace
- (c) Muffle furnace (d) Blast furnace.
- 75. In froth floatation process for the purification of ores, the particles of ore float because :
- (a) they bear electrostatic charge
- (b) they are insoluable
- (c) they are light
- (d) their surface is not easily wetted by water.
- 76\. In the equation :
4M + 8CN‒ + 2H2O + O2→ 4\[M(CN)2\]‒ + 4OH‒
identify the M :
- (a) Platinum (b) Titanium
- (c) Vanadium (d) Gold.
- 77. Heating an ore in the absence of air below its melting point is called :
- (a) calcination (b) smelting
- (c) roasting (d) leaching.
- 78. A basic lining is given to a furnance by using :
- (a) limestone (b) calcined dolomite
- (c) silica (d) haematite.
- 79\. For which ore of the metal, froth floatation method is used for concentration ?
- (a) Horn silver (b) Bauxite
- (c) Cinnabar (d) Haematite.
- 80\. The process of extracting the metal from its ore is called :
- (a) refining (b) concentration
- (c) leaching (d) metallurgy.
- 81\. A sulphide ore of a metal is converted into metal oxide by the process of :
- (a) smelting (b) roasting
- (c) calcination (d) bessemerization.
- 82\. When the ore limestone is heated, carbon dioxide is given off. This operation in metallurgy is known as :
- (a) smelting (b) ore dressing
- (c) calcination (d) roasting.
- 83\. Zone refining is a method to obtain :
- (a) very high temperature (b) ultrapure metal oxides
- (c) ultrapure metals (d) ultrapure aluminium.
- 84\. The process of ore dressing is carried out to :
- (a) remove the siliceous materials
- (b) add flux to the mineral
- (c) convert the ore to oxide
- (d) remove the poisonous impurities.
- 85\. Which method of purification is represented by the equations ?
Ti + 2I2 —50—0 —K (Impure)
(a) Cupellation
(c) Van Arkel
→ TiI4
16—75—K→ Ti + 2I2 (Pure)
(b) Poling
(d) Zone refining.
Answers
1\. (b)
2\. (c)
3\. (a)
4\. (a)
5\. (d)
6\. (d)
7\. (c)
8\. (b)
9\. (b)
10\. (a)
11\. (b)
12\. (c)
13\. (b)
14\. (c)
15\. (d)
16\. (a)
17\. (b)
18\. (d)
19\. (d)
20\. (a)
21\. (d)
22\. (b)
23\. (b)
24\. (a)
25\. (b)
26\. (b)
27\. (c)
28\. (b)
29\. (c)
30\. (d)
31\. (c)
32\. (d)
33\. (d)
34\. (b)
35\. (a)
36\. (d)
37\. (b)
38\. (b)
39\. (b)
40\. (a)
41\. (a)
42\. (a)
43\. (c)
44\. (d)
45\. (c)
46\. (b)
47\. (b)
48\. (d)
49\. (a)
50\. (d)
51\. (c)
52\. (c)
53\. (b)
54\. (c)
55\. (d)
56\. (d)
57\. (a)
58\. (d)
59\. (b)
60\. (d)
61\. (d)
62\. (c)
63\. (c)
64\. (b)
65\. (c)
66\. (a)
67\. (b)
68\. (b)
69\. (c)
70\. (c)
71\. (c)
72\. (b)
73\. (d)
74\. (b)
75\. (d)
76\. (d)
77\. (a)
78\. (b)
85\. (c).
79\. (c)
80\. (d)
81\. (b)
82\. (c)
83\. (c)
84\. (a)
Section ‘B’
Short Answer Type Questions
Q. 1. Whether zinc will displace copper from its salt solution or not ?
(E°Zn2+/Zn = ‒ 0∙76 Volt; E°Cu2+/Cu = + 0∙34 Volt)
Ans. E°cell = E°Cu2+/Cu ‒ E°Zn2+/Zn
= 0∙34 – (– 0∙76) = + 1∙10 Volt
Since E° is positive, zinc will displace copper from its salt solution.
Q. 2. Describe van Arkel method for the purification of metals.
Ans. By this method small amounts of very pure metal can be produced. In this method the impure metal (Ti, Zr etc.) is heated in an evaculated vessel with iodine. The gaseous tetraiodide (MI4) so formed is then decomposed on a white hot tungsten filament.
Impure Ti + 2I 50-250°C TiI 1400°C Ti + 2I
2 > 4 > 2
Tungsten Filament (Pure)
Q. 3. What is slag ?
Ans. Slag is a substance which is produced during smelting by reaction of the flux with impurities (gangue). It is immiscible with the metal and has a low melting point. Where the metal is liquid at the temperature of the furnace, the slag, having a lesser density, floats on the metal and protects it from oxidation. Because of the layering due to difference in density and immiscibilites, the two can be tapped off separately.
Q. 4. Explain the terms roasting and smelting.
Ans. Roasting is a process in which ore is heated alone or with some other material usually in presence of air below its fusion temperature in order to bring about some desired chemical changes like oxidation etc. It is generally done in a reverberatory furnace or in a blast furnace.
Smelting is a process in which a metal is extracted by fusing the ore in a suitable furnace in presence of a reducing agent such as carbon.
Q. 5. Whether Ag metal will reduce Sn2+ or not ?
(E°Ag+/Ag = + 0∙80 Volt; E°Sn2+/Sn = ‒ 0∙14 Volt)
Ans. Ag ^ Ag+ + e-: E° = - 0-8 Volt
AG° = - 1(- 0-8) = 0-8 F
or 2Ag ^ 2Ag+ + 2e- ; E° = - 0-8 volt;
[AG° = 1^6 F(i)](#bookmark129)
Sn2+ + 2e ^ Sn; E° = - 0-14 Volt;
[AG° = - 2(- 0-14) F = 0^28 F(ii)](#bookmark130)
On adding (i) and (ii)
2Ag + Sn2+ ^ 2Ag+ + Sn; AG° = + 1-6F + 0^28 F = 1^88 F
AG° 1^88 F
[E°=—- = —— = - 0-94 Volt - n F- 2 F](#bookmark131)
Since the value of E° is negative, the above reaction is not feasible i. e., Ag will not reduce Sn2+ to Sn.
Q. 6. Write a note on froth floatation process.
Ans. Froth floatation process is usually employed for the concentration of sulphide ores. In this process powdered ore is mixed with water and a small amount of pine oil or eucalyptus oil (frothing agent). Then a strong current of air is passed. On passing air, the sulphide-oil mixture is carried to the surface due to surface tension forces by air-bubbles surrounded by films of oil while impurities settle down to the bottom.
Q. 7. The EMF diagram for Fe is given as :
EA° (volts) FeO42“ ———^ Fe3+ ——77 Fe2+---0—45> Fe°
\+ 1.08
Determine the value of E°FeO 2-/Fe2+.
Ans. The half-reaction showing FeO42- ^ Fe3+ and Fe3+ ^ Fe2+ changes are
FeO42- + 8H+ + 3e- ^ Fe3+ + 4H2O (n = 3); E°A= + 2.20 V
Fe3++ e- ^ Fe2+ (n =1); E°A= + 0.77 V
On adding the above two equations we get
FeO42- + 8H+ + 4e- ^ Fe2+ + 4H2O (n = 4) which represents FeO42- ^ Fe2+ change
2.20 x 3 + 0.77 x1
FeÛ42-/Fe2+ = 3 ^ 1
= + 1.84 V.
Q. 8. Explain what is meant by flux.
Ans. Flux is a substance which is added to the ore during smelting to assist in the removal of earthy impurities (gangue) as slag.
Flux + Impurities\_\_\_\_> Slag
The selection of flux depends upon the nature of the impurities present. If the impurities are acidic (sand etc.) a basic flux like lime is used and if the impurities are basic (lime etc.) an acidic flux like sand is used.
Q. 9. Whether Ag will displace H2 gas from HCl or not ?
(E°Ag+/Ag = 0∙80 Volt; E° H+/H2 = 0∙0 Volt)
Ans. The probable reaction will be
Ag + HCl> AgCl + 2h
E°cell(0∙0 – 0∙80) = – 0∙80 Volt
Since the value of E° is negative, the above reaction is not feasible i.e., Ag will not react with HCl to liberate H2 gas.
Section ‘C’
Long Answer Type Questions
Q. 1. Compare the following properties of elements of second and third transition series with those of first series :
- (a) Oxidation states (b) Ionic radii
- (c) Magnetic properties (d) Stereochemistry
Ans. (a) Oxidation states
The elements of second and third transition series show a pronounced tendency towards higher oxidation states. Whereas the +2 state is known for all elements of the first transition series it is relatively unimportant for the heavier metals. Cd is nearly resricted to +2 oxidation state and Hg (II), Pd (II) and Pt (II) are the only other important dipositive species. Although Co is known as both Co(II) and Co(III) its congeners Rh and Ir are essentially limited to +3 state or higher. Cr (III) is the most stable oxidation state of Cr but both Mo and W are strongly reducing in that oxidation state with the +6 oxidation state being much more important. In general the stability of the highest possible oxidation state is considerably greater in heavier metals. Thus \[ReO4\]– unlike \[MnO4\]– is not a strong oxidizing agent. The trend is extended further along the series as well culminating in RnO4 and OsO4. Further examples are the stabilities of Pd(IV), Pt(IV) and Au (III) relative to lighter congeners. Gold is even able to achieve the unexpected oxidation state of +5.
- (b) Ionic radii
The radii of the heavier metals and ions are larger than those of the first series. Because of the lanthanide contraction, the radii of the third series show a little difference from those of the second series, despite the increased atomic number and total number of electrons.
- (c) Magnetic properties
The heavier elements tend to give low-spin compounds. Ions with an even number of electrons are often diamagnetic. Even where there is an odd number of d electrons, there is frequently only one unpaired electron. The simple interpretation of magnetic moments that is usually possible for first row paramagnetic species can seldom be made because of complications due to spin-orbit coupling. The spin-pairing can be attributed to the greater extension of 4d and 5d orbitals in space. Double occupancy of an orbital produces less inter electronic repulsion than in the smaller 3d orbitals. The electronic absorption spectra are also more difficult to interpret. It may be noted that a given set of orbitals produce splitting in the order 5d > 4d > 3d.
- (d) Stereochemistry
In general the coordination number of the elements of second and third transition series tend to be greater than for the first series because the ionic radii are larger by about 0.15 to 0.20 Å for corresponding species. Thus tetrahedral coordination is considerably less frequent although observed in species such as OsO4, \[WO4\]2 – and \[ReO4\]–. The square planar coordination is found in d8 species such as Pt(II), Au(III), Pd(II) and Rh(I). The octahedral species are quite common and the occurrence of coordination number 7, 8, 9 and 10 is fairly common.
Q. 2. The reduction potential diagram (Latimer diagram) for Cu in acid solution is :
Cu2+
+0∙15 Volt +0∙50 Volt
\--------> Cu+---------» Cu
E° = x Volt
Calculate x. Does Cu+ ion disproportionate in solution
Ans. Cu2+ + e ^Cu+; E° = + 045 ;
AG° = - n FE° = - 1 x 045 F = - 045 F
Cu+ + e ^ Cu ; E° = 0^50 ;
AG° = - 1 x 0^50 F = - 0^50 F
On adding
Cu2+ + 2 e ^ Cu; AG° = - 0^65 F
ag°
-0-65F
–2F
E° Cu2+/Cu =
\- n F
= 0∙325 Volt
a x = 0325 Volt
Further
Cu+ ^ Cu2+ + e; AG° = - n FE°
= – 1 × (– 0∙15) F
= 0∙15 F
Cu+ + e ^ Cu; AG° = - 1 x 0^50 x F
= – 0∙50 F
On adding
2Cu+ ^ Cu2+ + Cu; AG° = 0-15F + (- 0^50 F)
= – 0∙35 F
AG° -0-35 F
- • E°=----=------= + 0-35 F
- • • E - nF - F
Since E° of the above reaction is possible, the reaction is feasible i.e., Cu+ ion disproportionates.
Q. 3. Discuss in brief the various steps involved in the extraction of metals from their ores.
Ans. The various steps involved in the extraction of pure metals from their respective ores are as follows :
- 1\. Dressing or concentration of the ore : The ores are usually associated with large amounts of unwanted impurities e. g. earthly matter, rocky matter, sand etc. known as gangue or matrix. The removal of these impurities from the ore is known as dressing of the ore. In this process the percentage of the metal gradually goes on increasing. Hence the process is also called as concentration. The ore dressing operation may include one or more of the following operations according to the nature and quantity of the impurities present and also according to the nature of the ore.
- (a) Hand picking : The lumps of ores are broken to smaller pieces and if possible ores are separated from rocky and sandy pieces by hand picking. The picked up ore is obtained in sufficient degree of purity by breaking away the adherent rocky material with a hammer or a chisel.
- (b) Gravity separation : In this process an advantage is taken of the difference in the specific gravity of the ore particles and gangue. The ore is crushed and finely ground and sieved. The sieved ore is either subjected to dry centrifugal separation or is placed in big shallow tanks when a current of water flows. Heavy ore particles settle down to the bottom rapidly whereas lighter gangue particles are carried away by the current.
- (c) Magnetic concentration : In some cases when one of the impurities is magnetic in nature e.g. tin stone containing wolframite (FeMn)WO4 (magnetic)
and in the case of ferro-magnetic ores of iron separation is affected by means of this method. The powdered ore is dropped on a belt moving over electromagnetic rollers. The magnetic particles are attracted by the magnet and form a heap near the magnet while the non-magnetic particles form a heap a little away from it. The process may also be used for separating other transition metal ores such as magnetite (Fe3O4), chromite (FeO.Cr2O3) and pyrolusite (MnO2) from unwanted gangue.
- (d) Electrostatic separation : In this process an electrically charged surface is used. On account of electrical attraction and repulsion the easily chargeable metallic particles of the ore are thrown out of the whole mass and thus separated from the non-metallic portions.
- (e) Froth floatation process : This process is widely used for sulphide ores. A mixture of very finely ground ore and water is placed in a wooden tank or a cell to which a small amount of pine or eucalyptus oil (frothing agent) is added. Now a strong current of air is passed. On passing air sulphide-oil mixture is carried to the surface due to surface tension forces by air-bubbles surrounded by films of oil while impurities settle down at the bottom. The sulphide ores of copper, zinc, lead etc. are concentrated by this process.
- (f) Leaching : It is a chemical method for the concentration of the ores. In this process the ore is treated with a suitable reagent as to make it soluble while impurities remain insoluble. e.g. during the extraction of aluminium from bauxite (Al2O3.2H2O) the finely divided ore is treated with hot conc. solution of NaOH. Alumina present in the ore dissolves forming soluble sodium metaaluminate while impurities are left behind and are filtered off. Leaching method is also used to concentrate silver and gold ores.
- 2\. Calcination : It is a process in which ore is heated in absence of air below its fusion temperature in order to expel all organic matters and free moisture present in the ore as well as water from a hydrate or hydroxide or CO2 from a carbonate e.g.
CaCO3 ^ CaO + CO2
It renders the ore porous and easily workable in subsequent stages.
- 3\. Roasting : It is process in which ore is heated alone or with some other material usually in presence of air below its fusion temperature in order to bring about some desired chemical change like oxidation etc. It is carried out either in a reverberatory furnace or in a blast furnace.
S + O2 ^ SO2
2ZnS + 3O2 ^ 2ZnO + 2SO2
2PbS + 3O2 ^ 2PbO + 2SO2
4 As + 3O2 ^ 2As2O3
Sometimes during roasting, small amounts of sulphides are also oxidized to sulphates.
PbS + 2O2→ PbSO4
- 4\. Reduction to free metal : The calcined or roasted ore is then reduced to metal. Some of the methods commonly used for the purpose are as follows :
- (a) Smelting : In the process the ore is mixed with suitable quantity of carbon (coal or coke) and is heated to a high temperature above the melting point of the metal in a furnace. Carbon and CO produced by incomplete combustion of carbon reduce the oxide to free metal e.g.
ZnO + C → Zn + CO
SnO2 + 2C → Sn + 2CO
Fe2O3 + 3CO → 2Fe + 3CO2
During roasting a substance known as flux is added. It combines chemically with the residual gangue and forms a slag which has a low melting point and forms a separate layer being immiscible with the molten metal. The choice of flux depends upon the impurities present in the ore. e.g. if the ore contains acidic impurities such as SiO2 or P2O5 the basic fluxes like CaO, magnetic (MgCO3) etc.are used :
SiO2 + CaO → CaSiO3 (Slag)
On the other hand if the ore contains basic impurities such as FeO, CaO etc. then acidic fluxes like sand etc. are used :
FeO + SiO2→ FeSiO3 (Slag)
Smelting is generally carried out in a reverberatory furnace or in a blast furnace. It may be noted that in case of volatile metals e.g. Zn, reduction of the oxide ore is carried out in fire clay retorts.
- (b) Reduction by heating in air : In case of less active metals e.g. Hg whose oxides are unstable towards heat only roasting in air is enough for the separation of the metal :
2HgS + 3O2→ 2HgO + 2SO2
2HgO → 2Hg + O2
- (c) Reduction by aluminium : Certain oxides e.g. Cr2O3, Mn3O4 etc. are not easily reduced by carbon. In such cases aluminium is used as a reducing agent. A mixture of metallic oxide and Al powder (Thermite) is ignited in a closed crucible by means of a lighted Mg ribbon :
Cr2O3 + 2Al → 2Cr + Al2O3
3Mn3O4 + 8Al → 9Mn + 4Al2O3
- (d) Electrometallurgy : The oxides of very active metals like alkali metals, alkaline earth metals, Al etc. are not easily reduced by chemical reducing agents. In all such cases hydrometallurgy is not possible and extraction by smelting
process (pyrometallurgy) is unsuitable as the reduction requires extreme temperatures and under these circumstances the metal combines with carbon to form a carbide. Such metals are usually obtained by the electrolysis of their fused salts.
- (e) Hydrometallurgy : In this proces the concentrated ore is leached by aq. soln. of some suitable chemical reagent to extract the metal in the form of its soluble salt. The metal is then recovered from the salt solution either by electrolysis or by the use of suitable precipitating agent. e.g. silver ores are leached with dilute aqueous solution of NaCN or KCN is presence of air to extract the metal in the form of cyanide complex from which Ag is precipitated by adding reducing agent like Al or Zn.
Ag2S + 4NaCN ==== 2Na\[Ag(CN)2\] + Na2S 4Na2S + 5O2 + 2H2O ^ 2Na2SO4 + 4NaOH + 2S 2Na\[Ag(CN)2\] + Zn ^ 2Ag + Na2\[Zn(CN)4\]
- (f) Amalgamation : This process is used for the extraction of Ag and Au from their native ores. The mixture of finely powdered ore and water is allowed to flow over copper or brass plates coated with Hg and arranged in a slanting position. The metal particles form an amalgam with Hg and are retained on these plates. The amalgam from the plates is scrapped off and distilled in iron retorts when Hg distils over leaving the free metal behind.
- 5\. Purification or refining : The metals obtained by any of the methods described above are not pure. Hence they are subjected to certain purifying processes which depend upon the metal under treatment and nature of the impurities to be removed. The usual refining processes are discussed below :
- (a) Liquation : This method is applicable in the case of metals having low melting point such as tin and lead. The crude metal is kept on a sloping hearth and heated. The metal melts down and drains away leaving the solid impurities behind.
- (b) Poling : In this process molten metal is stirred by means of logs of green wood. Torrents of reducing gas containing CH4 bubbles up through the metal and reduce any oxide present as impurity in the metal. Copper and tin are refined by this process.
- (c) Distillation : This process is used in the case of metals such as Zn, Hg etc. which are easily vaporized at low temperatures. The impure metal is heated in a retort and its vapours are separately condenced in a receiver. The pure metal distils over while the non-volatile impurities are left behind in the retort.
- (d) Electrolytic refining : This method is widely used for the purification of a number of metals e.g. Cu, Ag, Au, Al, Ni etc. The crude metal is made the
anode and thin sheets of pure metal are made the cathode. The electrolyte is generally an aq. soln. of a suitable salt (simple or complex) of the metal with some corresponding acid if necessary. On passing electric current pure metal deposits on the cathode through the solution. The low electropositive impurities settle down at the bottom and are removed as anode mud while the more electropositive impurities pass into the solution.
- (e) Oxidation process : This process is used when the impurities have a greater affinity for oxygen than the metal itself. The impure metal is fused in a suitable furnace is stirred well by passing air thereby impurities are oxidized and begin to float on the surface of molten metal and are skimmed off. The various oxidation processes used for different metals bear different names e.g. cupellation, puddling etc.
- (f) Vapour phase refining : This is examplified by Mond process.
Ni + 4CO—5—0 —C → Ni(CO)4 1—80—C—→ Ni + 4CO
Volatile Pure
Another process given by van Arkel is very similar to the above in which Ti or Zr is heated in an evacuated vessel with iodine. The gaseous tetraiodide so formed is then decomposed on white hot tungsten filament.
50 – 250°C 1400°C
Impure Ti + 2I2 –––———→ TiI4 ———→ Ti + 2I2 Tungsten Pure Filament
- (g) Zone refining : By this method ultrapure Ge, silicon, boron, indium etc. are obtained. In this method the impure element is taken in the form of a rod. A narrow region is melted at one end. The molten region is then progressively transferred from one end of the rod to the other by slowly moving the source of heat. Impurities collect in the molten part or zone and are progressively swept to one end of the rod where they can be cut off and discarded. This method is based upon the fact that impurieties are more soluble in the melt than in the pure metal.
Q. 4. Discuss the stability of various oxidation states of an element in aqueous solution.
Ans. There are three sources of thermodynamic instability for a particular oxidation state of an element in aqueous solution.
- (i) The element may reduce hydrogen in water or hydronium ions.
- (ii) It may oxidize oxygen in water or hydroxide ion.
- (iii) It may disproportionate.
From the EMF values for reduction of hydrogen in water, we can determine
the minimum oxidation EMF necessary for a species to effect reduction of hydrogen : 1M acid, E° > 0.000 V; neutral solution, E > + 0.414 V; 1M base, E° > +0.828 V For manganese the only oxidation state that is unstable in this way is Mn(0) which readily reacts with acid.
Mn (s) → Mn2+ (aq) + 2e– E° = +1.18 V
2H+ (aq) + 2e–→ H2(g) E° = 0.00 V
Mn (s) + 2H+ (aq) → Mn2+ (aq) + H2 (g) E° = +1.18 V
Similarly from the EMF values for oxidation of oxygen in water we can determine the minimum reduction EMF necessary for a species to effect oxidation of oxygen : 1M acid, E° > +1.229 V; neutral solution, E > +0.815 V; 1M base, E° > + 0.401 V. There are several oxidation states of manganese that are reduced by water but the protonated manganate ion is typical.
HMnO4– (aq) + 3H+ (aq) + 2e–→ MnO2(s) + 2H2O E° = + 2.09 V
H2O → 12 O2(g) + 2H+ (aq) + 2e– E° = – 1.23 V
HMnO4–(aq) + H+ (aq) → MnO2(s) + H2O + 12 O2(g) E° = + 0.86 V
The species that reduce or oxidize water can be spotted rapidly in EMF (Latimer) diagrams e.g. the EMF diagram for manganese is as forward :
\+ 1.51
+0.90 +2.09 +0.90 +1.56 –1.18
MnO4– —→ HMnO4– —→ MnO2 —→ Mn3+ —→ Mn2+ —→ Mn
\+ 1.70 + 1.23
It may be noted that in acid solution all negative EMFs result in reduction of H+ ion by the species to the right of that EMF value. All values more positive than +1.23 V result in oxidation of the water by the species to the left of that value. The examination of the above manganese diagram for acid solution reveals that the following species are unstable : Mn° (oxidized to Mn2+), Mn3+ (reduced to Mn2+) and MnO4– (reduced to MnO2). One should also examine the skip-step EMF values for possible reactions leading to unstability. Thus, although water will not reduce MnO4– to HMnO4–, the skip-step EMF for MnO4– to MnO2 ( +1.70 V) is sufficiently large to make the reaction proceed.
2MnO4– (aq) + 2H+ (aq) + 2e–→ 2HMnO4– (aq) E° = + 0.90 V
H2O → 12 O2 (g) + 2H+ (aq) + 2e– E° = –1.23 V
2MnO4– (aq) + H2O → 2 HMnO4– (aq) + 12O2 (g) E° = – 0.33 V
2MnO4– (aq) + 8H+(aq) + 6e–→ 2MnO2 (s) + 4H2O E° = +1.70 V
3H2O → 32 O2(g) + 6H+ (aq) + 6e– E° = – 1.23 V
2MnO4– (aq) + 2H+ (aq) → 2MnO2(s) + 32O2 (g) + H2O E° = + 0.47 V
The disproportion occurs when a species is both a good oxidizing agent and a good reducing agent e.g. in basic solution Cl2 disproportionates to Cl– and OCl– ions.
- 12 Cl2 (aq) + e–→ Cl–(aq) E° = +1.40 V
- 12 Cl2 (g) + 2OH– (aq) → OCl– (aq) + H2O + e– E° = – 0.89 V
Cl2 (g) + 2OH– (aq) → Cl–(aq) + OCl– (aq) + H2O E° = +0.51 V
The species that are susceptible to disproportionation are readily picked out from EMF diagram. The normal behaviour of an element is for the EMF values to decrease steadily from left to right. Good reducing agents are on the right, good oxidizing agent are on the left and stable species are towards the middle. Whenever this gradual change from more positive to more negative is broken disproportionation occurs. For Mn in acid solution such breaks occur at two species : Mn3+ and HMnO4– Both these ions are unstable because they are reduced by water.
Mn3+ (aq) + e– → Mn2+ (aq) E° = + 1.56 V
Mn3+ (aq) + 2H2O → MnO2 (s) + 4H+ (aq) + e– E° = – 0.90 V
2Mn3+ (aq) + 2H2O → Mn2+ (aq) + MnO2 (s) + 4H+ (aq) E° = + 0.66 V
Unit-III
Section ‘A’
Objective Type Questions
- 1. According to IUPAC nomenclature sodium nitroprusside is named as :
- (a) sodium pentacyanonitrosylferrate (III)
- (b) sodium nitroferricyanide
- (c) sodium nitroferrocyanide
- (d) sodium pentacyanonitrosylferrate (II)
- 2\. Which one of the following will show paramagnetism corresponding to
- 2 unpaired electrons ?
(Atomic Numbers : Ni= 28, Fe =26)
- (a) \[NiCl4\]2– (b) \[Ni(CN)4\]2–
- (c) \[FeF6\]3– (d) \[Fe(CN)6\]3–.
- 3\. Among the following identify the species having both tetrahedral and diamagnetic attributes :
- (a) \[Ni(CO)4\] (b) \[Ni(CN)4\]2–
- (c) \[FeCl4\]– (d) \[(C6H5)4P\]+.
- 4\. The number of unpaired electrons in the complex ion \[CoF6\]3‒ is :
- (a) 0 (b) 2
- (c) 3 (d) 4
(Atomic No. Co = 27)
- 5\. Which of the following ligands forms a chelate ?
- (a) Acetate (b) Oxalate
- (c) Cyanide (d) Ammonia.
- 6\. The geometry of the complex \[Cu(NH3)4\]2+ is :
- (a) square planar (b) tetrahedral
- (c) triangular bipyramid (d) octahedral.
- 7\. One mole of the complex compound Co(NH3)5Cl3 gives 3 moles of ions
on dissolution in water. One mole of the same complex reacts with two moles of AgNO3 solution to yield two moles of AgCl(s). The structure of the complex is :
- (a) \[Co(NH3)4Cl\]∙Cl2∙NH3 (b) \[Co(NH3)5Cl\]Cl2
- (c) \[Co(NH3)3Cl3\]∙2NH3 (d) \[Co(NH3)4Cl2\]∙Cl∙NH3
- 8\. Pick out the complex compound in which the central metal atom obeys EAN rule strictly :
- (a) K3Fe(CN)6 (b) K4\[Fe(CN)6\]
- (c) \[Cu(NH3)4\]SO4 (d) \[Cr(H2O)6\]Cl3
- 9\. Which one of the following has largest number of isomers ?
- (a) \[Ru(NH3)4Cl2\]+ (b) \[Co(NH3)5Cl2+
- (c) \[Ir(PR3)2H(CO)\]2+ (d) \[Co(en)2Cl2\]+
(R = alkyl group, en = ethylenediamine)
- 10\. The ligand called π acid is :
- (a) CO (b) NH3
- (c) C2O42– (d) ethylenediamine.
- 11\. The compound EDTA (ethylenediaminetetraacetic acid) combines with cations to form :
- (a) chelates (b) clathrates
- (c) polymers (d) ion exchange resines.
- 12\. K4\[Fe(CN)6\] is an octahedral (d2sp3 hybridization) complex and is :
- (a) paramagnetic (b) diamagnetic
- (c) ferromagnetic (d) antimagnetic
\[Note : Outer electronic configuration of Fe is \[Ar\]3d64s2\]
- 13. In the compound lithium tetrahydridoaluminate, the ligand is :
- (a) H+ (b) H
- (c) H– (d) none of these.
- 14. The effective atomic number (EAN) of cobalt in Co2(CO)8 is :
- (a) 35 (b) 36 (c) 70 (d) 54.
- 15. \[Pt(NH3)4Cl2\]Br2 and \[Pt(NH3)4Br2\]Cl2 are related to each other as :
- (a) optical isomers (b) coordinate isomers
- (c) ionization isomers (d) linkage isomers.
- 16. The number of geometrical isomers of the complex \[Co(NO2)3 (NH3)3\] is :
- (a) 2 (b) 4 (c) 3 (d) 0.
- 17. Coordination number of Ni in \[Ni(C2O4)3\]4‒is :
- (a) 3 (b) 6 (c) 4 (d) 2.
- 18. Which one is an example of octahedral complex ?
- (a) \[FeF6\]3– (b) \[Zn(NH3)4\]2+
- (c) \[Ni(CN)4\]2– (d) \[Cu(NH3)4\]2+
- 19. IUPAC name of \[Pt (NH3)3 (Br) (NO2) Cl\] Cl is :
- (a) triamminechlorobromonitroplatium (IV) chloride
- (b) triamminebromonitrochloroplatinum (IV) chloride
- (c) triamminebromochloronitroplatinum (IV) chloride
- (d) triamminenitrochlorobromoplatinum (IV) chloride
- 20. The correct nomenclature for Fe4\[Fe(CN)6\]3 is :
- (a) ferroso ferric cyanide (b) ferric ferrous hexacyanate
- (c) iron (III) hexacyanoferrate (II) (d) hexacyanoferrate(III-II).
- 21. In which there is outer orbital hybridization ?
- (a) \[Fe(CN)6\]3– (b) \[Cr(NH3)6\]3+
- (c) \[CoF6\]3– (d) \[Co(H2O)6\]3+.
- 22. The pair of coordination compounds \[Co(NH3)5Br\]SO4 and \[Co(NH3)5SO4\]Br is an example for :
- (a) geometrical isomerism (b) coordination isomerism
- (c) structural isomerism (d) ionization isomerism.
- 23. Which of the following will give maximum number of isomers ?
- (a) \[Co(NH3)4Cl2\] (b) \[Ni(en)(NH3)4\]2+
(c) \[Ni(C2O4)(en)2\]2– (d) \[Cr(SCN)2(NH3)4\]+
- 24. Which one of the following octahedral complexes will not show geometric isomerism ? (A and B are monodentate ligands) :
- (a) \[MA5B\] (b) \[MA2B4\]
- (c) \[MA3B3\] (d) \[MA4B2\]
- 25\. In which of the following pairs both the complexes show optical isomerism ?
- (a) Cis- \[Cr(C2O4)2Cl2\]3 –, Cis-\[Co(NH3)4Cl2\]
- (b) \[Co(en)3\]Cl3, cis-\[Co(en)2Cl2\]Cl
- (c) \[PtCl(dien)\]Cl, \[NiCl2Br2\]2–
- (d) \[Co(NO3)3(NH3)3\], cis-\[Pt(en)2Cl2\]
- 26\. A coordination complex compound of cobalt has the molecular formula containing five ammonia molecules, one nitro group and two chlorine atoms for one cobalt atom. One mole of this compound produces three mole ions in an aqueous solution. On reacting this solution with excess of AgNO3 solution, we get two moles of AgCl precipitate. The ionic formula for this complex would be :
- (a) \[Co(NH3)4 (NO2) Cl\] \[(NH3)Cl\]
- (b) \[Co(NH3)5 Cl\] \[Cl(NO2)\]
- (c) \[Co(NH3)5 (NO2)\] Cl2
- (d) \[Co(NH3)5\] \[(NO2)2Cl2\].
- 27\. CN‒ is a strong field ligand. This is due to the fact that :
- (a) it carries negative charge
- (b) it is a pseudohalide
- (c) it can accept electrons from metal species
- (d) it forms high spin complexes with metal species.
- 28\. \[Cr(H2O)6\]Cl3 (at. no. of Cr = 24) has a magnetic moment of 3.83 BM. The correct distribution of 3d electrons in the chromium of the complex is :
- (a) 3dxy1, 3d(x2 – y2)1, 3dyz1 (b) 3dxy1, 3dyz1, 3dxz1
- (c) 3dxy1, 3dyz1, 3dz21 (d) 3d(x2 – y2)1, 3dz21 , 3dxz1
- 29\. In nitroprusside ion the iron and NO exist as FeII and NO+ rather than FeIII and NO. These forms can be differentiated by :
- (a) estimating the concentration of iron
- (b) measuring the concentration of CN–
- (c) measuring the solid state magnetic moment
- (d) thermally decomposing the compound.
- 30\. The hybridization of Ag in the linear complex \[Ag(NH3)2\]+ is :
- (a) dsp2 (b) sp (c) sp2 (d) sp3.
- 31\. Amongst \[Ni(CO)4\], \[Ni(CN)4\]2‒ and \[NiCl4\]2‒ :
- (a) \[Ni(CO)4\] and \[NiCl4\]2– are diamagnetic and \[Ni(CN)4\]2– is paramagnetic
- (b) \[NiCl4\]2– and \[Ni(CN)4\]2– are diamagnetic and \[Ni(CO)4\] is paramagnetic
- (c) \[Ni(CO)4\] and \[Ni(CN)4\]2‒ are diamagnetic and \[NiCl4\]2‒ is paramagnetic
- (d) \[Ni(CO)4\] is diamagnetic and \[NiCl4\]2‒ and \[Ni(CN)4\]2‒ are paramagnetic.
- 32. Which one of the following forms with excess of CN‒ (cyanide) a complex having coordination number two ?
- (a) Cu+ (b) Ag+
- (c) Ni2+ (d) Fe2+.
- 33. Which one has the highest paramagnetism ?
- (a) Ni(CO)4 (b) \[Ni(NH3)4\]Cl2
- (c) \[Ni(NH3)6\]Cl2 (d) \[Cu(NH3)4\]Cl2
- 34. Among the following the paramagnetic one is :
- (a) Ni(CO)4 (b) \[Ni(CN)4\]2‒
- (c) \[NiCl4\]2‒ (d) \[Ag(NH3)2\]+
- 35. In the complex \[Ni(CO)4\] the hybridized orbitals of Ni are :
- (a) sp2 (b) sp3
- (c) dsp2 (d) d2sp3.
- 36. Which of the following does not exhibit geometrical isomerism ?
- (a) PtCl4 (b) Pt(NH3)2Cl2
- (c) \[Pt(NH3)4Cl2\]Cl (d) \[Co(NH3)4\]Cl2.
- 37. Which one of the following complexes is an outer orbital complex ?
(At. Nos. : Mn = 25, Fe = 26, Co = 27, Ni = 28)
- (a) \[Fe(CN)6\]4‒ (b) \[Mn(CN)6\]4‒
- (c) \[Co(NH3)6\]3+ (d) \[Ni(NH3)6\]2+.
- 38. The correct order for the wavelength of absorption in the visible region
is :
- (a) \[Ni(NO2)6\]4‒ < \[Ni(NH3)6\]2+ < \[Ni(H2O)6\]2+
- (b) \[Ni(NO2)6\]4‒ < \[Ni(H2O)6\]2+< \[Ni(NH3)6\]2+
- (c) \[Ni(H2O)6\]2+< \[Ni(NH3)6\]2+< \[Ni(NO2)6\]4‒
- (d) \[Ni(NH3)6\]2+< \[Ni(H2O)6\]2+< \[Ni(NO2)6\]4‒.
- 39. An example of hexadentate ligand is :
- (a) 2, 2’ ‒dipyridyl (b) ethylenediamine
- (c) ethylenediamine tetraacetate ion (d) dimethylglyoximato.
- 40. The coordination number of a metal in coordination compounds is :
- (a) same as primary valency
- (b) sum of primary and secondary valencies
- (c) same as secondary valency
- (d) none of the above.
- 41\. According to Lewis the ligands are :
- (a) acidic in nature
- (b) basic in nature
- (c) neither acidic nor basic
- (d) some are acidic and others are basic
- 42\. The total number of possible isomers for the complex compound \[CuIINH3)4\] \[PtIICl4\] is :
- (a) 3 (b) 6 (c) 5 (d) 4.
- 43\. The hybridization state of Fe in \[Fe(H2O)5NO\] SO4 is (At. no. of Fe =
- 26\) :
- (a) dsp2 (b) sp3d (c) sp3d2 (d) d2sp3.
- 44\. IUPAC name of \[Co (NH3)3 (H2O)2 Cl\] Cl2 is :
- (a) diaquochlorodiamminecobalt (III) chloride
- (b) triamminediaquochlorocobalt (III) chloride
- (c) chlorodiammine diaquocobalt (III) chloride
- (d) diamminediaquochlorocobalt (III) chloride
- 45\. Pick out from the following complex compounds, a poor electrolytic conductor in solution :
- (a) K2\[PtCl6\] (b) \[Co(NH3)3(NO2)3\]
- (c) K4\[Fe(CN)6\] (d) \[Cu(NH3)4\]SO4.
- 46\. Coordination number of Cu in complex \[Cu(H2O4)4\]2+ is :
- (a) 4 (b) 3 (c) 2 (d) 1.
- 47\. Generally, a group of atoms can function as a ligand if :
- (a) they are positively charged ions
- (b) they are free radicals
- (c) they are either neutral molecules or negatively charged ions
- (d) none of the above.
- 48\. Which of the following is not a coordination compound ?
- (a) Cuprammonium sulphate (b) Potassium ferrocyanide
- (c) Potash alum (d) Potassium ferricyanide.
- 49\. Coordination isomerism is caused by the interchange of ligands between the :
- (a) cis and trans structure
- (b) complex cation and complex anion
- (c) inner sphere and outer sphere
- (d) low oxidation and higher oxidation states.
- 50\. The coordination number of a central metal atom in a complex is determined by :
- (a) the number of ligands around a metal ion bonded by sigma bonds.
- (b) the number of ligands around a metal ion bonded by pi-bonds.
- (c) the number of ligands around a metal ion bonded by sigma and pi-bonds both
- (d) the number of only anionic ligands bonded to the metal ion.
- 51. The number of isomers possible for square planar complex K2\[PbClBr2(SCN)\] is :
- (a) 2 (b) 3 (c) 4 (d) 6.
- 52\. Ni (Z = 28) combines with a uninegative monodentate ligand X‒ to form a paramagnetic complex \[NiX4\]2‒. The number of unpaired electron (s) in the nickel and geometry of this complex ion are respectively :
- (a) one, square planar (b) two, square planar
- (c) one, tetrahedral (d) two, tetrahedral.
- 53. The complexes \[Co(NH3)6\] \[Cr(C2O4)3\] and \[Cr(NH3)6\] \[Co(C2O4)3\] are the example of :
- (a) linkage isomerism (b) geometrical isomerism
- (c) coordination isomerism (d) ionization isomerism.
- 54. The concept of effective atomic number (EAN) was given by :
- (a) Werner (b) Sidgwick
- (c) Pauling (d) Morgan.
- 55. How many EDTA (ethylene diaminetetracetic acid) molecules are required to make an octahederal complex with a Ca2+ ion ?
- (a) One (b) Two (c) Six (d) Three.
- 56. The IUPAC name for the complex \[Co(NO2)(NH3)5\]Cl2 is :
- (a) pentaammine nitrito-N-cobalt (II) chloride
- (b) pentaammine nitrito-N-cobalt (III) chloride
- (c) nitrito-N-pentaammine cobalt (III) chloride
- (d) nitrito-N-pentaammine cobalt (II) chloride.
- 57. The proper name for \[Co(NH3)6\]Cl3 is :
- (a) cobalt hexaamminechloride
- (b) hexaamminecobalt (III) chloride
- (c) hexaammoniacobalt trichloride
- (d) hexaamminecobalt trichloride.
- 58. Which of the following will exhibit maximum ionic conductivity ?
- (a) K4\[Fe(CN)6\] (b) \[Co(NH3)6\]Cl3
- (c) \[Cu(NH3)4\]Cl2 (d) \[Ni(CO)4\].
- 59. The pair in which both species have same magnetic moment (spin only value) is :
- (a) \[Cr(H2O)6\]2+, \[CoCl4\]2‒ (b) \[Cr(H2O)6\]2+\[Fe(H2O)6\]2+
- (c) \[Mn(H2O)6\]2+, \[Cr(H2O)6\]2+ (d) \[CoCl4\]2‒, \[Fe(H2O)6\]2+
- 60\. Which of the following complexes exhibits coordination isomerism ?
- (a) \[Co(en)2Cl2\]+ (b) \[Cr(NH3)6\]Cl3
- (c) \[Cr(NH3)6\] \[Co(CN)6\] (d) \[Pt(NH3)2Cl2\].
- 61\. Which one of the following has tetrahedral geometry ?
- (a) \[Co(NH3)6\]3+ (b) \[Ni(CN)4\]2‒
- (c) Fe(CO)5 (d) \[NiCl4\]2‒.
- 62\. Among the following ions which one has the highest paramagnetism ?
- (a) \[Cr(H2O)6\]3+ (b) \[Fe(H2O)6\]2+
(c) \[Cu(H2O)6\]2+ (d) \[Zn(H2O)6\]2+.
- 63\. \[Co(H2O)6\]2+ has unpaired electrons (At No. of Co = 27) :
- (a) 5 (b) 2
- (c) 3 (d) 4.
- 64\. Which one of the following will give a white precipitate with AgNO3 in aq. medium ?
- (a) \[Co(NH3)5Cl\] (NO2)2 (b) \[Pt(NH3)2Cl2\]
- (c) \[Pt(en)Cl2\] (d) \[Pt(NH3)4\]Cl2.
- 65\. The hybridization state of Fe in \[Fe(CN)6\]4‒ is :
- (a) sp3d (b) sp3 (c) d2sp3 (d) sp3d2.
- 66\. In which there is inner orbital hybridization ?
- (a) \[Mn(H2O)6\]3+ (b) \[Fe(H2O)6\]2+
- (c) \[Cr(H2O)6\]2+ (d) \[Co(H2O)6\]3+.
- 67\. Optical isomerism is not shown by the complex :
- (a) \[Cr(ox)3\]3‒ (b) \[Co(en)2Cl2\]+ (cis form)
- (c) \[Co(en)2Cl2\]+ (trans form) (d) \[Cr(en)3\]3+.
(OX = oxalate, en = ethylenediamine)
- 68\. The possible number of isomers for the complex \[MCl2Br2\]SO4 is :
- (a) 1 (b) 2 (c) 4 (d) 5.
- 69\. Which type of hybridization is involved in the metal ion of \[Fe(CN)6\]3‒ complex ?
- (a) sp3 (b) dsp2 (c) d2sp3 (d) sp3d2
- 70\. The type of isomerism exhibited by the compounds when both the cationic part and the anionic part are complexes and there is an interchange of ligands between the two complex is :
- (a) linkage isomerism (b) coordination isomerism
- (c) geometrical isomerism (d) ionization isomerism.
- 71\. Which of the following ions form most stable complex compound ?
- (a) Cu2+ (b) Ni2+ (c) Fe2+ (d) Mn2+.
- 72\. Which of the following is not true for ligand-metal complex ?
- (a) Larger the ligand, the more stable is the metal-ligand bond
- (b) Highly charged ligand forms stronger bond
- (c) Larger the permanent dipole moment of ligand, the more stable is the bond
- (d) Greater the ionization potential of central metal, the stronger is the bond.10. The strongest ligand in the following is :
(a) CN‒ (b) Br‒ (c) HO‒ (d) F‒.
73\.
The term chelate ligands was coined by :
(a) Pauling (b) Sidgwick (c) Morgan (d) Werner.
74\.
\[Co(NH3)4Cl2\]NO2 and \[Co(NH3)4ClNO2\] Cl are .............. isomers.
- (a) geometrical (b) optical
- (c) linkage (d) ionization.
75\.
In the case of platinum complexes
\[Pt(NH3)6\]Cl4 \[Pt(NH3)Cl\]Cl3
I II
\[Pt(NH3)4Cl2\]Cl2
III
the molar electrical conductance (in water) is expected to follow the order :
- (a) I > II > III (b) I < II < III
- (c) I = II = III (d) I ≥ II ≥ III.
76\.
The shape of \[Cu(NH3)4\]2+ is square planar. Cu2+ in this complex is :
- (a) sp3 hybridized (b) dsp2 hybridized
- (c) sp3d hybridized (d) sp3d2 hybridized.
77\.
Which of the following is paramagnetic ?
(a) \[Ni(CO)4\] (b) \[Co(NH3)6\]3+
(c) \[Ni(CN)4\]2‒ (d) \[NiCl4\]2‒.
78\.
Which of the following is diamagnetic ?
\[At. Nos. : Cu = 29, Ni = 28, Co = 27\]
(a) \[Ni(CN4)2‒ (b) \[NiCl4\]2‒
(c) \[Cu(NH3)4\]2+ (d) \[CoCl4\]2‒.
79\.
Which statement is incorrect ?
- (a) Ni(CO)4 tetrahedral, paramagnetic
- (b) \[Ni(CN)4\]2‒ square planar, diamagnetic
- (c) Ni(CO)4 tetrahedral, diamagnetic
- (d) \[NiCl4\]2‒ tetrahedral, paramagnetic.
80\.
The incorrect match for the possible hybridization in the case of nickel complex is shown by :
(a) Ni(CO)4
3d 4s 4p
1111111111\] 111111111 Diamagnetic
sp3
- (b) \[NiCl4\]2‒
3d 4s 4p
Iu |u hi In In Ihi In hi hi I Paramagnetic sp3
- (c) \[Ni(CN)4\]2‒
3d 4s 4p
|ii|ii|ii|ii|iilMn|ii| I Diamagnetic dsp2
- (d) none of the above.
- 81. Which of the following is diamagnetic ?
\[At. Nos. Cu = 29, Ni = 28, Co = 27\]
- (a) \[Ni(CN)4\]2‒ (b) \[NiCl4\]2‒
- (c) \[Cu(NH3)4\]2+ (d) \[CoCl4\]2‒.
- 82. Which of the following compounds shows optical isomerism ?
- (a) \[Co(CN)6\]3‒ (b) \[Cr(C2O4)3\]3‒
- (c) \[ZnCl4\]2‒ (d) \[Cu(NH3)4\]2+.
- 83\. Which one of the following is not a transition metal complex ion by definition ?
\[At. Nos. : Co = 27, Ni = 28, Zn = 30, Mn = 25\]
- (a) \[CoF6\]3‒ (b) \[Ni(H2O)6\]2+
(c) \[Zn(H2O)6\]2+ (d) \[Mn(H2O)6\]2+.
- 84. Which one of the isomeric pairs shows ionization isomerism ?
- (a) \[Co(NH3)6\] \[Cr(CN)6\] and \[Cr(NH3)6\] \[Co(CN)6\]
- (b) \[Cr(H2O)6Cl3 and \[Cr(H2O)5Cl\] Cl2∙H2O
- (c) \[Pt(NH3)2Cl2\] and \[Pt(NH3)4\] \[PtCl4\]
- (d) \[Co(NH3)5Br\]SO4 and \[Co(NH3)5SO4\]Br.
- 85. Which of the following ligands is expected to be bidentate ?
- (a) CH3NH2 (b) C2O42‒
- (c) CH3C≡N (d) Br‒.
- 86. Which of the following complexes is likely to show optical activity ?
- (a) Trans \[Co(NH3)4Cl2\]+
- (b) \[Cr(H2O)6\]3+
- (c) Cis\[Co(NH3)2(en)2\]3+ (where en is ethylenediamine)
- (d) Trans \[Co(NH3)2(en)2\]3+.
- 87. Geometrical isomerism is found in coordination compounds having coordination number :
- (a) 2 (b) 3
- (c) 4 (tetrahedral) (d) 6.
- 88. The value of the ‘spin only’ magnetic moment for one of the following configurations is 2∙84 BM. The correct one is :
- (a) d5 (in strong ligand field)
- (b) d3 (in weak as well as in strong fields)
- (c) d4 (in weak field ligand)
- (d) d4 (in strong field ligand)
- 89. The structure of the \[Cu(H2O)4\]2+ ion is :
- (a) square planar (b) tetrahedral
- (c) distorted rectangle (d) octahedral.
- 90. The coordination number of copper in cuprammonium sulphate is :
- (a) 2 (b) 3 (c) 4 (d) 6.
- 91. The complex ion \[Co(NH3)6\]3+ is formed by sp2d3 hybridization. Hence the ion should possess :
- (a) octahedral geometry (b) tetrahedral geometry
- (c) square planar geometry (d) tetragonal geometry.
- 92. When monodentate ligand in the complex compound contains more than one donor atom, the isomerism exhibited by it will be :
- (a) optical isomerism (b) geometrical isomerism
- (c) linkage isomerism (d) ionization isomerism.
- 93. \[Co(en)2Cl2)+ ion has the following forms :
- (a) 2 (b) 3 (c) 4 (d) 5.
- 94. What is the correct name for \[Co(NH3)4Cl2\]Cl :
- (a) trichlorotetraamminecobalt chloride
- (b) tetraamminecobalt (III) chloride
- (c) dichlorotetraammine cobalt (III) chloride
- (d) chlorotetraamminecobalt (III) chloride
- 95. The proper name for K2\[PtCl6\] is :
- (a) potassium platinumhexachloride
- (b) potassium hexachloroplatinum (IV)
- (c) potassium hexachloroplatinate (IV)
- (d) potassium hexachloroplatinum.
- 96. Which of the following coordination compounds would exhibit optical isomerism ?
- (a) pentamminocobalt (III) chloride
- (b) diaminedichloroplatinum (II)
- (c) trans-dicyano bish (ethylene diamine) chromium (III) chloride
- (d) tris-(ethylene diaminecobalt (III) bromide.
- 97. IUPAC name of complex K3 \[Al (C2O4)3\] is :
- (a) potassium aluminooxalate
- (b) potassium trioxalatoaluminate (III)
- (c) potassium aluminium (III) oxalate
- (d) potassium trioxalatoaluminate (VI).
- 98. In which of the following complexes oxidation state of metal is zero ?
- (a) \[Pt (NH3)2 Cl2\] (b) \[Cr(CO)6\]
- (c) \[Cr(NH3)3Cl3\] (d) \[Cr(en)2 Cl2\].
- 99. The effective atomic number of Cr (Atomic No. = 24) in \[Cr(NH3)6\]Cl3 is :
- (a) 27 (b) 33 (c) 35 (d) 36.
- 100\. In which there is outer orbital hybridization ?
- (a) \[Zn(NH3)6\]2+ (b) \[Co(NH3)6\]3+
- (c) \[Cr(NH3)6\]3+ (d) \[V(NH3)6\]3+.
- 101\. Which of the following compounds exhibits linkage isomerism ?
- (a) \[Co(en)3\]Cl3 (b) \[Co(NH3)6\]\[Cr(CN)6\]
- (c) \[Co(en)2(NO2)Cl\]Br (d) \[Co(NH3)5Cl\]Br2
- 102\. The complex compound \[Co(NH3)3 NO2ClCN\] is named as :
- (a) triamminechlorocyanonitro cobalt (III)
- (b) triamminenitrochlorocyano cobalt (III)
- (c) cyanonitrochlorotriammine cobalt (III)
- (d) tiamminenitrochlorocyano cobalt (III).
- 103\. Which of the following complexes shows ionization isomerism ?
- (a) \[Cr(NH3)6\]Cl3 (b) \[Cr(en)2Cl2\]
- (c) \[Cr(en)3\]Cl3 (d) \[Co(NH3)5Br\]SO4.
- 104\. The number of isomers possible in the square planar complex \[PtClBrI(NH3)\]‒ would be :
- (a) 2 (b) 3 (c) 4 (d) 6.
- 105\. Which of the following statement is incorrect ?
- (a) In \[Cu(NH3)4\]SO4, the ligand has satisfied only the secondary valancy of copper.
- (b) In K4\[Fe(CN)6\] the ligand has satisfied both primary and secondary valencies of ferric ion
- (c) In K3\[Fe(CN)6\] the ligand has satisfied both primary and secondary valencies of ferric ion
- (d) In K3\[Fe(CN)6\] the ligand has satisfied only the secondary valency of ferric ion.
- 106\. The IUPAC name of K3\[Ir(C2O4)3\] is :
- (a) potassium trioxalatoiridium (III)
- (b) potassium trioxalatoiridate (III)
- (c) potassium tris(oxalato)iridium (III)
- (d) potassium tris(oxalato)iridate (III)
- 107\. Which of the following will have three isomeric forms
- (i) \[Co(en)2ClBr\]
- (iii) \[Cr(NO3)3(NH3)3\]
(en = ethylenediamine)
- (a) (i) and (ii)
(c) (ii) and (iv)
(ii) K3\[Co(C2O4)2 Cl2\]
(iv) K3\[Co(C2O4)3\]
- (b) (i) and (iii)
(d) (iii) and (iv).
- 108\. The number of geometrical isomers for \[Pt(NH3)2Cl2\] is :
- (a) two (b) one (c) three (d) four.
- 109\. Among the following the compound that is both paramagnetic and coloured is :
- (a) K2Cr2O7 (b) (NH4)2(TiCl6)
- (c) VOSO4 (d) K3\[Cu(CN)4\].
- 110\. Which is the correct IUPAC name for K4\[Fe(CN)6\] ?
- (a) Potassium ferricyanide
- (b) Potassium ferrocyanide
- (c) Potassium hexacyanoferrate (II)
- (d) Potassium hexacyanoferrate (III).
- 111\. Which of the following has square planar structure ?
- (a) \[NiCl4\]2‒ (b) \[Ni(CO)4\]
- (c) \[Ni(CN)4\]2‒ (d) none of these.
- 112\. Pick the correct name of \[Co(NH3)5Cl\]Cl2 :
- (a) Chloropentaammine cobaltate (III)
- (b) Pentaammine cobalt (III) Chloride
- (c) Chloropentaammine cobalt (III) Chloride
- (d) Chloropentaammine cobalt (II) Chloride.
- 113\. The complex ion which has no d electrons in the central metal atom is :
- (a) \[MnO4\]‒ (b) \[Co(NH3)6\]3+
- (c) \[Fe(CN)6\]3‒ (d) \[Cr(H2O)6\]3+
- 114\. The number of ions per mole of a complex \[CoCl2.5NH3\] in aqueous solution will be :
- (a) nine (b) four (c) three (d) two.
- 115\. Which of the following complexes will have four isomers ?
- (a) \[Co(en)2Cl2\]Cl (b) \[Co(en)(NH3)2Cl2\]Cl
(c) \[Co(PPh3)2(NH3)2Cl2\]Cl (d) \[Co(en)3\]Cl3
(en = ethylenediamine)
- 116\. In tris (ethylenediamine) cobalt (III) chloride the coordination number of cobalt is :
- (a) 3 (b) 4 (c) 6 (d) 7.
- 117. Dichlorodiammineplatinum (II) complex has the formula :
- (a) Pt \[Cl2 (NH3)2\] (b) Pt \[R (NH2)2\] Cl2
- (c) \[PtCl2 (NH3)2\] (d) \[Pt R (NH2)2\] Cl2.
- 118\. IUPAC name for the complex compound K3\[Fe(CN)6\] is :
- (a) potassium hexacyanoferrate (III)
- (b) potassium ferrocyanide ion (III)
- (c) potassium hexacyanoferrate (II)
- (d) potassium cyanohexaferrate (II).
- 119\. The formula of dichlorobis (urea) copper(II) is :
- (a) \[Cu {O = C (NH2)2}\] Cl2
- (b) \[CuCl2 {O = C (NH2)2}2\]
- (c) \[Cu {O = C (NH2)2} Cl\] Cl
- (d) \[CuCl2\] \[O = C (NH2)2H2\].
- 120\. \[Co(NH3)4 (NO2)2\]Cl exhibits :
- (a) linkage isomerism, ionization isomerism and geometrical isomerism
- (b) ionization isomerism, geometrical isomerism and optical isomerism
- (c) linkage isomerism, geometrical isomerism and optical isomerism
- (d) linkage isomerism, ionization isomerism and optical isomerism.
- 121\. The hypothetical complex chloro diaquotriamminecobalt (III) chloride can be represented as :
- (a) \[CoCl(NH3)3(H2O)2\]Cl2 (b) \[Co(NH3)3(H2O)Cl3\]
- (c) \[Co(NH2)3(H2O)2Cl\] (d) \[Co(NH3)3(H2O)3\]Cl3
- 122\. Ethylenediamine is an example of a ..................... ligand.
- (a) monodentate (b) bidentate
- (c) tridentate (d) hexadentate.
- 123\. The pair of complex compounds \[Cr(H2O)6\]Cl3 and \[Cr(H2O)5Cl\] Cl2H2O is an example of :
- (a) linkage isomerism (b) ionization isomerism
- (c) coordination isomerism (d) hydrate isomerism.
- 124\. The coordination number of Pt in \[Pt(NH3)4Cl2\]2+ ion is :
- (a) 2 (b) 4 (c) 6 (d) 8.
- 125\. The scientist who explained structure and isomerism in the coordination complexes is :
- (a) Sidgwick (b) Pauling
- (c) Powell (d) Werner.
- 126\. The most stable complex among the following is :
- (a) K3\[Al(C2O4)3\] (b) Pt(en)2\]Cl2
- (c) \[Ag(NH3)2Cl (d) K2\[Ni(EDTA)\]
- 127\. Which one of the following will not show geometrical isomerism ?
- (a) \[Cr(NH3)4Cl2\]Cl (b) \[Co(en)2Cl2\]Cl
- (c) \[Co(NH3)5(NO2)\]Cl2 (d) \[Pt(NH3)2Cl2\]
- 128\. Which one of the following cyano complexes would exhibit the lowest value of paramagnetic behaviour ?
- (a) \[Co(CN)6\]3‒ (b) \[Fe(CN)6\]3‒
(c) \[Mn(CN)6\]3‒ (d) \[Cr(CN)6\]3‒
(At. nos. : Cr = 24, Mn = 25, Fe = 26, Co = 27)
- 129\. The usual coordination number of the electoral (nucleus) atom in complex salts is :
- (a) 2 (b) 3 (c) 4 (d) 6
- 130\. Which one of the following is not a chelating ligand ?
- (a) NH2‒CH2‒CH2‒NH2 (b) CH3COO‒
- (c) EDTA (d) C2O42‒.
- 131\. Stability of complexes of Cu2+, Ni2+, Co2+ and Fe2+ varies in the
order :
- (a) Cu2+ > Ni2+ > Co2+ > Fe2+ (b) Cu2+ > Fe2+ > Ni2+ > Co2+
- (c) Ni2+ > Co2+ > Fe2+ > Cu2+ (d) Cu2+ > Ni2+ > Co2+ > Fe2+
- 132\. IUPAC name of \[Co(ONO)(NH3)5\]Cl2 is :
- (a) pentaamminenitrocobalt (III) chloride
- (b) pentaamminenitritocobalt (III) chloride
- (c) pentamminenitrosocobalt (III) chloride
- (d) pentammineoxo-nitrocobalt (III) dichloride.
- 133\. Considering H2O as a weak field ligand, the number of unpaired electrons in \[Mn(H2O)6\]2+ will be (At. No. of Mn = 25).
- (a) three (b) five
- (c) two (d) four.
- 134\. Which one of the following is expected to exhibit optical isomerism ?
(en = ethylene diamine)
- (a) trans- \[Co(en)2Cl2\]+ (b) trans-\[Pt(NH3)2Cl2\]
- (c) cis- \[Pt(NH3)2Cl2\] (d) cis-\[Co(en)2Cl2\]+
- 135\. The number of ions in tetraammine copper(II) hydroxide is :
- (a) four (b) seven (c) two (d) three.
- 136\. Atomic numbers of Cr and Fe are respectively 25 and 26. Which of the following is paramagnetic with the spin of electron ?
- (a) \[Cr(CO)6\] (b) \[Fe(CO)5\]
(c) \[Fe(CN)6\]4‒ (d) \[Cr(NH3)6\]3+
- 137\. Amongst the following the lowest degree of paramagnetism per mole of the compound at 298 K will be shown by :
(a) MnSO4.4H2O (b) CuSO4.5H2O
(c) FeSO4.6H2O (d) NiSO4.6H2O.
- 138\. The correct order of magnetic moments (spin only value in BM) among
is :
- (a) \[MnCl4\]2‒ > \[CoCl4\]2‒ > \[Fe(CN)6\]4‒
- (b) \[MnCl4\]2‒ > \[Fe(CN)6\]4‒ > \[CoCl4\]2‒
- (c) \[Fe(CN)6\]4‒ > \[MnCl4\]2‒ > \[CoCl4\]2‒
- (d) \[Fe(CN)6\]4‒ > \[CoCl4\]2‒ > \[MnCl4\]2‒
(Atomic Nos. : Mn = 25, Fe = 26, Co = 27)
- 139\. Which of the following is considered to be an anticancer species
_files/AU20B.20Sc.20II20Chemistry20I(OK)-12.png)CH2 ||
CH2
Cl
(c)
_files/AU20B.20Sc.20II20Chemistry20I(OK)-13.png)Cl
Cl
_files/AU20B.20Sc.20II20Chemistry20I(OK)-14.png)Cl
Pt
NH3
H3N Cl
Pt
H3N / \\ Cl
- 140\. Among the following the species having square planar geometry for the central atom are :
- (i) XeF4 (ii) SF4 (iii) \[NiCl4\]2‒ (iv) \[PdCl4\]2‒
- (a) (i) and (iv) (b) (i) and (ii)
- (c) (ii) and (iii) (d) (iii) and (iv).
- 141\. Amongst TiF62‒CoF62‒Cu2Cl2 and NiCl42‒ the colourless species are : (At. Nos. Ti = 22, Co = 27, Cu = 29, Ni = 28)
- (a) CoF63‒ and NiCl42‒ (b) TiF62‒ and CoF63‒
- (c) Cu2Cl2 and NiCl42‒ (d) TiF62‒ and Cu2Cl2.
- 142\. The example of coordination isomerism is :
- (a) \[Co(NH3)6\]\[Cr(CN)6\] and \[Cr(NH3)6\] \[Co(CN)6\]
- (b) \[Co(NH3)5Br\]SO4 and \[Co(NH3)5SO4\]Br
- (c) \[Co(NH3)5NO3\]SO4 and \[Co(NH3)5SO4\]NO3
- (d) \[Pt(NH3)4Cl2\]Br2 and \[Pt(NH3)4Br2\]Cl2
- 143\. Which compound is zerovalent metal complex ?
- (a) Ni(CO)4 (b) K3\[Fe(CN)6\]
- (c) \[Cu(NH3)4\]SO4 (d) \[Pt(NH3)2Cl2\]
- 144\. In the coordination compound, K4\[Ni(CN)4\], the oxidation state of nickel is :
- (a) +2 (b) ‒ 1
- (c) 0 (d) + 1.
- 145\. Which one of the following is an inner orbital complex as well as diamagnetic in behaviour ?
- (a) \[Cr(NH3)6\]3+ (b) \[Co(NH3)6\]3+
(c) \[Ni(NH3)6\]2+ (d) \[Zn(NH3)6\]2+
\[Atomic Numbers : Zn = 30, Cr = 24, Co = 27, Ni = 28\]
- 146\. The diamagnetic species is :
- (a) \[Ni(CN)6\]2‒ (b) \[NiCl4\]2‒
- (c) \[CoCl4\]2‒ (d) \[CoF6\]2‒.
- 147\. Among Ni(CO4), \[Ni(CN)4\]2‒, \[NiCl4\]2‒ species the hybridization states at the Ni atom are respectively :
- (a) sp3, dsp2, dsp2 (b) sp3, dsp2, sp3
- (c) sp3, sp3, dsp2 (d) dsp2, sp3, sp3
(At. No. of Ni = 28)
148\. The ligands in anticancer drug cisplatin are :
(a) NH3, Cl
(b) NH3, H2O
(c) Cl, H2O
(d) NO, Cl.
149\. The number
of possible isomers
of an octahedral
complex
\[Co(C2O4)2(NH3)2\]‒ is :
(a) 1
(b) 2
(c) 3
(d) 4.
150\. The strongest ligand in the following is :
(a) CN‒
(b) Br‒
(c)
HO‒
(d) F‒.
Answers
1\. (d)
2\. (a)
3\. (a)
4\. (d)
5\. (b)
6\. (a)
7\. (b)
8\. (b)
9\. (d)
10\. (a)
11\. (a)
12\. (b)
13\. (c)
14\. (b)
15\. (c)
16\. (a)
17\. (b)
18\. (a)
19\. (c)
20\. (c)
21\. (c)
22\. (d)
23\. (d)
24\. (a)
25\. (b)
26\. (c)
27\. (c)
28\. (b)
29\. (c)
30\. (b)
31\. (c)
32\. (b)
33\. (c)
34\. (c)
35\. (b)
36\. (a)
37\. (d)
38\. (a)
39\. (c)
40\. (c)
41\. (b)
42\. (d)
43\. (c)
44\. (b)
45\. (b)
46\. (a)
47\. (c)
48\. (c)
49\. (b)
50\. (a)
51\. (a)
52\. (d)
53\. (d)
54\. (b)
55\. (a)
56\. (b)
57\. (b)
58\. (a)
59\. (b)
60\. (c)
61\. (d)
62\. (b)
63\. (c)
64\. (d)
65\. (c)
66\. (d)
67\. (c)
68\. (c)
69\. (c)
70\. (b)
71\. (a)
72\. (a)
73\. (c)
74\. (d)
75\. (a)
76\. (b)
77\. (d)
78\. (a)
79\. (a)
80\. (b)
81\. (b)
82\. (b)
83\. (c)
84\. (d)
85\. (b)
86\. (c)
87\. (b)
88\. (d)
89\. (a)
90\. (c)
91\. (a)
92\. (c)
93\. (b)
94\. (d)
95\. (c)
96\. (d)
97\. (b)
98\. (b)
99\. (b)
100\. (a)
101\. (c)
102\. (a)
103\. (d)
104\. (b)
105\. (d)
106\. (b)
107\. (a)
108\. (a)
109\. (c)
110\. (c)
111\. (c)
112\. (c)
113\. (a)
114\. (d)
115\. (b)
116\. (c)
117\. (c)
118\. (a)
119\. (b)
120\. (a)
121\. (a)
122\. (b)
123\. (d)
124\. (c)
125\. (d)
126\. (d)
127\. (c)
128\. (b)
129\. (d)
130\. (b)
131\. (a)
132\. (b)
133\. (b)
134\. (d)
135\. (d)
136\. (d)
137\. (b)
138\. (a)
139\. (c)
140\. (a)
141\. (d)
142\. (a)
143\. (a)
144\. (a)
145\. (b)
146\. (a)
147\. (b)
148\. (a)
149\. (c)
150\. (a).
Section ‘B’
Short Answer Type Questions
Q. 1. What is effective atomic number ? Calculate the effective atomic number of the central metal atom in the following complex compounds :
- (i) \[Cr(en)3\]Cl3 (ii) K3\[Co(C2O4)2Cl2\]
- (iii) Ni(CO)4 (iv) K3\[Cu(CN)4\]
Ans. The concept of effective atomic number (EAN) was given by sidgwick. According to him in the formation of complex compound the central metal atom tends to reach its EAN i.e. atomic number of next higher insert
Complex
Central metal atom
Atomic number of central metal atom (Z)
Oxidation state of central metal atom (X)
Coordination number of entral metal atom (n1)
Number of electrons donated by ligands
EAN of central atom \[Z‒X)+n1\]
(i) \[Cr(en)3)Cl3
Cr
24
\+ 3
6
4 x 3 = 12
24 ‒ 3 +
12 = 33
(ii) K3\[Co(C2O4)2Cl2\]
Co
27
\+ 3
6
4 x 2 + 2
x 2 = 12
27 ‒ 3 +
12 = 36
(iii) Ni(CO)4
Ni
28
0
4
2 x 4 = 8
28 ‒ 0 + 8 = 36
(iv) K3\[Cu(CN)4\]
Cu
29
\+ 1
4
2 x 4 = 8
29 ‒ 1 + 8 = 36
gas.The effective atomic number (EAN) of a metal in a complex is equal to atomic number of that metal minus the number of electrons lost in ion formation plus the number of electrons gained by coordination e.g., in K4\[Fe(CN)6\] the EAN of Fe is 26 (atomic number of iron) ‒2 (electrons lost in the formation of ferrous ion) +12 (electrons gained from six CN‒ ions) = 36 which is the atomic number of the next higher inert gas krypton. It may be noted that the attainment of stable noble gas configuration is not a necessary condition for complex formation.
Q. 2. Explain why \[Co(NH3)6\]3+ is diamagnetic and octahedral ?
Ans. In \[Co(NH3)6\]3+ cobalt is present as Co3+ ion whose outer electronic configuration is :
3d
_files/AU20B.20Sc.20II20Chemistry20I(OK)-15.png)
4s
_files/AU20B.20Sc.20II20Chemistry20I(OK)-16.png)
4p
_files/AU20B.20Sc.20II20Chemistry20I(OK)-17.png)
Co3+ ion (¿5
(ground state)
In presence of strong ligand NH3 molecules, rearrangement takes place and the electrons are paired up against the Hund’s rule. Thus, the configuration of Co3+ ion in presence of NH3 ligands becomes
3d
_files/AU20B.20Sc.20II20Chemistry20I(OK)-18.png)
4s
_files/AU20B.20Sc.20II20Chemistry20I(OK)-19.png)
4p
_files/AU20B.20Sc.20II20Chemistry20I(OK)-20.png)
Co after rearrangement in presence of NH3 ligands
The six electron pairs donated by six NH3 molecules enter the hybrid orbitals formed by hybridizing two 3d, one 4s and three 4p orbitals. Thus, the configuration of \[Co(NH3)6\]3+ complex ion becomes
3d
4s
4p
\[Co(NH3)6\]3+ ion
/
/
/
x x
x x
x x
x x
x x
x x
(Six </sp hybrids having 12 electrons donated by six ligands, 6NH3 molecules)
Since there is d2sp3 hybridization and no unpaired electron, \[Co(NH3)6\]3+ complex ion should be diamagnetic and octahedral in nature.
Q. 3. Indicate ‘primary’ and ‘secondary’ valencies of central metal atom in the following complexes.
- (a) K4\[Fe(CN)6\] (b) \[Co(NH3)6\]Cl3
Ans. Primary valency is equal to oxidation state of central metal atom
while secondary valency is equal to coordination number of central metal atom. Hence in
- (a) K4\[Fe(CN)6\] Primary valency = 2
Secondary valency = 6
- (b) \[Co(NH3)6\]Cl3 Primary valency = 3
Secondary valency = 6
Q. 4. Write the IUPAC name of the following complex compounds :
(a) K2\[PtCl6\]
(b)Na3\[Ag(S2O3)2\]
- (c) K3\[Co(C2O4)2Cl2\]
(d)\[Ag(NH3)2\]Cl
(e)K4\[Fe(CN)6\]
- (f) \[Cr(H2O)4 Cl2\]NO2
(g)\[Cr(en)3\]Cl3
(h)\[Co(NH3)5Cl\]Cl2
- (i) \[Cu(NH3)4\]SO4
- (j) Na2\[CrOF4\]
(k)\[Co(NH3)6\] \[CuCl5\]
- (l) \[Cr(C6H6)2\]
NH2
- (m) \[(NH3)5Co Co(NH3)5\] (NO3)5
(n)NH4\[Cr(NH3)2(NCS)4\]
(o)K3\[Co(NO2)6\]
- (p) Fe(CO)5
- (q) LiAlH4
Ans. (a) Potassium hexachloroplatinate (IV)
- (b) Sodium bis(thiosulphato) argentate (I)
- (c) Potassium dichlorodioxalatocobaltate (III)
- (d) Diamminesilver (I) chloride
- (e) Potassium hexacyanoferrate (II)
- (f) Tetraaquodichlorochromium (III) nitrite.
- (g) Tris (ethylenediamine) chromium (III) chloride
- (h) Pentamminechlorocobalt (III) chloride
- (i) Tetramminecopper (II) sulphate
- (j) Sodium tetrafluorooxochromate (IV)
- (k) Hexamminecobalt (III) pentachloro cuprate (II)
- (l) Bis(benzene) chromium (o)
- (m) µ-amido bis(pentaammine cobalt (III)) nitrate
- (n) Ammonium diamminetetrathiocyanato-N chromate (III)
- (o) Potassium hexanitrocobaltate (III)
- (p) Pentacarbonyl iron (o)
- (q) Lithium tetrahydrido aluminate (III)
Q. 5. Write the possible isomers of the following complex compounds and explain in brief the isomerism involved :
- (i) \[Co(NH3)6\] \[Cr(CN)6\]
- (ii) \[Cr(H2O)6\]Cl3
- (iii) \[Pt(NH3)2Cl2\]
- (iv) \[Pt(NH3) (Py)Cl Br\]
- (v) \[Co(NH3)5Br\]SO4
Ans. (i) The possible isomer is \[Cr(NH3)6\] \[Co(CN)6\] and the isomerism involved is coordination isomersim because the distribution of coordination
groups in the cations and anions of the two compounds of identical composition is different.
- (ii) The possible isomers are \[Cr(H2O)5Cl\]Cl2.H2O and \[Cr(H2O)4Cl2\]Cl∙ 2H2O and the isomerism involved is hydrate isomerism because the number of water molecules in the coordination sphere is different.
- (iii) The possible isomers are \[Pt(NH3)4\]\[PtCl4\] and \[Pt(NH3)3Cl\]2 \[PtCl4\] and the isomerism involved is polymerization isomerism because they are
polymers of the first.
(iv) The possible isomers are :
_files/AU20B.20Sc.20II20Chemistry20I(OK)-21.png)
_files/AU20B.20Sc.20II20Chemistry20I(OK)-22.png)Br
Py
and the isomerism involved is geometrical isomerism because NH3 is trans to each of the three groups.
- (v) The possible isomer is \[Co(NH3)5SO4\] Br and the isomerism involved is ionization isomerism because the position of coordinating groups within and outside the coordination sphere is different.
Q. 6. Name the type of isomerism involved in the following complex compounds, giving reason :
- (i) \[Co(NH3)5Br\]SO4 and \[Co(NH3)5SO4\]Br
OH
- (ii) \[(NH3)4Co Co(NH3)2Cl2\]SO4
OH
OH
and \[Cl(NH3)3Co zCo(NH3)3Cl\]SO4
OH
- (iii) \[Pd(dipy)(SCN)2\] and \[Pd(dipy)(NCS)2\]
- (iv) \[Co(NH3)6\]\[Cr(CN)6\] and \[Cr(NH3)6\]\[Co(CN)6\]
NO2
H3N 2 NO2
(v) 3 Co 2
H3N NH3
NH3
NO2
H3N NH3
and
H3N NH3
NO2
(vi)
en
_files/AU20B.20Sc.20II20Chemistry20I(OK)-23.png)
_files/AU20B.20Sc.20II20Chemistry20I(OK)-24.png)en
(vii) \[Cr(H2O)5Cl\]Cl2.H2O and \[Cr(H2O)4Cl2\]Cl.2H2O
Ans. (i) The isomerism involved is ionization isomerism because the position of coordinating groups within and outside the coordination sphere is different.
- (ii) The isomerism involved is coordination position isomerism because the distribution of coordinating groups between two coordination centres is different.
- (iii) The isomerism involved is linkage isomerism because the mode of linking of a coordinating group to a central atom is different.
- (iv) The isomerism involved is coordination isomerism because the distribution of coordinating groups in the cations and anions of the two compounds of identical composition is different.
- (v) The isomerism involved is geometrical isomerism because the two coordinating groups differing from the others occur in edgewise and axial positions respectively.
- (vi) The isomerism involved is optical isomerism because two isomers are mirror-image forms of the compound.
- (vii) The isomerism involved is hydrate isomerism because the number of water molecules in the coordination sphere is different.
Q. 7. Write a note on chelation and its application.
Ans. Whenever a ligand occupies two or more coordination positions on the same central metal ion, a complex possessing a closed ring structure is formed. Such ligands are called polydentate ligands and the cyclic complexes
so formed are known as chelate compounds or usually chelates. Of these ligands the most common are the bidentate e.g.
(i) Ethylenediamine (en in short)
CH2 ‒ H2N :
|
CH2 ‒ H2 N :
(ii) Oxalate ion
:O—C = O |
:O C-O
The examples of complexes formed by these bidentate ligands are given below :
CH-H,N:
Cu + + 2|
CH,—H,N:
'CH—H,N |
.CH,—HjN
Cu
2+
|
NH2—CH,J
bis (ethylenediamine) copper (II) ion
O = C—O: O-C CK /O C O
Fe +3 | —► | ^Fe |
O = C—O: O = C—O'/ ^O—C = O
O O
c—c
II II
\_ o o \_
Trioxalatoferrate(III) ion
Importance of chelates. Chelating agents are used (i) as indicators in volumetric estimations of metals (ii) in softening water and estimation of hardness of water i.e. estimation of Ca2+ and Mg2+ ion by EDTA (iii) in the separation of ions by solvent extraction method (iv) in food preservation (v) in the elimination of harmful radioactive metals from the body speedily (vi) In qualitative analysis for the identification of nickel, aluminium ions etc. (vii) In gravimetric analysis for the quantitative precipitation of nickel ion etc.
Q. 8. Write the formulae of the following complex compounds :
- (a) Potassium hexacyanoferrate (III)
- (b) Ammonium trioxalatocobaltate (III)
- (c) Bis (acetyl acetanto) oxovanadium (IV)
- (d) Potassium dichlorodioxalatocobaltate (III)
- (e) Ammonium diamminetetrathiocyanato-S chromate (III)
- (f) Potassium tetracyanonickelate (o)
- (g) Triamminechlorocyanonitrocobalt (III)
- (h) Tris (ethylenediamine) chromium (III) chloride
- (i) Tetrammineplatinum (II) tetrachloroplatinate (II)
- (j) Dichlorobis (urea) copper (II)
- (k) Potassium tetracyanonickelate (II)
- (l) Sodium bis(thiosulphato) argentate (I)
- (m) Dichlorobis(ethylenediamine) cobalt (III) chloride
- (n) Tetraaquodichlorochromium (III) nitrite
- (o) µ-amido-µ-hydroxo bis \[tetraammine cobalt (III)\] bromide
- (p) µ-amido bis (pentaammine cobalt (III)) nitrate
Ans. (a) K3\[Fe(CN)6\]
- (b) (NH4)3 \[Co(C2O4)3\]
- (c) \[VO(acac)2\]
- (d) K3\[Co(C2O4)2Cl2\]
- (e) NH4\[Cr(NH3)2(SCN)4\]
- (f) K4\[Ni(CN)4\]
- (g) \[Co(NH3)3(NO2)(CN)Cl\]
- (h) \[Cr(en)3\]Cl
- (i) \[Pt(NH3)4\] \[PtCl4\]
- (j) \[CuCl 2{O = C (NH2)2}2\]
- (k) K2\[Ni(CN)4\]
- (l) Na3\[Ag(S2O3)2\]
- (m) \[Co(en)2Cl2\]Cl
- (n) \[Cr(H2O)4Cl2\]NO2
NH2
- (o) \[(NH3)4Co Co(NH3)4\]Br4
OH
nh2
- (p) «NH^gZ XCo(NH3)5\](NO3)5
Q. 9. Explain hydrate isomerism with example.
Ans. Hydrate isomerism is due to the variation in the number of water molecules in the coordination sphere. e.g., there are following three different hexahydrates of chromic chloride with an emperical formula of CrCl3.6H2O
\[Cr(H2O)6\]Cl3 Hexaaquochromium (III)
\[Cr(H2O)5Cl\]Cl2.H2O Pentaaquochloro-
chloride; chromium (III) chloride
and \[Cr(H2O)4Cl2\]Cl.2H2O
Tetraaquodichlorochromium (III) chloride.
Q. 10. Explain why \[Ni(CN)4\]2– ion is diamagnetic and square planar ?
Ans. In Ni(CN)4\]2‒ nickel is present as Ni2+ ion whose outer electronic configuration is : 3d 4s 4p
Ni2+ion (3d8) I I .
(ground state) Ik Ik Ik I I
In presence of strong ligand CN‒ ion, rearrangement takes place and the electrons are paired up against the Hund’s rule. Thus, the configuration of Ni2+ in presence of CN‒ ligand becomes
3d 4s 4p
_files/AU20B.20Sc.20II20Chemistry20I(OK)-25.png)
Ni2+after rearrangement in presence of CN~ ligands
The four electron pairs donated by four CN‒ enter the hybrid orbitals
formed by hybridizing one 3d one 4s and two 4p orbitals. Thus, the configuration of \[Ni(CN)42‒ \]complex ion becomes
4s
x x
3d
\[Ni(CN)4\]2-ion
f
/
x x
4p
x x
x x
(Four dsp2 hybrids having 8 electrons donated by four ligands, 4CN‒ ions)
Since there is sp3 hybridization and no unpaired electron, this \[Ni(CN)42‒\] complex ion is diamagnetic and tetrahedral in nature.
Q. 11. Explain why Cr3+ ions always form paramagnetic complexes.
Ans. In the electronic configuration of Cr3+ ion there are three d electrons. The distribution of these electrons in the ground state according to Hund’s rule will be as shown below :
Thus, three of the orbitals will be partially filled each having one unpaired electron. The other two orbitals will be empty. Hence Cr3+ ion will always form paramagnetic complexes.
Q. 12. Explain which one of the following has maximum magnetic moment ?
\[Co(H2O)6\]3+, \[CoF6\]3– and \[Co(CN)6\]3–
Ans. Electronic configuration of \[Co(H2O)6\]3+
3d 4s 4p
_files/AU20B.20Sc.20II20Chemistry20I(OK)-26.png)d2sp3 hybridization
Electronic configuration of \[CoF6\]3‒
3d 4s 4p 4d
11/
XX
XX
XX
XX
XX
sp3d2 hybridization
Electronic configuration of \[Co(CN)6\]3‒
_files/AU20B.20Sc.20II20Chemistry20I(OK)-27.png)d2sp3 hybridization
Since the \[CoF6\]3‒ complex ion contains 4 unpaired electrons it has maximum magnetic moment.
Q. 13. Explain ionization isomerism with example.
Ans. Ionization isomerism is due to different position of ligands within and outside the coordination sphere and as such the compounds yield different
ions in solution, e.g., \[Co(NH3)5Br\]SO4 Pentaamminebromo cobalt (III) sulphate
and \[Co(NH3)5SO4\]Br
Pentaamminesulphato-cobalt (III) bromide
The first compound gives \[Co(NH3)5Br\]2+ and SO42‒ ions while the second compound gives \[Co(NH3)5SO4\]+ and Br‒ ions.
Q. 14. Explain why K3\[Fe(CN)6\] is paramagnetic ?
Ans. In this complex compound \[Fe(CN)6\]3‒ is the complex ion. In this complex ion Fe is present as Fe3+ ion whose outer electronic configuration is :
4s
Feî+ ion (d5)
(ground state)
_files/AU20B.20Sc.20II20Chemistry20I(OK)-28.png)
4p
In presence of strong ligand CN‒ ions, rearrangement takes place and the electrons are paired up against the Hund’s rule. Thus, the configuration of Fe3+ ion in presence of CN‒ ligand becomes
4s
Fe3+after rearrangement in presence of CbT ligands
3d
_files/AU20B.20Sc.20II20Chemistry20I(OK)-29.png)
4p
The six electron pairs donated by six CN‒ ions enter the hybrid orbitals formed by hybridizing two 3d, one 4s and three 4p orbitals. Thus, the configuration of \[Fe(CN)6\]3‒ complex ion becomes
3d
Pe(CN)6\] ion
_files/AU20B.20Sc.20II20Chemistry20I(OK)-30.png)
4s 4p
x x x x x x x x
(Six d1sp hybrids having 12 electrons donated by six ligands, 6CN ions)
Since the ion \[Fe(CN)6\]3‒ has one unpaired electron, K3\[Fe(CN)6\] is paramagnetic in nature.
Q. 15. Ni(CO)4 is diamagnetic and tetrahedral. Explain why
Ans. In Ni(CO)4 nickel as present as Ni(o) whose outer electronic configuration is :
3d 4s 4p
Ni(o) (ground state)
_files/AU20B.20Sc.20II20Chemistry20I(OK)-31.png)
t
In presence of strong ligand CO molecules, rearrangement takes place
and the electrons are paired up against the Hund’s rule. Thus, the configuration
4s
4p
of Ni(o) in presence of CO ligand becomes 3d
Ni(o) after rearrangement
1
11
in presence of CO ligands
The four electron pairs donated by four CO molecules enter the hybrid orbitals formed by hybridizing one 4s and three 4p orbitals. Thus, the configuration of Ni(CO)4 becomes
3d
Ni(CO)4 t t ^ ^ 1
4s 4p
x x x x x x x x
(Four sp3 hybrids having eight electrons donated by four ligands, 4CO molecules)
Since there is sp3 hybridization and no unpaired electron, the complex Ni(CO)4 is diamagnetic and tetrahedral in nature.
Section ‘C’
Long Answer Type Questions
Q. 1. Explain the geometry of the following on the basis of valence bond theory :
- (i) \[Fe(CN)6\]3–
- (ii) \[Co(NH3)6\]3+
- (iii) \[Ni(CN)4\]2–
Ans. (i) \[Fe(CN)6\]3– : In this complex ion Fe is present as Fe3+ ion whose outer electronic configuration is :
3d 4s 4p
Fe3+ ion (d5)
(ground state)
In presence of strong ligand CN‒ ion, rearrangement takes place and the electrons are paired up against the Hund’s rule. Thus, the configuration of Fe3+ in presence of CN‒ ligand becomes
3d
4s
Fe3+after rearrangement
in presence of CbT ligands
4p
The six electron pairs donated by six CN‒ ions enter the hybrid orbitals
formed by hybridizing two 3d, one 4s and three 4p orbitals. Thus, the
configuration of \[Fe(CN) \]3‒ complex ion becomes 6 3d 4s
\[Fe(CN)6\] ion
/
x x
x x
x x
4p
x x
x x
x x
(Six cfsp3 hybrids having 12 electrons donated by six ligands, 6CN ions)
Since there is d2sp3 hybridization and one unpaired electron, the \[Fe(CN)6\]3‒ complex ion is paramagnetic and octahedral in nature.
(ii) \[Co(NH3)6\]3+ : In this complex ion cobalt is present as Co3+ ion whose outer electronic configuration is :
3d 4s 4p
Codioni/) 'll 4 4 4 4
(ground state) I I I I I I
In presence of strong ligand NH3 molecules, rearrangement takes place and the electrons are paired up against the Hund’s rule. Thus, the configuration of Co3+ ion in presence of NH3 ligands becomes
3d 4s 4p
Co3+ after rearrangement in presence of NH3 ligands
The six electron pairs donated by six NH3 molecules enter the hybrid orbitals formed by hybridizing two 3d, one 4s and three 4p orbitals. Thus, the configuration of \[Co(NH3)6\]3+ complex ion becomes
3d 4s
\[Co(NH3)6\] ion
/
f
x x
x x
4p
x x
x x
x x
x x
(Six d2sp hybrids having 12 electrons donated by six ligands, 6NH3 molecules)
Since there is d2sp3 hybridization and no unpaired electron, the
\[Co(NH3)6\]3+ complex ion is diamagnetic and octahedral in nature.
(iii) \[Ni(CN)4\]2- : In this complex ion Ni is present as Ni2+ ion whose outer electronic configuration is :
3d
4s
Ni2+ ion (3d8)
(ground state)
_files/AU20B.20Sc.20II20Chemistry20I(OK)-32.png)
/I
4p
In presence of strong ligand CN‒ ion, rearrangement takes place and the electrons are paired up against the Hund’s rule. Thus, the configuration of Ni2+ ion in presence of CN‒ ligands becomes
4s 4p
3d
_files/AU20B.20Sc.20II20Chemistry20I(OK)-33.png)
Ni2+after rearrangement in presence of CN~ ligands
The four electron pairs donated by four CN‒ ions enter the hybrid orbitals formed by hybridizing one 3d, one 4s and two 4p orbitals. Thus, the configuration of \[Ni(CN)4\]2‒ complex ion becomes
3d 4s 4p
\[Ni(CN)4\]2" ion
/
!
f
/
!
/
!
x x
x x
x x
x x
Four dsp2 hybrids having 8 electrons donated by four ligands, 4CN‒ ions
Since there is dsp2 hybridization and no unpaired electron, this \[Ni(CN)4\]2‒ complex ion is diamagnetic and square planar in nature.
Q. 2. Explain whether the following complexes are low spin or high spin :
- (i) \[Co(NH3)6\]Cl3
- (ii) K3\[Fe(CN)6\]
- (iii) \[CoF6\]3–
Ans. (i) \[Co(NH3)6\]Cl3 : In this complex compound, \[Co(NH3)6\]3+ is the complex ion. In this complex ion cobalt is present as Co3+ ion whose outer electronic configuration is :
4s
3d
Co3+ ion (d6) (ground state)
_files/AU20B.20Sc.20II20Chemistry20I(OK)-34.png)
4p
In presence of strong ligand NH3 molecules, rearrangement takes place
and the electrons are paired up against the Hund’s rule. Thus, the configuration
of Co3+ ion in presence of NH3 ligands becomes 3d
4s
CoW after rearrangement
in presence of NH3 ligands
4p
The six electron pairs donated by six NH3 molecules enter the hybrid orbitals formed by hybridizing two 3d, one 4s and three 4p orbitals. Thus, the
configuration of \[Co(NH3)6\]3+complex ion becomes
3d 4s
\[Co(NH3)6\]3+ ion
f
1
<
x x
x x
x x
4p
x x
x x
x x
(Six <tsp hybrids having 12 electrons donated by six ligands, 6NH3 molecules)
Since in this complex the electron distribution of the complexed metal ion is not similar to that found in the gaseous metal ion, it is a low spin complex.
(ii) K3\[Fe(CN)6\] : In this complex compound \[Fe(CN)6\]3“ is the complex ion. In this complex ion Fe is present as Fe3+ ion whose outer electronic configuration is :
3d
4s
_files/AU20B.20Sc.20II20Chemistry20I(OK)-35.png)
Fe3+ ion (çt) >
(ground state) |\_.
In presence of strong ligand CN‒ ions, rearrangement takes place and the electrons are paired up against the Hund’s rule. Thus, the configuration of Fe3+ ion in presence of CN‒ ligand becomes
_files/AU20B.20Sc.20II20Chemistry20I(OK)-36.png)
_files/AU20B.20Sc.20II20Chemistry20I(OK)-37.png)
_files/AU20B.20Sc.20II20Chemistry20I(OK)-38.png)
_files/AU20B.20Sc.20II20Chemistry20I(OK)-39.png)
4p
_files/AU20B.20Sc.20II20Chemistry20I(OK)-40.png)
3d
_files/AU20B.20Sc.20II20Chemistry20I(OK)-41.png)
4s
_files/AU20B.20Sc.20II20Chemistry20I(OK)-42.png)
4p
_files/AU20B.20Sc.20II20Chemistry20I(OK)-43.png)
Fe3+after
rearrangement
in presence of CK ligands
The six electron pairs donated by six CN‒ ions enter the hybrid orbitals formed by hybridizing two 3d, one 4s and three 4p orbitals. Thus, the
configuration of \[Fe(CN)6\]3‒ complex ion becomes
3d 4s
CFe(CN)6\] ion
f
/
x x
x x
x x
4p
x x
x x
x x
(Six d1sp hybrids having 12 electrons donated by six ligands, 6CN ions)
Since in this complex the electron distribution of the complexed metal ion is not similar to that found in the gaseous metal ion, it is a low spin complex.
(iii) \[CoF6\]3- : In this complex ion cobalt is present as Co3+ whose outer
configuration is : 3d
4s
4p
4d
Co3+(d6)
(ground state)
The six electron pairs donated by six F‒ ions enter the hybrid orbitals formed by hybridizing one 4s, three 4p and two 4d orbitals.
Thus the configuration of \[CoF6\]3‒ complex ion becomes
3
3d
4s
4p
4d
ion
(Six sp3d2 hybrids having 12 electrons donated by six ligands, 6F‒ ions)
Since in this complex ion the electron distribution of the complexed metal ion is similar to that found in gaseous metal ion, it is a high spin complex.
- Q. 3. Give a detailed account of Werner’s coordination theory.
Ans. The fundamental postulates of Werner’s coordination theory may be summarised as follows :
- (i) Metals possess two types of valencies :
- (a) Primary, ionizable or principal valency (which is equal to oxidaiton state of central metal atom) and
- (b) Secondary, non-ionizable or auxillary valency.
- (ii) Each metal has a fixed number of secondary valencies also termed as the coordination number of the metal cation. The coordination number varies from 2 to 9. Generally it is six. The coordination number of some metal ions are given below :
Metal ion Coordination Number
[Ag+, Cu+2](#bookmark267)
[Cu2+, Ni2+, Au3+, Pt2+, Co2+4](#bookmark268)
[Fe2+, Co3+, Pt4+, Cr3+, Fe3+6](#bookmark269)
The ligands are coordinated to the central ion through these secondary valencies and this results in the formation of a complex ion.
- (iii) The primary valencies are satisfied by negative ions which are written outside the coordination sphere whereas the secondary valencies are satisfied either by negative ions or by neutral molecules. In certain cases a given negative group may satisfy both and is written inside the coordination sphere. It may be noted that in every case fulfilment of the coordination number of the metal ion is essential. Modern studies have shown that secondary valencies may also be satisfied by positive ions.
- (iv) The primary valencies are non-directional while secondary valencies are directed in space about the central cation. Six such valencies are regarded to be arranged at the six corners of a regular octahedron with the complex former at its center whereas four such valencies are either arranged at the four corners of a regular tetrahedron with the complex former at its center or at the four corners of a square, the center of which is occupied by the complex former. This postulate predicted the existence of different types of isomerism in coordination complex. Werner after a period of 19 years became successful in resolving a strictly inorganic compound into its optically active isomers.
Applications of Werner’s theory : Representing primary valencies with solid lines and secondary valencies with dotted line, Werner gave the following structure to complex compounds like CoCl3∙6NH3; CoCl3∙5NH3∙H2O; CoCl3.5NH3 and CoCl3.4NH3.
_files/AU20B.20Sc.20II20Chemistry20I(OK)-44.png)NH
Co
Cl
NH3
NH3
NH3
NH3
NH
NH3
_files/AU20B.20Sc.20II20Chemistry20I(OK)-45.png)Co
NH3
NH3
NH3 Cl
NH3 NH3 Cl
NH3
; Co Cl
/ i V-
z 1 \\ NH3
NH3 H2O Cl
NH3 Cl
\\ I NH3
Co Cl
^ NH3
NH3 Cl
It may be noted that in writing the formula of complex compounds, the central ion and the ligands attached to it are written within square brackets. Thus, the compounds considered above are written as :
\[Co(NH3)6\]Cl3 \[Co(NH3)5H2O\]Cl3
\[Co(NH3)5Cl\]Cl2 \[Co(NH3)4Cl2\]Cl
In these complexes the entire portion within square brackets behaves as an ion and forms the inner coordination sphere while the portion outside the brackets forms the outer coordination sphere and is ionizable.
Q. 4. Discuss the various types of isomerism shown by complex compounds with suitable examples.
Ans. The various types of isomerism met within coordination compounds are discussed below :
- (i) Ionization isomerism. This isomerism is due to different position of ligands within and outside the coordination sphere and as such the compounds yield different ions in solution, e.g.,
\[Co(NH3)5Br\]SO4 and \[Co(NH3)5SO4\]Br
The first compound gives \[Co(NH3)5Br\]2+ and SO42‒ ions while the second compound gives \[Co(NH3)5SO4\]+ and Br‒ ions.
- (ii) Coordination isomerism. This isomerism arises from the difference in the distribution of the ligands in the cations and anions of the two compounds of identical composition.
\[Co(NH3)6\]\[Cr(CN)6\] and \[Cr(NH3)6\]\[Co(CN)6\]
In the first compound cobalt occurs in the cation while in the second it is in the anion. The reverse is the case with chromium.
- (iii) Polymerization isomerism. In this isomerism, the isomers have the same stoichiometric composition but whose molecular compositions are multiples of the simplest stoichiometric arrangement. e.g.,
\[Pt(NH3)2Cl2\]; \[Pt(NH3)4\]\[PtCl4\] and \[Pt(NH3)3Cl\]2\[PtCl4\]
- (I) (II) (III)
Here the second and third compounds are said to be the polymers of the first.
- (iv) Hydrate isomerism. This isomerism is due to the variation in the number of water molecules in the coordination sphere. e.g., there are following three different hexahydrates of chromic chloride with an empirical formula of CrCl3.6H2O
\[Cr(H2O)6\]Cl3, \[Cr(H2O)5Cl\]Cl2.H2O and \[Cr(H2O)4Cl2\]Cl.2H2O
- (v) Salt or linkage isomerism. This isomerism results from the different manners of linking of a ligand to a central atom e.g.,
\[Co(NH3)5(NO2)\]Cl2 and \[Co(NH3)5(ONO)\]Cl2
- (vi) Coordination position isomerism. In this isomerism the distribution of ligands between two coordination centres is different, e.g.
OH
C1(NH3)3 Co,
:Co(NH3)3C1
Cl
and
OH
_files/AU20B.20Sc.20II20Chemistry20I(OK)-46.png)Co(NH3)2Cl2
Cl
- (vii) Stereoisomerism. In this isomerism, the two substances of the same composition and constitution differ only in the relative position of ligands in space.
- (a) Stereoisomerism in Octahedral Complexes
- (i) Geometrical isomerism. Geometrical isomerism is not possible for a complex of the type \[Ma6\] and \[Ma5b\]. However, for complex of the type \[Ma4b2\] and \[Ma4bc\] two isomers—cis and trans—are possible. In cis isomer the two coordinating groups differing from the others occur in the edgewise positions while in the trans isomer they occur in the axial positions as shown in the figure. Thus, there are following two forms of tetraamminedinitrocobalt (III) nitrate \[Co(NH3)4(NO2)2\]NO3.
_files/AU20B.20Sc.20II20Chemistry20I(OK)-47.png)
_files/AU20B.20Sc.20II20Chemistry20I(OK)-48.png)It should also be noted that following two isomers-fac and mer are possible
for a complex of the general \[Ma3b3\]. In fac-isomer the three a’s occupy 1, 2 and 3 positions which in mer-isomer the three a’s occupy 1, 2 and 6 positions.
_files/AU20B.20Sc.20II20Chemistry20I(OK)-49.png)Cis Trans
- (ii) Optical isomerism. Optical isomerism is shown by octahedral complexes of the type \[M(AA)3\] where AA represents a symmetrical bidentate ligand such as oxalate ion, C2O42‒ and ethylenediamine, H2N.CH2.CH2.NH2 (en in short). The figure given below represents the configuration for the ion, \[Cr(C2O4)3\]3‒.
_files/AU20B.20Sc.20II20Chemistry20I(OK)-50.png)
_files/AU20B.20Sc.20II20Chemistry20I(OK)-51.png)
mirror plane
The above two isomers are mirror-image forms of the compound. Optical isomerism also appears in complexes of the type \[Ma2b2c2\], \[Mabcdef\], \[M(AA)2a2\], \[M(AA)a2b2\], \[M(AA)2ab\], \[M(AA)2(BB)\], \[M(AB)3\] and \[M(AA)2(AB)\] where AA and BB are two different symmetrical bidentate ligands and AB is unsymmetrical bidentate ligand.
- (b) Stereoisomerism in Square planar Complexes
Square planar complexes exhibit geometrical isomerism. e.g., for a square planar complex of the type \[Ma2b2\] or \[Ma2bc\] following two isomers—cis and trans are possible. e.g., diamminedichloroplatinum (II), \[Pt(NH3)2Cl2\].
_files/AU20B.20Sc.20II20Chemistry20I(OK)-52.png)
_files/AU20B.20Sc.20II20Chemistry20I(OK)-53.png)
(when similar groups are in adjacent positions)
(when similar groups are on opposite sides)
_files/AU20B.20Sc.20II20Chemistry20I(OK)-54.png)
_files/AU20B.20Sc.20II20Chemistry20I(OK)-55.png)
_files/AU20B.20Sc.20II20Chemistry20I(OK)-56.png)
However three isomers are possible for a square planar complex of the type \[Mabcd\] corresponding to a trans to each of the three groups.
_files/AU20B.20Sc.20II20Chemistry20I(OK)-57.png)
It may be noted that optical isomerism rarely occurs in square planar complexes.
- (c) Stereoisomerism in Tetrahedral Complexes
Optical isomerism is expected in tetrahedral complexes of the type \[Mabcd\], e.g., the following are the mirror-image isomers of As3+ ion complex, \[AsIII (CH3) (C2H5)(S) (C6H4COO)\]2+.
_files/AU20B.20Sc.20II20Chemistry20I(OK)-58.png)
It may be noted that geometrical isomerism is not possible with this configuration.
Q. 5. Discuss valence bond theory of metal-ligand bonding with suitable examples. What are the limitation of this theory ?
Ans. Valence bond theory of metal-ligand bonding is mainly due to Pauling. The main assumptions of this theory are as follows :
- 1\. The central metal ion makes available a number of empty s, p and d atomic orbitals (equal in number to its coordination number) for the formation of coordinate bonds with suitable ligands.
- 2\. These vacant orbitals hyridize to give new orbitals called hybrid orbital equal in number to the mixing orbitals and having same shape and energy.
Some of the common type of hybridization and the geometry of the complexes formed by them are given in the table.
- 3\. The d orbitals involved in the hybridization may be inner viz. (n ‒ 1) d orbital or the outer viz. nd orbital. The complexes formed in these two ways are referred to as inner orbital or outer orbital complexes respectively.
- 4\. Each ligand has atleast one sigma orbital containing a lone pair of electrons.
- 5\. The non-bonding metal electrons present in inner orbitals don’t participate in the hybridization.
- 6\. The vacant hybrid orbitals of metal ion overlap with the filled sigma orbital of ligand forming ligand→metal coordinate bond. The number of such bonds depends upon the number of vacant orbitals made available by the central ion.
Coordination Hybridized
Number Orbitals
Molecular Geometry
Examples
2
sp
Linear
\[Ag(NH3)2\]+, \[Ag(CN)2\]‒
3
sp
Trigonal planar \[HgI3\]‒
4
sp
Tetrahedral
Ni(CO)4, \[Zn(NH3)4\]2+
4
dsp2
\[The d orbital involved is dx2–y2 of the inner i.e., (n ‒ 1)th shell\]
Square planar
\[Ni(CN)4\]2‒, \[Pt(NH3)4\]2+
5
dsp3
\[The d orbital involved is dx2 – y2 of the inner i.e.,
Square pyramidal
Fe(CO)5 , \[SbCl5\]2‒
5
(n ‒ 1)th shell\] sp3d
\[The d orbital involved is
dz2 of the outer i.e., the
Trigonal bipyramidal
\[CuCl5\]3‒
6
valency shell\]
d2sp3 of sp3d2 \[The d orbitals involved are dz2 and dx2 – y2 of inner i.e., (n ‒ 1)th shell in the first case and of the outer shell in the second case\]
Octahedral
\[Co(NH3)6\]3+, \[PtCl6\]2‒, \[CoF6\]3‒. It may be noted that in the case of \[Co(NH3)6\]3+ d orbitals of a lower energy level are involved.
In the case of \[CoF6\]3‒ however, d orbitals used are of the same principal energy level as the s- and p-orbitals.
- 7\. If the complex contains unpaired electrons, it is paramagnetic in nature otherwise diamagnetic.
- 8\. Under the influence of a strong ligand, the electrons can be forced to pair up against the Hund’s rule of maximum multiplicity.
\[Note : Complexes in which some of the unpaired electrons of the gaseous metal ion have been forced to pair are called low spin complexes. In high spin complexes the electron distribution of the complexed metal ion is similar to that found in the gaseous ion.\]
Now let us take a few examples to illustrate the following points.
- (i) Example of square planar complexes. The formation of square planar complexes by VBT can be explained by considering \[Ni(CN)4\]2‒ complex ion. In this complex ion Ni is present as Ni2+ ion whose outer electronic configuration is :
4s
4p
3d
Ni2+ ion (3/) (ground state)
_files/AU20B.20Sc.20II20Chemistry20I(OK)-59.png)
In presence of strong ligand CN‒ ion, rearrangement takes place and the electrons are paired up against the Hund’s rule. Thus, the configuration of Ni2+ ion in presence of CN‒ ligands becomes
3d 4s 4p
Ni2+ after
rearrangement
_files/AU20B.20Sc.20II20Chemistry20I(OK)-60.png)
in presence of CN~ ligands
The four electron pairs donated by four CN‒ ions enter the hybrid orbitals
formed by hybridizing one 3d, one 4s and two 4p orbitals. Thus, the configuration of \[Ni(CN)4\]2‒ complex ion becomes
3d
\[Ni(CN)4\]^ion
/
f
t
f
f
x x
4s
x x
4p
x x
x x
Four dsp2 hybrids having 8 electrons donated by four ligands, 4CN‒ ions
Since there is dsp2 hybridization and no unpaired electron, this \[Ni(CN)4\]2‒ complex ion should be diamagnetic and square planar in nature. These observations have in fact been proved by experimental evidence.
(ii) Example of tetrahedral complexes. The formation of tetrahedral complexes by VBT can be explained by considering the formation of \[ZnCl4\]2‒ complex ion. The configuration of Zn2+ ion and that of \[ZnCl4\]2‒ ion is :
_files/AU20B.20Sc.20II20Chemistry20I(OK)-61.png)
(Four sp3 hybrids having 8 electrons donated by four ligands, 4Cl‒ ions)
Since there is sp3 hybridization and no unpaired electron, this \[ZnCl4\]2‒ ion should be diamagnetic and tetrahedral in nature.
(iii) Example of octahedral complexes. Octahedral complexes are the most common and have been studied most extensively. As an example we may consider the formation of \[Co(NH3)6\]3+ complex ion on the basis of VBT. In this complex ion cobalt is present as Co3+ ion whose outer electronic
configuration is :
3d
Co3+ ion (¿f6) (ground state)
_files/AU20B.20Sc.20II20Chemistry20I(OK)-62.png)
_files/AU20B.20Sc.20II20Chemistry20I(OK)-63.png)
4s
4p
In presence of strong ligand NH3 molecules, rearrangement takes place and the electrons are paired up against the Hund’s rule. Thus, the configuration of Co3+ ion in presence of NH3 ligands becomes
4s
3d
Co3\* after rearrangement
_files/AU20B.20Sc.20II20Chemistry20I(OK)-64.png)
in presence of NH3 ligands
4p
The six electron pairs donated by six NH3 molecules enter the hybrid orbitals formed by hybridizing two 3d, one 4s and three 4p orbitals. Thus, the configuration of \[Co(NH3)6\]3+ complex ion becomes 3d 4s
\[Co(NH3)6\] ion
1
/
7
f
x x
x x
4p
x x
x x
x x
x x
(Six d2sp3 hybrids having 12 electrons donated by six ligands, 6NH3 molecules)
Since there is d2sp3 hybridization and no unpaired electron, the \[Co(NH3)6\]3+ complex ion should be diamagnetic and octahedral in nature. These observations have in fact been proved by experimental evidence.
Limitations of Valence Bond Theory :
- 1\. This theory could not explain the magnetic and spectral properties of complexes. This theory shows the number of unpaired electrons. From this the magnetic moment can be calculated. However, it does not explain why the magnetic moments change with temperature.
- 2\. VBT does not explain the behaviour of complexes having d8 central ion (e.g. Ni2+, Pb2+ etc.) in forming the expected 5-coordinated complexes. Moreover, it is not clear why this theory prefers only square planar geometry of complexes to other possible geometrics such as tetrahedral and trigonal bipyramidal with coordination number 5.
- 3\. VBT cannot explain why square planar complexes of Cu2+ ion (d9 system) like \[Cu(NH3)4\]2+ are not reducing agents like inner orbital octahedral complexes of Co2+ ion (d7 system) although in both the cases promotion of a non-bonding d electron to some higher energy level is required.
- 4\. In this theory, too much stress has been laid on the central ion while the importance of ligand is not properly stressed.
- 5\. VBT cannot explain reaction rates and mechanism of reactions.
Unit IV
Section ‘A’
Objective Type Questions
- 1. Rare earth elements usually :
- (a) form colourless ions (b) have multiple valencies
- (c) are inert (d) are diamagnetic
- (e) most reactive of the elements.
- 2. Which lanthanide is most commonly used ?
- (a) Lanthanum (b) Nobelium
- (c) Thorium (d) Cerium.
- 3. The basic character of hydroxides of lanthanide elements :
- (a) increase from La to Lu
- (b) decreases from La to Lu
- (c) shows no regular trend
- (d) increases from La to Gd and then decreases from Gd to Lu.
- 4. Lanthanide for which + II and + III oxidation states are common is :
- (a) La (b) Nd
- (c) Ce (d) Eu.
- 5. Which elements are chemically most similar ?
- (a) Na, Al (b) Cu, S
- (c) Zr, Hf (d) Ti, Zr.
- 6. Of the following an example of an actinide element is :
- (a) selenium (b) osmium
- (c) curium (d) cerium
- (e) astatine
- 7. The lanthanide contraction is responsible for the fact that :
- (a) Zr and Y have about the same radius
- (b) Zr and Nb have similar oxidation state
- (c) Zr and Hf have about the same radius
- (d) Zr and Zn have the same oxidation state.
- 8. The degree of complex ion formation in actinides decreases in the order :
- (a) M4+ > M3+ > MO22+ > MO2+
- (b) M4+ > MO2+ > M3+ > MO22+
- (c) MO22+ > MO2+ > M4+ > M3+
- (d) M4+ > MO22+ > M3+> MO2+
- 9. The lanthanides have electronic configuration of 6s2 in common but with variable occupation of :
- (a) 6p level (b) 5p level
- (c) 4p level (d) 5d level
- (e) 4f level.
- 10. The chemist who has helped in the discovery of the maximum number of transuranic elements is :
- (a) Sir Robert Thomson
- (b) Sir J. J. Robinson
- (c) Prof. G. T. Seaborg
- (d) Sir C. N. Hinshelwood.
- 11. Cerium (Z = 58) is an important member of the lanthanides. Which of the following statements about cerium is incorrect ?
- (a) The common oxidation states of cerium are + 3 and + 4
- (b) The + 3 oxidation state of cerium is more stable than the + 4 oxidation state
- (c) The + 4 oxidation state of cerium is not known in solutions
- (d) Cerium (IV) acts as an oxidizing agent.
- 12. Which group of elements is analogous to the lanthanides ?
- (a) Halides (b) Actinides
- (c) Carbides (d) Borides.
- 13. Actinide elements have vacant :
- (a) s orbitals (b) p orbitals
- (c) d orbitals (d) f orbitals
- 14. The name of the element with atomic number 100 was adopted in honour of :
- (a) Alfred Nobel (b) Enrico Fermi
- (c) Dimitri Mendeleev (d) Albert Einstein.
- 15. Which of the following is a man-made element ?
- (a) Np (b) C-14
- (c) Ra (d) U
- 16. The complexing power of the following mononegative ions is in the order : NO3‒ ClO3‒ F‒ Cl‒
- (a) NO3‒ > F‒ > Cl‒ > ClO3‒ (b) ClO3‒ > F‒ > NO3‒ > Cl‒
- (c) F‒ > NO3‒ > Cl‒ > ClO3‒ (d) Cl‒ > F‒ > ClO3‒ > NO3‒
- 17. Which of the following forms a stable + 4 oxidation state ?
- (a) Lanthanum (b) Cerium
- (c) Europium (d) Gadolinium
- 18. Which of the following is not an actinide ?
- (a) Thorium (b) Protactinium
- (c) Californium (d) Cerium
- 19. Transuranium elements are those which are :
- (a) heavier than uranium
- (b) lighter than uranium
- (c) of lower atomic number than uranium
- (d) of same atomic number as uranium.
- 20\. The radius La3+ (atomic number of La = 57) is 1∙06 Å. Which one of the following given values will be closest to the radius of La3+ (Atomic Number of La = 71) ?
- (a) 0∙85 Å (b) 1∙60Å
- (c) 1∙40 Å (d) 1∙06 Å.
- 21. The lanthanide contraction refers to :
- (a) ionic radius of that series
- (b) valence electrons of the series
- (c) the density of the series
- (d) nuclear mass of that series
- (e) the electronegativity.
- 22. Which of the following factors may be regarded as the main cause of lanthanide contraciton ?
- (a) Greater shielding of 5d electrons by 4f electrons
- (b) Poorer shielding of 5d electrons by 4f electrons
- (c) Effective shielding of one of 4f electron by another in the sub-shell (d) Poor shielding of one of 4f electron by another in the sub-shell.
- 23. The main reason for larger number of oxidation states exhibited by the actinides than the corresponding lanthanides is :
- (a) more energy difference between 5f and 6d orbitals than between 4f and 5d orbitals
- (b) lesser energy difference between 5f and 6d orbitals than between 4f and 5d orbitals
- (c) larger atomic size of actinides than the lanthanides
- (d) greater reactive nature of the actinides than the lanthanides.
- 24\. In lanthanides the effective magnetic moment of most of the M3+ions is given by the equation :
- (a) μ = g J BM (b) μ = gVJ(J + 1) BM
- (c) μ = g a/ J(J + 2) BM (d) μ = gVJ(J + 3) BM .
- 25. Which of the following ions will be colourless ?
- (a) La3+ (b) Nd3+
- (c) Pr3+ (d) Ti3+
- 26. Lanthanides are :
- (a) 14 elements in the sixth period (atomic no. = 90 to 103) that are filling
4f sub-level
- (b) 14 elements in the seventh period (atomic no. = 90 to 103) that are filling 5f sub-level
- (c) 14 elements in the sixth period (atomic no. = 58 to 71) that are filling 4f sub-level
- (d) 14 elements in the seventh period (atomic no. = 58 to 71) that are filling
4f sub-level.
- 27. Lanthanum is grouped with f-block elements because :
- (a) it has partially filled f orbitals.
- (b) it is just before Ce in the periodic table
- (c) it has both partially filled f and d orbitals
- (d) the properties of lanthanum are very similar to the elements of 4f-block.
- 28. Which of the following statements concerning lanthanide elements is false ?
- (a) Lanthanides are separated from one another by ion exchange methods
- (b) The ionic radii of trivalent lanthanides steadily increases with increase in the atomic number
- (c) All lanthanides are highly dense metals
- (d) More characteristic oxidation state of lanthanide elements is +3.
- 29. The separation of lanthanides by the ion exchange method is based on :
- (a) the solubility of their nitrates
- (b) size of the hydrated ions
- (c) basicity of the hydroxides
- (d) size of the unhydrated ions.
- 30. Typical oxidation state of all lanthanides is :
- (a) + 2 (b) + 3
- (c) + 4 (d) + 2, + 4
- 31. The correct order of ionic radii of Y3+, La3+, Eu3+ and Lu3+ is :
- (a) La3+ < Eu3+ < Lu3+ < Y3+ (b) Y3+ < La3+ < Eu3+ < Lu3+
- (c) Y3+ < Lu3+ < Eu3+ < La3+ (d) Lu3+ < Eu3+ < La3+ < Y3+
(Atomic Nos. Y = 39, La = 57, Eu = 63, Lu = 71)
- 32. Which of the following statement is not correct ?
- (a) Ca(OH)2 is less basic than Ln(OH)3
- (b) In lanthanide series ionic radius of Ln3+ ions decreases
- (c) La is actually an element of transition series rather lanthanides
- (d) Atomic radii of Zr and Hf are same because of lanthanide contraction
- 33. The lanthanide contraction is related with :
- (a) sharp decrease in atomic size in lanthanide series
- (b) slow or gradual decrease in atomic size in lanthanide series
- (c) constancy in atomic size
- (d) all the above.
- 34. Of the following the elements which is not a lanthanide is :
- (a) holmium (b) hafnium
- (c) cerium (d) lutecium
- (e) europium.
- 35. A reduction in atomic size with increase in atomic number is a characteristic of elements of :
- (a) radioactive series (b) high atomic masses
- (c) d-block (d) f- block.
Answers
1\. (b)
2\. (d)
3\. (b)
4\. (d)
5\. (c)
6\. (c)
7\. (c)
8\. (d)
9\. (e)
10\. (c)
11\. (c)
12\. (b)
13\. (d)
14\. (b)
15\. (a)
16\. (c)
17\. (b)
18\. (d)
19\. (a)
20\. (a)
21\. (a)
22\. (d)
23\. (b)
24\. (b)
25\. (a)
26\. (c)
27\. (d)
28\. (b)
29\. (b)
30\. (b)
31\. (c)
32\. (a)
33\. (b)
34\. (b)
35\. (d)
86
Vikas, 2009 (A. U.)
Section ‘B’
Short Answer Type Questions
Q. 1. Write a note on lanthanide contraction.
Ans. In lanthanide series there is a steady decrease in the size of the atoms and ions with increasing atomic number. This steady decrease in the atomic and ionic radii is called lanthanide contraction. The cause of this lanthanide contraction is the imperfect shielding of one electron by another in the same sub-shell. As one proceeds through La to Lu the nuclear charge and the number of 4f electrons increase by one at each step. The mutual shielding of 4f electrons is very imperfect so that at each increase the effective nuclear charge experienced by each 4f electron increases. This causes a reduction in the size of the entire 4 fn sub-shell. The accumulation of these successive contractions is the total lanthanide contraction.
Q. 2. Discuss in brief the magnetic and spectroscopic properties of actinides.
Ans. The magnetic properties of the actinide ions are considerably harder to interpret than those of the lanthanide ions, although there are similarities. The experimental magnetic moments are usually lower than the values calculated by using Russell-Saunders coupling and this appears to be due to (i) ligand field effects similar to those operating in d-transition series and (ii) inadequacy of this coupling scheme. Since 5f sub-shells can participate to some extent in covalent bonding, ligand effects are to be expected.
The absorption spectra of actinide ions in aqueous solution and in crystals contain like those of the lanthanides, narrow bands in the visible, near ultraviolet and near infra red regions of the spectrum. There is much evidence to show that these bands arise from the electronic transitions between energy states within the 5f electrons sub-shell. In general the absorption bands of the actinide series are about 10 times more intense than that of the lanthanide ions.
Q. 3. Discuss the various oxidation states of lanthanides.
Ans. All the lanthanides form ions in oxidation state +3. Certain lanthanides show +2 or +4 states but these are always less stable than the +3 state. Sc, Y and La form only M3+ ions because the removal of three electrons gives the stable 4f 14and 4f 7 configurations.
The most stable M2+ and M4+ ions are formed by those elements which by doing so attain f 0, f 7 or f 14 configurations, e.g., Ce4+ and Tb4+ attain f 0 and f7 configurations respectively whereas Eu2+ and Yb2+ acquire f 7 and f 14 configurations respectively. However, this argument does not hold good when we note that Sm and Tm also form M2+ ions having f 6 and f 13 configurations
but no M+ ion whereas Pr and Nd give M4+ ions with configurations f 1 and f2 but no M5+ or M6+ ions.
Q. 4. Discuss complex ion formation in actinides.
Ans. The electronic configuration of an ion is important in determining the tendency of an actinide ion towards complex ion formation. Since the electronic configuration of actinides is so nearly the same, tendency of complex ion formation of actinide ions is largely determined by ionic size and charge. Hence the maximum tendency of complex ion formation occurs with small highly charged M4+ ions e.g., Pu4+ forms such strong anion complexes that the uncomplexed ion is the exception. The degree of complex ion formation decreases in the following order :
M4+ > MO22+ > M3+ > MO2+
The relatively high tendency of MO22+ ions towards complex ion formation is due to high concentration of positive charge on the metal atom.
It may be noted that binegative ions form stronger complexes than do the mononegative ions. The complexing power of different binegative ions with the actinides is in the following order :
CO32‒ > C2O42‒ > SO42‒
and of mononegative ions is in the following order :
F‒ > NO3‒ > Cl‒ > ClO3‒
The actinide ions also form complex ions with a large number of organic susbtances which render them soluble in these substances. Of special interest are the chelate compounds especially those formed with TTA (Thenoyl Trifluoro Acetate) because they have been proved very effective in a number of separtion procedures.
Section ‘C’
Long Answer Type Questions
Q. 1. Discuss the various methods that have been used for the separation of lanthanides.
Ans. Since lanthanide ions are typically trivalent (M3+) and are of about the same size, their properties are almost identical. This makes the separation of lanthanides exceedingly difficult. Recently methods like fractional crystallization, ion exchange etc. have been used.
- 1\. Fractional crystallization. This method is based on slight differences in the solubility of various simple salts like nitrates, sulphates, bromates, oxalates etc. as well as double salts like 2Ln(NO3)3.3Mg(NO3)2.24H2O, Ln(NO3)3.2NH4NO3.4H2O etc. It may be noted that the solubility of simple and double salts decreases from La to Lu.
- 2\. Precipitation. The basic character of the hydroxides of lanthanide elements decreases from La to Lu. Hence if OH‒ is added to a solution containing nitrates of lanthanide elements, Lu(OH)3 which is the weakest base is precipitated first while the strongest base La(OH)3 is precipitated last.
- 3\. Fractional thermal decomposition of oxy-salts. This method is based on the fact that the rate of thermal decomposition of oxy-salts such as nitrate, sulphates and acetates of lanthanide elements at a given temperature decreases from La to Lu. Thus, if a mixture containing the nitrates of lanthanide elements is fused and then leached with water, ytterbium group lanthanides (Gd64 to Lu71) which are less basic are rapidly concentrated and their separation from the more basic cerium group lanthanides (La57 to Eu63) is effected.
- 4\. Change of oxidation states. Some lanthanides also exhibit +2 and +4 oxidation states in addition to +3 state which is most characteristic of all the elements of the family. Moreover, the properties of M2+ and M4+ ions are different from those of the usual M3+ ions. This forms the basis of separating specific lanthanides e.g., cerium can be separated from other lanthanides by its oxidation with alkaline KMnO4 to Ce4+ which is smaller and hence less basic and is precipitated as Ce(OH)4 by the addition of a small amount of alkali leaving all other M3+ ions in solution.
- 5\. Solvent extraction. Ce4+ can be readily extracted from nitric acid solutions by tributyl phosphate dissolved in kerosene or other inert solvent and can thus be readily separated from +3 lanthanide ions.
- 6\. Complex formation. The oxalates of lanthanides are insoluble in water but their complexes with EDTA (ethylenediamine tetra-acetic acid) are soluble. These complexes differ in their stability. Hence on adding an acid, the least stable complex is decomposed first and parent oxalate is precipitated.
- 7\. Ion exchange method. This is the most effective method now available for the separation of lanthanides. This method utilizes a column of synthetic ion exchange resin with acid functional groups such as —COOH or —SO3H. An aqueous solution containing a mixture of tripositive lanthanide ions is allowed to pass down the column when lanthanide ions replace H+ ions of COOH or —SO3H group of the resin and thus get fixed on the resin.
M3+ + 3H (resin) → M ‒ (resin)3 + 3H+
Since Lu3+ (aq) is largest in size and La3+ (aq) is smallest, La3+ (aq) is bound most tightly while Lu3+ (aq) least. The H+ ions flow down the column and move out through the tap below. In order to remove M3+ ions fixed as M-(resin)3 on the resin, the column is eluted with a complexing agent in aqueous solution like a buffer solution of citric acid—ammonium citrate. During elution process NH4+ ions of the eluting agent replace M3+ ions from M-(resin)3 to give M3+ ions which react with citrate ion to form M-citrate complex.
M-(resin)3 + 3NH4+ → 3NH4 - resin + M3+ M3+ + citrate ion → M-citrate complex
Since La3+ (aq) is bound to the resin most tightly while Lu3+ (aq) least, Lu-citrate complex comes out of the column first while La-citrate complex comes out last.
Q. 2. Discuss the chemistry of the separation of Np, Pu and Am from U.
Ans. The methods used for the separation of Np, Pu and Am from U are based on the following principles.
- 1\. Stabilities of oxidations states : The stabilities of the major ions involved are UO22+ > NpO22+ > PuO22+ > AmO22+ ;
Am3+ > Pu3+ > Np3+ > U4+.
Hence it is possible by choice of suitable oxidizing or reducing agents to obtain a solution containing the elements in different oxidation states. Then they can be separated by solvent extraction or precipitation. e.g. Pu can be oxidized to PuO22+ while Am remains as Am3+. The former can then be removed by solvent extraction or the latter by precipitation of AmF3.
- 2\. Precipitation reactions : It may be noted that only M3+ and M4+ give insoluble phosphates or fluorides from acid solutions. The higher oxidation states give either no precipitate or can be presented from precipitation by complex formation with sulphate or other ions.
- 3\. Extractability into organic solvents : The MO22+ ions are extracted from nitrate solutions into ethers. The M4+ ions are extracted into tributyl phosphate in kerosene from 6M nitric acid solution. Similarly M3+ ions are extracted from 10 to 16 M nitric acid and neighbouring actinides can be separated by a choice of conditions.
- 4\. Ion exchange methods : These methods have been indispersible in the separation of the elements following Am (often called the trans americium elements) and also for trace quantities of Np, Pu and Am.
To illustrate briefly the practical employment of these principles on a large scale, the following processes may be mentioned.
- 1\. The hexone \[MeC(O)CH2CHMe2\] method in which both U and Pu are oxidized to AnO22+ ions which are then extracted into an organic solvent by the hexone, leaving the fission products in the aqueous layer. Uranium is then separated from Pu by selectively (SO2) reducing the latter to Pu4+ and solvent extracting only the UO22+
- 2\. The tributyl phosphate extraction method where control of the concentration of nitric acid from 6 to 16 M allows the control of the transfer of various actinide ions into kerosene containing 30% (C4H9O)3PO.
3\. The Lanthanium fluoride cycle in which hexone extraction along with LaF3 as a carrier precipitate are used.
Q. 3. Compare the oxidation states of actinides and lanthanides.
Ans. The known oxidation states of actinides and lanthanides are summerized in the below table in such a way to bring out the analogy between the two group of elements.
La
Ce
Pr
Nd
Pm
Sm
Eu
Gd
Tb Dy
3
3
3
3
3
2
2
3
33
4
(4)
3
3
(4)
Ac
Th
Pa
U
Np
Pu
Am
Cm
Bk Cf
3
(3)
(3)
3
3
3
3
3
33
4
4
4
4
4
(4)
(4)
4
5
5
5
5
5
6
6
6
6
7
7
The unstable oxidation states are designated with parenthesis. None of these unstable states can exist in aqueous solutions but have been produced only in solid compounds although Pa(III) may be an exception. The (IV) state of curium has been observed only in solid CmO2 and CmF4 showing the extremely strong tendency of curium to retain the III state with its stable halffilled 5f sub-shell.
Although the dominant oxidation state of both the actinide and lanthanide elements is the III state, this oxidation state does not become important in the actinide series until uranium is reached. All the actinides elements upto americium display several oxidation states in aqueous solutions and there is no similar example among the lanthanides. This is due to the nearness of the energies of 7s, 6d and 5f sub-shells in this region. Thorium does not display a III state in aqueous solutions while the evidence for a III state in protactinium is rare. It may be noted that with the increase in atomic number the higher oxidation states become increasingly less stable. In americium the III state is so firmly established that it is the only state in acidic aqueous solutions. In curium the III state is only one that can exist in aqueous solution.
In berkelium both the III and IV states are stable in aqueous solutions indicating that berkelium has eight 5f electrons, one more than the stable structure of half filled 5f sub-shell.
Q. 4. What are actinides ? Discuss their electronic configuration.
Ans. Actinium (Z = 89) and the succeeding elements upto lawrencium (Z = 103) are called actinides or actinons and constitute the second inner transtion series. Just as in lanthanides the antepenultimate 4f sub-shell is filled successively by the addition of one electron at a time similarly in actinides the antepenultimate 5f sub-shell is filled successively by the addition of one electron at each step. Just as lanthanum is prototype of the lanthanides, actinium is the prototype of the actinides.
The electronic configuration of the neutral gaseous atoms of these elements is give in the below table.
Element
Symbol
Atomic Number Electronic Configuration
Actinium
Ac
89
\[Rn\] 6d1 7s2
Thorium
Th
90
\[Rn\] 6d2 7s2
Protactinium
Pa
91
\[Rn\] 5f2 6d1 7s2or
5f1 6d2 7s2
Uranium
U
92
\[Rn\] 5f3 6d17s2
Neptunium
Np
93
\[Rn\] 5f5 7s2
Plutonium
Pu
94
\[Rn\] 5f6 7s2
Americium
Am
95
\[Rn\] 5f7 7s2
Curium
Cm
96
\[Rn\] 5f7 6d17s2
Berkelium
Bk
97
\[Rn\] 5f8 6d17s2or 5f 9 7s2
Californium
Cf
98
\[Rn\] 5f10 7s2
Einsteinium
Es
99
\[Rn\] 5f11 7s2
Fermium
Fm
100
\[Rn\] 5f12 7s2
Mendelevium
Md
101
\[Rn\] 5f13 7s2
Nobelium
No
102
\[Rn\] 5f14 7s2
Lawrencium
Lr
103
\[Rn\] 5f14 6d17s2
From the above table it is clear that there is still uncertanity in allocation in a few cases. In the actinide elements the fourteen 5f electrons are added formally, though not necessarily actually from thorium (Z = 90) onwards and the 5f sub-shell is complete at element 103.
Since f sub-shells are being filled, the actinides have a close relation with lanthanides. Thus Ac and La occupy corresponding positions as prototypes for the two series of elements. Although important similarities do exist between the actinides and lanthanides, very important differences also occur. The difference arises mainly from the relatively lower binding energies and less effective shielding by the outer of the 5f as compared to the 4f electrons. It
appears from spectroscopic, chemical and other data that the 5f sub-shell becomes progresively lower in energy compared to 6d with increasing atomic number. It may be noted that 5f electrons are not present in thorium or probably in protactinium but for uranium and succeeding elements 5f electrons are present.
In actinides the energies of 5f, 6d, 7s and 7p sub-shell are comarable over a range of atomic numbers (especially U-Am). Hence the energies involved in an electron shifting from 5f to 6d lie within the range of chemical binding energies. Thus, the electronic configuration of the element in a given oxidation state may vary between compounds and in solution be dependent on the nature of the ligands. It is also often impossible to say which sub-shells, are being utilized in bonding or to decide meaningfully whether the bonding is covalent or ionic.
Q. 5. What are lanthanides and why are they also known as ‘rare earths’. Discuss the following properties of these elements:
- (a) Electronic configuration
- (b) Colour
- (c) Atomic and Ionic radii
- (d) Magnetic and Spectral properties.
Ans. The elements with electronic configuration (n ‒ 2) f 0‒14 (n ‒ 1)d0‒2 ns2 are known as inner transition elements or f-block elements. They are further classified as 4f and 5f block elements corresponding to the filling of 4f and 5f sub-shells of the antepenultimate shell. 4f block elements (Z = 58 to 71) along with lanthanium (Z = 57) are known as lanthanides or lanthanons. They were originally called ‘rare earths’. The word ‘earth’ was used because they occurred as oxides (or in old usage, earth) and the word ‘rare’ was used because their occurrence was believed to be very scarce.
- (a) Electronic configuration. In lanthanides there is filling of f-sub-shell of the antepenultimate shell as is clear from the electronic configuration of these elements.
Element Symbol Atomic Electronic Configuration
Number
Lanthanium
La
57
2, 8, 18
4s2p6d10
5s2p6d1
6s2
Cerium
Ce
58
2, 8, 18
4s2p6d10 f1
5s2p6d1
6s2
Praseodymium
Pr
59
2, 8, 18
4s2p6d10 f3
5s2p6d0
6s2
Neodymium
Nd
60
2, 8, 18
4s2p6d10 f4
5s2p6d0
6s2
Promethium
Pm
61
2, 8, 18
4s2p6d10 f5
5s2p6d0
6s2
Samarium
Sm
62
2, 8, 18
4s2p6d10 f6
5s2p6d0
6s2
Europium
Eu
63
2, 8, 18
4s2p6d10 f7
5s2p6d0
6s2
Gadolinium
Gd
64
2, 8, 18
4s2p6d10 f7
5s2p6d1
6s2
Terbium
Tb
65
2, 8, 18
4s2p6d10 f9
5s2p6d0
6s2
Dysprosium
Dy
66
2, 8, 18
4s2p6d10 f10
5s2p6d0
6s2
Holmium
Ho
67
2, 8, 18
4s2p6d10 f11
5s2p6d0
6s2
Erbium
Er
68
2, 8, 18
4s2p6d10 f12
5s2p6d0
6s2
Thulium
Tm
69
2, 8, 18
4s2p6d10 f13
5s2p6d0
6s2
Ytterbium
Yb
70
2, 8, 18
4s2p6d10 f14
5s2p6d0
6s2
Lutetium
Lu
71
2, 8, 18
4s2p6d10 f14
5s2p6d1
6s2
From the above table it is clear that lanthanides have similar configuration in the outermost shell. Hence they have very similar physical and chemical properties.
- (b) Colour. The colours of M3+ ions are given in the table :
Ion
No. of unpaired electrons in 4f sub-shell
Colour
Ion
No. of unpaired electrons in 4f sub-shell
Colour
La3+
0
Colourless
Lu3+
0
Colourless
Ce3+
1
Colourless
Yb3+
1
Colourless
Pr3+
2
Green
Tm3+
2
Green
Nd3+
3
Lilac
Er3+
3
Lilac
Pm3+
4
Pink-Yellow
Ho3+
4
Pink-
Yellow
Sm3+
5
Yellow
Dy3+
5
Yellow
Eu3+
6
Pale Pink
Tb3+
6
Pale Pink
Gd3+
7
Colourless
Gd3+
7
Colourless
From this table it is clear that the colour sequence in the La-Gd series is repeated in the series Lu-Gd. It can thus be concluded that the colours depend upon the number of unpaired electrons in 4f orbitals but the fact that M2+ ions (e.g., Sm2+) and M4+ ions (e.g., Ce4+) which are isoelectronic with M3+ ions have different colours indicates that the situation is somewhat more complex. It may be noted that the colours are due to f-f transitions.
- (c) Atomic and Ionic radii. The size of atoms and ions (M3+) are given in the table below :
Element
Symbol
Atomic number
Atomic radius
Radius of M3+ ion
Lanthanum
La
57
1∙88
1∙061
Cerium
Ce
58
1∙82
1∙034
Praseodymium
Pr
59
1∙83
1∙013
Neodymium
Nd
60
1∙82
0∙995
Promethium
Pm
61
—
(0∙979)
Samarium
Sm
62
1∙80
0∙964
Europium
Eu
63
2∙04
0∙950
Gadolinium
Gd
64
1∙80
0∙938
Terbium
Tb
65
1∙78
0∙923
Dysprosium
Dy
66
1∙77
0∙908
Holmium
Ho
67
1∙77
0∙894
Erbium
Er
68
1∙76
0∙881
Thulium
Tm
69
1∙75
0∙869
Ytterbium
Yb
70
1∙94
0∙858
Lutetium
Lu
71
1∙73
0∙848
From the above table it is clear that there is a steady decrease in the size of the atoms and ions with increasing atomic number. This steady decrease in the atomic and ionic radii is called lanthanide contraction.
- (d) Magnetic and Spectral properties. The paramagnetism is generally caused by the presence in the substance of ions, atoms or molecules having unpaired electrons. Thus, both La3+ (4f 0, 5d0, 6s0) and Lu3+ (4f 14, 5d0, 6s0) ions are diamagnetic. All other M3+ ions exhibit paramagnetism. But since f-sub-shell are very effectively shielded from the influence of external forces by the 5s2 and 5p6 sub-shells, it is not possible to explain their magnetic moments in terms of the number of unpaired electrons alone. The following equation :
jn = -J n (n + 2)BM
which holds good in the case of d-block elements is not valid in the case of f-block elements. The orbital contribution which was ignored in the case of d-block elements, cannot be ignored in the case of f-block elements. Hence, the effective magnetic moments of M3+ ions with the exception of Sm3+ and Eu3+ ions are given by the equation :
^ = gJJ+I) BM
where g = Lande splitting factor and J = total angular momentum.
When f-f transitions occur, the absorption bands of M3+ ions (except Ce3+and Yb3+ ions) are extremely sharp. They are similar to those for free atoms but are quite unlike the broad bands observed for d-d transitions. Virtually all the absorption bands are line-like and become even narrower as the temperature is lowered.
Unit ‘V’
Section ‘A’
Objective Type Questions
- 1\. Which of the following is a Lewis acid ?
- (a) OH‒ (b) CH3COO‒
- (c) NH3 (d) BF3
- 2\. Cl– is the conjugate base of :
- (a) HClO4 (b) HCl
- (c) HOCl (d) HClO3
- 3\. The conjugate acid of CO32– is :
- (a) H2O (b) H2CO3
- (c) OH‒ (d) HCO3‒
- 4\. Which equilibrium can be described as an acid-base reaction using the Lewis acid-base definition but not using the Bronsted-Lowry definition ?
- (a) NH3 + CH3COOH === CH3COO‒ + NH4+
- (b) H2O + CH3COOH === H3O+ + CH3COO‒
- (c) 4NH3 + \[Cu(H2O)4\]2+===\[Cu(NH3)4\]2+ + 4H2O
- (d) 2NH3 + H2SO4=== 2NH4+ + SO42‒
- 5\. The conjugate acid of a strong base is a :
- (a) strong acid (b) weak acid
- (c) strong base (d) weak base
- 6\. Which of the following is a Lewis acid ?
- [(a) NH3 (b)H](#bookmark353)
- [(c) BF3 (d)Cl](#bookmark354)
- 7\. Water acts as an acid in the presence of :
- [(a) NH3 (b)H](#bookmark355)
- [(c) C6H6 (d)HCl](#bookmark356)
- 8\. The conjugate acid of N3– (azide) ion is :
- (a) NH3 (b) HN3
- (c) NH2‒ (d) N2
- 9\. The conjugate base of OH– is :
- (a) H+ (b) H2O
- (c) OH2‒ (d) O2‒
10\.
The conjugate acid of (CH3)2NH is :
(a) (CH3)2N‒COOH (b) (CH3)2N‒
(c) (CH3)2N+ (d) (CH3)2NH2+
11\.
Water is a :
- (a) protophobic solvent (b) protophilic solvent
- (c) amphiprotic solvent (d) aprotic solvent
12\.
The conjugate base in the following reaction is : H2SO4 + H2O → H3O+ + HSO4–
- (a) H2O (b) HSO4‒
- (c) H3O+ (d) SO42‒
13\.
Of the given anion the strongest Bronsted base is :
- (a) ClO‒ (b) ClO2‒
- (c) ClO3‒ (d) ClO4‒
14\.
The conjugate base of ammonium ion is :
(a) OH‒ (b) NH3
(c) NH4OH (d) NH4Cl
15\.
The conjugate base in HCO3– is :
(a) CO32‒ (b) H2CO3
(c) H2O (d) CO2
16\.
Which one of the following substances has the highest proton affinity
(a) PH3 (b) H2O
(c) H2S (d) NH3
17\.
According to Lewis concept which one of the following is not a base ?
- (a) OH‒ (b) H2O
- (c) Ag+ (d) NH3
18\.
According to Bronsted concept water is a :
- (a) base (b) acid
- (c) both an acid and base (d) salt
19\.
According to Bronsted concept, base is a substance which is :
- (a) a proton donor (b) an electron-pair acceptor
- (c) a proton acceptor (d) an electron-pair donor
20\.
Which of the following is Lewis base ?
(a) NaCl (b) NH4Cl
(c) MgCl2 (d) BF3
21\.
Conjugate acid of NH3 is :
(a) NH2‒ (b) NH4+
(c) HCl (d) none of these.
- 22. What is the decreasing order of strength of the base : OH-, NH2-, H - C s C- and CH3 - CH2- ?
- (a) CH3 - CH2- > NH2- > H-C = C- > OH-
- (b) H - C s C- > CH3 - CH2- > NH2- > OH-
- (c) OH- > NH2- > H - C = C- > CH3 - CH2-
- (d) NH2- > H - C C > OH- > CH3 - CH2-
- 23\. In the following reaction :
HC2O4- + PO43- ^ HPO42- + C2O42-
which are the two Bronsted bases ?
- (a) HC2O4‒ and PO43‒ (b) HPO42‒ and C2O42‒
- (c) HC2O4‒ and HPO42‒ (d) PO43‒ and C2O42‒
- 24\. In the equilibrium :
CH3COOH + HF ^ CH3COOH2+ + F-
- (a) F‒ is the conjugate acid of CH3COOH
- (b) F‒ is the conjugate base of HF
- (c) CH3COOH is the conjugate acid of CH3COOH2+
- (d) CH3COOH2+ is the conjugate base of CH3COOH
- 25. BF3 is acid according to :
- (a) Arrhenius (b) Lewis
- (c) Bronsted and Lowry (d) all of these.
- 26. Which one of the following species can act both as an acid and as a base ?
- (a) SO42‒ (b) HSO4‒
- (c) H3O+ (d) CO32‒
- 27. Amongst the following the one having characteristics of Lewis acid is :
- (a) ClF3 (b) BF3
- (c) NCl3 (d) BrF3
- 28. Which of the following is a Lewis base ?
- (a) HCl (b) H2O
- (c) BF3 (d) CH4
- 29. Which of the following is not a Lewis base ?
- (a) NH3 (b) H2O
- (c) CO (d) BF3
- 30. Which of the following statement about water is wrong ?
- (a) It can act as a base
- (b) It can act as an acid
- (c) It can act both as an acid and a base
- (d) It cannot act either as acid or as base.
31\.
Which of the following is not a Lewis acid ?
(a) CO (b) SiCl4
(c) SO3 (d) Zn2+
32\.
Which of the following is the strongest conjugate base ?
(a) Cl‒ (b) CH3COO‒
(c) SO42‒ (d) NO2‒
33\.
Which of the following is not a Lewis base ?
(a) OH‒ (b) AlCl3
(c) ROH (d) NH3
34\.
Which among following cannot be a Bronsted acid ?
(a) NH4+ (b) HSO3‒
(c) H2O (d) SO2
35\.
Which one of the following compounds is Lewis acid ?
(a) NaCl (b) AlCl3
(c) CCl4 (d) NH3
36\.
Strongest Lewis base among the following is :
- (a) CH3‒ (b) F‒
- (c) NH2‒ (d) OH‒
37\.
The conjugate base of a strong acid is a :
(a) strong base (b) strong acid
(c) weak acid (d) weak base
38\.
Which one of the following is least likely to act as Lewis base ? (At. No.
of P =15, S = 16, Cl =17, I = 53)
- (a) PCl3 (b) SCl2
- (c) I‒ (d) I+
39\.
Ionic dissociation of acetic acid is represented as :
CH3COOH + H2O == H3O+ + CH3COO–
According to Lowry and Bronsted, we have in this equation :
- (a) an acid and three bases (b) only one acid and one base
- (c) two acids and two bases (d) three acids and a base
40\.
Which of the following is a better electron pair donor ?
(a) P(CH3)3 (b) PH3
(c) N(CH3)3 (d) NH3.
41\.
Among the following, the weakest Lewis base is :
(a) H‒ (b) OH‒
(c) Cl‒ (d) HCO3‒
42\.
HNO3 in liquid hydrogen fluoride behaves :
- (a) as an acid (b) as a base
- (c) neither as a base nor as an acid (d) as a base and as an acid
43\.
Which of the following can act both as Bronsted acid and a Bronsted base ?
(a) Na2CO3 (b) OH‒
(c) HCO3‒ (d) NH3
44\.
In HS–, I–, R–NH2, NH3 order of proton excepting tendency will be :
(a) I‒ > NH3 > R‒NH2 > HS‒ (b) NH3 > R‒NH2 > HS‒ > I‒
(c) R‒NH2 > NH3 > HS‒ > I‒ (d) HS‒ > R‒NH2 > NH3 > I‒
45\.
The species which can act both as an acid and a base is :
(a) SO42‒ (b) HSO4‒
(c) HCl (d) PO43‒
46\.
Which of the following are not Lewis bases ?
(a) Ag+ (b) H2O
(c) CN‒ (d) CH4
47\.
In the reaction
2H2O → H3O+ + OH– water is :
- (a) a weak base
- (b) a weak acid
- (c) both a weak acid and a weak base
- (d) neither acid nor a base
48\.
Which one is Lewis acid ?
(a) Cl‒ (b) Ag+
(c) C2H5OH (d) S2‒
49\.
Which of the following is not a Lewis acid ?
(a) BF3 (b) SnCl2
(c) CCl4 (d) AlCl3
50\.
Which of the following is Lewis acid ?
(a) BF3 (b) SnCl4
(c) CCl4 (d) PbCl2
51\.
Conjugate base of H2PO4– is :
(a) PO43‒ (b) P2O5
(c) H3PO4 (d) HPO42‒
52\.
BF3 molecule is :
(a) Lewis acid (b) Lewis base
(c) Neutral salt (d) None of these
53\.
The conjugate base of NH3 is :
(a) NH4OH (b) NH2‒
(c) NH2‒ (d) N2H2
100
Vikas, 2009 (A. U.)
54\.
Among the following, the weakest base is :
(a) H‒ (b) CH3‒
(c) CH3O‒ (d) Cl‒
55\.
Ammonium ion is :
(a) a conjugate acid (b) a conjugate base
(c) neither an acid nor a base (d) both an acid and a base
56\.
The conjugate acid of S2O82– is :
(a) HSO4‒ (b) H2SO4
(c) H2S2O8 (d) HS2O8‒
57\.
AlCl3 is :
- (a) a Lewis acid (b) a Lewis base
- (c) a Bronsted-Lowry acid (d) an Arrhenius acid
58\.
The strongest conjugate base results from :
(a) formic acid (b) benzoic acid
(c) acetic acid (d) acetylene
59\.
Which of the following can act as an acid and as a base ?
(a) HCO3‒ (b) H2PO4‒
(c) HS‒ (d) All of these.
60\.
In the reaction I2 + I– → I3–, the Lewis base is :
- (a) I2 (b) I‒
- (c) I3‒ (d) none of these
61\.
Which of the following is a Lewis acid ?
(a) PCl3 (b) AlCl3
(c) NCl3 (d) AsCl3
62\.
The increasing basic strength order of ClO4–, Cl– and CH3COO– is :
- (a) ClO4‒ < Cl‒ < CH3COO‒
- (b) ClO4‒ < Cl‒ < CH3COO‒
- (c) Cl‒ < ClO4‒ < CH3COO‒
- (d) CH3COO‒ < Cl‒ < ClO4‒
63\.
Which of the following statements is correct ?
- (a) BCl3 and AlCl3 are both Lewis acids and BCl3 is stronger than AlCl3
- (b) Both BCl3 and AlCl3 are not Lewis acids
- (c) BCl3 and AlCl3 are both equally strong Lewis acids
- (d) BCl3 and AlCl3 are both Lewis acids and AlCl3 is stronger than BCl3
64\.
Which of the following is not Lewis acid ?
(a) BF3 (b) AlCl3
(c) FeCl3 (d) NH3
- 65. Which of the following will act as Lewis base, nucleophilic reagent and as well as ligand in complex compound ?
- (a) H2O (b) H3O+
- (c) AlCl3 (d) CH3+
- 66. According to lewis concept, an acid is :
- (a) an electron-pair donor (b) an electron-pair acceptor
- (c) a proton donor (d) a proton acceptor
- 67. H+ is a :
- (a) Lewis acid (b) Lewis base
- (c) Bronsted and Lowry base (d) None of these.
- 68. Which one of the following compounds is not a protonic acid ?
- (a) SO2(OH)2 (b) B(OH)3
- (c) PO(OH)3 (d) SO(OH)2.
- 69. Which one of the following can be classified as a Bronsted base ?
- (a) CH3COOH (b) NH4+
- (c) NO3‒ (d) H3O+
- 70. Which of the following is a conjugate acid-base pair :
- (a) HCl, NaOH (b) NH4Cl, NH4OH
- (c) H2SO4, HSO4‒ (d) KCN, HCN
- 71. Which one of the following orders of acid strength is correct ?
- (a) RCOOH > HC=CH > HOH > ROH
- (b) RCOOH > ROH > HOH > HC = CH
- (c) RCOOH > HOH > ROH > HC = CH
- (d) RCOOH > HOH > HC = CH > ROH.
- 72. Ammonia gas dissolves in water to give NH4OH. In this reaction water acts as :
- (a) an acid (b) a base
- (c) a salt (d) a conjugate base
- 73. The conjugate acid of NH2–is :
- (a) NH3 (b) NH2OH
- (c) NH4+ (d) N2H4
- 74. The conjugate base in HBr is :
- (a) H2Br+ (b) H+
- (c) Br‒ (d) Br+
- 75. According to Lewis concept a base is a substance that :
- (a) accepts an electron pair (b) donates a proton
- (c) donates an electron pair (d) accepts a proton
- 76. According to Lowry and Bronsted system, the chloride ion (Cl–) in aqueous solution is a :
- (a) weak base (b) strong base
- (c) weak acid (d) strong acid.
- 77. Which of the following statements is true ?
- (a) The conjugate base of a strong acid is a strong base
- (b) The conjugate base of a weak acid is a strong base
- (c) The conjugate base of a weak acid is a weak base
- (d) The base and its conjugate acid react to form a neutral solution.
- 78. Which of the following is not used as a Lewis acid ?
- (a) SnCl4 (b) FeCl3
- (c) KCl (d) BF3
- 79. Which of the following is an example of Lewis acid ?
- (a) C2H5ONa (b) FeCl3
- (c) HCl (d) HCOOH
- 80\. In the reaction
NH3 + BF3 = NH3 ^ BF3
BF3 is :
- (a) a Lewis acid
- (b) a Lewis base
- (c) neither a Lewis acid nor a Lewis base
- (d) a Lewis acid and also a Lewis base
- 81. In the equilibrium
HC1O4 + H2O ====== H3O+ + ClO4-
- (a) HClO4 is the conjugate acid of H2O
- (b) H2O is the conjugate acid of H3O+
- (c) H3O+ is the conjugate base of H2O
- (d) ClO4‒ is the conjugate base of HClO4.
- 82. H2O can act as either an acid or a base. Which of the following reactions best illustrates the behaviour of water as a base ?
- (a) HCl + H2O ^ H3O+ + Cl"
- (b) HCl + NaOH ^ NaCl + H2O
- (c) H2O + NH2 ^ NH3 + OH"
- (d) H2O + NH3 ^ NH4+ + OH"
- 83. The conjugate acid of HPO42– is :
- (a) PO43‒ (b) H2PO4‒
- (c) H3PO4 (d) H2PO4+
- 84. Which of the following is a Lewis acid ?
- (a) NaCl (b) MgO
- (c) BF3 (d) PCl3
- 85\. The following equilibrium is established when hydrogen chloride is dissolved in acetic acid
HCl + CH3COOH == Cl- + CH3COOH2+
The set that characterises the conjugate acid-base pairs is :
- (a) (HCl, CH3COOH) and (CH3COOH2+, Cl‒)
- (b) (HCl, CH3COOH2+) and (CH3COOH, Cl‒)
- (c) (CH3COOH2+, HCl) and (Cl‒, CH3COOH)
- (d) (HCl, Cl‒) and (CH3COOH2+, CH3COOH)
- 86\. Which one of the following is not a Lewis acid ?
- (a) ZnCl2 (b) BF3
- (c) Ag+ (d) H2O
- 87\. Which anion is the weakest base ?
- [(a) C2H5O‒ (b)NO](#bookmark412)
- (c) F‒ (d) CH3COO‒
- 88\. The compound that is not a Lewis acid is :
- [(a) BF3 (b)AlCl](#bookmark413)
- [(c) H2O (d)SnCl](#bookmark414)
- 89\. Which of the following is a Lewis base ?
- (a) CH4 (b) C2H5OC2H5
- (c) Acetone (d) Sec. amine
- 90\. According to Lowry and Bronsted a base must :
- (a) contain a metal (b) readily give up proton
- (c) contain oxygen (d) readily unite with a proton
- 91\. Which one of the following compounds is Lewis acid ?
- (a) PCl3 (b) BCl3
- (c) NCl3 (d) CHCl3
- 92\. The compound that is not Lewis acid is :
- (a) BF3 (b) C2H5OH
- (c) AlCl3 (d) BCl3
- 93\. Amphoteric behaviour is shown by :
- (a) H2O and H3O+
- (b) HCO3‒ and H2CO3
- (c) H2O and CO32‒
- (d) none of these.
- 94\. Conjugate base of HPO42– is :
- (a) PO43‒ (b) H2PO4‒
- (c) H3PO4 (d) H4PO3
- 95. Lowry and Bronsted theory recognizes ammonia molecule as a base because :
- (a) it dissolves in water to form ammonium hydroxide
- (b) it combines with hydrogen ion to form the ammonium ion
- (c) it turns wet red litmus paper blue
- (d) it forms fumes of ammonium chloride
- 96. The conjugate base of NH2– is :
- (a) NH3 (b) NH2‒
- (c) NH4 (d) N3
- 97. Which of the following can act both as Bronsted acid and a base ?
- (a) Cl‒ (b) HCO3‒
- (c) H3O+ (d) SO42‒
- 98. The conjugate base of a weak acid is :
- (a) a strong base (b) a weak base
- (c) a neutral species (d) may be a weak or strong base.
- 99. Among the following reactions, the Lewis acid-base concepts is best illustrated by :
- (a) Zn + 2HCl → ZnCl2 + H2 (b) NH3 + BF3→ H3N.BF3
- (c) AgNO3 + HCl → AgCl + HNO3 (d) CaCO3→ CaO + CO2
- 100\. Which of the following is a Lewis acid ?
- (a) HCOO‒ (b) H2SO4
- (c) SiF4 (d) H2S
- 101\. Which one of the following is not an amphoteric substance ?
- (a) NH3 (b) H2O
- (c) HCO3‒ (d) HNO3
- 102\. Which of the following represents the conjugate acid-base pair for the equilibrium reaction HCl + H2O === H3O+ + Cl– ?
- (a) HCl, H2O (b) HCl, H3O+
- (c) HCl, Cl‒ (d) Cl‒, H2O
- 103\. In the reaction HCl + H2O === H3O+ + Cl– :
- (a) H2O is the conjugate base of HCl acid
- (b) Cl‒ is the conjugate base of HCl acid
- (c) H3O+ is the conjugate base of HCl acid
- (d) Cl‒ is the conjugate acid of H2O base
Answers
1\. (d) 2. (b) 3. (d) 4. (c) 5. (b) 6. (c) 7. (a)
8\. (b) 9. (d) 10. (d) 11. (c) 12. (b) 13. (a) 14. (b)
15\. (a)
16\. (d)
17\. (c)
18\. (c)
19\. (c)
20\. (b)
21\. (b)
22\. (a)
23\. (d)
24\. (b)
25\. (b)
26\. (b)
27\. (b)
28\. (b)
29\. (d)
30\. (d)
31\. (a)
32\. (b)
33\. (b)
34\. (d)
35\. (b)
36\. (a)
37\. (d)
38\. (d)
39\. (c)
40\. (c)
41\. (c)
42\. (b)
43\. (c)
44\. (c)
45\. (b)
46\. (a)
47\. (c)
48\. (b)
49\. (c)
50\. (a)
51\. (d)
52\. (a)
53\. (b)
54\. (d)
55\. (a)
56\. (d)
57\. (a)
58\. (d)
59\. (d)
60\. (b)
61\. (b)
62\. (a)
63\. (a)
64\. (d)
65\. (a)
66\. (d)
67\. (a)
68\. (b)
69\. (c)
70\. (c)
71\. (c)
72\. (a)
73\. (a)
74\. (c)
75\. (c)
76\. (a)
77\. (b)
78\. (c)
79\. (b)
80\. (a)
81\. (d)
82\. (a)
83\. (b)
84\. (c)
85\. (d)
86\. (b)
87\. (b)
88\. (c)
89\. (d)
90\. (d)
91\. (b)
92\. (b)
93\. (d)
94\. (a)
95\. (b)
96\. (b)
97\. (b)
98\. (a)
99\. (a)
100\. (c)
101\. (d)
102\. (c)
103\. (b)
Section ‘B’
Short Answer Type Questions
Q. 1. Give equations for the auto-ionization of liquid ammonia and liquid sulphur dioxide.
Ans. Auto ionization of liquid ammonia :
NH3 + NH3 === NH4+ + NH2‒
Ammonium ion Amide ion
Auto-ionization of liquid sulphur dioxide :
SO2 + SO2 === SO2+ + SO32‒
Thionyl ion Sulphite ion
Q. 2. Describe what will happen if CH3COOH is dissolved in liquid ammonia.
Ans. When CH3COOH is dissolved in liquid ammonia, the stronger protophilic nature of liquid ammonia greatly increases the dissociation of CH3COOH which then behaves as a strong electrolyte.
CH3COOH + NH3→ CH3COO‒ + NH4+
Acid Base Base Acid
(strong) solvent
Q. 3. Liquid ammonia is a better solvent for organic compounds than water.
Ans. The dielectric constant of liquid ammonia is 22(at ‒33.3°C) as compared with 78.5 (at 25°C) for water. Hence liquid ammonia is a better solvent for organic compounds and generally a poorer one for ionic substances. Thus
sulphates, sulphites, carbonates, phosphates, arsenates, oxides and sulphides are insoluble in liq. NH3, Metal hydroxides are likewise insoluble. All the metal amides except those of alkali metals are insoluble. Lithium amide is also insoluble. The metal salts which are in general most soluble in liq. NH3 include thiocyanates, perchlorates, nitrates, nitrites and many of the iodides. Most ammonium salts are soluble in liq. NH3.
Q. 4. Discuss in brief Lux and Flood concept of acids and bases.
Ans. Lux and Flood proposed a theory to explain acid-base concept of oxide system. According to this concept an acid is an oxide ion acceptor while a base is an oxide ion donor. e.g.
CaO (Base)
\+ CO2 (Acid)
^
CaCO3
CaO (Base)
\+ SiO2 (Acid)
^
CaSiO3
This concept has a direct relation to aqueous chemistry of acids and bases.
Q. 5. Classify with reasons the following as Lewis acids and bases : ROH, BF3, NH3, AlCl3, SnCl4, H3O+, Ag+, HCl, H2O, M and L in a complex compound \[ML6\] where M is a metal ion and L is any ligand, CO and Fe in Fe(CO)5.
Ans. ROH is a Lewis base because oxygen has two unshared pair of electrons and therefore, it can donate an electron pair.
BF3 is a Lewis acid because the central atom (B) has incomplete octet and therefore, it can accept an electron pair.
NH3 is a Lewis base because the central atom (N) has one unsharped pair of electrons and, therefore, it can donate an electron pair.
AlCl3 is a Lewis acid because the central atom (Al) has incomplete octet and, therefore, it can accept an electron pair.
SnCl4 is a Lewis acid because the central atom (Sn) has vacant d orbitals and therefore, it can accept an electron pair.
H3O+ is a Lewis acid because being a cation it can accept an electron pair.
Ag+ is a Lewis acid because being a cation it can accept an electron pair.
HCl is a Lewis acid because it can accept a lone pair of electrons from a base such as water though this is followed by ionization.
H2O + HCl ^ \[H2O ^ HCl\] ^ H3O+ + CF
H2O is a Lewis base because the central atom (O) has two unshared pair of electrons and, therefore, it can donate an electron pair.
In the complex compound \[ML6\] the metal ion M is Lewis acid because it accepts electrons while the ligand L is Lewis base because it donates electrons.
In Fe(CO)5, Fe is Lewis acid because it accepts electrons while CO is Lewis base because it donates electrons.
Q. 6. Write a note on classification of solvents.
Ans. The solvents have been classified in a number of ways depending on their physical and chemical properties.
- (I) This is a common classification based on proton-donor and protonacceptor property of solvents. On this basis the solvents may be divided into three types :
- (A) Protic or Protonic solvents : These solvents have hydrogen atom in their formulae and are of two types :
- (i) Acidic or Protogenic solvents : They have strong tendency to donate protons e.g. H2SO4, HF, CH3COOH.
- (ii) Basic or Protophilic solvents : They have strong tendency to accept protons e.g. NH3.
- (B) Aprotic or non-protonic solvents : They may or may not have hydrogen in their formulae and neither donate nor accept protons e.g. C6H6, CHCl3, CCl4 etc.
- (C) Amphiprotic or Amphoteric solvents : They have hydrogen in their formulae and donate or accept protons i.e. these can act both as acids and bases and hence are amphoteric in nature e.g. H2O, CH3COOH.
- (II) Ionizing and Non-ionizing solvents : It is based upon polar or nonpolar nature of solvents e.g. H2O, NH3, HF etc. are ionizing while C6H6, CCl4 etc. are non-ionizing.
- (III) Aqueous and Non-aqueous solvents : The solvents other than water are called non-aqueous solvents e.g. liquid NH3.
Q. 7. Discuss Arrhenius concept of acids and bases. What are the main limitations of this concept.
Ans. According to Arrhenius an acid is defined as a hydrogen compound which in water solution gives hydrogen ions and a base is a hydroxide compound which in aqueous solution yields OH‒ ions. The examples of Arrhenius acids are HCl, HNO3, CH3COOH etc. while of Arrhenius bases are NaOH, KOH etc.
The main limitations of this theory are as follows :
- 1\. It defines acids and bases in terms of aqueous solutions and not in terms of the substances themselves.
- 2\. It also fails to explain the properties of a salt such as NH4NO3 in liquid ammonia.
- 3\. This concept restricts bases merely to hydroxides. Thus even metal oxides would not by regarded as bases.
- 4\. This theory has an impossible idea of a bare proton existing in aqueous solution.
- 5\. In this concept an artificial explanation is necessary to explain the basicity of an aqueous solution of ammonia. According to Arrhenius theory ammonia must react with water to produce ammonium hydroxide which in turn ionizes to produce OH‒ ions.
NH3 + H2O === NH4OH === NH4+ + OH‒
There is no proof for the existence of the compound designated as ammonium hydroxide and the assumption of such an ionization is entirely misleading.
Q. 8. Solutions of alkali metals in liquid ammonia are coloured.
Ans. Solutions of alkali metals in liquid ammonia are coloured. Dilute solutions are blue and the blue colour is due to ammoniated electrons.
M → M
+
(ammoniated)
\+ e‒
(ammoniated)
As the concentration of alkali metal increases, metal ion cluster formation takes place and at very high concentration the solution becomes coppercoloured (metallic bronze like appearance).
Q. 9. Discuss Lewis concept of acids and bases. Give merits and demerits of this concept.
Ans. Lewis concept : This concept is based upon the electronic theory of valence. According to this concept an acid is any species (atom, ion or molecule) which may accept an electron pair to form a coordinate bond and a base is any species that has a pair of electrons which it can donate to establish a coordinate bond e.g.
F
H
FH
|
F ‒ B
+
|
: N ‒ H
→
| |
F ‒ B → N ‒ H
| F
|
H
| |
FH
(acid)
(Base)
(Coordinate bond)
It may be noted that Lewis bases and the Bronsted bases are same because any substance which can add a proton must possess an unbonded electron pair. Thus Lewis and Bronsted theories are identical as far as bases are concerned except in the working of definitions.
Merits of this concept :
- 1\. This concept also includes those reactions in which no protons are
involved.
- 2\. It is more general than Bronsted-Lowry concept in that acid-base behaviour is not dependent on the presence of any particular element or on the presence or absence of a solvent.
Demerits of this concept : In this concept the base and acid strengths are not fixed because the order of base strength of a series of Lewis bases may change when the type of the acid which they are allowed to combine changes.
Q. 10. Arrange the members in the following groups of acids or bases in order of decreasing acidity or basicity.
- (i) HCl, HClO, HClO2, HClO3
- (ii) NH3, PH3, AsH3
- (iii) HF, HCl, HBr, HI
- (iv) H2O, H3O+, OH–
- (v) NH3, NH2–, NH4+
- (vi) HCl, HOBr, HOI
Ans. (i) HClO3 > HClO2 > HClO > HCl
- (ii) NH3 > PH3 > AsH3
- (iii) HI > HBr > HCl > HF
- (iv) OH‒ > H2O > H3O+
- (v) NH2‒ > NH3 > NH4+
- (vi) HCl > HOBr > HOI
Q. 11. Write the conjugate base of the following acids :
CH4, C2H5OH, NH4+, H2O, HCl, H3O+, HCO3–, \[Al(H2O)6\]3+ Ans.
Acid
Conjugate base
CH4
CH3‒
C2H5OH
C2H5O‒
NH4+
NH3
H2O
OH‒
HCl
Cl‒
H3O+
H2O
HCO3‒
CO32‒
\[Al(H2O)6\]3+
\[Al(H2O)5OH\]
2+
Q. 12. What are non-aqueous solvents ? Mention the advantages of liquid ammonia over water as a solvent.
Ans. The solvents other than water are called non-aqueous solvents. Some important non-aqueous solvents are liquid ammonia, liquid SO2, liquid HF and glacial acetic acid.
Advantages of using liquid ammonia over water as a solvent :
There are following advantages of using liquid ammonia over water as a solvent :
- 1\. Liquid NH3 dissolves alkali metals without apparent chemical action and the dissolved metals may be recovered unchanged by simply allowing the solvent to evaporate.
- 2\. Liquid NH3 has a lesser tendency than water to undergo solvolysis reactions with dissolved solutes.
- 3\. Reducing agents stronger than hydrogen liberate H2 from water whereas solutions of alkali and alkaline earth metals in liquid NH3 provide stronger reducing systems in which the solvent is unaffected.
- Q. 13. Discuss in brief the solvent system concept of acids and bases.
Ans. The concept of solvent system as a means of explaining acid-base behaviour had its origin with Franklin. This was extended by Cady and Elsey. Consider any solvent which undergoes self-ionization. e. g.
2NH3 → NH4+ + NH2‒
2H2O → H3O+ + OH‒
Acids are substances which increase the concentration of positive ions characteristic of self-ionization of solvent (H3O+ in the case of water and NH4+ in the case of liquid ammonia). On the other hand bases are substances which increase the concentration of negative ions characteristic of self-ionization of the solvent (OH‒ in the case of water and NH2‒ in the case of liquid ammonia).
Section ‘C’
Long Answer Type Questions
Q. 1. Discuss Bronsted-Lowry concept of acids and bases. Give merits and demerits of this concept.
Ans. Bronsted-Lowry concept : In an attempt to overcome some of the limitations of the Arrhenius concept, a more general definition for acids and bases was proposed almost simultaneously by Bronsted and Lowry. According to them an acid is defined as any molecule or ion that can release a proton whereas a base is any molecule or ion that can accept a proton. In short, acids are protogenic while bases are protophilic.
Bronsted-Lowry acids : HCl, HCO3‒, HNO3, \[Al(H2O)6\]3+ etc.
Bronsted-Lowry bases : HCO3‒, OH‒, Pyridine, \[Al(H2O)5OH\]2+
Consider the following reaction :
HCl + H2O === H3O+ + Cl‒
In this reaction HCl donates a proton to H2O and is, therefore, an acid. H2O on the other hand, accepts a proton from HCl and is, therefore, a base. In the reverse reaction H3O+ ion donates a proton to Cl‒ ion. Hence H3O+ is an acid and Cl‒ is a base. Acid-base pairs such as HCl and Cl‒ or H3O+ and H2O the members of which can be formed from each other mutually by the gain or loss of proton are called ‘conjugate acid-base pairs’. It may be noted that stronger acid has the weaker conjugate base and the stronger base has weaker conjugate acid.
According to this theory a strong acid is one which has a strong tendency to donate a proton and whose conjugate base is weak. On the other hand, a weak acid is one which does not readily release a proton and whose conjugate base is strong. Similar consideration can be applied to strong and weak bases also.
Those substances which can accept as well as donate a proton are called amphiprotic e.g., H2O, NH3, HCO3‒ etc.
Merits of this concept :
- 1\. It defines acid and base in terms of the substances themselves and not in terms of their ionization in aqueous solution.
- 2\. The idea of hydrolysis is not essential to explain the acidic character of NH4+ ion or the basic character of CO32‒ ion.
- 3\. It is neither restricted to nor dependent upon any particular solvent.
- 4\. According to this concept the basic character of NH3 is due to the ability of NH3 molecule to accept proton in forming the acid NH4+.
- 5\. This concept is also useful in accounting for the hydrolysis of salt solution e.g. aq. solution of Na2CO3 is alkaline because proton acceptor ability of CO32‒ ion exceeds the proton donor ability of hydrated sodium ion.
Demerits of this concept : It lays excessive emphasis on the proton transfer. Although it is true that most common acids are protonic in nature yet there are many which are not. Moreover, there are a number of acid-base reactions in which no proton transfer take place.
Q. 2. Discuss with examples the behaviour of liquid ammonia as non-aqueous solvent with reference to the following :
- (i) Precipitation reactions
- (ii) Ammonolysis
- (iii) Redox reactions
- (iv) Complex formation reactions
- (v) Neutralization reactions
Ans. (i) Precipitation reactions : Some of the reactions of this type are mentioned below :
- (a) Precipitation of LiCl from liq. NH3 solutions of amm.chloride and lithium nitrate.
NH4C1 + LiNO3 ^ LiCl + NH4NO3
Similarly precepitation of KCl from liq. NH3 solutions of KI and NH4Cl.
KI + NH4C1 ^ KCl + NH4I
- (b) Precipitation of sulphides of Zn, Cd, Ba, Ag, Pb etc. by amm.sulphide dissolved in liq. NH3.
Ba(NO3)2 + (NH4)2S ^ BaS + 2NH4NO3
2AgNO3 + (NH4)2S ^ Ag2S + 2NH4NO3
- (c) Precipitation of barium chloride from solutions of silver chloride and barium nitrate in liq. NH3.
2AgCl + Ba(NO3)2 ^ BaCl2 + 2AgNO3
- (d) Precipitation of bromides or iodides from solutions of various nitrates (e.g., Sr(NO3)2, Zn(NO3)2 and ammonium bromide or ammonium iodide in liq. NH3
Zn(NO3)2 + 2NH4I ^ ZnI2 + 2NH4NO3
Sr(NO3)2 + 2NH4Br ^ SrBr2 + 2NH4NO3
- (e) Precipitation of many insoluble halides as complex amines.
2NaCl + Ca(NO3)2 + 8NH3 ^ \[Ca(NH3)8\]Cl2 + 2NaNO3
2AgBr + Ba(NH3)2 + 8NH3 ^ \[Ca(NH3)8\]Br2 + 2AgNO3
- (ii) Ammonolysis : It is also usually referred to as solvolytic or ammonolytic reactions. In these reactions an atom, ion or radical from the compound undergoing ammonolysis is replaced by—NH2, > NH or = N group, e.g.
M'H + NH3 ^ M+NH2" + H2
NH3(l)
Hg2Cl2> Hg(NH2)Cl
NH3(l) rB (NH2)3 ’
BCl3 > B2(NH)3
NH3(l)
SiCl4> Si(NH2)4
NH (l) 0°C
PC13 > P(NH2)3-------> P2(NH)3
NH (l)
VOC13-----^ VO(NH2)3
- (iii) Redox reactions : They are those rections in which the nitrogen or hydrogen atom of the ammonia molecule undergoes a change in oxidation
state. Example of reaction in which ammonia acts as a reducing agent include the catalytic oxidation of NH3 to NO
Pt
4NH3 + 5O2------→ 4NO + 6H2O
and the reduction of metal oxide with ammonia at high temperatures.
3CuO + 2NH3------→ 3Cu + 3H2O + N2
In its reaction with active metals, NH3 acts as an oxidizing agent.
2Na + 2NH3------→ 2NaNH2 + H2
3Mg + 2NH3------→ Mg3N2 + 3H2
Ammonia is oxidized by some non-metals e.g. Cl2 oxidizes NH3 to chloramine in the gas phase, in aqueous solution or in liq. NH3.
Cl2 + 2NH3------→ NH2Cl + NH4Cl
It may be noted that if ammonia is not in excess dichloramine (NHCl2), nitrogen trichloride or nitrogen are formed.
- (iv) Complex formation reactions : Many metal amides, imides and nitrides dissolve in solutions of KNH2 in liquid NH3 forming complexes.
liq. NH3
AgNH2 + KNH2------→ K\[Ag(NH2)2\]
liq. NH3
Zn(NH2)2 + 2KNH2------→ K2\[Zn(NH2)4\]
- (v) Neutralization reactions : The auto-ionization of liquid NH3 is as follows :
NH3 + NH3 === NH4+ + NH2‒
Hence any substance which gives NH4+ ions in liq. NH3 (e.g. NH4Cl) is an ammono acid while the substance which gives NH2‒ (amide), NH2‒ (imide) or N3‒ (nitride) ions (e.g. NaNH2, PbNH and BiN) is called ammono base.
It may be noted that the reaction of an ammono acid with an ammono base to give a salt called ammono salt and the solvent liquid NH3 is called neutralization reaction. The following reactions illustrate the neutralization reaction in liq. NH3.
liq. NH3
Ammono + Ammono —————→ Ammono + Solvent
acid
base
liq. NH3
salt
NH4NO3
+
KNH2
——→
KNO3
+
2NH3
2NH4I
+
PbNH
liq. NH3
——→
PbI2
+
3NH3
3NH4I
+
BiN
liq. NH3
——→
BiI3
+
4NH3
liq. NH
NH2CO.NH2 + KNH2-->3 K+\[H2NC=ONH\]- + NH3
Q. 3. Discuss with examples the various types of reactions in liquid sulphur dioxide.
Ans. Following five types of reactions have been observed in liquid sulphur dioxide.
- (i) Complex formation reactions : The mode of formation of many complex ions in liquid SO2 is similar to the process occurring in aqueous solutions. Typical behaviour of this kind is the solubility of a precipitate in an excess of its own anion e. g.
2AlCl3 + 3\[(CH3)4N\]2SO3 ========== Al2(SO3)3 + 6\[(CH3)4.N\]Cl
Insoluble
Al2(SO3)3 + 3\[(CH3)4N\]2SO3 ======== 2\[(CH3)4N\]3\[Al(SO3)3\]
Soluble
Another example of complex formation not involving a precipitate is the interaction between antimony trichloride and potassium chloride to give the soluble complex K3SbCl6.
SO2
SbCl3 + 3KCl ======= KJSbCy
- (ii) Redox reactions : Following redox systems have been investigated in liq. SO2.
- (a) Oxidation of soluble iodides in liq. SO2 by SbCl5 or FeCl3.
6KI + 3SbCl5> 2K3\[Sba6\] + 3I2 + SbCl3
- (b) Reduction of iodine by sulphite in liq. SO2.
- (iii) Solvolytic reactions : These reactions in liq. SO2 are much less common and much more complex. PCl3, WCl6, NbCl5 etc. readily undergo solvolytic reaction in liq. SO2.
PC15 + SO2( l)> POC13 + SOC12
70°C
WCl6 + SO2( l)--> WOC14 + SOC12
70°C
NbCl5 + SO2( l)--> NbOCl3 + SOC12
Other examples are :
2NH4(OOCCH3) + 2SO2( l) > (NH4)2SO3 + (CH3COO)2SO
II
(CH3CO)2O + SO2
Zn(C2H5)2 + 2SO2( l) > ZnSO3 + (C2H5)2SO
- (iv) Acid-base reactions : The auto-ionization of liq. SO2 is as follows :
SO2 + SO2> SO2+ + SO32-
Hence it is clear that comounds containing or making available SO32‒ ions (e.g. alkali metal sulphites) will act as typical bases in liq. SO2. Similarly the compounds containing or making available SO2+ ions (e.g. SOCl2 etc.) will behave as acids in liq. SO2. Some typical examples of acid-base reactions are as follows :
Acid
+
Base
→
Salt
+
Solvent
SOCl2
+
Cs2CO3
→
2CsCl
+
2SO2
SO(SCN)2
+
K2SO3
→
2KSCN
+
2SO2
SOBr2
+
\[(CH3)4N\]2SO3
→
2\[(CH3)4N\]Br
+
2SO2
(v) Precipitation reactions. A few examples of precipitation reactions are
given below :
liq. SO2
BaI2
+
Zn(CNS)2
\----→
Ba(CNS)2
↓
+
ZnI2
liq. SO2
AlCl3
+
3NaI
3NaCl
↓
+
AlI3
liq. SO2
SbCl3
+
3Lil
SbI3
↓
+
3LiCl
liq. SO2
PbF2
+
Li2SO4
PbSO4
↓
+
2LiF
liq. SO2
↓
2Ag(OOCCH3) +SOCl2
2AgCl
+
SO(OOCCH3)2
liq. SO2
2KBr
+
SOCl2
2KCl
↓
+
SOBr2
liq. SO2
2KI
+
SOCl2
2KCl
↓
+
SOI2
Very Short Answer Type Questions
Q. 1. What are d-block elements ?
Ans. Elements in which last electron enters in the d-orbital are known as d-block elements.
Q. 2. Give the general electronic configuration of d-block elements.
Ans. General electronic configuration of d-block elements is :
(n ‒ 1)d1‒10,ns1‒2
Q. 3. Give the electronic configuration of Cr24 and Cu29.
Ans. Cr24 = 1s2, 2s22p6, 3s23p63d5, 4s1
Cu29= 1s2, 2s22p6, 3s23p63d10, 4s1
Q. 4. Which one of the following Mn2+, Fe2+, Co2+, Ni2+ will represents highest para-magnetism ?
Ans. Number of unpaired electrons in Mn2+, Fe2+, Co2+ and Ni2+ are 5, 4, 3 and 2 respectively. Hence, magnetic moment for Mn2+ ion will be maximum because number of unpaired electrons are related with magnetic moment by the rotation.
^ = y\]n(n + 2) where n ^ no. of unpaired electrons
Q. 5. What is the oxidation state of iron in potassium
hexacyan ferrate ?
Ans. The oxidation state of iron in potassium hexacyan ferrate is +3.
Q. 6. Write down the structure formula of the complex bis
(dimethyal glyoxamato) nickel (II).
Ans.
O --- H ‒ O
CH ‒ C = N N = C ‒ CH
33
| Ni |
CH3‒ C = N N = C ‒ CH3
O ‒ H ---- O
Q. 7. Give the geometry of tetramine cupric sulphate.
Ans. The geometry of tetramine cupric sulphate is tetrahedral.
Q. 8. Which transition series contains the following
elements
W, Re, Os
Ans. Third transition series (La — Hg) contains W, Re and Os.
Q. 9. Which of the following transition metal represents highest oxidation state ?
Ru, Os, Mn, Re
Ans. Osmium (Os) represents highest oxidation state (+ 8).
Q. 10. Give the electronic configuration of Pd and Ag.
Ans. Pd46 = 1s2, 2s22p6, 3s23p63d10, 4s24p64d10, 5s0
Ag47 = 1s2, 2s22p6, 3s23p63d10, 4s24p64d10, 5s1
Q. 11. What is slag ?
Ans. Slag is a substance which is produced during smelting by reaction of the flux with impurities.
SiO2 + CaO ^ CaSiO3
Flux Impurity Slag
Q. 12. What do you mean by a flux ?
Ans. Flux is a substance which is added to the ore during smelting to assist in the removal of earthy impurities as slag.
Q. 13. What do you mean by calcination ?
Ans. It involves heating of the ore below its fusion temperature is absence of air.
Q. 14. Explain the term smelting.
Ans. Smelting is a process in which a metal is extracted by fussion the ore is a suitable furnace in presence of a reducing agent such as carbon.
Q. 15. What do you mean by Bassemerisation ?
Ans. The process in which a impure metal is heated in a furnace and a blast of compress air is brown through the molten mass by which the impurities get oxidised is called bassemerisation.
Q. 16. Which metal is extracted by Mac Aurthur Forrest cyanide process ?
Ans. Silver metal is extracted by Mac Aurthur Forrest cyanide process.
Q. 17. Give one example each of double and complex salt.
Ans. Double salt Alum K2SO4.Al2(SO4)3.24H2O
Complex Salt.Potassiumferrocyanide K4Fe(CN)6
Q. 18. What is EAN ?
Ans. EAN is defined as the resultant number of electrons with the metal atom or ion after gaining electrons from the donor atoms of the ligands.
Q. 19. What is the effective atomic number of coabalts in \[Co(NH3)6\]Cl3 ?
Ans. Effective atomic number of Co in \[Co(NH3)6\]Cl3 is = (27 ‒ 3) + 6 × 2 = 24 + 12 = 36
Q. 20. What do you mean by co-ordination number ?
Ans. The number of atoms of the ligands then are directly bound to the central metal atom or ion by co-ordinate bonds is known as co-ordination number of the metal atom or ion.
Q. 21. Define ligands.
Ans. The neutral molecules or anions which are directly linked with the central metal atom or ion in a complex ion are called ligands.
Q. 22. How ligands are attached to central metal ion or atoms ?
Ans. The ligands are attached to the central metal ion or atoms through co-ordinate bonds.
Q. 23. Give one examples each of bidentate and hexadentate ligand.
Ans. CH2 ‒ NH2
HOOCCH2 HOOCCH2
N ‒ CH2 ‒ CH2‒ N
HOOCCH2 HOOCCH2
Ethylene diamine tetraacetic acid (EDTA) (Hexadentate ligand)
|
CH2 ‒ NH2
Ethylene diamine
(Bidentate)
Q. 24. Give the IUPAC names of the following complexes.
- (a) \[Cr(NH3)6\]\[Co(CN)6\]
- (b) \[Pt(NH3)6\]Cl4
Ans. (a) Hexamine Chromium (II) hexacyanocobaltate (II)
- (b) Heamine platinum (IV) chloride.
Q. 25. Give an example of ionisation isomerism.
Ans. \[Co(NH3)4Cl2\]NO2 dichlorotetraminecobalt (III) nitrite and \[Co(NH3)4ClNO2\]Cl chloronitrotetramine cobalt (III) chloride are ionisation isomers.
Q. 26. Write the possible isomers of the complex compound \[Pt(NH3)2Cl2\] and explain in brief the isomerism involved.
Ans. The possible isomers are \[Pt(NH3)4\]\[PtCl4\] and \[Pt(NH3)3Cl\]2\[PtCl4\] and the isomerism involved is polymerisation isomerism because they are polymers of the first.
Q. 27. What do you mean by hydrate isomerism.
Ans. This type of isomerism arises when different number of water molecule are present inside and outside the co-ordination sphere e.g.
\[Cr(H2O)6\]Cl3, \[Cr(H2O)5Cl\]\[Cl2.H2O and \[Cr(H2O)4Cl2\]Cl.2H2O
Q. 28. What is formed when ethylene diamine tetra acetic acid combine with a cation ?
Ans. When ethylene diamine combine with cation formation of cholase takes place.
Q. 29. Give the general electronic configuration of lanthanides.
Ans. General electronic configuration of lanthanides is (n ‒ 2)f1‒14(n ‒ 1)d0 ‒ 2 ns2
Q. 30. How many elements are in lanthanide series.
Ans. Elements from lerium (58) to Lutetium (71) along with lanthanum (57) there are fifteen elements in lanthanide series.
Q. 31. Which oxidation state is more stable in case of lanthanide ions ?
Ans. +3 oxidation state is most stable oxidation form in case of Ln ions.
Q. 32. What is lanthanide contraction ?
Ans. Steady decrease in atomic and ionic radii with increasing atomic number is called lanthanide contraction.
Q. 33. Why lantanide ions are coloured ?
Ans. Lanthanide ions are coloured due to f + transitions.
Q. 34. Give two methods by which lanthanides are extracted from monazite sand.
Ans. (i) Solvent extraction method
- (ii) Ion exchange method.
Q. 35. In actinides which sub shell is filled successively ?
Ans. In actinides 5f-sub shell is filled successively.
Q. 36. How many elements are there in actinide series.
Ans. Elements from thorium (Th90) to law rencium (Lw103) along with actinium (AC89), there are fifteen elements inactinide series.
Q. 37. Which oxidation state is most stable incase of actinide series ?
Ans. +3 oxidation state is most. Stable in actinide series.
Q. 38. Which oxidation state is more stable in Thorium ?
Ans. +4 oxidation state is more stable in case of thorium.
Q. 39. Which among the following elements belongs to actinide series ?
W, Ru, Rh, Zr, Gd, Pu
Ans. Plutonium (Pu) belongs to actinide series.
Q. 40. Give Arrhenius concept of acids and bases ?
Ans. According to Arrhenius theory of ionisation all substances which give hydrogen ions when dissolved in water, could be called acids while those which ionise in water to form hydroxyl group, could be called bases.
Q. 41. Give two examples each of Arrhenius acids or bases.
Ans. Arrhenius Acids HCl, H2SO4
Arrhenisus Bases NaOH, NH4OH
Q. 42. What do you mean by proton-donor accepter concept ?
Ans. An acid is any hydrogen contained substances (a molecule or a cation or an anion) where as a base is any substance (a molecule or a cation or an anion) thats can accept a proton from any other substance.
Q. 43. What do you mean by conjugate acid-base pairs ?
Ans. Pairs of substances which are formed from one another by the gain or loss of a proton, are called conjugate acid-base pairs.
Q. 44. What do you meant by levelling effect of the solvent.
Ans. Phenomenon in which the strenght of all the acids becomes equal to than of H3O+ ion, is kown as levelling effect of the solvent.
Q. 45. Give the Cady Elsey acid-base concept.
Ans. Substance which increase the concentration of the cation characteristics of the solvent, are acids whereas substances which increase the concentration of the anion characteristics of the solvent are bases.
Q. 46. Write a note on Lewis concept of acids and bases.
Ans. According to Lewis a base is a substance that has an unshared electron pair with which it can forma covalent bond with an atom, molecule or ion an acid is a substance that can form a covalent bond by accepting an electron pair from a base.
Q. 47. Out of BF3 and NH3 which is an acid and which is a base ?
Ans. BF3 is an acid because it accept lone electron pair and NH3 is a base because it gives an electron pair.
Q. 48. The following equilibrium is established when hydrogen chloride is dissolved in acetic acid.
HCl + CH3COOH ====== Cl– + CH3COOH2+ Give the set than characterises conjugate acid-base pairs. Ans. The set that characterises acid-base piar are
HCl, Cl‒ and CH3COOH2+, CH3COOH
Q. 49. Why Lowry and Bronsted recognize ammonia molecule as a base ?
Ans. Lowry and Bronsted recognize ammonia molecule as above because is combines with hydrogen ion to form the ammonium ion.
Q. 50. Give the equation for the auto-ionisation of liquid NH3 and H2O.
Ans. Auto-ionisation of liquid NH3
NH3 + NH3 ======== NH4+ + NH2‒
Auto ionisation of water
H2O + H2O ======== H3O+ + OH‒
Q. 51. Give the names of the reactions with occur in liquid ammonia.
Ans. Following types of reactions are occured in liquid ammonia :
- (a) Precipitation Reactions (b) Ammonolysis
- (c) Redox Reactions (d) Complex formation reactions
- (e) Neutralisation reactions.
Q. 52. Give the names of the general properties of ionisating solvents.
Ans. (a) Dipolemoment.
- (b) Dielectric Constant
- (c) Electrical Conductance
- (d) Viscosity
- (e) Proton Offinity.
Q. 53. Give equation for the auto-ionisation of liquid sulphurdi-oxide.
Ans. Auto ionisation of liquid sulphur dioxide.
SO2 + SO2 ======= SO2+ + SO32‒
Thionyl ion Sulphite ion
Q. 54. Mention one advantage of liquid ammonix over water as a solvent.
Ans. Liquidammonia dissolves alkalimetals without apperent chemical action and the dissolved metals may be recovered unchange by simply allowing the solvent to evaporate.
B. Sc. II, Chemistry I
121