C H A P T E R

09
Coordination
Compounds
A Quick Recapitulation of the Chapter
1. Coordination compounds are those in which a central 9. Werner’s Theory
metal atom or ion is attached with a fixed number of
(i) According to Werner’s theory of coordination
groups or molecules (ligands) through coordinate compounds, metals possess two types of valencies,
bonds. called primary or ionisable valency (oxidation number)
2. Double salts when dissolved in water dissociate and secondary or non-ionisable valency (coordination
almost completely into ions in the same way in which number).
the individual components of the double salts (ii) Every metal atom must satisfy both of its valencies.
combine. Primary valencies are non-directional.
3. In complex compounds, the central metal atom or ion Secondary valencies are directed in space towards
can act as Lewis acid and ligands act as Lewis base. fixed position giving a definite shape to the complex.
4. Molecules which can donate a lone pair of electron 10. The positive part of a coordination compound is
to the central atom or ion are called ligands. named first and is followed by the name of negative
5. Chelating ligand is a neutral molecule, free radical or part.
ion with two or more lone pairs of electrons on 11. The ligands are named first followed by the central
different atoms. metal. The prefixes di, tri etc., are also used to
6. The charge present on central metal atom or ion is indicate number of each kind of ligand present.
called oxidation number of central atom. The oxidation state of central metal is indicated in
7. Complexes in which a metal is bound to roman numbers in parantheses.
(i) only one kind of ligands → homoleptic complexes, If a complex species has negative charge, the name
e.g. [Co(NH3 )6 ]3+ . of the central metal ends in -ate.
(ii) more than one kind of donor group → heteroleptic 12. Isomerism geometrical isomerism arises when two
complexes, e.g. [Co(NH3 )4 Cl2 ]+ . identical ligands occupy adjacent positions to each
8. (i) The part of the coordination which is present other (cis-isomer) or placed opposite to each other
outside the square brackets is known as ionisation (trans-isomer)
sphere. Square planar complexes of the type MA2 X 2,
(ii) The central atom/ion and the ligands present in MABX 2, MA2 XY and MABXY and octahedral
the square brackets are collectively known as complexes of the type MA4 X 2 , MA3 X 3 show
coordination sphere. geometrical isomerism.
13. Compounds of type MA2 X 2Y 2, MA2 X 2YZ , MA2 XYZL, 21. Crystal Field Theory It is a more appropriate
MABXYZL, M ( AA) 3 , M ( AA) 2 X 2 shows optical isomerism. theory than VBT. According to CFT, under the
influence of ligand field, degeneracy of the
14. Ionisation isomerism arises when complex compounds with d-orbital is destroyed and it splits into two or more
same molecular formula give different ions in solution. energy levels.
15. Linkage isomerism arises when a complex contains a
22. The extent of splitting depends on the strength of
monodentate ligand with more than one donor atoms.
ligand. A strong ligand causes greater splitting
16. Coordination isomerism arises when there is exchange of while a weak ligand causes smaller splitting.
ligands between complex cation and complex anion.
23. The difference of energy between the two sets of
17. Hydrate isomerism arises when different number of water d-orbital (in octahedral complexes) is called
molecules present in the coordination sphere. crystal field splitting energy (CFSE) or ∆ . The
18. Valence Bond Theory (VBT) According to valence bond magnitude of ∆o depends upon the nature of the
theory (VBT), the metal atom or ion under the influence of ligands.
ligands can use (n − 1) d ,ns, np, nd orbitals for 24. The increasing order of ∆o is given below:
hybridisation to yield square planar, tetrahedral or
I− < Br − < S2 − < SCN− < Cl− < F − < OH− < C 2O24 –
octahedral geometry. These hybridised then overlap with
ligand orbitals which can donate an electron pair for < H2O < NCS− < NH3 < en < NO−2 < CN− < CO
bonding. This series is called spectrochemical series.
19. A complex compound may be inner orbital complex or outer 25. Bonding in metal carbonyls The metal carbon
orbital complex depending upon whether d-orbitals of the bond in metal carbonyl possess both σ and
inner shell or outer shell are involved in hybridisation. Inner π-character. The M − C, σ-bond is formed by the
orbital complex is generally high spin complex while outer donation of lone pair of electrons on the
orbital complex is generally high spin complex. carbonyl carbon into a vacant orbital of the
20. If no unpaired electron is present, the complex is metal.
diamagnetic. If unpaired electron/s is/are present, 26. Haemoglobin is a complex of iron, vitamin B12 is a
complex is paramagnetic and its magnetic moment complex of Co, chlorophyll is a complex of Mg.
µ = n(n + 2) where, n is the number of unpaired electrons Wilkinson catalyst is used for the hydrogenation
present. of alkenes.

Objective Questions Based on NCERT Text

Topic 1
Werner’s Theory of Coordination Compounds
1. Primary valency of binary compounds CrCl 3 , CoCl 2 and 3. Which one of the following complexes will
PdCl 2 are consume more equivalents of aqueous solution of
AgNO 3 ? (JEE Main 2016)
(a) 2, 2 and 3 respectively (b) 2, 3 and 2 respectively
(c) 3, 3 and 2 respectively (d) 3, 2 and 2 respectively (a) Na 2 [CrCl 6 ]
(b) [Cr(H2 O)5 Cl]Cl 2
2. Given, the molecular formula of some hexa coordinated
complex (c) [Cr(H2 O)6 ]Cl 3
A. CoCl 3 ⋅ 6NH3 B. CoCl 3 ⋅ 5NH3 d) Na 2 [CrCl 5 (H2 O)]
C. CoCl 3 ⋅ 4NH3 4. According to Werner’s theory, the primary
If the number of coordinated NH 3 molecules in A, B and C valencies of the central metal atom
respectively are 6, 5 and 4, the primary valency, in (A), (B) (a) are satisfied by negative ions or neutral molecule
and (C) respectively (b) are satisfied by negative ions
(a) 6, 5 and 4 (b) 3, 2 and 1 (c) are equal to its coordination number
(d) decide the geometry of the complex
(c) 0, 1 and 2 (d) 3, 3 and 3
 5. Werner proposed the term secondary valency 8. The primary and secondary valencies of chromium in
for the the complex ion, dichlorodioxalato chromium (III)
(a) number of groups bound indirectly to the metal ion respectively are
(b) number of groups bound directly to the metal ion (a) 3 and 4 (b) 4 and 3 (c) 6 and 3 (d) 3 and 6
(c) total number of groups bound directly as well as
9. What is/are true about the double salt and a complex
indirectly to the metal ion
ion?
(d) None of the above
(a) Both are formed by the combination of two or more
6. According to postulates of Werner’s theory for stable compounds in stoichiometric ratio
coordination compounds, which of the following is (b) Double salts do not dissociate into simple ions when
true? dissolved in water
(a) Primary valencies are ionisable (c) Complex ion dissociate into simple ions when
dissolved in water
(b) Secondary valencies are ionisable
(d) All of the above are true
(c) Only primary valencies are non-ionisable
(d) Primary and secondary valencies are non-ionisable 10. Categorise the following compounds into double salts
3+ – and complex ions.
7. Consider the formula [Co(NH 3 ) 6 ] (3Cl ).
I. [Fe(CN) 6 ] 4– II. [KCl ⋅ MgCl 2⋅ 6H2 O]
In this formula the species within the square
bracket and the ions outside the square bracket III. FeSO4⋅ (NH4 ) 2 SO4⋅ 6H2 O IV. KAl(SO4 ) 2 ⋅ 12H2 O
respectively Choose the correct option.
(a) Coordination entities and counterions (a) I, II – complex ions; III, IV – double salts
(b) Counterions and coordination entities (b) II, III – complex ions; I, IV – double salts
(c) Counterions and counter entities (c) I – complex ion; II, III, IV – double salts
(d) Coordinated ions and counter entities (d) I, II, IV – complex ions; III – double salt

Topic 2
Some Important Terms to Coordination Compounds
11. One mole of the complex compound Co(NH 3 ) 5 Cl 3 , 14. Which of the following is hexadentate ligand?
gives 3 moles of ions on dissolution in water. One (a) Ethylene diamine
mole of the same complex reacts with 2 moles of (b) Ethylene diamine tetra acetic acid
AgNO 3 solution to yield 2 moles of AgCl( s). The
(c) 1,10-phenanthroline
structure of the complex is
(d) Acetyl acetonate
(a) [Co(NH3 )5 Cl]Cl 2 (b) [Co(NH3 )3 Cl 2 ] ⋅ 2NH3
(c) [Co(NH3 )4 Cl 2 ]Cl ⋅ NH3 (d) [Co(NH3 )4 Cl]Cl 2 ⋅ NH3 15. Classify the following structures into
3+ 4+ nitrito–N(A), nitrito–O(B), thiocyanato (C), and
12. Both Co and Pt have a coordination number of isothiocyanato (D)
six. Which of the following pairs of complexes will
show approximately the same electrical conductance I. M ← O — N == O
for their 0.001 M aqueous solution? II. M ← SCN
(a) CoCl 3 ⋅ 4NH3 and PtCl 4 ⋅ 4NH3 III. M ← NCS
(b) CoCl 3 ⋅ 3NH3 and PtCl 4 ⋅ 5NH3 O
(c) CoCl 3 ⋅ 6NH3 and PtCl 4 ⋅ 5NH3 IV. M ← N
(d) CoCl 3 ⋅ 6NH3 and PtCl 4 ⋅ 3NH3 O
13. Ligands can be small ions small molecules, larger Choose the correct option.
molecules etc. The correct example if are
(a) I – A; II – B; III – C; IV – D
(a) simple ions like Cl − , small molecules like H2 O
(b) I – B; II – C; III – D; IV – A
(b) larger molecules like H2 NCH2 CH2 NH2
(c) macromolecules like proteins (c) I – D; II – C; III – B; IV – A
(d) All of the above (d) I – C; II – A; III – D; IV – B
16. The correct structure of ethylenediaminetetraacetic 23. The coordination number of a central metal atom in a
acid (EDTA) is complex is determined by
HOOCCH2 CH2 COOH (a) the number of ligands around a metal ion bonded by
(a) N — CH == CH — N σ-bonds
(b) the number of ligands around a metal ion bonded by
HOOCCH2 CH2 COOH
π-bonds
HOOC COOH (c) the number of ligands around a metal ion bonded by
(b) N — CH2 — CH2 — N σ and π bonds both
HOOC COOH (d) the number of only anionic ligands bonded to the metal ion

HOOCCH2 CH2 COOH 24. A chelating agent has two or more than two donor
(c) N — CH2 — CH2 — N atoms to bind to a single metal ion. Which of the
HOOCCH2 CH2 COOH following are chelating agents?
COOH I. Oxalato
H2C II. Glycinato
(d) HOOC—H2C H III. Ethane-1,2-diamine
N—CH — CH—N
H CH2 — COOH III. Thiosulphato
CH2
HOOC Choose the correct option.
(a) I, II and III (b) II, III and IV
17. Among the properties ( A ) reducing, ( B ) oxidising, (c) I, III and IV (d) I, II and IV
(C ) complexing, the set of properties shown by 25. Coordination number of Fe in [Fe(CN) 6 ]4− and
CN − ion towards metal species is [Fe(CN) 6 ]3− are respectively
(a) B and C (b) A and C (c) A, B and C (d) A and B (a) 2 and 3 (b) 6 and 6
18. Coordination number (CN) of a metal ion in a (c) 6 and 3 (d) 6 and 4
complex can be defined as 26. In the complex K 4 [Fe(CN) 6 ] identify the
(a) number of ligand donor atoms coordination sphere and the counter ion.
(b) number of ligand donor atoms to which the metal is
(a) K + coordination sphere; [Fe(CN)6 ]4− counter ion
indirectly bonded
(c) number of ligand donor atoms to which the metal is (b) [Fe(CN)6 ]4− coordination sphere; K + counter ion
directly bonded (c) (CN)6 coordination sphere; K + counter ion
(d) None of the above (d) None of the above
19. The most stable system for a chelate is 27. The oxidation number of the central atom in a
(a) five fused cyclic system complex is defined as the charge it would carry
(b) four fused cyclic system (a) if all the ligands are removed along with the electron
(c) three fused cyclic system pairs that are donated by the central atom
(d) two fused cyclic system (b) if all the ligands are removed along with the electron
pairs that are shared with the central atom
20. A ligand can also be regarded as (c) if all the ligands are removed from the central metal atom
(a) Bronsted base (b) Bronsted acid
(d) if all the electrons pairs are shared with the central atom
(c) Lewis base (d) Lewis acid
21. Coordination number of Pt and Ni in [PtCl 6 ]2– and 28. The oxidation state of Fe in the brown ring complex
[Ni(NH 3 ) 4 ] 2+
are respectively. [Fe(H 2O) 5 NO]SO 4 is
(a) +3 (b) 0 (c) +2 (d) +1
(a) 4 and 6 (b) 6 and 4 (c) 4 and 4 (d) 6 and 6
29. A complex compound in which the oxidation number
22. In the complex ions, [Fe(C 2O 4 ) 3 ]3– and [Co(en) 3 ]3+
of a metal is zero, is
the coordination number of Fe and Co is 6. Why?
(a) [Ni(CO)4 ] (b) [Pt(NH3 )4 ]Cl 2
(a) Because C2 O2–4 and ethane-1,2-diamine are bidentate
ligands (c) K 3 [Fe(CN)6 ] (d) K 4 [Fe(CN)6 ]
(b) Because C2 O2–
4 and ethane-1,2-diamine are ambidentate 30. The pair of compounds having metals in their highest
ligands oxidation states is
(c) Because C2 O2–
4 and ethane-1,2-diamine are polydentate (a) [MnO4 ]– , CrO2 Cl 2 (b) [NiCl 4 ]2– , [CoCl 4 ]–
ligands
(d) None of the above (c) MnO2 , FeCl 3 (d) [Fe(CN)6 ]3– , [Co(CN)6 ]−3
31. The equation which is balanced and represents the 32. [Co(NH 3 ) 6 ]3+ and [Co(NH 3 ) 4 Cl 2 ]+ are the
correct product (s) is (JEE Main 2014) examples of
(a) Li 2 O + 2KCl → 2LiCl + K 2 O (a) homoleptic complexes
(b) [ CoCl(NH3 )5 ]+ + 5H+ → Co 2 + + 5NH+4 + Cl − (b) heteroleptic complexes
(c) [Co(NH3 )6 ]3+ homoleptic complex; [Co(NH3 )4 Cl 2 ]+
(c) [Mg(H2 O)6 ]2 + + (EDTA)4 − →
Excess NaOH
heteroleptic complex
[ Mg (EDTA)]2 + + 6H2 O (d) [Co(NH3 )6 ]3+ heteroleptic complex; [Co(NH3 )4 Cl 2 ]+
(d) CuSO4 + 4 KCN → K 2 [ Cu(CN)4 ] + K 2 SO4 homoleptic complex

Topic 3
Nomenclature of Coordination Compounds
33. Which of the following statement is incorrect for (b) [PtCl 2 (en)2 ](NO3 )2 is the molecular formula of
nomenclature of coordination compounds? dichlorido bis- (ethane-1,2-diammine) platinum (IV)
(a) The cation is named first in both positively and nitrate
negatively charged coordination entities (c) Molecular formula of iron (III) hexacyanoferrate (II) is
(b) The ligands are named in an alphabetical order before Fe4 [Fe(CN)6 ]3
the name of central atom/ion (d) If complex name starts with metal, it is an anionic complex
(c) Names of anionic ligands with electron 39. In the given compound [Pt(NH 3 ) 2 Cl(NH 2 ) (CH 3 )] Cl
(d) Prefixes mono, di, tri etc., are used to indicate the identify the metal ( A ), counter ion ( B ) and ligands (C ).
number of the individual ligands in coordination (a) A → Cl; B → Pt; C → NH3
34. The neutral ligands NH 3 , CO and NO respectively (b) A → Pt; B → Cl; C → NH3 ,Cl,NH2 and CH3
are named is (c) A → Pt; B → Cl; C → NH3 , Cl
(a) ammine, carbonyl and nitrosyl (d) A → Cl; B → Pt; C → NH2 ,CH3
(b) amine, carbonyl and nitrito
(c) amine, carboxyl and nitrito OH2
(d) amine, carboxyl and nitrosyl H 3N NH3
Cl
35. If the complex ion is anion, the name of the metal ends 40. Give the name of Cl Co
Cl
with the suffix-ate, for cationic complex ion, H 3N NH3
(a) the metal is named same as the element OH2
(b) the name of the metal ends with the seffix -ate (a) tetraammineaquacobalt chloride
(c) the metal is named ends with the suffix −O (b) tetraamminediaquacobalt (III) chloride
(d) None of the above (c) tetraamminediaquacobalt (IV) chloride
(d) None of the above
36. Name the compound [Co(en) 2 Cl 2 ]2 SO 4
41. The IUPAC name of the complex [Co(NH 3 ) 4 Cl 2 ]Cl is
(a) dichloroethylenediamminecobalt (III) sulphate
(b) dichlorobisethylenediamminecobalt (III) sulphate (a) tetraammine dichlorocobalt (III) chloride
(b) tetraammine dichlorocobalt (IV) chloride
(c) 2-bisdichloroethylenediamminecobalt (III) sulphate
(c) tetraammine dichlorocobalt (II) chloride
(d) dichlorobisethylenediamminecobalt (II) sulphate
(d) dichloro tetraammine cobalt (II) chloride
37. Which among the following will be named as 42. Ammonia gas does not evolve from the complex
dibromidobis (ethylenediammine) chromium (III) FeCl 3 ⋅ 4NH 3 but it gives white precipitate with
bromide ? (AIEEE 2012) aqueous solution of AgNO 3 . Coordination number of
(a) [ Cr(en)3 ] Br3 (b) [Cr(en)2 Br2 ]Br central metal ion in above complex is six.
(c) [ Cr(en)Br4 ]− (d) [ Cr(en)Br2 ]Br Give IUPAC name of the complex.
(a) Ammonium trichloro triammine iron (III)
38. Identify the false statement. (b) Tetraammine iron (III) chloride
(a) Molecular formula of tetraamminediaquacobalt (III) (c) Dichlorotetraammine ferrate (II) chloride
chloride is [Co(NH3 )4 (H2 O)2 Cl 3 ] (d) Dichlorotetraammine iron (III) chloride
43. The IUPAC name of compound K 3 [Fe(CN) 5 NO] is 49. Write the IUPAC name of K 2 [Zn(OH) 4 ] .
(a) pentacyanonitrosylpotassium ferrate (II) (a) Potassium tetraaquazincate (III)
(b) potassiumcyanopentanitrosyl ferrate (II) (b) Potassium tetrahydroxidozincate (III)
(c) potassiumpentacyanonitrosyl ferrate (III) (c) Potassium tetraaquazincate (II)
(d) potassiumpentacyanonitrosyl ferrate (II) (d) Potassium tetrahydroxidozincate (II)
44. The IUPAC name of [Ni(PPh 3 ) 2 Cl 2 ] is 50. In [Co(H 2 NCH 2CH 2 NH 2 ) 3 ]2 (SO 4 ) 3 , the counterion is
−
(a) bis-dichloro (triphenylphosphine) nickel (II) (a) CH, NH (b) CH3 (c) NH2 (d) SO4
(b) dichlorobis (triphenylphosphine) nickel (II)
51. The name of the complex [Al(OH) 2 (H 2O) 4 ]SO 4 is
(c) dichlorotriphenylphosphine nickel (II)
(d) triphenylphosphine nickel (II) dichloride (a) dihydroxotetrahydridealuminium (III) sulphate
(b) tetraaquadihydroxyaluminium (III) sulphate
45. According to IUPAC nomenclature sodium
(c) tetraaquadihydroxyaluminium (IV) sulphate
nitroprusside is named as
(d) None of the above
(a) sodium pentacyanonitrosyl ferrate (II)
(b) sodium pentacyanonitrosyl ferrate (III) 52. What is the IUPAC name of Hg[Co(SCN) 4 ] ?
(c) sodium nitroferricyanide (a) Mercury tetrathiocyanato cobalt (II)
(d) sodium nitroferrocyanide (b) Mercury tetrathiocyanato cobalt (III)
46. IUPAC name of Na 3 [Co(NO 2 ) 6 ] is (c) Mercury tetrathiocyanato cobaltate (III)
(d) Mercury tetracyanato cobaltate (III)
(a) sodium hexanitrito cobaltate (II)
(b) sodium hexanitro cobaltate (III) 53. Write the formula for the coordination compund
(c) sodium hexanitrito cobaltate (II) dichloridobis (ethane-1, 2-diamine) platinum (IV) nitrate
(d) sodium cobaltinitrite (II) (a) [PtCl 2 (en)2 ](NO3 )2 (b) [PtCl 2 (NO3 )2 ](en)2
(c) [Pt(en)2 (NO3 )]Cl 2 (d) None of these
47. Write the formula for coordination compound
diamminesilver (I) dicyanoargentate (I). 54. The IUPAC name of coordination compound
(a) [Ag(NH3 )2 ][Ag(CN)2 ] (b) [Ag(NH3 )3 ][Ag(CN)2 ] K 3 [Al(C 2O 4 ) 3 ] is
(a) potassium trioxalatoaluminate (III)
(c) [Ag(NH3 )2 ][Ag(CN)3 ] (d) [Ag(NH2 )2 ][Ag(CN)2 ]
(b) potassium hexacarbonylaluminate (III)
48. Write the formula for coordination compound (c) trioxalato aluminium III potassium
named as tris (ethane-1,2-diammine) cobalt (III) (d) potassium trioxalato aluminate (II)
sulphate.
55. Write the IUPAC name of the coordination compound
(a) [Co (H2 N CH2 CH2 NH2 )3 ]2 (SO4 )3
Fe 4 [Fe(CN) 6 ]3 .
(b) [Co(H2 NCH2 CH2 NH2 )3 ]SO4 (a) Iron (II) hexacyanidoferrate (II)
(c) [Co(H2 NCH2 CH2 NH3 )2 ](SO4 )3 (b) Iron (III) hexacyanidoferrate (II)
(d) [Co(H2 NCH2 CH2 NH2 )3 ](SO4 )3 (c) Iron (II) hexacyanoferrate (III)
(d) Iron (II) hexacyanoiron (III)

Topic 4
Isomerism in Coordination Compounds
56. Choose the odd one out Choose the correct option.
(a) solvate isomerism (b) coordination isomerism (a) I and II (b) II and III
(c) ionisation isomerism (d) optical isomerism (c) I and III (d) II and IV

57. Identify the geometrical isomers of Pt[(NH 3 ) 2 Cl 2 ] 58. Which type of isomerism arises when bidentate ligands
(L–L) are present in complexes of formula
Cl NH3 Cl NH3 [ MX 2 ( L − L) 2 ]?
I. Pt ; II. Pt (a) Optical isomerism
Cl NH3 H3 N Cl
NH3 Cl (b) Geometrical isomerism
Cl Cl
III. Pt ; IV. Pt (c) Linkage isomerism
H3N NH3 Cl NH3 (d) Solvate isomerism
 – 3–
59. The existence of two different coloured complexes NH3
en
with the composition [Co(NH 3 ) 4 Cl 2 ]+ is due to NC NH3 Cl
(CBSE AIPMT 2012) 65. Fe Cr
(a) ionisation isomerism NC CN Cl
en
(b) linkage isomerism CN
(c) geometrical isomerism A B
(d) coordination isomerism 3– –
Cl NH3
60. Which one amongst the following, exhibit NC CN
geometrical isomerism? en Co en Fe
NC CN
(a) [Co(NH3 )5 Br]SO4
Cl NH3
(b) [Co(EDTA)]−
C D
(c) [Cr(SCN)6 ]3–
(d) [Pt(NH3 )2 Cl 2 ] In the above structures identify the cis- and
trans-isomers.
61. Identify the facial (fac) and meridional (mer) isomers (a) cis-isomers ⇒ B, C; trans-isomers ⇒ A, D
from the following. (b) cis-isomers ⇒ A, D; trans-isomers ⇒ B, C
NH3 NH3 (c) cis-isomers ⇒ A, B; trans-isomers ⇒ C, D
O2 N NH3 O2 N NO2 (d) cis-isomers ⇒ A, C; trans-isomers ⇒ B, D

Co Co 66. The complex, [Pt(py)(NH 3 )BrCl] will have how many

O2 N O2 N NO2 geometrical isomers? (CBSE AIPMT 2012)
NH3
NO2 NH3 (a) 2 (b) 3
(c) 4 (d) 0
A B
(a) mer- mer- (b) fac- fac- 67. Which of the following coordination compounds
(c) mer- fac- (d) fac- mer- would exhibit optical isomerism?
(a) Pentaamminenitrocobalt (III) iodide
62. Which one of the following has an optical isomer ?
(b) Diamminedinitroplatinum (II)
(en = ethylenediamine) (AIEEE 2011) (c) trans-dicyanobis (ethylenediamine)
(a) [Zn(en)(NH3 )2 ]2+ (d) Tris-(ethylenediamine) cobalt (III) bromide
(b) [Co(en)3 ]3+ 68. Which type of Ligand is contained in linkage
(c) [Co(H2 O)4 (en)]3+ isomerism?
(d) [Zn(en)2 ]2+ (a) Unidentate (b) Bidentate
(c) Polydentate (d) Ambidentate
63. The correct statement on the isomerism associated
with the following complex ions, (JEE Main 2016) 69. Coordination isomerism arises
2+ (a) when the counter ion in a complex salt is itself a
I. [Ni(H2 O) 5 NH3 ]
potential ligand
II. [Ni(H2 O) 4 (NH3 ) 2 ] 2+ (b) from the interchange of ligands between cationic and
III. [Ni(H2 O) 3 (NH3 ) 3 ] 2+ anionic entities of different metal ions present in a
(a) I and II show geometrical and optical isomerism complex
(b) II and III show only optical isomerism (c) Both (a) and (b) are correct
(c) II and III show only geometrical isomerism (d) None of the above
(d) II and III show geometrical and optical isomerism 70. The complexes [Co(NH 3 ) 6 ][Cr(CN) 6 ] and
64. In a coordination entity of the type [PtCl 2 (en) 2 ]2+ , [Cr(NH 3 ) 6 ][Co(CN) 6 ] are the examples of which
optical activity is shown by type of isomerism? (CBSE AIPMT 2012)
(a) only cis-isomer (a) Geometrical isomerism
(b) only trans-isomer (b) Linkage isomerism
(c) both cis- and trans-isomers (c) Ionisation isomerism
(d) None of the above (d) Coordination isomerism
71. The pair of [Co(SO 4 )(NH 3 ) 5 ]Cl and 73. Which one of the following complexes shows optical
[CoCl(NH 3 ) 5 ]SO 4 constitutes isomerism? (JEE Main 2016)
(a) optical isomers (b) linkage isomers (a) cis- [Co(en)2 Cl 2 ]Cl
(c) coordination isomers (d) ionisation isomers (b) trans- [Co(en)2 Cl 2 ]Cl
72. [Co(NH 3 ) 5 SO 4 ]Br and [Co(NH 3 ) 5 Br]SO 4 are (c) [Co(NH3 )4 Cl 2 ]Cl
(a) linkage isomers (b) coordination isomers (d) [Co(NH3 )3 Cl 3 ]
(c) ionisation isomers (d) solvate isomers

Topic 5
Bonding in Coordination Compounds
74. Which one of the following is an outer orbital (b) tetrahedral, square planar, octahedral
complex and exhibits paramagnetic behaviour? dsp 2 , sp 3 , sp 3 d 2 : 0, 5.9, 4.9
(CBSE AIPMT 2012) (c) square planar, tetrahedral, octahedral
dsp 2 , sp 3 , d 2 sp 3 : 5.9, 4.9, 0
(a) [ Ni(NH3 ) 6 ]2+ (b) [ Zn(NH3 ) 6 ]2+
(d) square planar, tetrahedral, octahedral
(c) [ Cr(NH3 ) 6 ]3+ (d) [ Co(NH3 ) 6 ]3+ dsp 2 , sp 3 , sp 3 d 2 : 0, 5.9, 4.9
75. Red precipitate is obtained when ethanol solution of 79. The pair having the same magnetic moment is
dimethylglyoxime is added to ammoniacal Ni(II). [atomic number, Cr = 24, Mn = 25, Fe = 26 and
Which of the following statements is not true? Co = 27] (JEE Main 2016)
(CBSE AIPMT 2010)
(a) [Cr(H2 O)6 ]2+ and [Fe(H2 O)6 ]2+
OH
H3C  C == N (b) [Mn(H2 O)6 ]2+ and [Cr(H2 O)6 ]2+
Dimethylglyoxime =  (c) [CoCl 4 ]2− and [Fe(H2 O)6 ]2+
H3C  C == N
OH (d) [Cr(H2 O)6 ]2+ and [CoCl 4 ]2−

(a) Red complex has a square planar geometry 80. For d 4 ions, the number of possible patterns of
(b) Complex has symmetrical H-bonding electron distribution arises is
(c) Red complex has a tetrahedral geometry (a) two (b) three
(d) Dimethylglyoxime functions as bidentate ligand (c) four (d) five
76. The d-electron configuration of Cr 2+ , Mn 2+ , Fe 2+ 81. For d 4 ions, the fourth electron enters one of the eg
2+ 4 5 6 7
and Co are d , d , d and d respectively. Which orbitals giving the configuration t 23g eg1 , when
one of the following will exhibit minimum (a) ∆ o > P (b) ∆ o < P
paramagnetic behaviour? (c) ∆ o = P (d) ∆ o ≥ P
(Atomic number of Cr = 24, Mn = 25, Fe = 26, Co = 27) 82. Which one of these statements about [Co(CN) 6 ]3− is
(CBSE AIPMT 2011) true? (NEET 2016)
(a) [Cr(H2 O)6 ] 2+
(b) [Mn(H2 O)6 ]2+ (a) [Co(CN)6 ]3− has no unpaired electrons and will be in a
low-spin configuration
(c) [Fe(H2 O)6 ]2+ (d) [Co(H2 O)6 ]2+
(b) [Co(CN)6 ]3− has four unpaired electrons and will be in a
77. Which of the following complex ions is diamagnetic low-spin configuration
in nature? (CBSE AIPMT 2011) (c) [Co(CN)6 ]3− has four unpaired electrons and will be in
3–
(a) [CoF6 ] (b) [NiCl 4 ]2− a high-spin cofiguration.
(c) [Ni(CN)4 ] 2−
(d) [CuCl 2 ]2 − (d) [Co(CN)6 ]3− has no nupaired electrons and will be in a
high-spin configuration.
78. [Ni(CN) 4 ]2− , [MnBr 4 ]2− and [CoF6 ]3− , geometry,
83. Among the ligands NH 3 , en, CN − and CO, the correct
hybridisation and magnetic moment of the ions
order of their increasing field strength, is (AIEEE 2011)
respectively, are
(a) tetrahedral, square planar, octahedral (a) CO < NH3 < en < CN− (b) NH3 < en < CN− < CO
−
sp 3 , dsp 2 , sp 3 d 2 : 5.9, 0, 4.9 (c) CN < NH3 < CO < en (d) en < CN− < NH3 < CO
84. Identify the correct trend given below 91. Cobalt (III) chloride forms several octahedral
(Atomic number, Ti= 22, Cr = 24 and Mo = 42) complexes with ammonia. Which of the following
(JEE Main 2016) will not give test for chloride ions with silver nitrate
(a) ∆ o of [Cr(H2 O)6 ]2 + < [Mo(H2 O)6 ]2 + at 25°C? (NEET 2016)
and ∆ o of [Ti(H2 O)6 ]3 + < [Ti(H2 O)6 ]2 + (a) CoCl 3 ⋅ 3NH3 (b) CoCl 3 ⋅ 4NH3
(c) CoCl 3 ⋅ 5NH3 (d) CoCl 3 ⋅ 6NH3
(b) ∆ o of [Cr(H2 O)6 ] 2+
> [Mo(H2 O)6 ] 2+

and ∆ o of (Ti(H2 O)6 )2 + > [Mo(H2 O)6 ]2 + 92. Which of the following statement(s) is/are not true when
EDTA solution is added to Mg 2+ ion solution?
(c) ∆ o of [Cr(H2 O)6 ]2 + > [Mo(H2 O)6 ]2 +
(a) pH of the solution decreases when Mg 2+ ion is present in
and ∆ o of [Ti(H2 O)6 ]2 + < [Mo(H2 O)6 ]2 + hard water
(d) ∆ o of [Cr(H2 O)6 ]2 + < [Mo(H2 O)6 ]2 + (b) Four coordinate sites of Mg 2+ occupied by EDTA and
and ∆ o of [Ti(H2 O)6 ]3+ > [Ti(H2 O)6 ]2 + remaining two sites occupied by water molecules
(c) Colourless [Mg-EDTA] 2− chelate is formed
85. Which of the following complex ion is not expected
(d) All six coordinate sites of Mg 2+ are occupied
to absorb visible light? (CBSE AIPMT 2010)
2+
(a) [Ni(H2 O)6 ] (b) [Ni(CN)4 ]2− 93. Which of the following complex ions has electrons
3+ 2+ that are symmetrically filled in both t 2g and
(c) [Cr(NH3 )6 ] (d) [Fe(H2 O)6 ]
eg orbitals? (JEE Main 2015)
86. When concentrated HCl is added to an aqueous (a) [FeF3 ]3− (b) [Mn(CN)6 ]4−
solution of CoCl 2 , its colour changes from reddish (c) [CoF6 ]3− (d) [Co(NH3 )6 ]2+
pink to deep blue. Which complex ion gives blue
colour in this reaction? (JEE Main 2015) 94. Metal-carbon bond in metal carbonyls possess
(a) [CoCl 4 ]2- (b) [CoCl 6 ]3- (a) only σ-character (b) only π-character
(c) [CoCl 6 ] 4-
(d) [Co(H2 O)6 ] 2+ (c) Both σ and π-character (d) Neither σ nor π-character

87. Crystal field stabilisation energy for high spin d 4 95. The M — C π-bond is formed by the
octahedral complex is (CBSE AIPMT 2010) (a) donation of a pair of electrons
(a) −0.6∆ o (b) −1.8∆ o (b) sharing of a pair of electrons
(c) −1.6 ∆ o + P (d) −1.2 ∆ o (c) receiving a pair of electrons
(d) None of the above
88. Atomic number of Mn, Fe and Co are 25, 26 and 27
respectively. Which of the following inner orbital 96. Consider the following figure.
octahedral complex ions are diamagnetic? π∗
π π
I. [Co(NH3 ) 6 ] 3+ II. [Mn(CN) 6 ] 3−
σ
III. [Fe(CN) 6 ] 4− IV. [Fe(CN) 6 ] 3− M C O
Choose the correct option. π
(a) II and III (b) I and IV
(c) I and III (d) II and IV Which type of bond formed between metal and
ligand?
89. A magnetic moment of 1.73 BM will be shown by
(a) synergic bond (b) σ-bond
one among the following. (CBSE AIPMT 2012)
(c) π-bond (d) None of these
(a) [ Cu(NH3 )4 ]2+ (b) [ Ni(CN )4 ]2−
97. Decacarbonyldimanganese(0) is made up of two
(c) TiCl 4 (d) [ CoCl 6 ]4− square pyramidal Mn(CO) 5 units joined by 2
90. The octahedral complex of a metal ion M 3 + with (a) Mn  Mn bond (b) Mn ≡≡ Mn bond
four monodentate ligands L1 , L2 , L3 and L4 absorb (c) Mn ≡≡ Mn bond (d) Mn  Mn bond
wavelengths in the region of red, green, yellow and 98. π-bond is available in
blue, respectively. The increasing order of ligand I. [NiCl 4 ] 2− II. Ni(CO) 4
strength of the four ligands is (NEET 2016)
III. [Co 2 (CO) 8 ] IV. Cr(CO) 6
(a) L4 < L3 < L2 < L1
The correct option is
(b) L1 < L3 < L2 < L4
(a) II, III and IV (b) I, II and III
(c) L3 < L2 < L4 < L1
(d) L1 < L2 < L4 < L3 (c) I, III and IV (d) I, II, III and IV
99. In Fe(CO) 5 , the Fe—CO bond possesses 101. Among the following metal carbonyls, C—O bond
(a) σ-character only order is lowest in
(b) π-character only (a) [Mn(CO)6 ]+ (b) [Fe(CO)5 ]
(c) Both (a) and (b) (c) [Cr(CO)6 ] (d) [V(CO)6 ]−
(d) ionic character
102. The total number of metal-metal bond present in
100. In emerald, Cr 3+ ions occupy octahedral sites [Co 2 (CO) 8 ] is
in the mineral beryl. The chemical formula of (a) 0 (b) 3 (c) 2 (d) 1
beryl is 103. In which compound synergic effect is present?
(a) Be2 AlSi 3 O9 (b) Be6 Al 4 Si12 O18 (a) [Ni(CO)4 ] (b) [NiCl 4 ]2–
(c) Be3 Al 2 Si 6 O18 (d) BeAISi 2 O6 (c) [CuCl 4 ]2− (d) [Mn(H2 O)6 ]2+

Topic 6
Stability Importance and Applications of Coordination Compounds
104. For the reaction of the type M + 4L s ML4 110. Estimation of calcium and magnesium is done by
(a) larger the stability constant, lower the proportion of ML4 (a) EDTA
that exists in solution (b) oxalate
(c) phosphate
(b) larger the stability constant, higher the proportion of ML4
that exists in solution (d) None of the above
(c) smaller the stability constant, higher the proportion of 111. Gold combines with cyanide in the presence of
ML4 that exists in solution oxygen and water to form coordination entity in
(d) None of the above aqueous solution. The coordination entity is
105. M + 4L s ML4 (a) [Au(CN)2 ]− (b) [Au(CN)2 ]2−
For this reaction, overall stability constant (β 4 ) is (c) [Au(CN)2 ]3− (d) [Au(CN)2 ]4−
expressed as
112. Coordination compounds have great importance in
(a) β 4 = [ ML] /[ ML3 ][ L] (b) β 4 = [ ML4 ]/[ ML3 ][ L]
biological systems. In this context which of the
(c) β 4 = [ ML4 ] /[ M ][ L]4 (d) β 4 = [ ML] /[ M ][ L] following statements is incorrect?
(a) Chlorophyll is green pigment in plants and contain
106. The reciprocal of the formation constant is called
calcium
(a) instability constant (b) dissociation constant
(b) Haemoglobin is the red pigment of blood and contains
(c) stability constant (d) Both (a) and (b)
iron
107. What is the overall equilibrium constant for the (c) Cyanocobalamine is vitamin B12 and contains cobalt
format ion [ ML4 ]2− ion, given that β 4 for this (d) Carboxypeptidase-A is an enzyme and contains zinc
complex is 2.5 ×1013 ? 113. The pigment responsible for photosynthesis,
(a) 2.5 × 1013 (b) 5 × 10−13 chlorophyll is a coordination compound of
(a) Cu (b) Zn (c) Mg (d) Ca
(c) 2.5 × 10−14 (d) 4.0 × 10−13
114. Which of the following statements is not correct?
108. Coordination compounds find their use in
(a) In oxyhaemoglobin Fe2+ is paramagentic
(a) qualitative chemical analysis
(b) During respiration the size of Fe2+ increases when it
(b) quantitative chemical analysis changes from diamagnetic to paramagnetic state
(c) qualitative physical analysis (c) Four heme groups are present in haemoglobin
(d) Both (a) and (b) (d) Heme is the prosthetic group and it is non-protein part
109. Hardness of water is estimated by simple titration 115. Cyanocobalamine, the antipernicious anemia factor, is
with Na 2 - EDTA because a coordination compound of
2+
(a) Ca ions form stable complexes with EDTA (a) Mn (b) Mo (c) Co (d) Cr
(b) Mg 2+ ions form stable complexes with EDTA 116. Wilkinson catalyst is used for the hydrogenation of
2+
(c) Ca ions form unstable complexes with EDTA (a) alkanes (b) alkynes
(d) Both (a) and (b) (c) alkenes (d) All (a) (b) and (c)
117. Articles can be electroplated with silver and gold 119. The excess of copper and iron are removed by
much more smoothly and evenly from solutions of the chelating ligands
complexes, respectively are (a) disferrioxime-B
(a) [Ag(CN)2 ]2− and [Au(CN)2 ]2− (b) D-penicillamine
(b) [Ag(CN)2 ]− and [Au(CN)2 ] (c) disferrioxime -B and D-penicillamine respectively
−
(c) [Au(CN)2 ] and [Ag(CN)]2 (d) D-penicillamine and disferrioxime -B respectively
(d) AgCN and AuCN
120. For lead-poisoning, the antidote used is
118. In photography, the use of Na 2S 2O 3 ⋅ 5H 2O is (a) white of an egg
(a) for converting AgBr into Ag 2 SO4 (b) cis-platin
(b) for converting AgBr into soluble thiosulphate complex (c) nickel
(c) for converting AgBr into silver thiosulphate (d) EDTA
(d) in reduction of Ag metal from AgBr

Special Format Questions

I. More Than One Correct Option 127. Which ions are not paramagnetic?
121. A bridging ligand possesses (a) Ni (NH3 )4 ] 2+ (b) [Ni (CO)4 ]
3+
(a) polydentate nature (c) [Co ( NH3 )6 ] (d) [Ni (CN)4 ] 2−
(b) two or more donor centres
128. Which is true in case of [Ni(CO) 4 ]?
(c) the tendency to get itself attached to two metal ions
(d) only one dentate site (a) Hybridisation of Ni is sp 3
(b) Tetrahedral shape
122. Diethylene triamine are (c) Diamagnetic
(a) chelating agent (b) polydentate ligand (d) Square planar
(c) tridentate ligand (d) not a ligand
129. Which of the following pairs do not represent linkage
123. A ligand having an unshared pair of electorns may isomers?
be a (a) [Cu(NH3 )4 ][PtCl 4 ] and [Pt(NH3 )4 ][CuCl 4 ]
(a) neutral molecule
(b) [Pd(PPh 3 )2 (NCS)2 ] and [Pd(PPh 3 )2 (SCN)2 ]
(b) positively charged ion
(c) [Co(NH3 )5 ]NO3 SO4 and [Co(NH3 )5 SO4 ]NO3
(c) negatively charged ion
(d) group containing a lone pair of electrons (d) [PtCl 2 (NH3 )4 ]Br2 and [PtBr2 (NH3 )4 ]Cl 2

124. The oxidation number of metal atom are zero in 130. Which complex is d 2 sp 3 hybridised?
(a) Ni( CO )4 (b) Fe( CO )5 (a) [ Co ( NH3 )6 ] 2+ (b) [ Fe(CN)6 ] 3−
(c) Na [Co(CO)4 ] (d) [Cu Cl 4 ] 2 − (c) [ Cu(NH3 )4 ] 2+ (d) None of the above
125. Which of the following do not have square planar
geometry? II. Statements Based Questions Type I
2 2−
(a) [CoCl 4 ] (b) [FeCl 4 ] ■ Directions (Q. Nos. 131-132) In the following
2− 2−
(c) [NiCl 4 ] (d) [PtCl 4 ] questions, a Statement I is followed by a corresponding
Statement II. Of the following Statements, choose the
126. Which of the following are true for ligands metal
correct one.
complex? (a) Both Statement I and Statement II are correct and
(a) Larger the ligand, the more stable is the metal ligand Statement II is the correct explanation of Statement I.
complex
(b) Both Statement I and Statement II are correct but
(b) Highly charged ligand forms stronger bonds Statement II is not the correct explanation
(c) Larger the permanent dipole moment of ligand, the more of Statement I.
stable is the bond (c) Statement I is correct but Statement II is incorrect.
(d) Greater the ionisation potential of central metal, the (d) Statement I is incorrect but Statement II is correct.
stronger the bond
131. Statement I When EDTA solution is added to Mg 2+ IV. Its structure is
ion solution then four coordinate site of Mg are 2+ CH2 COO–
occupied by EDTA and remaining two sites are H2 C — N
occupied by water molecules. CH2 COO–
Statement II EDTA is a hexadentate ligand.
CH2 COO–
132. Statement I Haemoglobin is the red pigment of blood H2 C — N
and contains iron.
CH2 COO–
Statement II Cynocobalamine is B12 and contains
The correct set of statements is
cobalt.
(a) I, II and III (b) II and IV
III. Statement Based Questions Type II (c) I, III and IV (d) I, II, III and IV

133. Consider the following statements, 137. Coordination number of the central atom/ion is
I. In coordination compounds metals show two types determined
of linkages (valencies). I. only by the σ-bonds formed by the ligand with the
II. The primary valencies are normally ionisable and central atom/ion.
are satisfied by negative ions. II. only by the π-bonds formed by the ligand with the
The correct statement(s) is/are central atom/ion.
(a) Only I Choose the correct option.
(b) Only II (a) Only II
(c) Both I and II (b) Both I and II
(d) Neither I nor II (c) Either I or II
(d) Only I
134. [CoCl 3 (NH 3 ) 3 ]
I. It is a coordination entity. 138. Which of the following complex can be ionised in
II. Cobalt ion is surrounded by three ammonia molecules. solution?
III. Cobalt ion is surrounded by three chloride ions. I. [CoCl 3 (NH3 ) 3 ] II. [Pt(NH3 ) 6 ]Cl 4
III. K 2 [PtF6 ] IV. K 4 [Fe(CN) 6 ]
The correct statements about [CoCl 3 (NH 3 ) 3 ] are
(a) I and II are correct Choose the correct option.
(b) II and III are correct (a) I, II and III (b) II, III and IV
(c) I and III are correct (c) I, III and IV (d) I, II, III and IV
(d) I, II and III are correct 139. Consider the following statements, while naming a
135. Consider the following statements. coordination compound.
I. When di or polydentate ligand uses its two or more I. The central metal atom is listed first.
donor atoms to bind a single ion, it is called chelate II. The ligands are then listed in alphabetical order.
ligand. III. The placement of a ligand in the list does not
II. When a ligand is bound to a metal ion through depend on its charge.
several donor atoms, the ligand is said to be a chelate The correct statement(s) is/are
ligand. (a) I and II
III. Ligand which can ligate through two different atoms is (b) II and III
called didentate ligand. (c) Only I
The correct statement(s) is/are (d) All I, II and III
(a) I and II (b) Only III 140. I. Polydentate ligands are not listed in alphabetical
(c) Only I (d) I and III order.
136. Which of the following statement(s) is/are true about II. In case of abbreviated ligand, the first letter of the
abbreviation is used to determine the position of the
[EDTA 4− ]?
ligand in the alphabetical order.
I. It is ethylene diaminetetracetate ion.
The correct option is
II. It is a hexadentate ligand. (a) Only I (b) Only II
III. It can bind through two nitrogen and four oxygen atoms (c) Both I and II (d) Neither I nor II
to a central metal ion.
141. Consider the following rules of naming coordination 147. The two forms dextro and laevo
compounds, I. depend upon the direction they rotate the plane of
I. The cation is named first in both positively and polarised light in a polarimeter.
negatively charged coordination entities. II. l-rotates to the right.
II. The ligands are named in an alphabetical order after III. d-rotates to the left.
the name of the central atom/ion. The correct statement(s) is/are
The incorrect statement(s) is/are (a) Both II and III (b) I, II and III
(a) Only I (b) Only II (c) Only I (d) None of these
(c) Both I and II (d) Neither I nor II
148. Consider the following statements about solvate
142. I. If the complex ion is a cation, the metal is named same
isomerism.
as the element.
I. It is also known as ‘hydrate isomerism’ where water is
II. Co in a complex ion is called cobalt and Pt is called
involved as a solvent.
platinum.
II. It is also similar to ionisation isomerism.
III. If the complex ion is an anion, the name of the metal
ends with the suffix–ate. III. Aqua complex[Cr(H2 O) 6 ]Cl 3 (violet) is an example of
solvate isomerism. Its solvate isomer is
The correct set of statements is [Cr(H2 O) 5 Cl]Cl 2 ⋅ H2 O.
(a) I and III (b) II and III
(c) I and II (d) I, II and III The correct set of statements is
(a) I and II (b) II and III
143. I. [Ag(NH3 ) 2 ][Ag(CN) 2 ] is named as (c) I and III (d) I, II and III
diamminesilver (I) dicyanoargentate (I).
149. In the octahedral complex [Co(NH 3 ) 6 ]3+
II. The molecular formula of tris (ethane-1,2-diammine)
cobalt(III) sulphate is I. the cobalt ion is in +3oxidation state.
[Co(H2 NCH2 CH2 NH2 ) 3 ] 2 (SO4 ) 3 . II. it is in 3d 6 electronic configuration.
The incorrect statement(s) is/are III. diamagnetism exists.
(a) Only I (b) Only II Which of these are true? Choose the correct option.
(c) Both I and II (d) Neither I nor II (a) I and II (b) II and III
(c) I and III (d) I, II and III
144. Consider the following structure and statements
regarding the structure 150. What is true about [MnCl 6 ]3− , [FeF6 ]3− and [CoF6 ]3− ?
[Co(H 2 NCH 2CH 2 NH 2 ) 3 ]2 (SO 4 ) 3 . I. Each of these are outer orbital complexes.
I. Counter anion in the molecule→ Sulphate II. Each of these have sp 3 d 2 hybridisation.
II. Charge on each complex cation→ + 2 III. Each of these are paramagnetic.
III. Neutral molecule in the compound→ IV. [MnCl 6 ] 3− ,[FeF6 ] 3− and [CoF6 ] 3− have four, five and
Ethane-1,2-diammine
four unpaired electrons respectively.
Which of the above statement(s) is/are true?
Choose the correct statements.
(a) Only II (b) Only III
(a) I, II and III (b) II, III and IV
(c) II and III (d) I and III
(c) I, III and IV (d) All four are true
145. Which of the following are the types of
151. Which of the following are the limitations of VBT?
stereoisomerism?
I. It does not give quantitative interpretation of magnetic
I. Linkage isomerism II. Optical isomerism data.
III. Geometrical isomerism IV. Ionisation isomerism II. It does not distinguish between weak and strong ligands.
Choose the correct pair of stereoisomers. III. It does not explain the colour exhibited by coordination
(a) I and II (b) II and III (c) I and IV (d) I, II and III compounds.
146. Optical isomers IV. It does not give a quantitative interpretation of the
I. are mirror images. thermodynamic or kinetic stabilities of coordination
II. cannot be superimposed on one another. compounds.
III. are also called as enantiomers. Identify the correct option from the choices given
The correct statement(s) is/are below.
(a) I and II (b) II and III (a) I, II, III and IV (b) II, III and IV
(c) I and III (d) I, II and III (c) I, III and IV (d) II, III and IV
152. Which of the following facts are related to CFT? 157. Decacarbonyldimanganese (0).
I. The five d-orbitals in an isolated gaseous metal I. Is made up of two square pyramidal Mn(CO) 5 units.
atom/ion have same energy, i.e. they are degenerate. II. These units are joined by a MnOMn bond.
II. If a spherically symmetrical field of negative charges III. Its structure is
surrounds the metal atom/ion, the degeneracy is
maintained. CO CO CO CO
III. When this negative field is due to ligands in a complex,
CO—Mn——Mn—CO
it becomes asymmetrical and the degeneracy of the
d-orbitals is lifted. CO CO CO CO
IV. It results in splitting of d-orbitals. This pattern of
splitting depends upon the nature of the crystal field. The fact(s) is/are
The correct set of statements is (a) I and II (b) II and III
(a) I, II, III and IV (b) II, III and IV (c) I and III (d) I, II and III
(c) I, II and III (d) I, III and IV
158. Consider the following statements.
153. I. [PtCl 4 ] 2− complex has a tetrahedral geometry. I. If ∆ o < P ,low spin state is more stable.
II. [Ni(CO) 4 ] complex has a square planar geometry. II. CO is a very weak ligand.
The incorrect statement(s) is/are III. Tetrahedral complexes exhibit nearly 50%
(a) Only I (b) Only II CFSE value than octahedral complexes.
(c) Both I and II (d) Neither I nor II IV. The colour of a complex depends on the nature of metal
154. Ruby, the gemstone ion.
I. is aluminium oxide (Al 2 O3 ) . The incorrect statements are
II. contains 0.5–1% Cr 3+ ions ( d 3 ) . (a) I, II and III (b) I, III and IV
III. Cr 3+ ions are randomly distributed in positions (c) I, II and IV (d) II, III and IV
normally occupied by Al 3+ . 159. Which of the following are false about [Cu(NH 3 ) 4 ]SO 4 ?
The true statement(s) is/are I. It is paramagnetic with one unpaired electron in the
(a) I and II (b) Only III d-subshell.
(c) II and III (d) I, II and III II. Its aqueous solution cannot conduct electricity.
155. Which of the following statements are correct for III. It gives white BaCl 2 solution.
[Fe(CN) 6 ]3− complex? IV. It is a square planar complex.
Choose the correct option.
I. It shows d 2 sp 3 hybridisation.
(a) I and II (b) II and III
II. It shows sp 3 d 2 hybridisation. (c) III and IV (d) I, II and IV
III. It is paramagnetic.
IV. It is diamagnetic. IV. Assertion-Reason Type Questions
Choose the correct option.
(a) I and III (b) II and III (c) III and IV (d) I and II
■ Directions (Q. Nos. 160-181) In the following
questions, a statement of Assertion (A) is followed by a
156. An aqueous pink solution of cobalt(II) chloride corresponding statement of Reason (R). Of the following
changes to deep blue on addition of excess of HCl. statements, choose the correct one.
This is because (a) Both A and R are correct; R is the correct explanation
I. [Co(H2 O) 6 ] 2+ is transformed into[CoCl 6 ] 4− . of A.
II. [Co(H2 O) 6 ] 2+ is transformed into[CoCl 4 ] 2− . (b) Both A and R are correct; R is not the correct explanation
of A.
III. Tetrahedral complexes have smaller crystal field
(c) A is correct; R is incorrect.
splitting than octahedral complexes.
(d) R is correct; A is incorrect.
IV. Tetrahedral complexes have larger crystal field
splitting than octahedral complex . 160. Assertion (A) Transition metals generally form
Choose the correct option. coordination compounds.
(a) II and III (b) I and IV Reason (R) These usually have partly filled d-orbitals of
(c) I, III and IV (d) II and IV the nth shell.
161. Assertion (A) Mohr’s salt gives NH +4 , Fe 2+ and SO 2−
4 ions 172. Assertion (A) The total number of isomers
in aqueous solution. shown by [Co(en) 2 Cl 2 ]+ complex ion is three.
Reason (R) Mohr’s salt is a double salt. Reason (R) [Co(en) 2 Cl 2 ]+ complex ion has an
162. Assertion (A) Primary and secondary are the two types of octahedral geometry.
linkages shown by coordination compounds. 173. Assertion (A) Octahedral geometry and
Reason (R) The ions/groups bound by the secondary diamagnetism exist in [Co(NH 3 ) 6 ]3+ .
linkages to the metal have characteristic spatial arrangements
corresponding to different coordination numbers. Reason (R) It has no unpaired electron.

163. Assertion (A) Toxic metal ions are removed by the chelating 174. Assertion (A) The ligands nitro and nitrito are
ligands. called ambidentate ligands.
Reason (R) Chelate complexes tend to be more stable. Reason (R) These ligands give linkage isomers.

164. Assertion (A) In the coordination compound 175. Assertion (A) Removal of water from
[Co ( H 2 NCH 2CH 2 NH 2 ) 3 ]2 , ethane-1,2-diammine is a [Ti(H 2O) 6 ]Cl 3 on heating renders it colourless.
neutral molecule. Reason (R) In the absence of ligand, crystal
Reason (R) Oxidation number of Co in the complex ion is field splitting does not occur.
+3. 176. Assertion (A) The coordination complexes
165. Assertion (A) Usually a sulphate ion is a bidentate ligand [Ni(CN) 4 ]2– and [NiCl 4 ]2− have not the same
but it can also act as monodentate in certain complexes. shape and similar magnetic behaviour.
Reason (R) Many times multidentate ligands do have Reason (R) Both are square planar in shape but
flexidentate character. [Ni(CN) 4 ]2– is diamagnetic but [Ni(Cl) 4 ]2− is
166. Assertion (A) Oxidation number of Cr in paramagnetic.
[Cr(NH 3 ) 3 (H 2O) 3 ]Cl 3 is same as the charge of the complex 177. Assertion (A) The stability of [Ni(en) 3 ]Cl 2 is
ion, +3.
lower than that of [Ni(NH 3 ) 6 ]Cl 2 .
Reason (R) All the ligands are neutral molecules in this
Reason (R) The geometry of Ni is trigonal
compound.
bipyramidal in [Ni(en) 3 ]Cl 2 .
167. Assertion (A) Isomers differ in one or more physical or
178. Assertion (A) Complexes are preferred in the
chemical properties.
electrolytic bath for electroplating.
Reason (R) These have different arrangement of atoms.
Reason (R) Complexes dissociate slowly and
168. Assertion (A) Geometrical isomerism arises in heteroleptic hence give a smooth and even deposit.
complexes.
179. Assertion (A) Potassium ferricyanide is
Reason (R) Different geometric arrangements of the ligands paramagnetic while potassium ferrocyanide is
are possible. diamagnetic.
169. Assertion (A) Tetrahedral complexes do not show Reason (R) Crystal field splitting in
geometrical isomerism. ferrocyanide ion is greater than that of
Reason (R) The relative positions of the unidentate ligands ferricyanide ion.
attached to the central metal atom are the same with respect 180. Assertion (A) As compared to non-chelated
to each other. complexes, chelated complexes are more
170. Assertion (A) Square planar complexes do not show optical stable.
isomerism. Reason (R) Labile complexes are the
Reason (R) These complexes do not possess chiral complexes which contain ligands those can be
structures. easily replaced by other ligands.
171. Assertion (A) All square planar complexes do not 181. Assertion (A) Metal carbonyls can also be
necessarily exhibit geometrical isomerism. called organometallics.
Reason (R) In such complexes metal generally assumes Reason (R) Metal carbonyls do contain metal
sp 3 -hybrid state. carbon bond.
V. Matching Type Questions 185. Match the columns and choose the correct option from
the codes given below.
182. Match the columns and choose the correct option
from the codes given below. Column I Column II
L
Column I Column II L L
(Compound) (Colour) A. M 1. Square planar
L L
A. CoCl 3 ⋅ 6NH3 1. Violet
L
B. CoCl 3 ⋅ 5NH3 2. Green L L
B. M 2. Square pyramidal
C. CoCl 3 ⋅ 4NH3 3. Purple L L
D. CoCl 3 ⋅ 3NH3 4. Yellow L
C. L
3. Octahedral
M
Codes L L
A B C D L
(a) 1 2 3 4 D. L L 4. Tetrahedral
(b) 2 1 4 3 M
L L
(c) 3 1 4 2
L
(d) 4 3 2 1
L Trigonal
183. Match the columns (formulation of cobalt III chloride E. L M 5.
L
bipyramidal
ammonia complexes) and choose the correct option L
from the codes given below.
Codes
Column III
Column I Column II A B C D E A B C D E
(Solution conductivity
(Colour) (Formula) (a) 1 2 3 4 5 (b) 5 4 3 2 1
corresponds to)
(c) 3 1 4 2 5 (d) 4 5 1 3 2
A. Yellow 1. [Co(NH3 )6 ]3+ 3Cl – I. 1 : 2 electrolyte
186. Match the following columns and choose the correct
B. Green 2. [CoCl(NH3 )5 ]2+ 2Cl – II. 1 : 1 electrolyte option from the codes given below.
C. Purple 3. [CoCl 2 (NH3 )4 ]+ Cl – III. 1 : 3 electrolyte Column I Column II
(Coordination compound) (Structure)
A. Potassium tetrahydroxidozincate (II) 1. K 3[Al(C2O4 )3 ]
Codes
B. Potassium trioxalatoaluminate (III) 2. [Ni(CO)4 ]
A B C
(a) 2(I) 1(II) 3(III)
C. Dichloridobis(ethane-1,2-diammine) 3. [CoCl 2 (en)2 ]+
cobalt (III)
(b) 3(II) 2(III) 1(I)
D. Tetracarbonyl nickel (0) 4. K 3[Zn(OH)4 ]
(c) 1(III) 3(II) 2(I)
(d) 2(III) 3(I) 1(II) Codes
A B C D A B C D
184. Match the following columns and choose the correct
(a) 1 2 3 4 (b) 2 4 1 3
option from the codes given below.
(c) 3 2 1 4 (d) 4 1 3 2
Column I Column II
187. Match the following columns and choose the correct
A. [Co(NH3 )6 ]3+ 1. Tetrahedral
option from the codes given below.
B. [Ni(CO)4 ] 2. Octahedral
Column I Column II
C. [PtCl 4 ]2− 3. Square planar (Structure) (Coordination compound)
A. Hg[Co(SCN)4 ] 1. Diamminechloridonitrito-N-
Codes platinum(II)
A B C B. [Co(NH3 )5 (CO)3 ]Cl 2. Pentaamminecarbonatocobalt
(a) 1 2 3 (III) chloride
(b) 2 1 3 C. [CoCl 2 (en)2 ]Cl 3. Dichloridobis
(c) 3 2 1 (ethane-1,2-diamine) cobalt (III)
(d) 3 1 2 chloride
D. [Pt(NH3 )2 Cl(NO2 )] 4. Mercury tetrathiocyanato
cobaltate(III)
 Codes 191. Match the following columns.
A B C D
Column II Column III
(a) 1 3 2 4 Column I
(Colour of light (Colour of
(b) 4 3 2 1 (Coordination entity)
absorbed) coordination entity)
(c) 4 2 3 1
A. [Ti(H2O)6 ]3+ I. Yellow 1. Blue
(d) 1 2 3 4
B. [Cu(H2O)4 ]2+ II. Blue 2. Violet
188. Match the following columns and choose the correct
option from the codes given below. C. [Co(NH3 )6 ]3+ III. Red 3. Yellow orange
2+
D. [CoCl(NH3 )5 ] IV. Blue green 4. Violet
Column I Column II
(Coordination compound) (Name) Codes
A K 3[Fe(C2O4 )3 ] 1. Amminebromidochloridonitrito A B C D
. -N-platinate (II)
(a) I(3) II(4) III(2) IV(1)
B. [Co(NH3 )5 Cl]Cl 2 2. Potassiumtrioxalato ferrate (III) (b) II(4) IV(3) I(2) III(1)
C. [Pt(NH3 )BrCl(NO2 )]− 3. Pentaamminechloridocobalt (c) III(1) I(2) IV(3) II(4)
(III) chloride
(d) IV(2) III(1) II(3) I(4)
Codes
192. Match Column I with Column II and choose the
A B C A B C
correct option from the codes given below.
(a) 2 3 1 (b) 3 1 2
(c) 1 2 3 (d) 3 2 1 Column II
Column I
(Central metal ion properties
189. Match the following columns and choose the correct (Complex) of complex ion)
option from the codes given below. A. [Ni(CN)4 ]2− 1. Ti 4+
Column I Column II B. Chlorophyll 2. sp3; paramagnetic
(IUPAC name) (Formula)
C. Ziegler-Natta catalyst 3. non-planar
A. Tetrahydroxoidozincate (II) 1. [Co(NH3 )5 (ONO)]2+
D. [NiCl 4 ]2− 4. Mg2+
B. Hexaammine platinum (IV) 2. [CuBr4 ]2–
E. Deoxyhaemoglobin 5. Planar
C. Tetrabromidocuprate (II) 3. [Zn(OH)4 ]2− 6. dsp2; diamagnetic
D. Pentaamminenitrito-O-cobalt (III) 4. [Pt(NH3 )6 ]4+ Codes
A B C D E
Codes
(a) 6 4 1 2 3
A B C D (b) 2 4 1 6 3
(a) 1 2 3 4 (c) 2 4 1 6 5
(b) 4 3 1 2 (d) 6 4 1 2 5
(c) 3 4 2 1
(d) 2 1 4 3 193. Match the following Column I with Column II
and choose the correct option from the codes given
190. Match the following columns and choose the correct below.
option from the codes given below.
Column I Column II
Column I Column II (Complex) (Structure and magnetic moment)
(IUPAC name) (Formula)
A. [Ag(CN)2 ]− 1. Square planar and 1.73 BM
A. Hexaamminenickel (II) 1. [Ni(NH3 )6 ]Cl 2
chloride B. [Cu(CN)4 ]3− 2. Linear and zero
B. Tris (ethane-1,2-diammine) 2. [Co(en)3 ]3+ C. [Cu(CN)6 ]3− 3. Octahedral and zero
cobalt (III) ion
D. [Cu(NH3 )4 ]2+ 4. Tetrahedral and zero
C. Diammine chlorido 3. [Pt(NH3 )2 Cl(NH2CH3 )]Cl
(methylammine) platinum E. [Fe(CN)6 ]4− 5. Octahedral and 2.86 BM
(II) chloride
Codes
D. Hexaaquatitanium (III) ion 4. [Ti(H2O)6 ]3+
A B C D E
Codes (a) 1 2 3 4 5
A B C D A B C D (b) 5 3 1 4 2
(a) 4 3 2 1 (b) 1 2 3 4 (c) 2 4 5 1 3
(c) 3 1 4 2 (d) 2 4 1 3 (d) 3 5 4 2 1
194. Match the Column I with Column II and choose the VI. Passage Based Questions
correct option from the codes given below.
■ Directions (Q. Nos. 197-198) Read the following
Column I Column II table and answer the following questions.
(Homoleptic carbonyls) (Structure) On the basis of the following observations made with
A. Tetracarbonyl nickel(0) 1. Tetrahedral
aqueous solutions.

B. Pentacarbonyl iron (0) 2. Trigonal bipyramidal Moles of AgCl precipitated

Formula per mole of the compounds
C. Hexacarbonyl chromium (0) 3. Octahedral with excess AgNO 3
I. PdCl 2 ⋅ 4NH3 2
Codes
A B C II. NiCl 2 ⋅ 6H2O 2
(a) 1 2 3 III. PtCl 4 ⋅ 2HCl 0
(b) 3 2 1 IV. CoCl 3 ⋅ 4NH3 1
(c) 2 3 1
(d) 2 1 3 V. PtCl 2 ⋅ 2NH3 0

195. Match the Column I with Column II and choose the

correct option from the codes given below.
197. The secondary valencies of I and III is
Column I Column II (a) 4 and 5 respectively (b) 4 and 6 respectively
(Coordination compound) (Uses in medicinal chemistry) (c) 6 and 4 respectively (d) 2 and 4 respectively

A. cis-platin 1. Removal of excess of Cu 198. Complexes having the secondary valence of 6 are
(a) I, II and III (b) I, II, III and IV
B. EDTA 2. Removal of excess of Fe
(c) II, III and IV (d) I, II, III, IV and V
C. Desferrioxime-B 3. Lead poisoning
■ Directions (Q. Nos. 199-201) Consider the
D. D-penicillammine 4. Tumours
following passage and answer the following questions.
Codes The coordination number of nickel (II) ion is 4.
A B C D NiCl 2 + KCN → X
(Cyano complex)
(a) 1 2 3 4 (Excess)
(b) 2 4 1 3 X + conc. HCl → Y
(c) 3 1 4 2 (Excess) (Chloro complex)
(d) 4 3 2 1 199. The IUPAC name for the complexes X and Y
respectively are
(a) potassium tetracyanonickel (II) and potassium
196. Match the following columns.
tetrachloronickel (II)
Column I Column II (b) tetracyanonickel (II) and tetrachloronickel (II)
(Metal ion configuration in (CFSE ∆ value)
o (c) tetracyano potassium nickelate (II) and tetrachloro
strong ligand field)
potassium nickelate (III)
A. d4 1. – 2.4 (d) potassium tetracyanonickelate (II) and potassium
B. d 5
2. – 2.0 tetrachloronickelate (II)
C. d 6
3. – 1.8 200. The hybridisation of X and Y are
D. d 7
4. – 1.6 (a) sp 3 d 2 , d 2 sp 3 (b) dsp 2 , sp 3 (c) dsp 2 , sp 3 (d) sp 3 , dsp 3
201. The magnetic nature of X and Y are
Codes (a) Both are diamagnetic
A B C D (b) Both are paramagnetic
(a) 4 2 1 3
(c) X is diamagnetic while Y is paramagnetic containing two
(b) 1 2 3 4
unpaired electrons
(c) 3 4 2 1
(d) X is diamagnetic while Y is paramagnetic containing one
(d) 2 3 4 1
unpaired electron
 NCERT & NCERT Exemplar Questions
NCERT 210. Due to the presence of ambidentate ligands
202. Which of the following species is not expected to be a coordination compounds show isomerism. Palladium
ligand? complexes of the type [Pd(C 6 H 5 ) 2 (SCN) 2 ] and
(a) NO (b) NH+4 [Pd(C 6 H 5 )2 (NCS) 2 ] are
(c) NH2 CH2 CH2 NH2 (d) CO (a) linkage isomers
(b) coordination isomers
203. A chelating agent has two or more than two donor (c) ionisation isomers
atoms to bind a single metal ion. Which of the (d) geometrical isomers
following is not a chelating agent?
(a) Thiosulphato (b) Oxalato 211. The compounds [Co(SO 4 )(NH 3 ) 5 ]Br and
(c) Glycinato (d) Ethane-1, 2-diammine [Co(SO 4 )(NH 3 ) 5 ]Cl represent
(a) linkage isomerism
204. When 0.1 mol CoCl 3 (NH 3 ) 5 is treated with excess of
(b) ionisation isomerism
AgNO 3 , 0.2 mol of AgCl is obtained. The
(c) coordination isomerism
conductivity of solution will correspond to
(d) no isomerism
(a) 1 : 3 electrolyte
(b) 1 : 2 electrolyte
(c) 1 : 1 electrolyte NCERT Exemplar
(d) 3 : 1 electrolyte 212. The stabilisation of coordination compounds due to
205. When 1 mole of CrCl 3 ⋅6H 2O is treated with excess of chelation is called the chelate effect. Which of the
AgNO 3 , 3 moles of AgCl are obtained. The formula following is the most stable complex species?
of the complex is (a) [Fe(CO)5 ] (b) [Fe(CN)6 ] 3−
(a) [CrCl 3 (H2 O)3 ] ⋅ 3H2 O (b) [CrCl 2 (H2 O)4 ]Cl ⋅ 2H2 O (c) [Fe(C2 O4 )3 ] 3− (d) [Fe(H2 O)6 ] 3+
(c) [CrCl(H2 O)5 ]Cl 2 ⋅ H2 O (d) [Cr(H2 O)6 ]Cl 3
213. The CFSE for octahedral [CoCl 6 ]4− is 18,000 cm −1 .
206. The correct IUPAC name of [Pt(NH 3 ) 2 Cl 2 ] is
The CFSE for tetrahedral [CoCl 4 ]2− will be
(a) Diamminedichloridoplatinum (II)
(b) Diamminedichloridoplatinum (IV) (a) 18000 cm−1 (b) 16000 cm−1
(c) Diamminedichloridoplatinum (0) (c) 8000 cm−1 (d) 20000 cm−1
(d) Dichloridodiammineplatinum (IV) 214. Which of the following complexes formed by
207. IUPAC name of [Pt(NH 3 ) 2 Cl(NO 2 )] is Cu 2+ ions is most stable?
(a) platinum diamminechloronitrite (a) Cu 2+ + 4NH3 → [Cu(NH3 )4 ]2+ , log K = 11.6
(b) chloronitrito-N-ammineplatinum (II) (b) Cu 2+ + 4CN− → [Cu(CN)4 ]2 − , log K = 27.3
(c) diamminechloridonitrito-N-platinum (II) (c) Cu 2+ + 2en → [Cu(en)2 ]2+ , log K = 15.4
(d) diamminechloronitrito-N-platinate (II)
(d) Cu 2+ + 4H2 O → [Cu(H2 O)4 ]2+ , log K = 8.9
208. Indicate the complex ion which shows geometrical
isomerism. 215. The colour of the coordination compounds depends
+ on the crystal field splitting. What will be the correct
(a) [Cr(H2 O)4 Cl 2 ] (b) [Pt(NH3 )3 Cl]
order of absorption of wavelength of light in the
(c) [Co(NH3 )6 ] 3 + (d) [Co(CN)5 (NC)] 3− visible region, for the complexes, [Co(NH 3 ) 6 ]3+ ,
209. What kind of isomerism exists between [Co (CN) 6 ]3− and [Co(H 2O) 6 ]3+ .
[Cr(H 2O) 6 ]Cl 3 (violet) and [Cr(H 2O) 5 Cl]Cl 2 ⋅ H 2O (a) [Co(CN)6 ] 3 − > [Co(NH3 )6 ] 3+ > [Co(H2 O)6 ] 3+
(greyish-green)?
(a) Linkage isomerism (b) [Co(NH3 )6 ] 3+ > [Co(H2 O)6 ] 3+ > [Co(CN)6 ] 3−
(b) Solvate isomerism (c) [Co(H2 O)6 ] 3+ > [Co(NH3 )6 ] 3+ > [Co(CN)6 ] 3 −
(c) Ionisation isomerism (d) [Co(CN)6 ] 3 − > [Co(NH3 )6 ] 3+ > [Co(H2 O)6 ] 3+
(d) Coordination isomerism
216. Identify the correct statements for the behaviour of ■ Directions (Q.Nos. 225 to 229) In the following
ethane-1, 2- diamine as a ligand. questions a statement of Assertion (A) followed by a
(a) It is a neutral ligand statement of Reason (R) is given. Choose the correct
(b) It is a didentate ligand answer out of the following choices.
(c) It is a chelating ligand (a) Assertion and Reason both are true, Reason is correct
(d) It is a unidentate ligand explanation of Assertion.
(b) Assertion and Reason both are true but Reason is not the
217. Which of the following complexes are homoleptic? correct explanation of Assertion.
(a) [Co(NH3 )6 ] 3+ (b) [Co(NH3 )4 Cl 2 ]+ (c) Assertion is true, Reason is false.
(c) [Ni(CN)4 ]2− (d) [Ni(NH3 )4 Cl 2 ] (d) Assertion is false, Reason is true.

218. Which of the following complexes are heteroleptic? 225. Assertion (A) Toxic metal ions are removed by the
3+ + chelating ligands.
(a) [Cr(NH3 )6 ] (b) [Fe(NH3 )4 Cl 2 ]
(c) [Mn(CN)6 )] 4− (d) [Co(NH3 )4 Cl 2 ]
Reason (R) Chelate complexes tend to be more
stable.
219. Identify the optically active compounds from the
following. 226. Assertion (A) Linkage isomerism arises in
coordination compounds containing ambidentate
(a) [Co(en)3 ] 3+
(b) trans − [Co(en)2 Cl 2 ]+
ligand.
(c) cis − [Co(en)2 Cl 2 ]+ (d) [Cr(NH3 )5 Cl]
Reason (R) Ambidentate ligand has two different
220. Which of the following complexes show linkage donor atoms.
isomerism?
227. Assertion (A) Complexes of MX 6 and MX 5 L type
(a) [Co(NH3 )5 (NO2 )]2+ (b) [Co(H2 O)5 CO]3+
(X and L are unidentate) do not show geometrical
(c) [Cr(NH3 )5 ]SCN2+ (d) [Fe(en)2 Cl 2 ]+
isomerism.
221. Atomic number of Mn, Fe and Co are 25, 26 and 27 Reason (R) Geometrical isomerism is not shown by
respectively. Which of the following inner orbital complexes of coordination number 6.
octahedral complex ions are diamagnetic?
228. Assertion (A) [Fe(CN) 6 ]3− ion shows magnetic
(a) [Co(NH3 )6 ] 3+ (b) [Mn(CN)6 ] 3−
moment corresponding to two unpaired electrons.
(c) [Fe(CN)6 ] 4− (d) [Fe(CN)6 ] 3−
Reason (R) Because it has d 2 sp 3 type hybridisation.

222. Atomic number of Mn, Fe, Co and Ni are 25, 26, 27 229. Assertion (A) [Cr(H 2O 6 )]Cl 2 and [Fe(H 2O) 6 ]Cl 2 are
and 28 respectively. Which of the following outer reducing in nature.
orbital octahedral complexes have same number of Reason (R) Unpaired electrons are present in their
unpaired electrons? d-orbitals.
(a) [MnCl 6 ] 3− (b) [FeF6 ] 3−
230. Match the compounds given in Column I with the
(c) [CoF6 ] 3− (d) [Ni(NH3 )6 ]2+ oxidation state of cobalt present in it (given in
223. Which of the following options are correct for
Column II) and assign the correct code.
[Fe(CN) 6 ]3− complex? Column I Column II
2 3 3 2 (Compound) (Oxidation state of Co)
(a) d sp hybridisation (b) sp d hybridisation
A. [Co(NCS)(NH3 )5 ](SO3 ) 1. +4
(c) Paramagnetic (d) Diamagnetic
B. [Co(NH3 )4 Cl 2 ]SO4 2. 0
224. An aqueous pink solution of cobalt(II) chloride
changes to deep blue on addition of excess of HCl. C. Na 4[Co(S2O3 )3 ] 3. +2
This is because D. [Co2 (CO)8 ] 4. +3
2+ 4−
(a) [Co(H2 O)6 ] is transformed into [CoCl 6 ]
Codes
2+
(b [Co(H2 O)6 ] is transformed into [CoCl 4 ]2− A B C D
(c) tetrahedral complexes have smaller crystal field splitting (a) 1 2 4 3
than octahedral complexes (b) 4 3 2 1
(d) tetrahedral complexes have larger crystal field splitting (c) 3 1 4 2
than octahedral complex (d) 4 1 3 2
231. Match the coordination compounds given in 233. Match the complex ions given in Column I with the
Column I with the central metal atoms given in colours given in Column II and assign the correct code.
Column II and assign the correct code. Column I Column II
Column I Column II (Complex ion) (Colour)
(Coordination compound) (Central metal atom) A. [Co(NH3 )6 ]3+ 1. Violet
A. Chlorophyll 1. Rhodium 3+
B. [Ti(H2O)6 ] 2. Green
B. Blood pigment 2. Cobalt
C. Wilkinson catalyst 3. Magnesium C. [Ni(H2O)6 ]2+ 3. Pale blue
D. Vitamin B12 4. Iron D. [Ni(H2O)4 (en)]2+ (aq) 4. Yellowish orange

Codes Codes
A B C D A B C D A B C D A B C D
(a) 3 4 1 2 (b) 3 4 5 1 (a) 1 2 4 5 (b) 4 3 2 1
(c) 4 3 2 1 (d) 3 4 2 1 (c) 3 2 4 1 (d) 4 1 2 3
232. Match the complex species given in Column I with 234. Match the complex ions given in Column I with the
the possible isomerism given in Column II and assign hybridisation and number of unpaired electrons given
the correct code. in Column II and assign the correct code.
Column I Column II Column II
(Complex species) (Isomerism) Column I
(Hybridisation, number
1. Optical (Complex ion)
A. [Co(NH3 )4 Cl 2 ]+ of unpaired electrons)

B. cis − [Co(en)2 Cl 2 ]+ 2. Ionisation A. [Cr(H2O)6 ] 3+ 1. dsp2, 1

C. [Co(NH3 )5 (NO2 )]Cl 2 3. Coordination B. [Co(CN)4 ]2− 2. sp3d 2, 5

D. [Co(NH3 )6 ][Cr(CN)6 ] 4. Geometrical C. [Ni(NH3 )6 ]2+ 3. d 2sp3, 3

Codes D. [MnF6 ]4− 4. sp3d 2 , 2

A B C D
(a) 1 2 4 3 Codes
(b) 4 3 2 1 A B C D A B C D
(c) 4 2 1 3 (a) 3 1 4 2 (b) 4 3 2 1
(d) 4 1 2 3 (c) 3 2 4 1 (d) 4 1 2 3

Answers
1. (a) 2. (b) 3. (c) 4. (b) 5. (d) 6. (a) 7. (a) 8. (d) 9. (a) 10. (c) 11. (a) 12. (c) 13. (d) 14. (b) 15. (b)
16. (c) 17. (b) 18. (c) 19. (a) 20. (c) 21. (b) 22. (a) 23. (a) 24. (b) 25. (b) 26. (b) 27. (b) 28. (d) 29. (a) 30. (a)
31. (b) 32. (c) 33. (c) 34. (a) 35. (a) 36. (b) 37. (d) 38. (a) 39. (b) 40. (b) 41. (a) 42. (d) 43. (d) 44. (b) 45. (b)
46. (b) 47. (a) 48. (a) 49. (d) 50. (d) 51. (b) 52. (c) 53. (a) 54. (a) 55. (b) 56. (d) 57. (a) 58. (b) 59. (c) 60. (d)
61. (d) 62. (b) 63. (c) 64. (a) 65. (c) 66. (b) 67. (d) 68. (d) 69. (b) 70. (d) 71. (d) 72. (c) 73. (a) 74. (a) 75. (c)
76. (d) 77. (c) 78. (d) 79. (d) 80. (a) 81. (b) 82. (a) 83. (b) 84. (b) 85. (b) 86. (a) 87. (a) 88. (c) 89. (a) 90. (b)
91. (a) 92. (b) 93. (a) 94. (c) 95. (a) 96. (a) 97. (d) 98. (a) 99. (c) 100. (c) 101. (b) 102. (d) 103. (a) 104. (b) 105. (c)
106. (d) 107. (a) 108. (d) 109. (c) 110. (a) 111. (a) 112. (a) 113. (c) 114. (a) 115. (c) 116. (c) 117. (b) 118. (b) 119. (d) 120. (d)
121. (ab) 122. (ac) 123. (ad) 124. (ab) 125. (abc) 126. (cd) 127. (bcd) 128. (a c) 129. (acd) 130. (ab) 131. (d) 132. (b) 133. (c) 134. (d) 135. (c)
136. (d) 137. (d) 138. (b) 139. (d) 140. (b) 141. (b) 142. (d) 143. (d) 144. (d) 145. (b) 146. (d) 147. (c) 148. (b) 149. (d) 150. (d)
151. (a) 152. (a) 153. (c) 154. (a) 155. (a) 156. (a) 157. (d) 158. (c) 159. (a) 160. (c) 161. (a) 162. (b) 163. (a) 164. (b) 165. (a)
166. (a) 167. (a) 168. (a) 169. (a) 170. (a) 171. (d) 172. (c) 173. (a) 174. (a) 175. (a) 176. (c) 177. (d) 178. (a) 179. (c) 180. (b)
181. (d) 182. (d) 183. (c) 184. (b) 185. (c) 186. (d) 187. (c) 188. (a) 189. (c) 190. (b) 191. (d) 192. (a) 193. (c) 194. (a) 195. (d)
196. (a) 197. (b) 198. (c) 199. (d) 200. (b) 201. (c) 202. (b) 203. (a) 204. (b) 205. (d) 206. (a) 207. (c) 208. (a) 209. (b) 210. (a)
211. (d) 212. (c) 213. (c) 214. (b) 215. (c) 216. (abc) 217. (ac) 218. (bd) 219. (ac) 220. (ac) 221. (ac) 222. (ac) 223. (ac) 224. (bc) 225. (a)
226. (a) 227. (b) 228. (d) 229. (b) 230. (d) 231. (a) 232. (d) 233. (b) 234. (a)
 Hints & Explanations
2. (b) The primary valency is ionisable valency. It corresponds to 14. (b) EDTA is a hexadentate ligand because it has six
oxidation state of metal. The primary valency is always satisfied donor centres.
by anion
[Co(NH3 )6 ]Cl 3 → [Co(NH3 )6 ]+ + 3Cl − −
OOC ⋅ CH2 CH2 COO–
•• ••
(A ) N— CH2 — CH2 — N
∴ Number of primary valency is 3. −
OOC ⋅ CH2 CH2 COO–
[Co(NH3 )5 Cl]Cl 2 → [Co(NH3 )5 Cl]2+ + 2Cl −
(B ) O
∴ Number of primary valency is 2. 15. (b) Nitrito –N ⇒ M ← N
[Co(NH3 )4 Cl 2 ]Cl → [Co(NH3 )4 Cl 2 ]+ + Cl − O
Nitrito – O ⇒ M ← O— N== O
∴ Number of primary valency is 1.
Thiocyanato ⇒ M ← SCN
3. (c) [Cr(H2 O)6 ]Cl 3 contains maximum number of ionisable anion Isothiocyanato ⇒ M ← NCS
(3Cl − ). Hence, it will consume more equivalents of aqueous 16. (c) EDTA structure is
solution of AgNO3 and forms three moles of AgCl (white ppt.)
HOOCH2 C CH2 COOH
Dissociation in aqueous N — CH2 — CH2 — N
Complex
solution of AgNO 3 HOOCH2 C CH2 COOH
(a) Na[CrCl 6 ] 3Na + + [CrCl 6 ] 3−
17. (b) CN− is a better complexing agent (C) as well as
(b) [Cr(H2O) 5 Cl] Cl 2 [Cr(H2O) 5 Cl] 2 + + 2Cl −
reducing agent (A).
(c) [Cr(H2O) 6 ]Cl 3 [Cr(H2O) 6 ] 3 + + 3Cl − Thus, properties of (A) and (C) are shown by CN.
(d) Na 2[CrCl 5( H2O)] 2Na + + [CrCl 5 (H2O)] 2− Property (C)
6. (a) According to postulates of Werner’s theory for Ni 2+ + 4 CN− → [Ni(CN)4 ]2 −
coordination compounds, metal atoms exhibit two types Property (A)
of valencies; i.e. primary valency and secondary valency. II I 1
The primary valency is ionisable while the secondary valency is Cu Cl 2 + 5KCN → K 3 [Cu (CN)4 ] + (CN)2 + 2KCl
2
non-ionisable.
(CN− reduces Cu 2+ to Cu + ).
8. (d) The primary and secondary valencies of chromium in the
18. (c) Coordination number (CN) of a metal ion in a
complex ion, dichlorodioxalato chromium (III) are 3 and 6
complex can be defined as the number of ligand donor
respectively.
atoms to which the metal is directly bonded.
10. (c) [Fe(CN)6 ]4– is a complex ion while rest of the compounds
19. (a) Greater is the number of chelate rings, greater is
are double salts. the stability of the chelate. Hence, five fused cyclic
11. (a) [Co(NH3 )5 Cl]Cl 2 s [Co(NH3 )5 Cl]2+ + 2Cl – system is most stable for a chelate.
1444424444 3
3 ions 21. (b) Coordination number of Pt and Ni in [PtCl 6 ]2− and
2Cl – + 2Ag + → 2AgCl [Ni(NH3 )4 ]2+ are 6 and 4 respectively.
2 mol 2 mol 2 mol
22. (a) In the complex ion, [Fe(C2 O4 )3 ]3− and
12. (c) Co and Pt = 6 coordination number.
3+ 4+
[Co(en)3 ]3+ , the coordination number of Fe and Co is 6
CoCl 3 ⋅ 6NH3 and PtCl 4 ⋅ 5NH3 because C2 O2− 4 and ethane –1, 2-diamine are bidentate
ligands.
In solution
[Co(NH3 )6 ]Cl 3 → [Co(NH3 )6 ] 3+ + 3Cl –
24. (b) Glycinato, ethane-1, 2-diamine, thiosulphato are
In solution chelating agents.
[PtCl(NH3 )5 ]Cl 3 → [PtCl(NH3 )5 ]3+ + 3Cl –
25. (b) The number of ligands attached to the central metal
Number of ionic species are same in the solution of both
ion is called the coordination number.
complexes, therefore their equimolar solutions will show same
conductance. So, coordination number of Fe in [Fe(CN)6 ]4− and
[Fe(CN)6 ]3− are 6 and 6 respectively.
13. (d) Ligands can be simple ions like Cl − , small molecules like
H2 O, larger molecules like H2 NCH2 CH2 NH2 and 26. (b) In the complex K 4 [Fe(CN)6 ], K + is the counter ion
macromolecules like proteins. while coordination sphere is [Fe(CN)6 ] 4− .
27. (b) The oxidation number of the central atom in a complex 45. (b) Sodium pentacyanonitrosyl ferrate (III)
is defined as the charge it would carry if all the ligands are −Na 2 [Fe(CN)5 NO]
removed along with the electron pairs that are shared with
56. (d) Optical isomerism is the type of stereoisomerism.
the central atom.
Cl NH3 Cl NH3
28. (d) Let the oxidation state of Fe in 57. (a) I. Pt II. Pt
[Fe(H2 O)5 NO]SO4 is x. [Fe(H2 O)5 NO]2+ Cl NH3 NH3 Cl
∴ x+ 0+ 1= 2 cis-isomer trans-isomer
⇒ x= +1
58. (b) Geometrical isomerism arises when bidentate ligands
Here, NO exists as a nitrosyl ion (NO+ ).
29. (a) CO is a neutral ligand, so the oxidation state of metal in (L–L) are present in complexes of formula [ MX 2 ( L − L )2 ].
metal carbonyl is always zero. 59. (c) The existence of two different coloured complexes is
[Ni(CO)4 ] due to geometrical isomerism as cis and trans forms are
x + (0 × 4 ) = 0 present.
x=0 II
60. (d) [Pt (NH3 )2 Cl 2 ] shows geometrical isomerism.
30. (a) In MnO2 and FeCl 3 oxidation states of Mn and Fe are
Cl NH3 Cl NH3
+4 and +3 respectively.
In [MnO4 ] − and CrO2 Cl 2 oxidation states of Mn and Cr are Pt Pt
+7 and +6 respectively. Cl NH3 H 3N Cl
In [Fe(CN)6 ] 3− and [Co(CN)3 ] oxidation states of Fe and Co cis trans
(orange yellow) (pale yellow)
are +3 and +3 respectively.
In [NiCl 4 ] 2− and [CoCl 4 ] − , oxidation states of Ni and Co 61. (d) A→ fac−; B → mer −
are +2 and +3 respectively. 62. (b) Complex [Co(en)3 ]3+ has no plane of symmetry and
31. (b) (a) Li 2 O + KCl → 2LiCl + K 2 O centre of symmetry that’s why it is optically active.
This is wrong equation, since a stronger base K 2 O cannot en en
be generated by a weaker base Li 2 O.
(b) [CoCl(NH3)5 ] + + 5H+ → Co2 + (aq) + 5NH+4 + Cl −
en Co3+ Co3+ en
This is correct. All ammine complexes can be destroyed
by adding H⊕ . Hence, on adding acid to [CoCl(NH3)5 ], it
gets converted to Co2 + (aq) NH+4 and Cl − . en en
Mirror
OH −
(c) [Mg(H2O)6 ] 2 + + EDTA4− →
Excess
[Mg(EDTA)] 2 + 63. (c) (II) and (III) show only geometrical isomerism.
+ 6H2O [Ni(H2 O)4 (NH3 )2 ]2+ show cis and trans-isomers. These
This is wrong, since the formula of complex must be are followings
[Mg(EDTA)]2 − . NH3 NH3
(d) The 4th reaction is incorrect. It can be correctly H2O NH3 H2 O NH3
represented as Ni Ni
2CuSO4 + 10 KCN → 2K3[Cu(CN)4 ] H2O H2O H2O H2O
+ 2K2SO4 + (CN)2 ↑ H 2O NH3
cis trans
33. (c) Names of anionic ligands with e− .
[Ni(H2 O)3 (NH3 )3 ]2+ show facial and meridional
37. (d) Two Br, two (en) and one Cr are parts of the complex.
Charge on the complex is geometrical isomerism.
NH3 NH3
2 (Br) = − 2 
 H2O NH3 NH3 H2O
2 (en) = 0  = + 1
Ni Ni
1 (Cr) = + 3 
H 2O NH3 H 2O H2O
Thus, complex ion is [Cr(en)2 Br2 ] + H 2O NH3
fac-isomer mer-isomer
Since, anion is bromide, thus complex is [Cr(en)2 Br2 ] Br.
39. (b) In the given compound [Pt(NH3 )2 Cl(NH2 )(CH3 )]Cl 64. (a) In a coordination entity of the type [PtCl 2 (en)2 ]2+ ,
the metal is Pt, counter ion is Cl and ligands are
(NH3 )2 Cl(NH2 )(CH3 ). optical activity is shown by cis-isomer only.
 [Co(NH3 )4 Cl 2 ]Cl can exist in both cis and trans forms that
– 3– are given below:
NH3 Cl
Cl + NH3 +
NC NH3 H3N NH3 H3N Cl
65. (c) Fe en Co en
NC CN Co Co
CN Cl
cis- trans- H3N NH3 H3N Cl
A C
3– isomers isomers Cl NH3
NH3 –
en trans-[Co(NH3)4Cl2]Cl cis-[Co(NH3)4Cl2]Cl
Cl (optically inactive) (optically inactive)
NC CN
Cr
Fe
Cl NC [Co(NH3 )3 Cl 3 ] exists in fac and mer-isomeric forms and
en CN
both are optically inactive.
NH3
B D NH3 NH3
Cl NH3 Cl NH3
66. (b) M ( ABCD ) type complex have three geometrical
Co Co
isomers as
py NH3 py Br py NH3 Cl NH3 Cl Cl
Pt Pt Pt Cl NH3
fac-isomer mer-isomer
Cl Br ; Cl NH3 ; Br Cl ; (optically inactive) (optically inactive)

67. (d) Tris-(ethylenediammine)cobalt (III)bromide,

[Co(en)3 ]Br3 exhibits optical isomerism. 74. (a) Outer orbital complex utilises d-orbitals for bonding and
exhibit paramagnetic behaviour, only if there present
3+ 3+
en en unpaired electrons.
(a) In [Ni(NH3 )6 ] 2+ , Ni 2+ = [Ar] 3d 8 4 s0 [Ni(NH3 )6 ] 2+ =
en Co en Co en
3d 4s 4p 4d

en en 123
→

d-form mirror l-form

Two unpaired sp 3d 2 hybridisation
68. (d) Linkage isomerism arises in a coordination compound electrons
containing ambidentate ligand. So, this is an outer orbital complex having paramagnetic
69. (b) Coordination isomerism arises from the interchange of character.
ligands between cationic and anionic entities of different (b) [Zn(NH3 )6 ] 2+ , sp3d 2 hybridisation, outer orbital complex
metal ions present in a complex. and diamagnetic.
70. (d) When the cation and anion both are complex ions, (c) [Cr(NH3 )6 ] 3+ , d 2sp3 hybridisation, inner orbital complex
the coordination compound exhibit coordination and paramagnetic.
isomerism. (d) [Co(NH3 )6 ] 3+ , d 2sp3 hybridisation, inner orbital complex
Thus, the given examples are of coordination isomerism. and diamagnetic.
71. (d) These examples are ionisation isomers because of 75. (c)
chloride and sulphate ion. It has no isomerism. CH3  C == NOH
72. (c) [Co(NH3 )5 SO4 ] Br and [Co(NH3 )5 Br]SO4 are + Ni2+
CH3  C == NOH
ionisation isomers. O– .........H  O
H 3C  
73. (a)
C == N N == C  CH3
Cl Cl
Cl
 Ni2+
C == N N == C  CH3
en Co en Co en
H 3C 
– .........
O HO
en
Cl
cis-[Co(en)2Cl2]Cl trans-[Co(en)2Cl2]Cl H-bond
(optically active) (optically inactive due
to plane of symmetry) DMG acts as a bidentate ligand.
76. (d) Smaller the number of unpaired electrons, smaller is the For [CoF6 ]3− , oxidation state of Co is +3
paramagnetic behaviour. F− = weak field ligand
Co 2 + → 3d 7 , 4 s0 In [CoF6 ]3−
, having minimum number
of unpaired electrons. (3) Co3+ =

Thus, Co 2+ has minimum paramagnetic behaviour. 4s 4p

4d
77. (c) Ni has dsp 2 hybridisation where CN− is a strong field ligand. ×× ×× ×× ×× ×× ××
dsp 2 hybridisation F– F– F– F– F– F–

sp 3d 2
3 2
sp d hybridisation, i.e. octahedral geometry four
CN – CN – CN – CN – unpaired electrons, i.e. magnetic moment is 4.91 BM.
Since, all the electrons are paired, it is diamagnetic. 79. (d) The complexes, in which metals ions have same
78. (d) For [Ni(CN)4 ]2– , oxidation state of Ni is +2. number of unpaired electrons will have same magnetic
moment.
CN− = strong field ligand
Ni 2+ (ground state) = Electronic
Number of unpaired
8 0 Complex ion configuration of
3d 4s electrons (n)
metal ion

[Cr(H2O) 6 ] 2 + Cr 2 + ; [Ar] 3 d 4 ;4
In [Ni(CN)4 ]2−
4s 4p
3d [Fe(H2O) 6 ] 2+ Fe2 + ; [Ar] 3 d 6 ; 4
Ni = 2+
×× ×× ×× ××

CN
–
CN CN CN
– – – [Mn(H2O) 6 ] 2+ Mn 2 + ; [Ar] 3 d 5 ;5

dsp2
[CoCl 4 ] 2− Co 2 + ; [Ar] 3 d 7 ;3
dsp 2 hybridisation, i.e. square planar geometry, zero unpaired
electron, i.e. zero magnetic moment.
For [MnBr4 ]2– , oxidation state of Mn is +2. 80. (a) For d 4 ions, the number of possible patterns of
electron distribution arises is two.
Br − = weak field ligand
81. (b) The fourth electron enters one of the eg orbitals
Mn 2+ (in ground state)
3d 4s giving the configuration t 23g e1g when ∆ o < P.

82. (a) [Co (CN)6 ]3−

2−
In [MnBr4 ] , Co 3+ = 1s2 , 2s2 , 2 p 6 , 3s2 , 3 p 6 , 3d 6
4s 4p
CN− is a strong field ligand and as it approaches the
3d
Mn
2+
= × × × × metal ion, the electron must pair up.
– – – –
Br Br Br Br The splitting of the d-orbitals into two sets of orbitals
sp 3 hybridisation, i.e, tetrahedral geometry, five unpaired in as octahedral complex [Co (CN)6 ] 3− may be
electrons, i.e. magnetic moment = 5.9 as [ µ = n ( n + 2 ) and represented as
n = 5] dx2-y2,dz2
Co 3+ in ground state = 3d eg orbitals
3d 4s 4p
∆ο
a

eg orbitals
a
a
a

dxy dyz dxz

6
Here, for d ions, three electrons first inter orbitals with
parallel spin out the remaining may pair up in t 2 g
orbital giving rise to low spin complex (strong ligand)
field.
∴ [Co (CN)6 ] 3− has no unpaired electrons and will be in
low spin configuration.
83. (b) Based on spectrochemical series, ligands arranged in (a) In [Cu(NH3 )4 ] 2+ , Cu 2+ = [Ar] 3d 9
increasing order of crystal field strength are as 3d 9
NH3 < en < CN− < CO
84. (b) For [Cr(H2 O)6 ] 2+ and Cr 2 + = [Ar],3d 4
(Although in the presence of strong field ligand NH3, the
unpaired electrons gets excited to higher energy level but
eg it still remains unpaired.)
0.6 ∆o [Thus, [Cu(NH3 )4 ] 2+ is the complex that exhibits a
Energy
magnetic moment of 1.73 BM.]
Average energy of –0.4 Do 90. (b) Ligand field strength ∝ Energy of light absorbed
d-orbitals in symmetrical
1
crystal field
t2g ∝
Wavelength of light absorbed
Splitting in an
octahedral crystal field Wavelength of absorbed light decreases
As H2 O is a weak field lignad, so pairing of electrons does λ L1 L3 L2 L4
not occur. Absorbed light Red Yellow Green Blue
CFSE for [Cr(H2 O)6 ]2+ = 3( −0.4∆ o ) + 1(0.6∆ o ) = −0.6∆ o As the energy of wavelengths absorbed increases, greater
Similarly, extent of crystal-field splitting, hence higher field strength
of the ligand.
Electronic
Complex Metal ion CFSE Increasing order
configuration
Energy L1 < L3 < L2 < L4
[Mo(H2O) 6 ] 2 + Mo 2 + t23g , e1g −0.6 ∆ o (Red) (Yellow) (Green) (Blue)

`[Ti(H2O) 6 ] 3 + Ti 3 + t12g , eg0 −0.4 ∆ o Ligand field strength L1 < L3 < L2 < L4

[Ti(H2O) 6 ] 2 + Ti 2 + t22g , eg0 −0.8 ∆ o 91. (a) [Co (NH3 )6 ] Cl 3 → [Co (NH3 )6 ]3+ + 3Cl −
[Co (NH3 )3 ] Cl 3 → [Co (NH3 )3 Cl 3 ]
∆ o ∝ Crystal Field Stabilisation Energy (CFSE)
[Co (NH3 )4 Cl 2 ]Cl → [Co (NH3 )4 Cl 2 ] + Cl −
∆ o depends on Zeff and for 3d-series, Zeff is less than
4d -series. [Co (NH3 )5 Cl] Cl 2 → [Co (NH3 )5 Cl]2+ + 2Cl −
Hence, ∆ o of [Cr(H2 O)6 ] 2 + > [Mo(H2 O)6 ]2 + So, [CO(NH3 )3 Cl 3 ] does not ionise so does not give test
for chloride ions.
From above table,
92. (b) Hard water has commonly high pH, when EDTA
∆ o of [Ti(H2 O)6 ]3 + > ∆ o of [Ti(H2 O)6 ]2 + solution is added to it, six coordinated Mg-EDTA
85. (b) [Ni(CN)4 ] 2− do not contain unpaired electrons, so complex is formed which lead to decrease in pH of
solution.
cannot absorb visible light.
93. (a) (a) [FeF3 ]2− oxidation state of Fe = +3
86. (a) When concentrated HCl is added in an aqueous
solution of CoCl 2 then it forms [CoCl 4 ]2− complex. So, Fe3 + = [Ar]3d 5 where, F− is weak field ligand.
[Co(H2 O)6 ]2 + + 4Cl − → [Co(Cl 4 )]2 − + 6H2 O 3d5
3 2
eg (Symmetrical field) [t 2g . eg ]
Pink colour Blue colour
complex complex

87. (a) CFSE = 3( −0.4) + 1(0.6) = −0.6 ∆O t2g

88. (c) [Co(NH3 )6 ]3+ and [Fe(CN)6 ]4– , inner orbital complex Hence, [FeF3 ]3− produces more symmetrical field.
ions are diamagnetic.
(b) [Mn(CN)6 ]4− oxidation state of Mn = +2
89. (a) Magnetic moment, µ is related with number of
So, Mn 2 + = [Ar]3d 5 where, CN− is strong field ligand,
unpaired electrons as so , it produce unsymmetrical field.
µ = n ( n + 2 ) BM
3d 5 eg
(1.73 )2 = n ( n + 2 ) [t 25g , eg0 ]

On solving n=1
a
a

Thus, the complex compound having one unpaired

electron exhibit a magnetic moment of 1.73 BM.
 (c) [CoF6 ]3− oxidation state of Co = +3 103. (a) The M—C π-bond in metal carbonyl which is formed
3+ by the donation of an electron pair from a filled d-orbital
So, Co = [Ar]3d 6
of metal into the vacant antibonding π-orbital of CO,
eg strengthens the M—C σ-bond.
3d 6
This is called synergic effect and is usually observed
in metal carbonyls. Thus, [Ni(CO)4 ] exhibits synergic
a

t2g effect.

a
104. (b) For the reaction of the type
Due to weak field ligand of F− , it produces
M + 4L s ML4 ,
unsymmetrical field
larger the stability constant, the higher the proportion of
i.e. t 42 g , e g2 ML4 that exists in solution.
(d) [Co(NH3 )6 ]2+ oxidation state of Co= +2 105. (c) For M + 4 L s ML4 ,
2+ the overall stability constant (β 4 ) is expressed as
So, Co = [Ar]3d 7

where, NH3 is strong field ligand, 106. (d) The reciprocal of the formation constant is called
instability constant or dissociation constant.
3d 7 eg
107. (a) β 4 for [ ML4 ]2− can be written as
[ ML4 ] 2 −
a
a

t2g β4 = = 2.5 × 1013

[ M 2+ ][ L−1 ] 4
a
a
a

t 62 g , e 1g (unsymmetrical field). The overall equilibrium constant for formation of

[ ML4 ]2 −
94. (c) Metal-carbon bond in metal carbonyls possess both σ and [ ML4 ] 2 − can be written as, K =
[ M 2+ ][ L−1 ]4
π-character.
K = β 4 = 2.5 × 1013 ; where, β 4 = [ ML4 ]/ [ M ][ L]4
95. (a) The M—C π-bond is formed by the donation of a pair of
electrons. 108. (d) Coordination compounds find use in both qualitative
96. (a) The metal-carbon bond in metal carbonyls possess both σ and quantitative chemical analysis.
and π-character. The metal to ligand bonding creates a 109. (c) Hardness of water is estimated by simple titration with
synergic effect which strengthens the bond between Co and Na 2 EDTA because Ca 2+ and Mg 2+ ions form stable
the meal. complexes with EDTA.
98. (a) The complexes which contain −Co ligand, will have
110. (a) Estimation of calcium and magnesium is done by
π-bond. All these three contain π-bond in their ligand.
EDTA.
99. (c) In Fe (CO)5 , the Fe—CO bond possesses both sigma and
111. (a) Gold combines with cyanide in the presence of
pi-bonds characters. oxygen and water to form the coordination entity
101. (b) Mn + = 3 d 5 , 4 s1 . In the presence of CO, effective [Au(CN)2 ]− in aqueous solution.
configuration = 3 d 6 , 4 s0 . 112. (a) Chlorophyll contains Mg, hence (a) is incorrect
Three lone pairs of back bonding with vacant orbital of C in statement.
CO
114. (a) Haemoglobin is prophyrin complex of ferrous iron
Fe0 = 3d 6 , 4 s2 . In the presence of CO, effective being coordinated to four nitrogen atoms and additionally
configuration = 3 d 8 . 4 lone pairs of back bonding with CO. coordinated to a water molecule. The water molecule
Cr 0 = 3 d 5 , 4 s1 . Effective configuration = 3 d 6 , 3 lone pairs appears to be replaceable reversible by a molecule of
or back bonding with CO. oxygen to give oxyhaemoglobin. Fe2+ is diamagnetic due
V− = 3 d 4 , 4 s2 . Effective configuration = 3 d 6 , 3 lone pairs to strong field ligands.
for back bonding with CO. 116. (c) Wilkinson catalyst is used for the hydrogenation of
Maximum back bonding in Fe(CO)5 , therefore, CO bond alkenes.
order is lowest here.
118. (b) AgBr + 2Na 2 S2 O3 → Na 3 [Ag(S2 O3 )2 ] + NaBr
102. (d) O
Soluble complex
OC C CO 119. (d) The excess of copper and iron are removed by the
OC—Co——Co—CO chelating ligands D-penicillamine and desterrioxime−B
OC C CO via the formation of coordination compounds.
O 120. (d) For lead poisoning, the antidote used is EDTA.
Total number of metal -metal bond = 1
125. (a,b,c) Cl − is a weak field ligand but Cl − causes the pairing 135. (c) When di- or polydentate ligand uses its two or more
of electron with large Pt 2+ and consequently give dsp 2 donor atoms to bind a single ion, it is called chelate ligand.
hybridisation with square planar geometry. 137. (d) Coordination number of the central atom/ion is
127. (b,c,d) The complexes which do not have unpaired determined only by the σ-bonds formed by the ligand with
electrons are diamagnetic. the central atom/ion.
(a) [Ni(NH3 )4 ]2+ 138. (b) Only [CoCl 3 (NH3 )3 ] complex does not ionised in
3d 4s 4p
solution, as no ion is present outside the coordination
sphere.
Ni =

a
a
a
a

3d 4s 4p
140. (b) Polydentate ligands are listed in alphabetical order.
Ni2+ = 141. (b) Ligands are named in a alphabetical order before the
a
a
a

3d 4s 4p
name of central metal ion.
[Ni(NH3)4 ]2+ = XX XX XX XX 142. (d) Co in a complex ion is called cobalt and Pt is called
a
a
a

3 platinum. If the complex ion is an anion the name of the

NH3,sp hybridisation
metal ends with the suffix-ate.
The resulting complex will be tetrahedral. It has two 144. (d) Charge on each complex cation in the given compound
unpaired electrons thus, paramagnetic. is +3.
(b) Ni(CO)4 146. (d) All the statements are true about optical isomers
3d 4s 4p
147. (c) The two forms, dextro and laevo, depend upon the
[Ni(CO)4 ]= XX XX XX XX
a
a
a
a
a

direction they rotate the plane of polarised light in a

CO CO CO CO
123 polarimeter. l-rotates to the left and d-rotates to the right.
sp3 hybridisation
149. (d) In the octahedral complex [Co(NH3 )6 ] 3+ , the cobalt ion
3+
(c) [Co(NH3 )6 ] is in +3 oxidation state. It is in 3 d 6 electronic configuration
3d 4s 4p and it is diamagnetic in nature.
Co =
a
a
a

3d 6 =
3d 4s 4p
NH3 being strong field ligand causes pairing resulting in a
3+
Co = diamagnetic complex.
a
a

3d 4s 4p 150. (d) All the statements are true.

[Co(NH3)6 ]3+ = XX XX XX XX XX XX 151. (a) All the statements are true.
a
a
a

123
NH3, d2sp3 hybridisation 152. (a) All the given facts are true related to CFT.
153. (c) [PtCl 4 ]2– complex has a square planar geometry.
(d) [Ni(CN)4 ]2−
[Ni(CO)4 ] complex has a tetrahedral geometry.
3d 4s 4p
155. (a) [Fe(CN)6 ] 3− complex has d 2 sp 3 hybridisation and it is
Ni2+ =
a
a
a

paramagnetic, due to pairing of electrons by strong field

3d 4s 4p ligand CN− .
[Ni(CN)4 ]2– = XX XX XX XX 156. (a) An aqueous pink solution of cobalt (II) chloride
a
a
a
a

123
changes to deep blue on addition of excess of HCl. This is
CN–, dsp2 hybridisation
because [Co(H2 O)6 ]2+ is transformed into [CoCl 4 ]2– and
In options (b),(c) and (d). no unpaired electrons are present. also the tetrahedral complexes have smaller crystal field
Thus, these are not paramagnetic. splitting than octahedral complexes.
129. (a,c,d) Linkage isomerism is caused due to presence of 157. (d) Decacarbonyldimanganese (0) is made up of two square
ambidentate ligands. In option (b), ambidentate ligand SCN pyramidal Mn(CO)5 units. These units are joined by a
is present. Thus, the pair mentioned in this option will Mn  Mn bond. Its structure is
show linkage isomerism.
CO CO
CO CO
133. (c) In coordination compounds metals show two types of
linkages (valencies). (i) Primary and (ii) Secondary CO—Mn——Mn—CO
valencies. The primary valencies are normally ionisable
CO CO
and are satisfied by negative ions. CO CO
 4
158. (c) Tetrahedral complexes have th of CFSE value than 181. (d) Reason is correct; Assertion is incorrect.
octahedral complexes. 9
182. (d) A → 4, B → 3, C → 2, D → 1
159. (a) [Cu(NH3 )4 ]SO4 is paramagnetic with one unpaired
183. (c) A → 1(III); B → 3(II); C → 2(I)
electron in the d-subshell. Its aqueous solution cannot
conduct electricity. 184. (b) A → 2, B → 1, C → 3
160. (c) Assertion is correct, reason is incorrect. Shape of [Co(NH3 )6 ]3+ is octahedral because it is d 2 sp 3
hybridised.
161. (a) Mohr’s salt gives NH+4 , Fe2+ and SO2−
4 ions in aqueous
Shape of [Ni(CO)4 ] is tetrahedral, as it is sp 3 hybridised.
solution because it is a double salt.
164. (b) Ethan-1, 2-diammine is a neutral molecule.as it Shape of [PtCl 4 ]2− is square planar, as it is dsp 2 hybridised.
carries no charge. Oxidation number of Co in the complex 185. (c) A → 3, B → 1, C → 4, D → 2, E → 5
ion is +3. 186. (d) A → 4, B → 1, C → 3, D → 2
165. (a) Usually a sulphate ion is a bidentate ligand but it can also 187. (c) A → 4, B → 2, C → 3, D → 1
act as a monodentate in certain complexes because many 188. (a) A → 2, B → 3, C → 1
times multidentate ligands do have flexidentate character. 189. (c) A → 3; B → 4; C → 2; D → 1
166. (a) Oxidation number of Cr in [Cr(NH3 )3 (H2 O)3 ]Cl 3 is 190. (b) A → 1, B → 2, C → 3, D → 4
same as the charge of the complex ion, i.e. +3 because all the 191. (d) A → IV (2), B → III (1), C → II (3), D → I (4)
ligands are neutral molecules in this compound. 192. (a) A → 6, B → 4, C → 1, D → 2, E → 3
167. (a) Isomers differ in one or more physical or chemical A. [Ni(CN)4 ]2−
properties because these have different arrangement of
Ni(28) = [Ar] 3 d 8 4 s2
atoms.
168. (a) Geometrical isomerism arises in heteroleptic complexes Ni 2 + = [Ar] 3 d 8 , 4 s0
because different geometric arrangements of the ligands are 3d 8
possible.
169. (a) Tetrahedral complexes do not show geometrical
isomerism because the relative positions of the unidentate Since, CN− is a strong field ligand, it causes pairing of
ligands attached to the central metal atom are same with electrons.
respect to each other. B. Chlorophyll contains Mg2+ ion.
171. (d) Square planar complexes necessarily exhibit geometrical C. Ziegler-Natta catalyst is Ti 4+ + (C2H5 )3 Al
isomerism. D. [NiCl 4 ]2−
172. (c) Assertion is correct, reason is incorrect. In this case, Cl − is a weak field ligand so doesn’t cause
173. (a) Octahedral geometry and diamagnetism exist in pairing. Hence,
3d 3s 3p
[Co(NH3 )6 ] 3+ as it has no unpaired electron.
∴ [Ni(CN)4]2– = [Ar] × × × ×
174. (a) The ligands nitro and nitrito are called ambidentate – – – –
CN CN CN CN
ligands. These give linkage isomers.
175. (a) Removal of water from [Ti(H2 O)6 ]Cl 3 on heating renders dsp2 hybridisation
(No unpaired electron
it colourless because in the absence of ligand, crystal field ⇒ diamagnetic)
splitting does not occur.
Due to the presence of unpaired electrons, it is
176. (c) The coordination complexes [Ni(CN)4 ]2– and [NiCl 4 ]2– paramagnetic.
have not the same shape also differ in magnetic behaviour also. E. Deoxyhaemoglobin is non-planar while
177. (d) [Ni(en)3 ]Cl 2 is more stable than [Ni(NH3 )6 ]Cl 2 . oxyhaemoglobin is planar.
193. (c) A → 2, B → 4, C → 5, D → 1, E → 3.
178. (a) Complexes are preferred in the electrolytic bath for
electroplating because complexes dissociate slowly and 194. (a) A → 1, B → 2, C → 3
hence give a smooth and even deposit. 195. (d) A → 4, B → 3, C → 2, D → 1
196. (a) A → 4, B → 2, C → 1, D → 3
179. (c) Crystal field splitting in ferrocyanide ion is smaller than
e g.
that of ferricyanide ion.
180. (b) As compared to non-chelated complexes, chelated d4
complexes are more stable.
Labile complexes are those complexes which contain ligands t2g
those can be easily replaced by other ligands. CFSE = −4 × 0.4 = − 1⋅ 6
 For d 5 , CFSE = − 5 × 0.4 = − 2.0 contain [Co(NH3 )5 Cl]2+ and two Cl − as constituent ions.
For d 6 , CFSE = − 6 × 0.4 = − 2.4 Thus, it is 1 : 2 electrolyte.
For d 7 , CFSE = − 6 × 0.4 + 1 × 0.6 [Co(NH3 )5 Cl]Cl 2 → [Co(NH3 )5 Cl]2+ ( aq ) + 2Cl − ( aq )
= − 2.4 + 0.6 = − 1.8 Hence, option (b) is the correct.
197. (b) Secondary valency means coordination number. So,
205. (d) 1 mole of AgNO3 precipitates one free chloride ion (Cl − ) .
the secondary valencies of I and III are 4 and 6
respectively. Here, 3 moles of AgCl are precipitated by excess of AgNO3 .
Hence, there must be three free Cl − ions.
198. (c) Complexes having the secondary valencies
(coordination number) of 6 are II, III and IV. So, the formula of the complex is [Cr(H2 O)6 ]Cl 3 and correct
choice is (d).
199. (d) NiCl 2 + 4KCN → K 2 [Ni(CN)4 ] + 2KCl
(excess) X 206. (a) The complex compound is [Pt(NH3 )2 Cl 2 ] .
Potassium tetra
cyanonickelate (II) The ligands present in the compound are
(i) NH3 − neutral ligand represented as ammine.
K 2 [Ni(CN)4 ] + 4HCl → K 2 [Ni(Cl)4 ] + 4HCN (ii) Cl − anion ligand (ending with-o-) represented as chlorido, di
Potassium prefixed to represent two ligands.
tetrachloronickelate (II)
The oxidation number of platinum in the compound is 2.
Hence, correct IUPAC name of [Pt (NH3 )2 Cl 2 ] is
200. (b) Hybridisation of X → K 2 [Ni(CN)4 ] is dsp 2
Diamminedichloridoplatinum (II).
Y → K 2 [NiCl 4 ] is sp 3 .
So, option (a) is correct.
201. (c) X contains paired electrons whereas Y contains 207. (c) Correct IUPAC name can be written as
unpaired electrons.
The ligands present in the given coordination compound are:
202. (b) Ligand must have an electron pair to donate metal (i) (NH3 ) represented as ammine
ion. These electron pair must be loosely bounded to (ii) Cl s represented as chlorido
ligand (iii) NOs2 represented as nitrito-N
•• •• ••
•
e.g. • N== O, NH2 CH2 CH2 NH2 ,•• CO, According to IUPAC rule, ligands are named in an
alphabetical order before central atom. Prefix di-will be used
Among NH+4 does not have any pair of electron.
to indicate the number of NH3 ligands present.
Hence NH+4 is not a ligand. Oxidation state of metal is indicated by Roman numeral in
203. (a) A chelating ligand has two or more binding donor parenthesis. So, IUPAC name will be
atoms to a single metal ion diamminechloronitrito-N-platinum (II)
e.g. Hence, option (c) is correct.
O
|| NH2 → 208. (a) [Cr(H2 O)4 Cl 2 ]+ shows geometrical isomerism because it
C—O → NH2 → CH2 is a MA 4 B2 type coordination compound which contains two
| CH2 |
C—O → C—O → CH2
set of equivalent ligands, four H2 O and 2Cl.
|| NH2 → Hence, the possible geometrical isomers are
O O
Oxalato Glycinato Ethan-1, 2-diamine Cl Cl
H 2O H 2O H2O Cl
Here (←) denotes binding site. Thiosulphato ) is (S2 O2−
3 Cr and Cr
not a chelating ligand because geometrically it is not H 2O H 2O H2O H 2O
favourable for S2 O2−
3 to chelate a metal ion.
Cl H 2O
S trans-isomer cis-isomer
||
S—O
Hence, correct choice is option (a).

O O
209. (b) Solvate isomerism is shown when two compounds having
same molecular formula differ by water or solvent molecule
204. (b) One mole of AgNO3 precipitates one mole of directly bonded to metal ion or present as free solvent
chloride ion. In the above reaction, when 0.1 mole of molecules in the crystal lattice.
CoCl 3 (NH3 )5 is treated with excess of AgNO3 , 0.2 mole When water is present as solvent and show this type of
of AgCl is obtained thus, there must be two free chloride isomerism is known as hydrate isomerism.
ions in the solution of electrolyte. Coordination compound [Cr(H2 O)6 ]Cl 3 and
So, molecular formula of complex will be [Cr(H2 O)5 Cl]H2 O ⋅ Cl 2 are solvate isomers, because water is
[Co(NH3 )5 Cl] Cl 2 and electrolytic solution must exchanged by chloride ion. This is why both of them show
different colour on exposure to sunlight.
210. (a) The ligand(s) which has two different bonding sites are 216. (a, b, c) Molecular formula of ethan-1, 2-diammine is
known as ambident ligands, e.g. NCS, NO2 etc. ••
CH2 — NH2
Here, NCS has two binding sites at N and S.
 ••
Hence, NCS (thiocyanate) can bind to the metal ion in two ways
CH2 — NH2
M ← NCS or M → SNC
Thus, coordination compounds containing NCS as a ligand can (a) Ethan-1, 2-diammine is a neutral ligand due to
show linkage isomerism, i.e.[Pd(C6 H5 )2 (SCN)2 ] and absence of any charge.
[Pd(C6 H5 )2 (NCS)2 ] are linkage isomers. (b) It is a bidentate ligand due to presence of two
Hence, correct choice is option (a). donor sites one at each nitrogen atom of amino
group.
211. (d) Compounds having same molecular formula but different
structural formula are known as isomers. [Co(SO4 )2 (NH3 )5 ]Br (c) It is a chelating, ligand due to its ability to chelate
and [Co(SO4 )(NH3 )5 ]Cl have not same molecular formula. with the metal.
Hence, they are not isomers. Hence, options (a), (b) and (c) are correct choices.
212. (c) Chelation (formation of cycle by linkage between metal ion 217. (a, c) Homoleptic complex The complex containing
and ligand) stabilises the coordination compound. The ligand only one species or group as ligand is known as
which chelates the metal ion are known as chelating ligand. homoleptic ligand.
Here, only [Fe(C2 O4 )3 ]3− is a coordination compound which e.g. [Co(NH3 )6 ]3+ , [Ni(CN)4 ]2−
contains oxalate ion as a chelating ligand. Hence, it stabilises Here, [Co(NH3 )6 ]3+ contain only NH3 as a ligand
coordination compound by chelating Fe3+ ion. and [Ni(CN)4 ]2− contain CN as a ligand. While other
213. (c) CFSE for octahedral and tetrahedral complexes are closely two complexes [Co(NH3 )4 Cl 2 ]+ and [Ni(NH3 )4 Cl 2 ]
4
related to each other by formula ∆ t = ∆ o . contain NH3 and Cl as ligands.
9
Hence, options (a) and (c) are correct choices.
where, ∆ o = CFSE for octahedral complex, ∆ t = CFSE for
tetrahedral complex 218. (b, d) Heteroleptic complexes Coordination
According to question, ∆ o = 18000 cm−1 complexes which contain more than one type of
ligands are known as heteroleptic complexes.
4 4
∴ ∆t = ∆ o = × 18000 cm−1 e.g. [Fe(NH3 )4 Cl 2 ]+ contain NH3 and Cl as a ligand
9 9
is as heteroleptic complex. Similarly, [Co(NH3 )4 Cl 2 ]
= 4 × 2000 cm−1 = 8000 cm−1 contain NH3 and Cl as ligand is also a heteroleptic
Hence, correct choice is option (c). complex.
214. (b) Greater the value of log K, greater will be stability of complex Hence, options (b) and (d) are correct choices.
compound formed. 219. (a, c) [Co(en)3 ]3+ and cis − [Co(en)2 Cl 2 ]+ are
For reaction, Cu 2+ + 4CN− → [Cu(CN)4 ]2 − optically active compounds because their mirror
[(Cu(CN)4 )2 − ] images are non-superimposable isomer.
K= and log K = 27.3
[Cu 2+ ][CN− ]4 en 3+
en 3+

For this reaction, log K has highest value among the given four
reactions. Hence, K will also be higher among these four en Co Co en
complexes, i.e. stability of the complexes with higher pH, will be
highest among these four complexes.
en en
215. (c) As we know that, strong field ligand splits the five degenerate
energy levels with more energy separation than weak field ligand, Non-superimposable isomers of [Co(en)3 ]3+
i.e. as strength of ligand increases crystal field splitting energy ]+ ]+
increases. en en
hc 1 1
Hence, ∆E = ⇒ ∆E ∝ ⇒ λ∝
λ λ ∆E en
Co en Co
As energy separation increases, the wavelength decreases.
Cl Cl
Thus, the correct order is
[Co(H2 O)6 ]3+ > [Co(NH3 )6 ]3+ > [Co(CN)6 ]3− Cl Cl

Here, strength of ligand increases, ∆E increases, CFSE increases Non-superimposable isomers of [Co(en)2 Cl 2 ]+
and λ absored decreases. Hence, options (a) and (c) are correct choices.
Hence, correct choice is option (c).
 3d 4s 4p
220. (a, c) Coordination compounds containing a ligand
with more than one non-equivalent binding position [18 Ar] ×× ×× ×× ×× ×× ××
(known as ambident ligand) show linkage isomerism.
e.g. [Co(NH3 )5 (NO2 )+ contains NO2 which have two CN CN CN CN CN CN
2
donor sites N and O can be shown by arrow (→) as d sp3 hybridisation
O Number of unpaired electron = 1
→N
O
→ Magnetic property = Paramagnetic
Thus, [Co(NH3 )6 ]3+ and [Fe(CN)6 ]4− are diamagnetic.
[Cr(NH3 )5 SCN]2+ contains SCN which have two Hence, correct choices are options (a) and (c).
different donor sites S and N can be shown by arrow 222. (a, c) Molecular orbital electronic configuration of Mn 3+ in
(→) as [MnCl 6 ]3− is
→ S — C ≡≡ N ←
3d 4s 4p 4d
Hence, [Co(NH3 )5 (NO2 )]2+ and [Cr(NH3 )5 SCN]2+
[18 Ar] ×× ×× ×× ×× ×× ××
show linkage isomerism. While [Co(H2 O)5 CO]3+
and [Fe(en)2 Cl 2 ]+ has no ambident ligand. So, these Cl Cl Cl Cl Cl Cl
two will not show linkage isomerism. sp d 3 2 hybridisation

Hence, options (a) and (c) are correct choices. Number of unpaired electrons = 4
221. (a, c) Molecular orbital electronic configuration of Magnetic property = Paramagnetic
Co 3+ in [Co(NH3 )6 ]3+ is Molecular orbital electronic configuration of Co 3+ in [CoF6 ]3− is
3d 4s 4p 3d 4s 4p 4d
[18 Ar] ×× ×× ×× ×× ×× ×× [18 Ar] ×× ×× ×× ×× ×× ××

NH3 NH3 NH3 NH3 NH3 NH3 F F F F F F

2
d sp 3 hybridisation sp 3 2d hybridisation
Number of unpaired electron = 0 Number of unpaired electrons = 4
Magnetic property = Diamagnetic Magnetic property = Paramagnetic
Molecular orbital electronic configuration of Mn 3+ in Molecular orbital electronic configuration of Fe3+ in [FeF6 ]3− is
[Mn(CN)6 ]3− 3p 4s 4p 3d
3d 4s 4p [18 Ar] ×× ×× ×× ×× ×× ××
[18 Ar] ×× ×× ×× ×× ×× ××
F F F F F F
CN CN CN CN CN CN
sp3 2 d hybridisation
d 2 sp 3 hybridisation
Number of unpaired electrons = 5
Number of unpaired electrons = 2 Magnetic property = Paramagnetic
Magnetic property = Paramagnetic Molecular orbital electronic configuration of Ni 2+ in
Molecular orbital electronic configuration of Fe2+ in [Ni(NH3 )6 ]2+ is
[Fe(CN)6 ]4− is 3d 4s 4p 4d
3d 4s 4p [18 Ar] ×× ×× ×× ×× ×× ×× ××
[18 Ar] ×× ×× ×× ×× ×× ××
NH3 NH3 NH3 NH3 NH3 NH3

CN CN CN CN CN CN sp 3 2
d hybridisation
2
d sp 3 hybridisation Number of unpaired electrons = 2
Number of unpaired electron = 0 Magnetic property = Paramagnetic
Magnetic property = Diamagnetic Thus, [MnCl 6 ]3− and [CoF6 ]3− are paramagnetic having four
Molecular orbital electronic configuration of Fe3+ in electrons each.
[Fe(CN)6 ]3− Hence, correct choices are (a) and (c).
223. (a, c) According to VBT, the molecular orbital electronic Hybridisation = d 2 sp 3
configuration of Fe3− in [Fe(CN)6 ]3− is n=1
3d 4s 4p 4d Hence, correct Assertion is [Fe(CN)6 ]3− ion shows
magnetic moment corresponding to one unpaired
[18 Ar] ×× ×× ×× ×× ×× ××
electron.
i.e., µ = n ( n + 2 ) = 1(1 + 2 )
CN CN CN CN CN CN
2 = 3 = 1.73 BM
d sp 3 hybridisation
Hybridisation = d 2 sp 3 229. (b) Assertion and Reason both are true but Reason is
Number of unpaired electron = 1 not correct explanation of Assertion.
Magnetic property = Paramagnetic Correct Reason is
Hence, correct choices are options (a) and (c). [Cr(H2 O6 )]Cl 2 and [Fe(H2 O)6 ]Cl 2 are reducing in
224. (b, c) Aqueous pink solution of cobalt (II) chloride is due to nature due to formation of more stable complex ion
electronic transition of electron from t 2g to e g energy level of after gaining of electron.
[Co(H2 O)6 ]2+ complex. When excess of HCl is added to this 230. (d) A → (4); B → (1); C → (3); D → (2)
solution Oxidation state of CMI (central metal ion) can be
(i) [Co(H2O)6 ]2+ is transformed into [CoCl 4 ]2− . calculated by considering the oxidation state of whole
(ii) Tetrahedral complexes have smaller crystal field splitting than molecule is equal to charge present on coordination
4 sphere.
octahedral complexes because ∆ t = ∆ o
9 A. [Co(NCS)(NH3 )5 ]SO3.
Hence, options (b) and (c) are correct choices. Let oxidation state of Co is x.
225. (a) Assertion and Reason both are correct and Reason is the x −1+ 5×0 = + 2
correct explanation of Assertion. x = + 2 + 1= + 3
Toxic metal ions are removed by chelating ligands. When a B. [Co(NH3 )4 Cl 2 ]SO4
solution of chelating ligand is added to solution containing toxic Let oxidation state of Co = x
metals ligands chelates the metal ions by formation of stable ⇒ x + 4 × 0 + 2 × (−1) = + 2
complex. ⇒ x−2=+2
226. (a) Assertion and Reason both are correct and Reason is correct x=+4
explanation of Assertion. Linkage isomerism arises in C. Na 4[Co(S2O3 )3 ]
coordination compounds containing ambidentate ligands because Let oxidation state of Co = x
ambidentate ligand has two different donor atoms. x + 3 × (−2) = − 4
e.g. SCN, NO2 etc. x − 6= −4
x = −4 + 6= + 2
227. (b) Assertion and Reason both are correct and Reason is not
D. [Co(CO)8 ]
correct explanation of Assertion.
Let oxidation state of Co = x
Complexes of MX 6 and MX 5 L type (X and L are unidentate) do x − 8×0 = 0
not show geometrical isomerism due to presence of plane of x=0
symmetry and necessary condition for showing geometrical
isomerism is that complex is must of MA 4 B2 type or Hence, correct choice is option (d).
[ M ( AB )2 X 2 ] type 231. (a) A → (3); B → (4); C → (1); D → (2)
…

|
L 232. (d) A → (4); B → (1); C → (2); D → (3)
X | X Isomerism in coordination compound is decided by
M
type of ligands and geometry of coordination and
X | X
X arrangement of ligands.
…

[MX5L] Plane of
symmetry
A. [Co(NH3 )4 Cl 2 ]+ shows geometrical isomerism due
to presence of two types of ligand whose
228. (d) Assertion is false but Reason is true. [Co(NH3 )4 Cl 2 ] + arrangement around central metal
According to VBT, MOEC of Fe3+ in [Fe(CN)6 ]3− is ion.
NH3 Cl
3d 4s 4p NH3 NH3 NH3 NH3
[18 Ar] ×× ×× ×× ×× ×× ×× Co Co
Cl NH3 H3N NH3
Cl Cl
CN CN CN CN CN CN
cis-isomer trans-isomer
2
d sp 3 hybridisation
 B. cis − [Co(en)2 Cl 2 ]+ shows optical isomer due to its According to VBT, hybridisation and number of unpaired
non-superimposable mirror image relationship. electrons of coordination compounds can be calculated as
A. [Cr(H2O)6 ]3+
2+ 2+
en en MOEC (Molecular orbital electronic configuration) of
Cr 3+ in [Cr(H2O)6 ]3+ is
Cl Cl
Co Co 3d 4s 4p
×× ×× ×× ×× ×× ××
Cl Cl
en en
H2O H2O H2O H2O H2O H2O
C. [Co(NH3 )5 (NO2 )]Cl 2 shows ionisation isomer due to Hybridisation = d sp 2 3
its interchanging ligand from outside the ionisation
n (number of unpaired electrons) = 3
sphere.
B. [Co(CN)4 ]2−
D. [Co(NH3 )6 ][Cr(CN)6 ] shows coordination isomer due to
interchanging of ligand in between two metal ions MOEC of Co2+ in [Co(CN)4 ]2− is
from one coordination sphere to another coordination
sphere. ×× ×× ×× ××
Hence, correct choice is option (d).
233. (b) A → (4); B → (3); C → (2); D → (1) CN CN CN CN
Colour of coordination compound is closely related to Hybridisation = dsp2 and n = 1
CFSE of coordination compound. Depending upon the C. [Ni(NH3 ) 6 ]2+
CFSE of given coordination compounds. Correct matching
will be as follows MOEC of Ni 2+ in [Ni(NH3 )6 ]2 + is

Column I Column II 3d 4s 4p 4d
(Complex ion) (Colour)
×× ×× ×× ×× ×× ××
A. [Co(NH3 ) 6 ] 3+ 4. Yellowish orange
B. [Ti(H2O) 6 ] 3+ 3. Pale blue NH3 NH3 NH3 NH3 NH3 NH3
2+
C. [Ni(H2O) 6 ] 2. Green
Hybridisation = sp d 3 2
and n = 2
D. [Ni(H2O) 4 (en)] 2+ (aq ) 1. Violet 4−
D. [MnF6 ]
Hence, correct choice is option (b). MOEC of Mn 2+ in [MnF6 ] 4− is
234. (a) A → (3); B → (1); C → (4); D → (2)
Formation of inner orbital complex and outer orbital 3d 4s 4p 4d
complex determines hybridisation of molecule which inturn
depends upon field strength of ligand and number of vacant
d-orbitals. F F F F F F
(i) Strong field ligand forms inner orbital complex with
hybridisation d 2sp3. Hybridisation = sp3d 2
(ii) Weak field ligand forms outer orbital complex with n=5
hybridisation sp3d 2. Hence, correct choice can be represented by (a).