Coordination Compounds

Solutions

Chapter 19

Coordination Compounds

SECTION - A

Objective Type Questions (One option is correct)

1. Of the following, which ligand does not posses the name suggested by IUPAC when it acts as a ligand in

complex?

(1) H2O, aqua (2) NH

3, ammonia (3) CO, carbonyl (4) F–, fluoro

Sol. Answer (2)

Fact.

2. Write the structural formula for aqua-bromo-bis (ethylenediamine) chromium (III) chloride

(1) [CrBr(H2O) (en)]Cl (2) [CrBr

2(H

2O) (en)]Cl

2

(3) [CrBr(H2O) (en)

2]Cl

2(4) [CrBr(H

2O) (en)

2]Cl

3

Sol. Answer (3)

2 2 2 2 2 2 2[CrBr(H O)(NH CH CH NH ) ]Cl

3. The complex cis [CoCl(NH3)(NH

2CH

2CH

2NH

2)2]2+ has an octahedral geometry. Name of this complex ion

according to IUPAC would be

(1) cis - chloroamminethylenediamine cobalt (II) ion

(2) cis - amminechloroethylenediamine cobalt (III) ion

(3) cis - amminechlorobis (ethylenediamine) cobalt (II) ion

(4) cis - amminechlorobis (ethylenediamine) cobalt (III) ion

Sol. Answer (4)

cis-amminechlorobis (ethylenediamine) cobalt (III) ion.

4. Which complex ion has cis and trans isomer?

(1) [PdCl3NH

3]– (2) [Pd(CN)

5NH

3]– (3) [PtCl

2(CN)

2]2– (4) [Pt(C

2O

4)2]2–

Sol. Answer (3)

For cis and trans isomer, complex should have [MA2B

2] or [MA

2BC] or [MA

4B

2] or [MA

2B

4] formula.

42 Coordination Compounds Solution of Assignment

5. Which of the following divalent metal ion form the most stable complexes?

(1) Mn2+ (2) Fe2+ (3) Ni2+ (4) Cu2+

Sol. Answer (4)

On the basis of Erwin William series, stability of complex increases as

Mn+2 < Fe+2 < CO+2 < Ni+2 < Cu+2 > Zn+2.

6. A six coordinate complex of formula CrCl3.6H

2O has green colour. A 0.1 M solution of the complex when treated

with excess of AgNO3 gave 28.7 g of white precipitate. The formula of the complex would be

(1) [Cr(H2O)

6]Cl

3(2) [Cr(H

2O)

5Cl]Cl

2.H

2O (3) [Cr(H

2O)

4Cl

2]Cl.2H

2O (4) [Cr(H

2O)

3Cl

3].3H

2O

Sol. Answer (2)

0.1 mole of complex gives 28.7 gm of AgCl.

Mole of AgCl formed 28.7

0.2108 35.5

So the formula is [Cr(H2O)

5Cl]Cl

2.H

2O.

7. If excess of AgNO3 solution is added to 100 ml of 0.024 M solution of dichlorobis (ethylene diamine) cobalt

(III) chloride, how many moles of AgCl will be precipitated?

(1) 0.0012 (2) 0.0016 (3) 0.0024 (4) 0.0048

Sol. Answer (3)

One Cl– ion is present outside the co-ordination sphere, so, mole of AgCl formed is equal to 1

10 × mole of

complex = 0.0024.

8. Which of the following complex has five unpaired electrons?

(1) [Mn(H2O)

6]2+ (2) [Mn(CN)

6]3– (3) [CrCl

3(H

2O)

3] (4) [Ag(NH

3)2]+

Sol. Answer (1)

2 5Mn : 3

d

H2O is a weak ligand, so, [Mn(H

2O)

5]+2 complex contain 5 unpaired e–.

9. Which of the following species is diamagnetic?

(1) An isolated, gas-phase V3+ ion (2) A high spin ocatahedral Fe+2 complex

(3) An isolated, gas phase Cu2+ ion (4) A low spin octahedral Co3+ complex

Sol. Answer (4)

Low spin complexes of Co+3 do not contain any unpaired e–.

10. The complex [Ni(CN)4]2– is diamagnetic and the complex [NiCl

4]2– is paramagnetic. What can you conclude

about their molecular geometries?

(1) Both complexes have square planar geometries

(2) Both complexes have tetrahedral geometries

(3) [NiCl4]2– has a square planar geometry while [Ni(CN)

4]2– has a tetrahedral geometry.

(4) [NiCl4]2– has a tetrahedral geometry while [Ni(CN)

4]2– has a square planar geometry

43Solution of Assignment Coordination Compounds

Sol. Answer (4)

2 2

4[Ni(CN) ] square planar dsp

2 3

4(NiCl ) Tetrahedral sp

11. [PtCl2(NH

3)2] [Electronic configuration of Pt : [Xe]5d86s2]

(1) A tetrahedral geometry and is paramagnetic in behaviour

(2) A tetrahedral geometry and is diamagnetic in behaviour

(3) A square planar geometry and is paramagnetic in behaviour

(4) A square planar geometry and is diamagnetic in behaviour

Sol. Answer (4)

Most of the complexes of Pt+2 are square planar and diamagnetic.

12. Which of the following complex is considered to be tetrahedral?

(1) [FeCl4]– (2) [Ni(CN)

4]2– (3) [PtCl

4]2– (4) [PdCl

2(NH

3)2]

Sol. Answer (1)

[FeCl4]– is a tetrahedral complex.

13. Which of the following complex ions is more likely to be colourless?

(1) [Co(H2O)

6]2+ (2) [Mn(CN)

6]3– (3) [CrCl

3(H

2O)

3] (4) [Ag(NH

3)2]+

Sol. Answer (4)

Ag+ have 3d10 electronic configuration. So, No d-d transition is allowed.

14. How many unpaired electrons are present in the high spin form of [CoF6]3– complex and which metal orbitals

are used in bonding?

(1) 0 unpaired electrons and 4s, 4p and 4d orbitals to give sp3d2 hybridisation

(2) 4 unpaired electrons and 4s, 4p and 4d orbitals to give sp3d 2 hybridisation

(3) 0 unpaired electrons and 3d, 4s and 4p orbitals to give d 2sp3 hybridisation

(4) 4 unpaired electrons and 3d, 4s and 4p orbitals to give d 2sp3 hybridisation.

Sol. Answer (2)

–

F is a weak ligand, so complex of Co+3 will be high spin

hybridization sp3d2

15. The EAN of Pt in potassium hexachloroplatinate (IV) is

(1) 46 (2) 86 (3) 36 (4) 84

Sol. Answer (2)

EAN = (78 – 4) + 2 × 6 = 86

16. What is the name of the complex [Ni(H2O)

4(NH

2CH

2CH

2NH

2)]SO

4.5H

2O as per IUPAC rules?

(1) Aquaethylenediamine nickel (II) sulphate-1-water

(2) Tetraaquaethylenediamine nickel (II) sulphate-5-water

(3) Tetraaqua bis (ethylene diamine) nickel (II) sulphate-5-water

(4) Tetraaqua bis (ethylene diamine) nickel (III) sulphate-5-water

Sol. Answer (2)

Tetraaquaethylenediamine nickel (II) sulphate-5-water.


17. Which pair of isomers illustrates the concepts of ionisation isomers?

(1) [Cr(SCN) (NH3)5]2+ and [Cr(NCS) (NH

3)5]2+ (2) [CoCl(NH

3)5] SO

4 and [Co(SO

4) (NH

3)5]Cl

(3) cis [PtCl2(NH

3)2] and trans [PtCl

2(NH

3)2] (4) (+) - [Co(en)

3]3+ and (–) - [Co(en)

3]3+

Sol. Answer (2)

3 5 4 4 3 5[CoCl(NH ) ]SO [Co(SO )(NH ) ]Cl give different ions in solution.

18. Which of the following coordination compounds is incapable of showing geometrical isomerism?

(1) [PtCl2(NH

3)2] (2) [CoCl

2(NH

3)4]+ (3) [Co(NO

2)3(NH

3)3] (4) [Co(en)

3]3+

Sol. Answer (4)

33[Co(en) ] does not show geometrical isomerism.

19. Which of the following complexes are diamagnetic?

[Mn(CN)6]4–, [Zn(NH

3)4]2+, [Fe(CN)

6]4–, [FeF

6]3–

(1) [Zn(NH3)4]2+ (2) [Zn(NH

3)4]2+ and [FeF

6]3–

(3) [Zn(NH3)4]2+ and [Fe(CN)

6]4– (4) [Zn(NH

3)4]2+, [Mn(CN)

6]3– and [Fe(CN)

6]4–

Sol. Answer (3)

2 43 4 6[Zn(NH ) ] and [Fe(CN) ] are diamagnetic.

20. Which metal ion is likely to form a square planar complex ion with CN–?

(1) Cu+1 (2) Ag+ (3) Ni2+ (4) Zn2+

Sol. Answer (3)

Ni+2 form square planar complex ion with CN–.

21. Which of the given metal ion form high tendency to complex with CO?

(1) Sc+3 (2) Ti+3 (3) Mn+2 (4) Fe+2

Sol. Answer (4)

Due to presence of high number of e– in d-orbital.

22.4

excess

CuSO KCN Complex

Correct regarding complex in this reaction is

(1) Octahedral, paramagnetic (2) Tetrahedral, paramagnetic

(3) Square planar, diamagnetic (4) Tetrahedral, diamagnetic

Sol. Answer (2)

23. Which of the following compounds exhibit geometrical isomer?

(1) Pt(NH3)2Cl

2(2) Zn(NH

3)2Cl

2(3) Pt(NH

3)3Cl (4) All of these

Sol. Answer (1)

Pb(NH3)2Cl

2


24. Effective atomic number of Fe in Fe2(CO)

9 is

(1) 35 (2) 36

(3) 37 (4) Cannot be calculated

Sol. Answer (2)

36

25. Which of the following is most stable complex?

(1) [Co(NH3)6]+2 (2) [Co(NH

3)6]+3 (3) [Co(en)

2(NH

3)2]+3 (4) [Co(en)

3]+3

Sol. Answer (4)

Due to Chelation Effect.

26. The colour of light absorbed by an aqueous solution of CuSO4 is [IIT-JEE-2012]

(1) Orange-red (2) Blue-green (3) Yellow (4) Violet

Sol. Answer (1)

Complementary colour of blue is (Orange-red)

27. The IUPAC name of [Ni(NH3)4][NiCl

4] is [IIT-JEE-2008]

(1) Tetrachloronickel (II)-tetraamminenickel (II) (2) Tetraamminenickel (II) - tetrachloronickel (II)

(3) Tetraamminenickel (II) - tetrachloronickelate (II) (4) Tetrachloronickel(II) - tetraamminenickelate (0)

Sol. Answer (3)

IUPAC name is tetraamminenickel(II)-tetrachloronickelate(II).

28. Among the following metal carbonyls, the C–O bond order is lowest in [IIT-JEE-2007]

(1) [Mn(CO)6]+ (2) [Fe(CO)

5] (3) [Cr(CO)

6] (4) [V(CO)

6]–

Sol. Answer (4)

29. Among the following, the coloured compound is [IIT-JEE-2008]

(1) CuCl (2) K3[Cu(CN)

4] (3) CuF

2(4) [Cu(CH

3CN)

4]BF

4

Sol. Answer (3)

In CuF2, copper is in +2 oxidation state

coloured. In other options copper is in +1 oxidation state therefore colourless.

30. Both [Ni(CO)4] and [Ni(CN)

4]2– are diamagnetic. The hybridisations of nickel in these complexes, respectively,

are [IIT-JEE-2008]

(1) sp3, sp3 (2) sp3, dsp2 (3) dsp2, sp3 (4) dsp2, dsp2

Sol. Answer (2)

[Ni(CO)4] ———— sp3 hybridization

[Ni(CN)4]–2 ———— dsp2 hybridization


31. The spin only magnetic moment value (in Bohr magneton units) of Cr(CO)6 is [IIT-JEE-2009]

(1) 0 (2) 2.84 (3) 4.90 (4) 5.92

Sol. Answer (1)

Cr(CO)6

Under the influenceof strong ligand

d sp2 3

hybridization

Cr in G.S.

Diamagnetic = 0

32. The correct structure of ethylenediaminetetraacetic acid (EDTA) is [IIT-JEE-2010]

(1) N CH CH N

HOOC CH2

HOOC CH2

CH2

COOH

CH2

COOH

(2) N CH2

CH2

N

HOOC

HOOC

COOH

COOH

(3) N CH2

CH2

N

CH2

COOH

CH2

COOH

HOOC CH2

HOOC CH2

(4) N CH CH N

H

CH2

COOH

HOOC CH2

HCH

2

HOOC

CH2

COOH

Sol. Answer (3)

N CH2

CH2

N

CH2

COOH

CH2

COOH

HOOC CH2

HOOC CH2

33. The ionization isomer of [Cr(H2O)

4Cl(NO

2)]Cl is [IIT-JEE-2010]

(1) [Cr(H2O)

4(O

2N)]Cl

2(2) [Cr(H

2O)

4Cl

2] (NO

2)

(3) [Cr(H2O)

4Cl(ONO)]Cl (4) [Cr(H

2O)

4Cl

2(NO

2)] H

2O

Sol. Answer (2)

Cr(H2O)

4Cl(NO

2)]Cl and [Cr(H

2O)

4Cl

2](NO

2) have same molecular formula but they give different ions on ionization.


34. The complex showing a spin-only magnetic moment of 2.82 B.M. is [IIT-JEE-2010]

(1) Ni(CO)4

(2) [NiCl4]2– (3) Ni(PPh

3)4

(4) [Ni(CN)4]2–

Sol. Answer (2)

Ni2+ is sp3 hybridized and metal ion is connected with weak ligands.

Ni2+ d8

sp3two unpaired electrons

s = 2 (2 2) B.M.

= 8 B.M. = 2.83 B.M.

35. Geometrical shapes of the complexes formed by the reaction of Ni2+ with Cl–, CN– and H2O, respectively are

[IIT-JEE-2011]

(1) Octahedral, tetrahedral and square planar (2) Tetrahedral, square planar and octahedral

(3) Square planar, tetrahedral and octahedral (4) Octahedral, square planar and octahedral

Sol. Answer (2)

Ni+2 + 4Cl– [NiCl4]–2 (tetrahedral)

Ni+2 + 4CN– Ni(CN)4

–2 (square planar)

Ni+2 + H2O Ni(H

2O)

6

+2 (octahedral)

36. Among the following complexes (K – P),

K3[Fe(CN)

6](K), [Co(NH

3)6]Cl

3(L),

Na3[Co(oxalate)

3]–3(M), [Ni(H

2O)

6]Cl

2(N), K

2[Pt(CN)

4](O) and [Zn(H

2O)

6](NO

3)2(P)

the diamagnetic complexes are [IIT-JEE-2011]

(1) K, L, M, N (2) K, M, O, P (3) L, M, O, P (4) L, M, N, O

Sol. Answer (3)

[Co(NH3)6]Cl

3d2sp3 diamagnetic

Na3[Co(OX)

3] d2sp3 diamagnetic

K2[Pt(CN)

4] dsp2 diamagnetic

[Zn(H2O)

6](NO

3)2

sp3d2 diamagnetic

37. As per IUPAC nomenclature, the name of the complex [Co(H2O)

4(NH

3)2]Cl

3 is [IIT-JEE-2012]

(1) Tetraaquadiaminecobalt (III) chloride (2) Tetraaquadiamminecobalt (III) chloride

(3) Diaminetetraaquacobalt(III) chloride (4) Diamminetetraaquacobalt (III) chloride

Sol. Answer (4)

38. NiCl2{P(C

2H

5)2(C

6H

5)}

2 exhibits temperature dependent magnetic behaviour (paramagnetic/diamagnetic). The

coordination geometries of Ni2+ in the paramagnetic and diamagnetic states are respectively [IIT-JEE-2012]

(1) Tetrahedral and tetrahedral (2) Square planar and square planar

(3) Tetrahedral and square planar (4) Square planar and tetrahedral


Sol. Answer (3)

The hybridisation of Ni2+ in paramagnetic complex is sp3 and hence the complex is tetrahedral. The diamagnetic

complex is square planar as the hybridisation of Ni2+ is dsp2.

39. Consider the following complex ions, P, Q and R.

P = [FeF6]3–, Q = [V(H

2O)

6]2+ and R = [Fe(H

2O)

6]2+

The correct order of the complex ions, according to their spin-only magnetic moment values (in B.M.) is

[JEE(Advanced) 2013]

(1) R < Q < P (2) Q < R < P (3) R < P < Q (4) Q < P < R

Sol. Answer (2)

The electronic configuration of central metal ion in complex ions P, Q and R are

3 3

6P [FeF ] ; Fe :

3d

Q = [V(H2O)

6]2+; V2+ :

3d

R = [Fe(H2O)

6]2+; Fe2+ :

3d

The correct order of spin only magnetic moment is Q < R < P

SECTION - B

Objective Type Questions (More than one options are correct)

1. Identify the colourless complexes

(1) Ti(NO3)4

(2) [Cu(NCCH3)4]+BF

4

– (3) [Cr(NH3)6]Cl

3(4) K

3[VF

6]

Sol. Answer (1, 2)

Ti+3 and Cu+ have no unpaired e– in d orbital. So, their compounds are colourless.

2. Which of the following complexes are paramagnetic?

(1) Ni(H2O)

6]2+ (2) [MnCl

4]2– (3) [Zn(NH

3)4]2+ (4) [Co(NH

3)6]3+

Sol. Answer (1, 2)

Ni+2 and Mn+2 have unpaired e–, so these are paramagnetic.

3. Identify the incorrect statements about [Cu(NH3)4]2+

(1) The complex is tetrahedral (2) The complex is square planar

(3) Cu2+ in the complex is dsp2 hybridised (4) Cu2+ in the complex is sp2 hybridised

Sol. Answer (1, 4)

23 4[Cu(NH ) ] is paramagnetic square planar complex in which Cu+2 show dsp2 hybridisation.

4. [Co(NH3)5NO

2]SO

4 shows

(1) Ionisation isomerism (2) Coordination isomerism

(3) Linkage isomerism (4) Position isomerism


Sol. Answer (1, 3)

[Co(NH3)

5NO

2]SO

4 can show linkage and ionization isomerism. Linkage isomerism is due to presence of

ambidient 2NO

ligand.

5. Which complexes show geometrical isomerism?

(1) Tetrahedral (2) Octahedral (3) Square planar (4) Linear

Sol. Answer (2, 3)

Octahedral and square planar complexes can show geometrical isomerism.

6.

H N3

H N3

Cl

Cl

and PtPt

H N3 Cl

NH3

Cl

represent

(1) Position isomerism (2) cis-trans isomerism

(3) Geometrical isomerism (4) Optical isomerism

Sol. Answer (2, 3)

[Pt(NH3)

2Cl

2] complex shows geometrical or cis-trans isomerism.

7. Optical isomers are

(1) Mirror images (2) Non-superimposable

(3) cis-trans (4) Chiral molecules

Sol. Answer (1, 2, 4)

Optical isomers are mirror images, non-superimposable and chiral molecules.

8. An example of coordination isomerism is

(1) [Cr(H2O)

6]Cl

3 and [Cr(H

2O)

5Cl]Cl

2.H

2O (2) [Co(NH

3)6] [Cr(CN)

6] and [Cr(NH

3)6] [Co(CN)

6]

(3) [Pt(NH3)4Cl

2] and [Pt(NH

3)2Cl

4] (4) [Co(NH

3)5Br]SO

4 and [Co(NH

3)5SO

4]Br

Sol. Answer (2)

Co-ordination isomerism is shown by [Co(NH3]6 [Cr(CN)

6] and [Cr(NH

3)]

6 [Co(CN)

6]

9. Which of the following compounds is paramagnetic?

(1) [Ni(CO)4] (2) K

2[Ni(CN)

4] (3) K

3[CoF

6] (4) [Cr(NH

3)6]Cl

3

Sol. Answer (3, 4)

K3[CoF

6] and [Cr(NH

3)6]Cl

3 are paramagnetic complex because Co+3 have 4 unpaired e– while Cr+3 have three

unpaired e–.

10. Identify the coordination compounds

(1) KCl.MgCl2.6H

2O (2) CuSO

4.4NH

3

(3) Fe(CN)2.4KCN (4) FeSO

4.(NH

4)2SO

4.6H

2O

Sol. Answer (2, 3)

3 4 4 4 6[Cu(NH ) ]SO , K [Fe(CN) ] .


11. [Cr(CN)6] [Co(NH

3)6] contains

(1) Cationic complex (2) Anionic complex (3) Neutral ligands (4) Anionic ligands

Sol. Answer (1, 2, 3, 4)

6 3 6[Cr(CN) ][Co(NH ) ] is a complex in which both cationic and anionic part are complex, CN– is a negative

ligand while NH3 is a neutral ligand.

12. Which of the following involves sp3 hybridisation?

(1) Ni(CO)4

(2) [ZnCl4]2– (3) [Ni(CN)

4]2– (4) [Cu(NH

3)4]2+


2

4 4[Ni(CO) ] and [ZnCl ] Ni and Zn involves sp3 hybridised. In [Cu(NH3)

4]+2, N involves sp3 hybridised.

13. Other theories explaining the bonding in coordination compounds are

(1) Ligand field theory (2) Molecular orbital theory

(3) VSEPR theory (4) Atomic orbital theory

Sol. Answer (1, 2)

Bonding in coordination compound is explained by

(1) Ligand field theory.

(2) M.O. Theory.

14. On mixing aqueous solution of copper sulphate with aqueous solution of ammonia a deep blue coloured solution

is obtained. Choose the correct options :

(1) The blue coloured solution shows presence of Cu2+ ions

(2) The blue coloured solution does not show presence of Cu2+ ions

(3) The blue coloured solution shows the presence of [Cu(NH3)4]2+ ions

(4) The blue coloured solution does not show the presence of [Cu(NH3)4]2+ ions

Sol. Answer (2, 3)

Blue colour is due to 2

3 4[Cu(NH ) ] complex which is square planar and dsp2 hybridised.

15. The formation of complexes which involves d-orbitals of outer shell are called

(1) Low spin complexes (2) High spin complexes

(3) Inner orbital complexes (4) Outer orbital complexes

Sol. Answer (2, 4)

Involvement of d-orbital of outer shell is called high spin complex and outer orbital complex.

16. Which of the following ligands are acceptor?

(1) NO (2) NO+ (3) PF3

(4) H2O


NO, NO+, PF3 from synergic bonding.

17. [Co(en)2NO

2Br]Cl can exhibit


(3) Linkage isomerism (4) Ionisation isomerism

Sol. Answer (1, 2, 3, 4)

Fact.


18. Which of the following complexes may be coloured due to d-d transition?

(1) Ni(Co)4

(2) K3[Fe(CN)

6] (3) Cu(H

2O)

4

+2 (4) K3[Cu(CN)

4]


Fact.

19. Stability of the complex may depend on

(1) Oxidation state (2) No. of d orbitals in metal ion

(3) Nature of ligands (4) Geometry of complex

Sol. Answer (1, 2, 3, 4)

Square planar.

20. Which of following ions favour square planar geometry?

(1) Au+3 (2) Ir+ (3) Pt+2 (4) Hg+2


Hg+2 favour tetrahedral geometry.

21. The compound(s) that exhibit(s) geometrical isomerism is(are) [IIT-JEE-2009]

(1) [Pt(en)Cl2] (2) [Pt(en)

2]Cl

2(3) [Pt(en)

2Cl

2]Cl

2(4) [Pt(NH

3)2Cl

2]

Sol. Answer (3, 4)

(C) Pt

Cl

en en

Cl

&

Cl

Cl

en

en

(D) Pt

NH3

ClNH3

Cl

&

NH3

Cl NH3

Cl

Pt

22. The pair(s) of coordination complexes/ions exhibiting the same kind of isomerism is(are)


(1) [Cr(NH3)5Cl]Cl

2 and [Cr(NH

3)4Cl

2]Cl (2) [Co(NH

3)4Cl

2]+ and [Pt(NH

3)2(H

2O)Cl]+

(3) [CoBr2Cl

2]2– and [PtBr

2Cl

2]2– (4) [Pt(NH

3)3](NO

3)Cl and [Pt(NH

3)3Cl]Br

Sol. Answer (2, 4)

The pair of complex ions [Co(NH3)

4Cl

2]+ and [Pt(NH

3)

2(H

2O)Cl]+ show geometrical isomerism. The pair of

complexes [Pt(NH3)

3(NO

3)]Cl and [Pt(NH

3)3Cl]Br show ionisation isomerism. The other pairs given do not have

same type of isomerism.


SECTION - C

Linked Comprehension Type Questions

Comprehension-I

Consider the following isomers of [Co(NH3)4Br

2]+. The black sphere represents Co, grey spheres represent NH

3 and

unshaded spheres represent Br.

(a) (b) (c) (d)

1. Which of the following are cis-isomers?

(1) Isomers (a) and (b) (2) Isomers (a) and (c)

(3) Isomers (b) and (d) (4) Isomers (c) and (d)

Sol. Answer (2)

cis isomers are (a) and (c)

Co

H

3 N

H

3 N

Br

NH

3

Br

Co

H

3 N

Br

NH

3NH

3

NH3

H

3 N

Br

2. Which of the following are trans-isomers?

(1) Isomers (a) and (b) (2) Isomers (a) and (c)

(3) Isomers (b) and (d) (4) Isomers (c) and (d)

Sol. Answer (3)

Trans isomers are (b) and (d)

Co

H

3 N

H

3 N NH

3

Br

Br

NH

3

Co

Br

H

3 N Br

NH3

NH

3

NH

3

3. Which structures are identical?

(1) None of the structures are identical

(2) Structure (a) = structure (b) and structure (c) = structure (d)

(3) Structure (a) = structure (c) and structure (b) = structure (d)

(4) Structure (a) = structure (d) and structure (b) = structure (c)

Sol. Answer (3)

Structure (a) is identical to (c) while structure (b) is identical to (d).


Comprehension-II

A coordination compound is a complex in which the numbers of ions or molecules attached to the central metal

atom or ion is beyond the number possible, on the basis of electrovalent or covalent bonding. The extra groups or

ions are linked to the metal by coordination bonds in which the linked group (L) is the donor and metal (M) is the

acceptor i.e. M L. Coordination compounds are mainly of two types:

(i) Neutral compounds like [Fe(CO)5]

(ii) Ionic compounds, which consists of ions

in which atleast one is complex ion. e.g. K4[Fe(CN)

6] complex ion : A complex ion is an electrically charged radical

which consists of a metal atom or ion surrounded by a group of ions or neutral molecules. For example, [Cu(NH3)4]2+,

[Fe(CN)6]3–. Ligands are defined as molecules or ions having atleast one pair of electrons for donation. Ligands are

also known as Lewis bases.

1. How many ions are produced by [Co(NH3)6]Cl

3 in solution?

(1) 6 (2) 4 (3) 3 (4) 2

Sol. Answer (2)

33 6 3 3 6[Co(NH ) ]Cl [Co(NH ) ] 3Cl .

2. Which of the following is non-ionizable?

(1) [Co(NH3)3Cl

3] (2) [Co(NH

3)4Cl

2]Cl (3) [Co(NH

3)5Cl]Cl

2(4) [Co(NH

3)6]Cl

2

Sol. Answer (1)

3 3 3[Co(NH ) Cl ] is non-ionizable.

3. AgCl dissolves in NH4OH due to the formation of

(1) [Ag(NH4)2Cl] (2) [Ag(NH

4)3]Cl (3) [Ag(NH

3)2]Cl (4) [Ag(NH

3)2OH]

Sol. Answer (3)

AgCl + NH4OH [Ag(NH

3)2]Cl

4. The complex CoCl3.3NH

3 ionizes to give

(1) 2 Cl– ions (2) 1 Cl– ion (3) 3 Cl– ions (4) 0 Cl– ion

Sol. Answer (4)

[Co(NH3)Cl

3] cannot give Cl– because all Cl– are present in coordination sphere.

Comprehension-III

X[Co(en) ClBr]NO

2 2

Isomer

3AgNO

A Yellow ppt

3AgNO

B White ppt

Complexes ‘A’ and ‘B’ are hexacoordinated complexes.

1. Compound A may show

(1) Geometrical isomerism (2) Linkage isomerism

(3) Optical isomerism (4) All of these

Sol. Answer (4)


2. How many total optically active forms of ‘A’ and ‘B’ are possible?

(1) 2 (2) 10 (3) 8 (4) 4

Sol. Answer (3)

SECTION - D

Assertion-Reason Type Questions

1. STATEMENT-1 : Zeise's salt contain C2H

4 molecule as one of the ligands.

and

STATEMENT-2 : Zeise's salt is an organometallic compound.

Sol. Answer (2)

The formula of Zeise salt is [Pt(C2H

4)Cl

3]–

2. STATEMENT-1 : Oxidation state of Fe in Fe(CO)5 is zero.

and

STATEMENT-2 : EAN of Fe in its complexes is always 36.

Sol. Answer (2)

CO does not carry any charge.

EAN = 26 + 2 × 5 = 36.

3. STATEMENT-1 : [Co(NH3)3Cl

3] does not give white ppt with AgNO

3 solution.

and

STATEMENT-2 : Chlorine is not present in the ionisable part of the given complex.

Sol. Answer (1)

[Co(NH3)

3Cl

2] does not ionize to give Cl–

4. STATEMENT-1 : Tetrahedral complexes with chiral structure exhibit optical isomerism.

and

STATEMENT-2 : They lack plane of symmetry.

Sol. Answer (1)

Tetrahedral complex with chiral structure are optically active because these are dissymmetric.

5. STATEMENT-1 : The IUPAC name of K4[Fe(C

2O

4)3] is potassiumtrioxalatoferrate (III).

and

STATEMENT-2 : Oxalate ion is a bidentate ligand.

Sol. Answer (4)

In K4[Fe(C

2O

4)

3], Fe is in + II oxidation state is a bidentate ligand.


6. STATEMENT-1 : Coordination isomerism occurs when both the cations and anions are complexes.

and

STATEMENT-2 : Oxidation state of central metal ion in both coordination spheres is always equal.

Sol. Answer (3)

Oxidation state of central metal ion in both coordination sphere may be equal or different.

7. STATEMENT-1 : The [Ni(en)3]Cl

2 has higher stability than [Ni(NH

3)6]Cl

2.

and

STATEMENT-2 : Ethylene diamine shows chelation with Ni+2 ion.

Sol. Answer (1)

Due to chelation stability increases.

8. STATEMENT-1 : Co+2 form octahedral stable complex with excess KCN.

and

STATEMENT-2 : CN– is stronger ligand than NH3.

Sol. Answer (4)

Co+2 does not form octahedral complex with KCN.

9. STATEMENT-1 : CN– may act as bridging ligand.

and

STATEMENT-2 : Lone pair resides on carbon as well as on nitrogen.

Sol. Answer (2)

In CN– both C & N have lone pair.

10. STATEMENT-1 : Co(NH3)6

+3 is more stable than (Co(en)2(NH

3)2)+3

and

STATEMENT-2 : Chelation is more in [Co(en)2(NH

3)2]+3 than [Co(NH

3)6]+3.

Sol. Answer (4)

Due to chelation, [Co(en)2(NH

3)

2] is more stable.

11. STATEMENT-1 : In square planar complexes, 2 2–x y

d is higher in energy than dxy.

and

STATEMENT-2 : Ligands approach along x and y axis.

Sol. Answer (1)

Fact.


12. STATEMENT-1 : CH3 MgCl is organometallic complex.

and

STATEMENT-2 : In ether RMgCl, co-ordination number of Mg is 6.

Sol. Answer (2)

CH3–Mg–Cl is bonding complex.

13. STATEMENT-1 : Fe(CO)5 has trigonal bipyramidal shape.

and

STATEMENT-2 : Fe(CO)5 is diamagnetic.

Sol. Answer (2)

Fe(CO)5 is paramagnetic.

14. STATEMENT-1 : Fe – CN bond length is smaller in Fe(CN)6

–4 than Fe(CN)6

–3.

and

STATEMENT-2 : Fe(CN)6

–3 is more stable than Fe(CN)6

–4.

Sol. Answer (2)

Due to synergic bond.

15. STATEMENT-1 : Ni(dmg)2 is square planar complex.

and

STATEMENT-2 : Chelation effect is present in Ni(dmg)2.

Sol. Answer (2)

Ni(dmg)2 is square planar complex.

16. STATEMENT-1 : [Fe(H2O)

5NO]SO

4 is paramagnetic.

and

STATEMENT-2 : The Fe in [Fe(H2O)

5NO]SO

4 has three unpaired electrons. [IIT-JEE-2008]

Sol. Answer (1)

[Fe(H2O)

5NO]SO

4 is paramagnetic in nature. Magnetic moment of complex is 3.9 BM. Hence it has 3 unpaired

electrons. In this complex, oxidation number of Fe is +1.

17. STATEMENT-1 : The geometrical isomers of the complex [M(NH3)4 Cl

2] are optically inactive.

and

STATEMENT-2 : Both geometrical isomers of the complex [M(NH3)4Cl

2] possess axis of symmetry.

[IIT-JEE-2008]

Sol. Answer (1)

[M(NH3)

4Cl

2] show cis-trans geometrical isomers and are not optically active due to presence of axis of

symmetry.


SECTION - E

Matrix-Match Type Questions

1. Match the column

Column I Column II

(A) [Ni(CN)4]2– (p) d 2sp3

(B) [Ni(CO)4] (q) sp3d 2

(C) [Co(en)3]3+ (r) dsp2

(D) [Co(Br)3Cl

3]3– (s) sp3

Sol. Answer A(r), B(s), C(p), D(q)

2 2

4[Ni(CN) ] dsp

34[Ni(CO) ]sp

3 2 33[Co(en) ] d sp

3 3 23 2[CoBr Cl ] sp d

2. Match the column

Column I Column II

(A) Glycinate ion (p) Bidentate

(B) EDTA (q) Hexadentate

(C) Oxalate (r) Monodentate

(D) Sulphate (s) Lewis base

Sol. Answer A(p, s), B(q, s), C(p, s), D(r, s)

Glycinate ion Bidentate, Lewis base

EDTA hexadentate, Lewis base

Oxalate Bidentate, Lewis base

Sulphate Monodentate, Lewis base

3. Match the complex with its oxidation state of central metal

Column I Column II

(A) [Fe(CN)6]4– (p) Oxidation number of central metal is +2

(B) [Co(NH3)6]3+ (q) Oxidation number of central metal is +3

(C) [NiCl4]2– (r) Diamagnetic

(D) [CoF6]3– (s) Paramagnetic


Sol. Answer A(p, r), B(q, r), C(p, s), D(q, s)

4 II 2 36[Fe(CN) ] Fe , d sp , diamagnetic

3 3 2 33 6[Co(NH ) ] Co , d sp , diamagnetic

2 2

4[NiCl ] Ni paramagnetic

3 36[CoF ] Co , paramagnetic

4. Match the given compound in column I to the properties given as per VBT in column II

Column I Column II

(A) [Zn(H2O)

4]2+ (p) sp3 hybridisation

(B) [Pt(NH3)

4] [PtCl

4] (q) dsp2 hybridisation

(C) Fe2(CO)

9(r) Paramagnetic

(D) [Cu(NH3)4]2+ (s) Diamagnetic

(t) Metal - metal bond

Sol. Answer A(p, s), B(q, s), C(s, t), D(q, r)

Fe2(CO)

9 contain M–M bond.

In [Cu(NH3)

4]2+, Cu has dsp2 hybridisation.

5. Match the compound given in column I to the properties given in column II

Column I Column II

(A) [Co(en)2ClBr]+ (p) Geometrical isomerism

(B) [Pt(NH3)2Cl

2] (q) Optical isomerism

(C) [Cu(NH3)6]+2 (r) All bond length are not equal

(D) [Ni(dmg)2] (s) Chelation effect is present

(t) Hydrogen bonding is present in complex

Sol. Answer A(p, q, s), B(p), C(r), D(s, t)

Fact.

6. Match the complexes in Column I with their properties listed in Column II. Indicate your answer by darkening

the appropriate bubbles of the 4 × 4 matrix given in the ORS. [IIT-JEE-2007]

Column I Column II

(A) [Co(NH3)4(H

2O)

2]Cl

2(p) geometrical isomers

(B) [Pt(NH3)2Cl

2] (q) paramagnetic

(C) [Co(H2O)

5Cl]Cl (r) diamagnetic

(D) [Ni(H2O)

6]Cl

2(s) metal ion with +2 oxidation state

Sol. Answer A(p, q, s), B(p, r, s), C(q, s), D(q, s)


7. Match each coordination compound in Column I with an appropriate pair of characteristics from Column II and

select the correct answer using the code given below the column. [JEE(Advanced)-2014]

{en = H2NCH

2CH

2NH

2; atomic numbers : Ti = 22, Cr = 24; Co = 27; Pt = 78}

Column I Column II

(A) [Cr(NH3)4Cl

2]Cl (p) Paramagnetic and exhibits ionisation isomerism

(B) [Ti(H2O)

5Cl](NO

3)2

(q) Diamagnetic and exhibits cis-trans isomerism

(C) [Pt(en)(NH3)Cl]NO

3(r) Paramagnetic and exhibits cis-trans isomerism

(D) [Co(NH3)4(NO

3)2]NO

3(s) Diamagnetic and exhibits ionisation isomerism

Code :

A B C D

(1) s q r p

(2) r p s q

(3) q p r s

(4) p r s q

Sol. Answer (2)

Magnetic character Isomerism

A. [Cr(NH3)4Cl

2]Cl Paramagnetic cis/trans

B. [Ti(H2O)

5Cl](NO

3)2

Paramagnetic ionization

C. [Pt(en)(NH3)Cl]NO

3Diamagnetic ionization

D. [Co(NH3)4(NO

3)2]NO

3Diamagnetic cis/trans

SECTION - F

Integer Answer Type Questions

1. How many sp3 hybridised atoms are there in one molecule of CuSO4.5H

2O?

Sol. Answer (6)

H2O and SO

4 have sp3 hybridisation.

2. How many type of geometrical isomers are possible for 3 2M NH BrCl H O⎡ ⎤⎣ ⎦ , where M is Pt+2?

Sol. Answer (3)

Fact.

3. The coordination number of Fe in 5

5 5 2Fe C H⎡ ⎤ ⎣ ⎦ is 2x. Then x will be________.

Sol. Answer (5)

Fact.


4. How many geometrical isomers are possible for [Zn(NH3)ClBrPPh

3]– ?

Sol. Answer (0)

It is tetrahedral complex.

5. How many geometrical isomers are possible for MA2B

2C

2?

Sol. Answer (5)

Fact.

6. How many rings are present in [MEDTA]+n?

Sol. Answer (5)

Due to Chelation.

7. How many maximum atoms are present in one plane of [PtCl3(n2C

2H

4)]–?

Sol. Answer (4)

C2H

4 is perpendicular to plane of and Cl– Pt+2

8. How many metal-metal bonds are present in Co4(CO)

12?

Sol. Answer (6)

Fact.

9. Total number of geometrical isomers for the complex [RhCl(CO)(PPh3)(NH

3)] is [IIT-JEE-2010]

Sol. Answer (3)

Rh Cl(CO)(PPh3)(NH

3)] is a square planar complex with four different ligands and hence it will have three

geometrical isomers

a b

cd

Rh

a b

dc

Rh

a c

bd

Rh

10. The volume (in mL) of 0.1 M AgNO3 required for complete precipitation of chloride ions present in 30 mL of

0.01 M solution of [Cr(H2O)

5Cl]Cl

2, as silver chloride is close to [IIT-JEE-2011]

Sol. Answer (6)

Applying equation N1V

1 = N

2V

2

0.1 × V = 2 × 0.01 × 30

2 30 106

100 1

V


11. EDTA4– is ethylenediaminetetraacetate ion. The total number of N—Co—O bond angles in [Co(EDTA)]1– complex

ion is [JEE(Advanced) 2013]

Sol. Answer (8)

So, bond angles are

(1) NI CoOI

Co

O – C = O

O – C = O

O

C CH2 CH

2

O

CH2

CH2

N

O N

C CH2

O

CH2

II

III

IV

I

I

II

(2) NI CoOII

(3) NI CoOIII

(4) NI CoOIV

(5) NII CoOI

(6) NII CoOII

(7) NII CoOIII

(8) NII CoOIV

So, total asked bond angles are 8.

12. For the octahedral complexes of Fe3+ in SCN– (thiocyanato-S) and in CN– ligand environments, the difference

between the spin-only magnetic moments in Bohr magnetons (when approximated to the nearest integer) is

[Atomic number of Fe = 26] [JEE(Advanced) 2015]

Sol. Answer (4)

Fe(26) [Ar]18

3d64s2

Fe3+ [Ar]18

3d54s0

SCN– is weak field ligand hence pairing will not occur.

Fe3+

Unpaired electrons = 5

Magnetic moment 5(5 2) B.M.

35 B.M. 5.92 B.M.

CN– is strong field ligand hence pairing will take place.

Fe3+

Unpaired electrons = 1

Magnetic moment 1(1 2)B.M. 3 B.M. 1.732

Difference = 5.92 – 1.732 = 4.188

Hence answer is (4).


13. In the complex acetylbromidodicarbonylbis (triethylphosphine) iron (II), the number of Fe–C bond(s) is


Sol. Answer (3)

CH – C3

–

Fe+2

Br–

O

Et P3 CO

COEt P3

14. Among the complex ions, [Co(NH2–CH

2–CH

2–NH

2)2Cl

2]+, [CrCl

2(C

2O

4)2]3–, [Fe(H

2O)

4(OH)

2]+, [Fe(NH

3)2(CN)

4]–,

[Co(NH2–CH

2–CH

2–NH

2)2(NH

3)Cl]2+ and [Co(NH

3)4(H

2O)Cl]2+, the number of complex ion(s) that show(s)

cis-trans isomerism is [JEE(Advanced) 2015]

Sol. Answer (6)

Co

N

N

N

N

Cl

Cl

Co

N

N

Cl

Cl

N

N

Cr

O

O

O

O

Cl

Cl

Cr

O

O

Cl

Cl

O

O

Fe

NC

NC

CN

CN

NH3

NH3

Fe

NC

NC

NH3

CN

NH3

CN

Co

NH2

NH2

NH2

NH2

NH3

Cl

Co

NH2

NH2

Cl

NH2

NH3

NH2

Co

H N3

H N3

NH3

NH3

OH2

Cl

Co

H N3

H N3

Cl

NH3

OH2

NH3

Fe

H O2

H O2

OH

OH

H2O

H O2

Fe

H O2

H O2

OH2

OH2

OH

OH

(All the six complexes will show cis-trans isomerism)


SECTION - G

Multiple True-False Type Questions

1. STATEMENT-1 : Tetrahedral complexes are always high spin complexes.

STATEMENT-2 : Crystal field splitting energy in tetrahedral complexes is 2/3 of the 0 (crystal field splitting

energy in octahedral complexes).

STATEMENT-3 : Tetrahedral complex n

MABCD

is optically active.

(1) TTT (2) TFT (3) F T F (4) T T F

Sol. Answer (2)

t 0

4

9

2. STATEMENT-1 : CO is stronger ligand than CN–.

STATEMENT-2 : CO and CN– both show synergic bonding with metal.

STATEMENT-3 : CO and N2 are isoelectronic ligands but N

2 is a weaker ligand than CO but stronger than NH

3.

(1) T F T (2) T T F (3) F F T (4) F T T

Sol. Answer (2)

N2 is weaker ligand than NH

3.

3. STATEMENT-1 : In LiAlH4, Al is sp3 hybridised.

STATEMENT-2 : LiAlH4 is a good reducing agent.

STATEMENT-3 : LiAlH4 complex is unstable in water.

(1) T T T (2) T T F (3) T F F (4) F T T

Sol. Answer (1)

LiAlH4 is very reactive with water.

4. STATEMENT-1 : In[Cr(NH3)6]Cl

3, the Werner primary valency is 3.

STATEMENT-2 : [Cr(NH3)6]Cl

3 is paramagnetic.

STATEMENT-3 : [Co(NH3)

6]+3 is a low spin complex.

(1) T F T (2) T T T (3) F T T (4) T F F

Sol. Answer (2)

Co(NH3)+3 is low spin complex.

5. STATEMENT-1 : [M(en)Cl2Br

2] is optically active complex.

STATEMENT-2 : en is a bidentate ligand.

STATEMENT-3 : In K[Cu(CN)2] co-ordination number of Cu is 3.

(1) T F T (2) T T T (3) T F F (4) F T T

Sol. Answer (4)

Fact.


SECTION - H

Aakash Challengers Questions

1. For Ag+ metal ion, correct sequence regarding ligand strength is

(1) F– > Cl– > Br– > I– (2) I– > Br– > Cl– > F–

(3) Cl– > Br– > F– > I– (4) F– > Br– > I– > Cl–

Sol. Answer (2)

Fact.

2. Choose the correct statement regarding complex CuF6

–4.

(1) All Cu – F bond lengths are equal (2) It is paramagnetic

(3) It is diamagnetic (4) Both (1) & (2)

Sol. Answer (2)

Two axial bond lengths are longer.

3. Which of the following ligands may be flexidentate?

(1) EDTA (2) CO3

–2 (3) NO3

– (4) All of these

Sol. Answer (4)

All these are flexidentate ligand.

4. In which complex C – O bond length is maximum?

(1) V(CO)6

– (2) Cr(CO)6

(3) Mn(CO)6

+ (4) Fe(CO)6

+2

Sol. Answer (1)

Due to back bonding.

5. Choose the incorrect regarding stability

(a) [Fe(phen)3]+2 > [Fe(phen)

3]+3 (b) [Fe(CN)

6]–4 > [Fe(CN)

6]–3

(c) Fe(dipy)3

+3 > Fe(dipy)3

+2

(1) a, b (2) b, c (3) a, c (4) a, b, c

Sol. Answer (1)

Due to synergic bonding.

6. Choose the correct statement

(1) Chelation always increase stability (2) CFSE have strong effect on stability

(3) Fe(dipy)3

+3 is unstable (4) All of these

Sol. Answer (2)

Fact.


7. Strongest C–O bond is present in

(1) Fe(CO)5

(2) Mn(CO)6

+ (3) Cr(CO)6

(4) V(CO)6

–

Sol. Answer (2)

Due to less back bonding.

8. (MA2B

2C

2)+n can exhibit


(3) Ionisation isomerism (4) Both (1) & (2)

Sol. Answer (4)

Fact.

9. The hybridisation of Co in (CoF6)–2 is

(1) sp3d2 (2) d2sp3 (3) sp3d3 (4) d3sp3

Sol. Answer (2)

It is an example of inner orbital complex.

10. Which of the following compound may be optical active?

(1) Co(en)2Cl

2(2) [Rh Cl Br NH

3 PH

3]–

(3) Cu(en)2Cl

2(4) All of these

Sol. Answer (1)

Cu(en)2Cl

2 does not form isomer.

11. Write the formula for each of the following complexes.

(i) Ammonium hepta fluorozirconate (IV)

(ii) Diamminesilver (I) hexacyanoferrate (II)

(iii) Dichlorobis (ethylenediamine) chromium (III) tetrachloropalladate (II)

(iv) Dicyanobis (ethylenediamine) cobalt (III) chlorate

(v) Potassium hexafluoronickelate (IV)

(vi) Bromotriammineplatinum (II) nitrite

Sol. (i) (NH4)3 [ZrF

7]

(ii) [Ag(NH3)2]4 [Fe(CN)

6]

(iii) [Cr(en)4Cl

2]2 [PdCl

4]

(iv) [Co(en)2(CN)

2]ClO

3

(v) K2[NiF

6]

(vi) [Pt(NH3)3Br]NO

2


12. Name the following complexes according to the IUPAC system of nomenclature

(i) [Co(NH3)4 (H

2O)Br] (NO

3)2

(ii) Na[Au(CN)2]

(iii) Na3[Fe(C

2O

4)3] (iv) [Co(en)

3]Cl

3

Sol. (i) Bromoaquatetraamminecobalt (III) nitrate

(ii) Sodiumdicyanoaurate (I)

(iii) Sodiumtrioxalatoferrate (III)

(iv) Tris (ethylenediamine) cobalt (III) chloride

13. Arrange the following compounds in order of increasing molar conductivity

(a) K[Co(NH3)2(NO

2)4] (b) [Cr(NH

3)3(NO

2)3]

(c) [Cr(NH3)5(NO

2)]

3 [Co(NO

2)6]2

(d) Mg[Cr(NH3) (NO

2)5]

Sol. The larger the number of ions and larger the charge on each, the larger the conductivity. The compounds from

lowest conductivity to highest conductivity are b < a < d < c.

14. Account for the following

(i) Co(II) is stable in aqueous solution but in the presence of strong ligand and air, it can get oxidized to Co(III).

(ii) [Ni(CN)4]2– is square planar and diamagnetic whereas [NiCl

4]2– is tetrahedral and paramagnetic.

Sol. (i) Co(II) has the configuration 3d+ i.e. it has three unpaired electrons. Water being a weak ligand. The unpaired

electrons do not pair up. In the presence of strong ligands, the two unpaired electrons in 3d pair up and

the third unpaired electron shifts to higher energy sub shell from where it can be easily lost and hence

shows an oxidation state of III.

(ii) CN– a strong field ligand forces the electron to pair up in d-orbitals so it shows dsp2 hybridisation due to

which shape is square planar. Further, no electron is left unpaired, so, it is diamagnetic, while, Cl– a weak

field ligand cannot force the electron to pair up showing sp3 hyrbridisation so shape is tetrahedral and due

to unpaired electron it is paramagnetic.

15. FeSO4 solution mixed with (NH

4)2SO

4 solution, (in the molar ratio 1 : 1) gives the test of Fe2+ ion but CuSO

4

solution mixed with liquid NH3 (the molar ratio 1 : 4) does not give the test of Cu2+. Explain why?

Sol. FeSO4 does not form any complex with (NH

4)

2SO

4, instead, they form a double salt FeSO

4(NH

4)

2SO

4.6H

2O

which dissociates completely into ions but CuSO4 combines with NH

3 to form the complex [Cu(NH

3)

4]SO

4 in

which complex ion, [Cu(NH3)

4]2+ does not dissociate to give Cu2+ ions.

16. (a) Square planar complexes with coordination number four exhibit geometrical isomerism where as tetrahedral

complexes do not why?

(b) Three geometrical isomers of the square planar complex [Pt(NH3)(H

2O) (py) (NO

2)]+ are possible. What

are they?


Sol. (a) Tetrahedral complexes do not show geometrical isomerism because the relative positions of the ligands

attached to the central metal atom are same with respect to each other.

(b)

17. A, B and C are three complexes of chromium (III) with empirical formula H12

O6Cl

3Cr. All the three complexes

have water and chloride ions are ligands. Complex A does not react with conc H2SO

4, where as complexes

B and C lose 6.75% and 13.5% of their original weight respectively, on treatment with conc. H2SO

4. Identify

A, B and C.

Sol. A = [Cr(H2O)

6]Cl

3

2 4H SO No reaction

(... All H2O molecules are present in coordination sphere)

B = [CrCl(H2O)

5]Cl

2.H

2O 2 4H SO one mole of H

2O is lost

Molecular weight of complex = 266.5

% loss = 18

100 6.75%266.5

C = [CrCl2(H

2O)

4]Cl

2.2H

2O 2 4H SO 2 moles of H

2O removed

% loss in weight =2 18

100 13.50%266.5

18. Classify each of the following complexes as either high or low spin. Explain your answers.

(i) [Co(H2O)

6]2+ = 3.87 B.M. [Co(CN)

6]3– = 0.0 B.M.

(ii) [Fe(NO2)6]4– = 0.0 B.M. [Fe(H

2O)

6]3+ = 5.94 B.M.

Sol. (i) Each contains Co2+ ions in an octahedral field. Obviously compound [Co(H2O)

6]2+ is a high spin having

magnetic moment value of 3.87 B.M. whereas [Co(CN)6]3– is low spin with a magnetic moment value of

0.0 B.M.

(ii) The complex [Fe(NO2)

6]4– contains Fe2+ which belongs to a d6 system. As the magnetic moment of the

complex is zero, it indicates that all 6 electrons are paired up hence the compound is low spin on the

other hand the magnetic moment of the 5.94 B.M for the complex [Fe(H2O)

6]3+ corresponds to five unpaired

electron in Fe3+. It indicates, there is no pairing of d-electrons of Fe3+ ion. Hence complex is high spin.

19. Write the IUPAC nomenclature of the given complex along with its hybridisation and structure

K2[Cr(NO)(NH

3)(CN)

4], = 1.73 B.M.

Sol. Potassiumamminetetracyanonitrosoniumchromium (I) Cr is in +1 state and d2sp

3 hybridisation

n(n 2) 3 1.73 B.M.


20. NiCl2 in the presence of dimethyl glyoxime (DMG) gives a complex which precipitates in the presence of

NH4OH, giving a bright red colour.

(i) Draw its structure and show H-bonding.

(ii) Give oxidation state of Ni and its hyrbidisation.

(iii) Predict whether it is paramagnetic or diamagnetic.

Sol.

oxidation state of nickel is +2 and hybridisation is dsp2.

sn(n 2) B.M.

n = 0

s = 0

� � �

Documents

Coordination Compounds