NCERT Solutions for Class 12 Chemistry Chapter 5 Coordination Compounds

NCERT
SOLUTIONS
CLASS - 12th

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Class : 12th
Subject : Chemistry
Chapter : 9
Chapter Name : Coordination Compounds

Q9.1 Write the formulas for the following coordination compounds:
(i) Tetraamminediaquacobalt(lll) chloride
(ii) Potassium tetracyanonickelate(ll)
(iii) Tris(ethane—1,2—diamine) chromium(lll) chloride
(iv) Amminebromidochloridonitrito-N-platinate(II)
(v) Dichloridobis (ethane-1,2-diamine)platinum (IV) nitrate
(vi) Iron(lll) hexacyanoferrate(ll)

Answer. (i)[CO(H O) (NH ) ] Cl 2 2 3 4 3

(ii) K [Ni(CN) ]
2 4

(iii)[Cr(en) ] Cl3 3

(iv)[Pt(NH) BrCl (NO )]
−
3 2

(v)[PtCl (en) ] (NO )
2 2 3 2

(vi)Fe [Fe(CN) ]
4 6 3

Page : 251 , Block Name : Intext Questions

Q9.2 Write the IIJPAC names of the following coordination compounds:
(i) [Co(NH ) ] Cl3 6 3

(ii) [Co(NH ) Cl] Cl
3 5 2

(iii) K [Fe(CN) ]
3 6

(iv) K [Fe(C O ) ]
3 2 4 3

(v) K [PdCl ]
2 4

(vi) [Pt(NH ) Cl (NH CH )] Cl
3 2 2 3

Answer. (i) Hexaamminecobalt(lll) chloride
(ii) Pentaamminechloridocobalt(lll) chloride
(iii) Potassium hexacyanoferrate(lll)
(iv) Potassium trioxalatoferrate(lll)
(v) Potassium tetrachloridopalladate(ll)
(vi) Diamminechloride(methylamine) platinum (II) chloride

Page : 251 , Block Name : Intext Questions

Q9.3 Indicate the types of isomerism exhibited by the following complexes and draw the structures for
these isomers:
(i) K [Cr(H O) (C O )
2 2 2 4 2

(ii) [Co(en) ] Cl
3 3

(iii) [Co(NH ) (NO )] (NO )
3 5 2 3 2

(iv) [Pt (NH ) (H O) Cl ]
3 2 2

Answer. (i) Both geometrical (cis-, trans-) isomers for K [Cr(H O) (C O ) ] can exist. Also, optical
2 2 2 4 2

isomers for cis-isomer exist.

Trans-isomer is optically inactive. On the other hand, cis-isomer is optically active.

(ii) Two optical isomers for [CO(en) ] Cl exist. 3 3

Two optical isomers are possible for this structure.

(iii) [CO(NH ) (NO )] (NO )
3 5 2 3 2

A pair of optical isomers:

It can also show linkage isomerism.
[CO(NH3 ) (NO2 )] (NO3 ) and [CO(NH3 ) (ONO)] (NO3 )
5 2 5 2

It can also show ionization isomerism.
[Co(NH3 ) (NO2 )] (NO3 ) [CO(NH3 ) (NO3 )] (NO3 ) (NO2 )
s 2 5

(iv) Geometrical (cis-, trans-) isomers of [Pt (NH ) (H O) Cl ] can exist. 3 2 2

Page : 254 , Block Name : Intext Questions

Q9.4 Give evidence that [Co(NH ) cl] SO and [Co(NH ) SO ] Cl are ionization isomers.
3 5 4 3 5 4

Answer. When ionization isomers are dissolved in water, they ionize to give different ions. These ions
then react differently with different reagents to give different products.
[CO(NH ) Cl] sO + Ag ⟶ no reaction
2+ ∗
[CO(NH ) Cl] sO + Ba
3 4 ⟶ ↓ 3 4
5 5
White precipitate

[CO(NH3 ) SO4 ] Cl + Ba
5
2+
⟶ no reaction
⟶ AgCl ↓
+
[CO(NH3 ) SO4 ] Cl + Ag
s
White precipitate

Page : 254 , Block Name : Intext Questions

Q9.5 Explain on the basis of valence bond theory that [Ni(CN) ] ion with square planar structure is
2−
4

diamagnetic and the [NiCl ] ion with tetrahedral geometry is paramagnetic.
2−
4

Answer. Ni is in the +2 oxidation state i.e., in d con guration.
8

There are 4 CN¯ ions. Thus, it can either have a tetrahedral geometry or square planar geometry. Since
CN¯ ion is a strong eld ligand, it causes the pairing of unpaired 3d electrons.

It now undergoes dsp hybridization. Since all electrons are paired, it is diamagnetic. In case of
2

ion is a weak eld ligand. Therefore, it does not lead to the pairing of unpaired 3d
2− −
[NiCl4 ] , Cl

electrons. Therefore, it undergoes sp hybridization. 3

Since there are 2 unpaired electrons in this case, it is paramagnetic in nature.

Page : 261 , Block Name : Intext Questions

Q9.6 [NiCl ] is paramagnetic while [Ni(CO) ] is diamagnetic though both are tetrahedral. Why?
2−
4 4

Answer. Though both [NiCl ] and [Ni(CO) ] are tetrahedral, their magnetic characters are different.
2−
4 4

This is due to a difference in the nature of ligands. Cl is a weak eld ligand and it does not cause the
−

pairing of unpaired 3d electrons. Hence, [NiCl ] is paramagnetic.
2−
4

But CO is a strong eld ligand. Therefore, it causes the pairing of unpaired 3d electrons.
Also, it causes the 4s electrons to shift to the 3d orbital, thereby giving rise to sp hybridization. Since 3

no unpaired electrons are present in this case, [Ni(CO) ] is diamagnetic. 4

Page : 261 , Block Name : Intext Questions

3+
Q9.7 [Fe(H O) ] Explain. is strongly paramagnetic where as [Fe(CN) ] is weakly paramagnetic.
3−
2 6 6

Explain.

3+
Answer. In both [Fe(H O) ] , fe exists in the +3 oxidation state i.e., in d
3− 5
2 and [Fe(CN)6 ]
6

con guration

Since CN is a strong eld ligand, it causes the pairing of unpaired electrons. Therefore, there is only
−

one unpaired electron left in the d-orbital.

Therefore,
μ = √n(n + 2)

= √1(1 + 2)

= √3

= 1.732BM

On the other hand, H O is a weak eld ligand. Therefore, it cannot cause the pairing of electrons. This
2

means that the number of unpaired electrons is 5. Therefore,
μ = √n(n + 2)

= √5(5 + 2)

= √35

= 6BM
2+
Thus , it is evident that [Fe(H O) ] is strongly paramagnetic, while [Fe(CN) ] is weakly
3−
2 6 6

paramagnetic.

Page : 261 , Block Name : Intext Questions

3+
Q9.8 Explain [Fe(H O) ] is an inner orbital complex whereas [Fe(CN) ] is an outer orbital
3−
2 6 6

complex.

Answer.

Page : 261 , Block Name : Intext Questions

Q9.9 Predict the number of unpaired electrons in the square planar [Pt(CN) ] ion.
2−
4

Answer.
2−
[Pt(CN)4 ]

In this complex, Pt is in the +2 state. It forms a square planar structure. This means that it undergoes
dsp hybridization. Now, the electronic con guration of pd(+2) is 5d .
2 8

CN
−
being a strong eld ligand causes the pairing of unpaired electrons. Hence, there are no unpaired
electron in [Pt(CN) ]
2−
4

Page : 261 , Block Name : Intext Questions

Q9.10 The hexaquo manganese(ll) ion contains ve unpaired electrons, while the hexacyanoion
contains only one unpaired electron. Explain using Crystal Field Theory.

Answer.

Hence, hexaaquo manganese (II) ion has ve unpaired electrons, while hexacyano ion has only one
unpaired electron.

Page : 261 , Block Name : Intext Questions

Q9.11 Calculate the overall complex dissociation equilibrium constant for the Cu(NH ) ion, given
2+
3 4

that β for this complex is 2.1 × 10 Ion,
13
4

Answer. β = 2.1 × 10
13
4

The overall complex dissociation equilibrium constant is the reciprocal of the overall stability constant,
β4
1 1
=
13
β4 2.1 × 10
−14
= 4.7 × 10

Page : 261 , Block Name : Intext Questions

Q9.1 Explain the bonding in coordination compounds in terms of Werner's postulates.

Answer. Werner's postulates explain the bonding in coordination compounds as follows:
(i) A metal exhibits two types of valencies namely, primary and secondary valencies. Primary valencies
are satis ed by negative ions while secondary valencies are satis ed by both negative and neutral ions.
(In modern terminology, the primary valency corresponds to the oxidation number of the metal ion
whereas the secondary valency refers to the coordination number of the metal ion. )
(ii) A metal ion has a de nite number of secondary valencies around the central atom.
Also, these valencies project in a speci c direction in the space assigned to the de nite geometry of the
coordination compound.
(iii) Primary valencies are usually ionizable, while secondary valencies are non-ionizable.

Page : 265 , Block Name : Exercise

Q9.2 FeSO solution mixed with (NH ) SO solution in 1:1 molar ratio gives the test of Fe
4 4 2 4
2+
ion but
solution mixed with aqueous ammonia in 1 : 4 molar ratio does not give the test of Cu ion.
2+
CuSO4

Explain why?

Answer.
⟶ FeSO4 ⋅ (NH4 ) SO4 ⋅ 6H2 O
2
(NH4 ) SO4 + FeSO4 + 6H2 O
2
Mohr's salt

CuSO4 + 4NH3 + 5H2 O ⟶ [Cu(NH3 ) ] SO4 ⋅ 5H2 O
4

tetraamminocopper (ii) sulphate

Both the compounds i.e. FeSO 4 ⋅ (NH4 ) SO4 ⋅ 6H2 O and [Cu(NH3 ) ] so4 ⋅ 5H2 O
2 4
fall under the
category of addition compounds with only one major difference i.e., the
former is an example of a double salt, while the latter is a coordination compound.
A double salt is an addition compound that is stable in the solid state but that which breaks up into its
constituent ions in the dissolved state. These compounds exhibit individual properties of their
constituents. For e.g. FeSO ⋅ (NH ) SO ⋅ 6H O break into Fe , NH , and SO ions. Hence, it
2+ 4+ 2−
4 4 2 4 2 4

gives a positive test for Fe ions. A coordination compound is an addition compound which retains its
2+

identity in the solid
as well as in the dissolved state. However, the individual properties of the constituents are lost. This
happens because [Cu(NH ) ] SO does not show the test for Cu The ions present in the
2+
3 4 ⋅ 5H2 O
4

2∗
solution of [Cu(NH ) ] so
2−
3 4 ⋅ 5H2 O are [Cu(NH3 ) ] and SO
4 4 4

Page : 265 , Block Name : Exercise

Q9.3 Explain with two examples each of the following: coordination entity, ligand, coordination
number, coordination polyhedron, homoleptic and heteroleptic.

Answer. (i) Coordination entity:
A coordination entity is an electrically charged radical or species carrying a positive or negative charge.
In a coordination entity, the central atom or ion is surrounded by a suitable number of neutral
molecules or negative ions ( called ligands). For example:
2+
cationic complex
4
[Ni(NH3 ) ] , [Fe(CN)6 ] =
6

anionic complex
2− −
[PI Cl4 ] , [Ag(CN)2 ] =

[Ni(CO) ] , [Co(NH ) Cl ] = neutral complex
4 3 2
4

(ii) Ligands
The neutral molecules or negatively charged ions that surround the metal atom in a coordination entity
or a coordinal complex are known as ligands. For example,
.

Ligands are usually polar in nature and possess at least one unshared pair of
−
N̈H3 , H2 O, cl , −OH

valence electrons.
(iii) Coordination number:
The total number of ligands (either neutral molecules or negative ions) that get attached to the central
metal atom in the coordination sphere is called the coordination number of the central metal atom. It is
also referred to as its ligancy.
For example:
(a) In the complex, K [PtCl ], there as six chloride ions attached to Pt in the coordinate sphere.
2 6

Therefore, the coordination number of Pt is 6.
(b) Similarly, in the complex [Ni(NH ) ] Cl the coordination number of the central atom (Ni) is 4.
3 4 2

(vi) Coordination polyhedron:

Coordination polyhedrons about the central atom can be de ned as the spatial arrangement of the
ligands that are directly attached to the central metal ion in the
coordination sphere. For example:

(v) Homoleptic complexes:
These are those complexes in which the metal ion is bound to only one kind of a donor
3+
group. For eg: [Co(NH ) ]
2−
3 , [PtCl4 ]
6

(vi) Heteroleptic complexes:
Heteroleptic complexes are those complexes where the central metal ion is bound to more than one
type of a donor group.
+
For example : [Co(NH ) Cl ] , [Co (NH ) , Cl]
2+
3 4 2 3

Page : 265 , Block Name : Exercise

Q9.4 What is meant by unidentate, didentate and ambidentate ligands? Give two examples for each.

Answer. A ligand may contain one or more unshared pairs of electrons which are called the donor sites
of ligands. Now, depending on the number of these donor sites, ligands can be classi ed as follows:
(a) Unidentate ligands: Ligands with only one donor sites are called unidentate
ligands. For e.g N̈H , Cl
3
−

(b) (b) Didentate ligands: Ligands that have two donor sites are called didentate ligands. For e.g.,

(c) Ambidentate ligands:
Ligands that can attach themselves to the central metal atom through two different atoms are called

ambidentate ligands. For example:

Page : 265 , Block Name : Exercise

Q9.5 Specify the oxidation numbers of the metals in the following coordination entities:
(i) [Co (H O) (CN)(en) ]
2+
2 2

(ii) [CoBr (en) ]
+
2 2

(iii) [PtCl ]
2−
4

(iv) K [Fe(CN) ]
3 6

(v) [Cr(NH ) Cl ]
3 3 3

Answer.

Page : 265 , Block Name : Exercise

Q9.6 Using IUPAC norms write the formulas for the following:
(i) Tetrahydroxozincate(ll)
(ii) Potassium tetrachloridopalladate(ll)

(iii) Diamminedichloridoplatinum(ll)
(iv) Potassium tetracyanonickelate(ll)
(v) Pentaamminenitrito-O-cobalt(III)
(vi) Hexaamminecobalt(lll) sulphate
(vii) Potassium , tri(oxalate) Chromate (III)
(viii) Hexaammineplatinum(IV)
(ix) Tetrabromidocuprate(ll)
(x) Pentaamminenitrito-N-cobalt(III)

Answer. (i) [Zn(OH] 2−

(ii) K [PdCl ]
2 4

(iii) [Pt(NH ) Cl ] 3 2 2

(iv) K [Ni(CN) ]
2 4

2+
(v) [Co(ONO)(NH ) ] 3 5

(vi) [Co(NH ) ] (SO ) 3 6 2
4 3

(vii) K [Cr(C O ) ]
3 2 4 3

4+
(viii) [Pt(NH ) ] 3 6

(ix) [Cu(Br) ]
2−
4

2+
(x) [Co [NO ) (NH ) ]2 3 5

Page : 265 , Block Name : Exercise

Q9.7 using IIJPAC norms write the systematic names of the following:
(i) [Co (NH ) 6] Cl 3 3

(ii) [Pt(NH ) Cl (NH CH )] Cl
3 2 2 3

(iii) [Ti (H O) 6]
3+
2

(iv) [Co (NH ) Cl (NO )] Cl
3 4 2

(v) [Mn(H O)6]
2+
2

(vi) [NiCl ]
2−
4

(vii) [Ni(NH ) ] Cl 3 6 2

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

(ix) [Ni(CO) ] 4

Answer. (i) Hexaamminecobalt(lll) chloride
(ii) Diamminechlorido(methylamine) platinum(ll) chloride
(iii) Hexaquatitanium(lll) ion
(iv) Tetraamminichloridonitrito-N-Cobalt(III) chloride
(v) Hexaquamanqanese(ll) ion
(vi) Tetrachloridonickelate(ll) ion
(vii) Hexaamminenickel(ll) chloride
(viii) Tris(ethane-l, 2-diammine) cobalt(lll) ion
(ix) Tetracarbonylnickel(o)

Page : 265 , Block Name : Exercise

Q9.8 List various types of isomerism possible for coordination compounds, giving an example of each.

Answer.

(a) Geometric isomerism :
This type of isomerism is common in heteroleptic complexes. It arises due to the
different possible geometric arrangements of the ligands. For example:

(b) Optical isomerism:
This type of isomerism arises in chiral molecules. Isomers are mirror images of each other and are non-
superimposable.

(c) Linkage isomerism:
This type of isomerism is found in complexes that contain ambidentate ligands. For example:
[Co(NH3 ) (NO2 )] Cl2 and [Co(NH3 ) (ONO)Cl2
5 5

Yellow form red form
(d) Coordination isomerism:
This type of isomerism arises when the ligands are interchanged between cationic and anionic entities
of differnet metal ions present in the complex.
[Co(NH3 ) ] [Cr(CN)6 ] and [Cr(NH3 ) ] [Co(CN)6 ]
6 6

(e) Ionization isomerism: This type of isomerism arises when a counter ion replaces a ligand within the
coordination sphere. Thus, complexes that have the same composition, but furnish different ions when
dissolved in water are called ionization isomers. For e.g.,
Co(NH3 ) SO4 )Br and Co(NH3 ) Br]SO4
5 5

(f) Solvate isomerism:
Solvate isomers differ by whether or not the solvent molecule is directly bonded to the metal ion or
merely present as a free solvent molecule in the crystal lattice.

[Cr[H2 O) ] C3 [Cr(H2 O) Cl] Cl2 ⋅ H2 O [Cr(H2 O) Cl2 ] Cl ⋅ 2H2 O
6 5 5

Violet Blue-green Dark green

Page : 265 , Block Name : Exercise

Q9.9 How many geometrical isomers are possible in the following coordination entities?
3−
(i)[Cr(C2 O4 ) ] ( ii) [Co(NH3 ) Cl3 ]
3 3

3−
Answer. (i) For [Cr(C O ) ] 2 4 3
no geometric isomer is possible as it is a bidentate ligand.

(ii) [Co(NH ) Cl ]3 3 3

Two geometrical isomers are possible,

Page : 265 , Block Name : Exercise

Q9.10 Draw the structures of optical isomers of:
3−
(i) [Cr (C O ) ]
2 4 3

(ii) [PtCl (en) ]
2+
2 2
+
(iii) [Cr(NH ) Cl (en)]
3 2 2

Answer.

Page : 265 , Block Name : Exercise

Q9.11 Draw all the isomers (geometrical and optical) of:
+
(i) [CoCl2 (en)2 ]

2+
(ii) [Co (NH3 ) Cl(en)2 ]

+
(iii) [Co(NH3 ) Cl2 (en)]
2

Answer.

Page : 266 , Block Name : Exercise

Q9.12 Write all the geometrical isomers of [Pt(NH )(Br)(Cl)(py)] and how many of these will exhibit
3

optical isomers?

Answer.
[Pt(NH3 )(Br)(Cl)(py)]

From the above isomers, none will exhibit optical isomers. Tetrahedral complexes rarely show optical
isomerization. They do so only in the presence of unsymmetrical chelating agents.

Page : 266 , Block Name : Exercise

Q9.13 Aqueous copper sulphate solution (blue in colour) gives:
(i) a green precipitate with aqueous potassium uoride, and
(ii) a bright green solution with aqueous potassium chloride
Explain these experimental results.

Answer. Aqueous CuSO exists as [Cu(H O) ] SO It is blue in colour due to the presence of
4 2 4 4

2+
[Cu[H2 O) ]
4
ions.
(i) When KF is added :
2+ 2−
[Cu(H2 O) ] + 4F → [Cu(F)4 ] + 4H2 O
4

(ii) When KCl is added :
2+ − 2−
[Cu(H2 O) ] + 4Cl ⟶ [CuCl4 ] + 4H2 O
4

In both these cases, the weak eld ligand water is replaced by the F ions.
− −
and Cl

Page : 266 , Block Name : Exercise

Q9.14 What is the coordination entity formed when excess of aqueous KCN is added to an aqueous
solution of copper sulphate? Why is it that no precipitate of copper sulphide is obtained when H S(g) is 2

passed through this solution?

Answer. CuSO 4(aq)
+ 4KCN(aq) ⟶ K2 [Cu(CN)4 ]
(aq)
+ K2 SO4(aq)

2+
I.e. , [Cu(H O) ]
− 2−
2 + 4CN ⟶ [Cu(CN)4 ] + 4H2 O
4

Thus, the coordination entity formed in the process is K [Cu(CN) ]. K [Cu(CN) ]very stable
2 4 2 4

complex, which does not ionize to give Cu ions when added to water. Hence, Cu ions are not
2+ 2+

precipitated when H S is passed through the solution.
2 (g)

Page : 266 , Block Name : Exercise

Q9.15 Discuss the nature of bonding in the following coordination entities on the basis of valence bond
theory:
4−
(i) [Fe(CN)6 ]

3−
(ii) [FeF6 ]

3−
(iii) [Co(C2 O4 )3]

3−
(iv) [CoF6 ]

Answer.

Page : 266 , Block Name : Exercise

Q9.16 Draw gure to show the splitting of d orbitals in an octahedral crystal eld.

Answer.

The splitting of the d orbitals in an octahedral eld takes palce in such a way that d 2
x −y
2 dz 2 experience a
rise in energy and form the e level, while d
g xy, dyz and dzx , experience a fall in energy and form the t 2g

level.

Page : 266 , Block Name : Exercise

Q9.17 What is spectrochemical series? Explain the difference between a weak eld ligand and a strong
eld ligand.

Answer. A spectrochemical series is the arrangement of common ligands in the increasing order of their
crystal- eld splitting energy (CFSE) values. The ligands present on the R.H.S of the series are strong
eld ligands while that on the C.H.S are weak eld ligands. Also, strong eld ligands cause higher
splitting in the d orbitals than weak eld ligands.
− 2− − − − − 2− − − −
I− < Br < S < SCN < Cl < N3 < F < OH < C2 O ∼ H2 O < NCS ∼ H < CN <
4

2− −
NH3 < en ∼ SO < NO < phen < cO
3 2

Page : 266 , Block Name : Exercise

Q9.18 What is crystal eld splitting energy? How does the magnitude of decide the actual con guration
of d-orbitals in a coordination entity?

Answer. The degenerate d-orbitals (in a spherical eld environment) split into two levels i.e., e and t g 2g

in the presence of ligands. The splitting of the degenerate levels due to the presence of ligands is called
the crystal- eld splitting while the energy difference between the two levels (e and t ) is called the g 2g

crystal- eld splitting energy. It is
denoted by Δ . o

After the orbitals have split, the lling of the electrons takes place. After electron
(each) has been lled in the three tag t orbitals, the lling of the fourth electron takes place in two
2g

ways. It can enter the e orbital (giving rise to t like electronic con guration) or the pairing of the
3 1
9 eg
2g

electrons can take place in the t-„orbitals (giving rise to t 4
29
e
0

9
eg like electronic con guration). If the Lo
value of a ligand is less than the pairing energy (P), then the electrons enter the eg orbital. On the other
hand, if the Ao value of a ligand is more than the pairing energy (P), then the electrons enter the t 2g

orbital.

Page : 266 , Block Name : Exercise

3+
Q9.19 [Cr(NH ) ] is paramagnetic while [Ni(CN) ] is diamagnetic . Explain why ?
2−
3 6 4

Answer.

Page : 266 , Block Name : Exercise

2+
Q9.20 A solution of [Ni(H O) ] is green but a solution of [Ni(CN) ] is colourless. Explain.
2−
2 6 4

2+
Answer. In [Ni(H O) ] is a weak eld ligand. Therefore, there are unpaired electrons in Ni .
¨ 2+
2 , H2 O
6

In this complex, the d electrons from the lower energy level can be excited to the higher energy level
i.e., the possibility of d—d transition is present. Hence, Ni(H O) ] is coloured. 2 6
2+

In [Ni(CN) ] the electrons are all paired as CN is a strong eld ligand. Therefore, d-d transition is
2− −
4

not possible in [Ni(CN) ] Hence, it is colourless.
2−
4

Page : 266 , Block Name : Exercise

Q9.21
2+
are of different colours in dilute solutions. Why ?
4−
[Fe(CN)6 ] and [Fe(H2 O) ]
6

Answer. The colour of a particular coordination compound depends on the magnitude of the crystal-
eld splitting energy, Δ. This CFSE in turn depends on the nature of the ligand. In case of
2+
the colour differs because there is a difference in the CFSE. Now, CN
4−
[Fe(CN)6 ] and [Fe(H2 O) ]
6

is a strong eld ligand having a higher CFSE value as compared to the CESE value of water. This means

that the absorption of energy for the intra d-d transition also differs. Hence, the transmitted colour also
differs.

Page : 266 , Block Name : Exercise

Q9.22 Discuss the nature of bonding in metal carbonyls.

Answer. The metal-carbon bonds in metal carbonyls have both o and n characters. A c bond is formed
when the carbonyl carbon donates a lone pair of electrons to the vacant orbital of the metal. A n bond is
formed by the donation of a pair of electrons from the lled metal d orbital into the vacant anti-bonding
n* orbital (also known as back bonding of the carbonyl group). The o bond strengthens the n bond and
vice-versa. Thus, a synergic effect is created due to this metal-ligand bonding. This synergic effect
strengthens the bond between CO and the metal.

Page : 266 , Block Name : Exercise

Q9.23 Give the oxidation state, d-orbital occupation and coordination number of the central metal ion
in the following complexes:
(i) K3 [Co(C2 O4 ) ]
3

(ii) cis − [Cr(en)2 Cl2 ] Cl

(iii) (NH4 ) [CoF4 ]
2

(iv) [Mn(H2 O) ] SO4
6

Answer. (i) K [Co(C O ) ]
3 2 4 3

The central metal ion is Co.
Its coordination number is 6.
The oxidation state can be given as:
x − 6 = −3

x = +3

The d orbital occupation for Co
3+ 6 0.
is t eg
2g

(ii) cis − [Cr(en) Cl ] Cl
2 2

The central metal ion is Cr.
The coordination number is 6.
The oxidation state can be given as:
x + 2(0) + 2(−1) = +1

x − 2 = +1

x = +3

3+ 3
The d orbital occupation for Cr is t .
2g

(iii) (NH ) [CoF ]
4 2 4

The central metal ion is Co.
The coordination number is 4.
The oxidation state can be given as:
x - 4 +-2
x = +2
The d orbital occupation for Co is e t
2+ 4 3.
g 2g

(iv) [Mn(H O) ] SO
2 6 4

The central metal ion is Mn
The coordination number is 6.
The oxidation state can be given as:
x + 0 = +2
x = +2
The d orbital occupation for Mn is t e
3 2
2g g

Page : 266 , Block Name : Exercise

Q9.24 Write down the IUPAC name for each of the following complexes and indicate the oxidation state,
electronic con guration and coordination number. Also give
stereochemistry and magnetic moment of the complex:
(i) K [Cr(H2 O) (C2 O4 ) ] ⋅ 3H2 O
2 2

(ii) [Co(NH3 ) Cl] Cl2
s

(iii) CrCl3 (py)3

(iv) Cs [FeCl4 ]

(v) K4 [Mn(CN)6 ]

Answer. (i) Potassium diaquadioxalatochromate (Ill) trihydrate.
Oxidation state of chromium = 3
Electronic con guration: 3d : t 3 3

29

Coordination number = 6
Shape: octahedral
Stereochemistry:

Magnetic moment, μ = √n(n + 2)

= √3(3 + 2)

= √15

∼ 4BM

(ii) [Co(NH ) Cl] Cl
3 5 2

IUPAC name: Pentaamminechloridocobalt(lll) chloride
Oxidation state of Co — +3
Coordination number 6
Shape: octahedral.
Electronic con guration: d : t .
6 6

29

Stereochemistry:

(iii) CrCl (py)
3 3

IUPAC name: Trichloridotripyridinechromium (Ill)
Oxidation state of chromium = +3
Electronic con guration ford d = t3 3

2g

Coordination number = 6
Shape: octahedral.
Stereochemistry:

Both isomers are optically active. Therefore, a total of 4 isomers exist.
Magnetic moment, μ = √n(n + 2)

= √3(3 + 2)

= √15

∼ 4BM

(iv) Cs [FeCl ]
4

IUPAC name: Caesium tetrachloroferrate (Ill)
Oxidation state of Fe +3
Electronic con guration of d = e t
6 2
g
3

2g

Coordination number = 4
Shape: tetrahedral
Stereochemistry: optically inactive
Magnetic moment:

μ = √n(n + 2)

= √5(5 + 2)

= √35 ∼ 6BM

(v) K [Mn(CN) ]
4 6

Potassium hexacyanomanganate(ll)
Oxidation state of manganese +2

Electronic con guration: d 5+
: t
5

2g

Coordination number = 6
Shape: octahedral.
Streochemistry: optically inactive
Magnetic moment, μ = √n(n + 2)

= √1(1 + 2)

= √3

= 1.732BM

Page : 266 , Block Name : Exercise

Q9.25 What is meant by stability of a coordination compound in solution? State the factors which
govern stability of complexes.

Answer. The stability of a complex in a solution refers to the degree of association between the two
species involved in a state of equilibrium. Stability can be expressed quantitatively in terms of stability
constant or formation constant.
M + 3L ⟷ ML3

[ML1 ]
Stability constant, β = 3
[M][L]

For this reaction, the greater the value of the stability constant, the greater is the
proportion of ML3 in the solution.
Stability can be of two types:
(a) Thermodynamic stability:
The extent to which the complex will be formed or will be transformed into another
species at the point of equilibrium is determined by thermodynamic stability.
(b) Kinetic stability:
This helps in determining the speed with which the transformation will occur to attain the state of
equilibrium.
Factors that affect the stability Of a complex are:
(a) Charge on the central metal ion: Thegreater the charge on the central metal ion,
the greater is the stability of the complex.
2.Basic nature Of the ligand: A more basic ligand will form a more stable
complex.
2. Presence Of chelate rings: Chelation increases the stability of complexes.

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Q9.26 What is meant by the chelate effect? Give an example.

Answer. When a ligand attaches to the metal ion in a manner that forms a ring, then the metal- ligand
association is found to be more stable. In other words, we can say that complexes containing chelate
rings are more stable than complexes without rings. This is known as the chelate effect.
For example:

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Q9.27 Discuss brie y giving an example in each case the role of coordination compounds in:
(i) biological system
(ii) medicinal chemistry
(iii) analytical chemistry'
(iv) extraction/metallurgy of metals

Answer. (i) Role Of coordination compounds in biological systems:
We know that photosynthesis is made possible by the presence of the chlorophyll pigment. This
pigment is a coordination compound of magnesium. In the human
biological system, several coordination compounds play important roles. For example, the oxygen-
carrier of blood, i.e., haemoglobin, is a coordination compound of iron.
(ii) Role of coordination compounds in medicinal chemistry: Certain coordination compounds of
platinum (for example, cis-platin) are used for
inhibiting the growth of tumours.
(iii) Role of coordination compounds in analytical chemistry: During salt analysis, a number of basic
radicals are detected with the help of the colour
changes they exhibit with different reagents. These colour changes are a result of the coordination
compounds or complexes that the basic radicals form with different ligands.
(iii) Role Of coordination compounds in extraction Or metallurgy Of metals: The process of extraction
of some of the metals from their ores involves the formation of
complexes. for example, in aqueous solution, gold combines with cyanide ions to form
[Au(CN) ] From this solution, gold is later extracted by the addition of zinc metal.
2

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Q9.28 How many ions are produced from the complex Co(NH ) Cl in solution?
3 6 2

(i) 6
(ii) 4
(iii) 3
(iv) 2

Answer.
(iii) The given complex can be written as [Co(NH3 ) ] Cl2
6

+ −
Thus, [Co(NH3 ) ] along with two Cl ions are produced.
6

Page : 266 , Block Name : Exercise

Q9.29 Amongst the following ions which one has the highest magnetic moment value?
3+
(i) [Cr(H2 O) ]
6

2+
(ii) [Fe(H2 O) ]
6

2+
(iii) [Zn(H2 O) ]
6

Answer.
3+
(i) No. of unpaired electrons in [Cr(H2 O) ] = 3
6

Then, μ = √n(n + 2)

= √3(3 + 2)

= √15

−4BM
2+
(ii) No. of unpaired electrons in [Fe(H2 O) ] = 4
6

Then, μ

= √24

= 5BM
2+
(iii) No. of unpaired electrons in [Zn(H2 O) ] = 0
6

2+
Hence, [Fe(H2 O) ] has the highest magnetic moment value.
6

Page : 266 , Block Name : Exercise

Q9.30 The oxidation number of cobalt in K [Co(CO) ] is 4

(i) + 1

(ii) + 3

(iii) − 1

(iv) − 3

Answer. We know that CO is a neutral ligand and K carries a charge of +1.
Therefore, the complex can be written as K [Co(CO) ] Therefore, the oxidation number
+ −
4

of Co in the given complex is —1. Hence, option (iii) is correct.

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Q9.31 Amongst the following, the most stable complex is

3+
(i) [Fe(H2 O) ]
6

3+
(ii) [Fe(NH3 ) ]
6

3+
(iii) [Fe(C2 O4 ) ]
3

3−
(iv) [FeCl6 ]

Answer. We know that the stability of a complex increases by chelation. Therefore, the most stable
3−
complex is [Fe(C O ) ] 2 4 3

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Q9.32 What will be the correct order for the wavelengths of absorption in the visible region for the
following:
4− 2+ 2+
[Ni(NO2 ) ] , [Ni(NH3 ) ] , [Ni(H2 O) ]
6 6 6

Answer. The central metal ion in all the three complexes is the same. Therefore, absorption in the
visible region depends on the ligands. The order in which the CFSE values of the ligands increases in the
spectrochemical series is as follows:
−
H2 O < NH3 < NO
2

Thus, the amount of crystal- eld splitting observed will be in the following order:

Hence, the wavelengths of absorption in the visible region will be in the order:
2+ 2+ 4−
[Ni(H2 O) ] > [Ni(NH3 ) ] > [Ni(NO2 ) ]
6 6 6

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