Class 12 - Chemistry - Coordination Compounds

Question 9.1.

Explain the bonding in coordination compounds in terms of Werner’s postulates.


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 satisfied by negative ions while secondary valencies are satisfied 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 definite number of secondary valencies around the central atom.

Also, these valencies project in a specific direction in the space assigned to the definite geometry of the coordination compound.

(iii) Primary valencies are usually ionizable, while secondary valencies are non-ionizable.


Question 9.2.

FeSO4 solution mixed with (NH4)2SO4 solution in 1:1 molar ratio gives the test of Fe2+ ion

but CuSO4 solution mixed with aqueous ammonia in 1:4 molar ratios does not give the test of Cu2+ ion. Explain why?


(NH4)2SO4 + FeSO4 +6H2O --> FeSO4 (NH4)2SO4.6H2O

                                             Mohr’s Salt

CuSO4 + 4NH3 + 5H2O --> [Cu (NH3)4] SO4.5H2O

                                        Tetraamineocopper (II) sulphate

Both the compounds i.e. FeSO4 (NH4)2SO4.6H2O and [Cu (NH3)4] SO4.5H2O 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. FeSO4 (NH4)2SO4.6H2O breaks into Fe2+, NH4+, and SO42− ions.

Hence, it gives a positive test for Fe2+ ions.

A coordination compound is an addition compound which retains its identity in the solid as well as in the dissolved state.

However, the individual properties of the constituents are lost.

This happens because [Cu (NH3)4] SO4.5H2O does not show the test for Cu2+ .

The ions present in the solution of [Cu (NH3)4] SO4.5H2O are [Cu (NH3)4] 2+ and SO42-


Question 9.3.

Explain with two examples each of the following: coordination entity, ligand, coordination number,

coordination polyhedron, homoleptic and heteroleptic.


(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:

[Ni (NH3)6]2+, [Fe (CN6)] 4+ = cationic complex

[PtCl4]2- , [Ag (CN) 2] -       = anionic complex

[Ni (CO) 4], [Co (NH3)4 Cl2] = neutral complex

(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: NH3, H2O: Cl - , -OH.

Ligands are usually polar in nature and possess at least one unshared pair of 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, K2 [PtCl6], there as six chloride ions attached to Pt in the coordinate sphere.

Therefore, the coordination number of Pt is 6.

(b) Similarly, in the complex [Ni (NH3)4] Cl2, the coordination number of the central atom (Ni) is 4.

(iv) Coordination polyhedron: Coordination polyhedrons about the central atom can be defined

as the spatial arrangement of the ligands that are directly attached to the central metal ion in the coordination sphere.

 For example:


 (b) Tetrahedral


(v) Homoleptic complexes: These are those complexes in which the metal ion is bound to only one kind of a donor group.

For Example: [Co (NH3)6]3+, [PtCl4]2- etc.

(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 e.g.:  [Co (NH3)4 Cl2] +, [Co (NH3)5 Cl] 2+


Question 9.4.

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


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 classified as follows:

(a) Unidentate ligands: Ligands with only one donor sites are called unidentate ligands. For e.g.:NH3, Cl - etc.

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

Example: Ethane-1, 2-diamine


Oxalate ion


Ambidentate ligands: These ligands which can attach them with the central metal atom by two different

atoms are called as ambidentate ligands.

For Example:



Question 9.5.

Specify the oxidation numbers of the metals in the following coordination entities:

  • [Co(H2O)(CN)(en)2]2+ ](ii) [CoBr2(en)2]+ (iii) [PtCl4] 2– (iv) K3[Fe(CN)6] (v) [Cr(NH3)3Cl3


  • [Co(H2O)(CN)(en)2]2+ ]

Let the oxidation number of Co be x.

The charge on the complex is +2.

[Co (H2O) (CN) (en) 2]2+] = (x + 0 + (-1) +2(0))

(x-1) =+2


  • [CoBr2(en)2]+

=(x + 2(-1) + 2(0))


(x-2) =+1


  • [PtCl4] 2–




  • K3[Fe(CN)6]

i.e. [Fe (CN) 6]3-

(x +6(-1)) =-3


  • [Cr(NH3)3Cl3

(x+3(0) +3(-1)) =0

(x-3) =0



Question 9.6.

Using IUPAC norms write the formulas for the following:

(i) Tetrahydroxozincate (II)

(ii) Potassium tetrachloridopalladate (II)

(iii) Diamminedichloridoplatinum (II)

(iv) Potassium tetracyanonickelate (II)

(v) Pentaamminenitrito-O-cobalt (III)

(vi) Hexaamminecobalt (III) sulphate

(vii) Potassium tri (oxalato) chromate (III)

(viii) Hexaammineplatinum (IV)

(ix) Tetrabromidocuprate (II)

(x) Pentaamminenitrito-N-cobalt (III)


(i) Tetrahydroxozincate (II) = [Zn (OH) 4)] 2-

(ii) Potassium tetrachloridopalladate (II) = K2 [PdCl4]

(iii) Diamminedichloridoplatinum (II) = [Pt (NH3)2Cl2]

(iv) Potassium tetracyanonickelate (II) = K2 [Ni (CN) 4]

(v) Pentaamminenitrito-O-cobalt (III) = [Co (ONO) (NH3)5]2+

(vi) Hexaamminecobalt (III) sulphate = [Co (NH3)6]2 (SO4)3

(vii) Potassium tri (oxalato) chromate (III) = K3 [Cr (C2O4)3]

(viii) Hexaammineplatinum (IV) = [Pt (NH3)6]4+

(ix) Tetrabromidocuprate (II) = [Cu (Br) 4]2+

(x) Pentaamminenitrito-N-cobalt (III) = [Co [(NO2) (NH3)5]] 2+


Question 9.7.

Using IUPAC norms write the systematic names of the following:

(i) [Co (NH3)6] Cl3

(ii) [Pt (NH3)2Cl (NH2CH3)] Cl

(iii) [Ti (H2O) 6]3+

(iv)  [Co (NH3)4Cl (NO2)] Cl

(v)  [Mn (H2O) 6]2+

(vi) [NiCl4]2–

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

 (ix) [Ni (CO) 4]


(i) Hexaamminecobalt (III) chloride

(ii) Diamminechlorido (methylamine) platinum (II) chloride

(iii) Hexaquatitanium (III) ion

(iv) TetraamminechloridonitritoCobalt –N-(III) chloride

(v) Hexaquamanganese (II) ion

(vi) Tetrachloridonickelate (II) ion

(vii) Hexaamminenickel (II) chloride

(viii) Tris (ethane-1, 2-diammine) cobalt (III) ion

(ix) Tetracarbonylnickel (0)


Question 9.8.

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


Isomers are the compounds which have same chemical formula but different arrangement of atoms in space.

There are principle two types of isomerism:

(i) Stereo isomerism:

(a) Geometrical isomerism

(b) Optical isomerism

(ii) Structural isomerism

(a) Ionisation isomerism

(b) Linkage isomerism

(c) Coordination isomerism

(d) Solvate isomerism

Geometrical isomerism comes into existence by the different spatial arrangement of groups around the central metal atom.

Similar groups may either be arranged on the same side or on opposite sides of the central metal atom.

This gives rise to two types of isomers called cis and trans isomers.

When groups under consideration are arranged on the same side of the central metal atom, isomers are called cis isomers

and when the groups under consideration are spatially placed on the opposite sides, isomers are called trans isomer.


Optical isomerism is exhibited by those compounds which possess chirality.

The presence of an element of symmetry makes a molecule symmetric and renders it optically inactive.

When molecule does not possess any element of symmetry its mirror image is non-superimposable with the molecule itself.

This makes the molecule optically active. Such an asymmetric molecule can show the phenomenon of optical .

The two forms of the molecule which are mirror images of each other are called enantiomers.

One form rotates the plane of plane polarized light in clockwise direction, while the other in anticlockwise direction.

The former is called the d-form, while the latter is termed as l-form.


Ionisation isomerism: In ionisation isomerism there is an exchange of ions inside and outside the coordination sphere.

Ionisation isomers have the same formula but produce different ions in solution.

It is also known as ion-ion exchange isomerism.

Ions present in the solution



[Co(NH3)5SO4]+ + Br-

[Co(NH3)5SO4] Br

Red Violet

[Co(NH3)5Br]+ + SO42-




In the two isomers (A) and (B), there is an exchange of ions inside and outside the coordination sphere.

The aqueous solution of (A) gives the precipitate of AgBr on treatment with AgNO3 as it contains Br, while that of (B)

gives precipitate of BaSO4, on treatment with BaCl2.


Linkage isomerism: In linkage isomerism the same ligand is bonded to central metal atom or ion through different atoms.

Linkage isomers have the same molecular formula but differ in the linkage of the ligand to central metal atom.

For Example: - [Co (NH3)5 (NO2)] Cl2 and [Co (NH3)5 (ONO) Cl2

Yellow form Red form

Coordination isomerism: this type of isomerism arises from different Complex ions having same molecular formula.

There is interchange of ligands between cationic and anionic entities of different metal ions present in the complex.

Example is [Co (NH3)6] [Cr (CN) 6] in which ammonia ligands are bounded to Cobalt and cyanide ligands to chromium ion.

In its coordination isomer [Cr (NH3)6] [Co (CN) 6] ammonia ligands are bounded to chromium ion and cyanide ligands to Cobalt ion.

Solvent isomerism or hydrate isomerism: in hydrate isomerism there is an exchange of water molecule inside and outside coordination sphere.

Hydrate isomers have the same molecular formula but differ in the number of molecules of water inside and outside the coordination sphere.

[Cr [H2O) 6]Cl3 [Cr (H2O) 5Cl] Cl2. H2O [Cr (H2O) 5Cl2] Cl. 2H2O

Violet Blue-green Dark green


Question 9.9.

How many geometrical isomers are possible in the following coordination entities?

(i) [Cr (C2O4)3] 3– (ii) [Co (NH3)3Cl3]


(i) For [Cr (C2O4)3]3−, no geometric isomer is possible as it is a bidentate ligand.


(ii) [Co (NH3)3Cl3]

Two geometrical isomers are possible.


Question 9.10.

Draw the structures of optical isomers of:

(i) [Cr (C2O4)3]3– (ii) [PtCl2 (en) 2]2+ (iii) [Cr (NH3)2Cl2(en)] +


Optical isomers are the one which rotate the plane polarized light by a certain angle when passed through them

and are mirror images of each other, also called as stereoisomer.

  • (C2O4)2- is a bidentate ligand so it can attach to central atom at two sites, forming a chelate ring.
  • In the complexes of this type, three symmetrical bidentate chelating ligands AA are coordinated to the central metal atom M.
  • Such complexes do not possess any element of symmetry and are optically active.
  • Moreover, these complexes can be resolved into optical isomers.


  • In this complex Cl is a unidentate ligand with one site of attachment whereas -en
  • (ethylenediamine) is bidentate ligand with two sites of attachment.
  • The complexes in which two symmetrical bidentate chelating ligands AA and two monodentate ligands a,
  • are coordinated to central metal atom M, exhibit the phenomenon of optical isomerism and can be resolved into their optical isomers.

An example of this type of complexes is given as shows both geometrical as well as optical isomerism.

Its cis form is unsymmetrical, while the trans form is symmetrical because it contains a plane of symmetry.

Hence, optical isomerism is shown by cis form.


  • In this there are three types of ligands. One is ammonia which is neutral (nitrogen donates the lone pair of electron to metal),
  • Cl - ligand is unidentate and -en is bidentate ligand. Coordination number is 6.
  • Two optical isomers are possible due to presence of 3 types of ligands.


Question 9.11.

Draw all the isomers (geometrical and optical) of:

(i) [CoCl2 (en) 2] + (ii) [Co (NH3) Cl (en) 2]2+ (iii) [Co (NH3)2Cl2(en)] +


(i) [CoCl2 (en) 2] +


In total, three isomers are possible.

  • [Co (NH3) Cl (en) 2] 2+


Trans-isomers are optically inactive.

Cis-isomers are optically active.

  • [Co (NH3)2Cl2(en)] +


Question 9.12.

 Write all the geometrical isomers of [Pt (NH3) (Br) (Cl) (py)] and how many of these will exhibit optical isomers?



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.


Question 9.13.

Aqueous copper sulphate solution (blue in colour) gives:

(i) A green precipitate with aqueous potassium fluoride and

(ii) A bright green solution with aqueous potassium chloride. Explain these experimental results.


Copper sulphate exists as [Cu (H2O) 4] SO4. It is blue in colour due to presence of the [Cu (H2O) 4]2+ ions.

When KF is added water is replaced by fluoride ion and green colour is due to [Cu (F) 4]2- ions

[Cu (H2O) 4]2+ + 4F- --> [Cu (F) 4]2- + 4H2O


When KCl is added water is replaced by chloride ion and bright green colour is due to presence of [CuCl4]2- ions.

[Cu (H2O) 4]2+ + 4Cl- --> [CuCl 4]2- + 4H2O

                                   (Bright Green)

In both the cases water, weak field ligand is replaced by fluoride and chloride ions.


Question 9.14.

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 H2S (g) is passed through this solution?


CuSO4 + KCN ⇒ K2 [Cu (CN) 4] + K2SO4

[Cu (H2O) 4]2+ + 4CN- ⇒ [Cu (CN) 4]2- + 4H2O

The coordination entity formed is K2 [Cu (CN) 4].

IUPAC name of the coordination entity is potassium tetracyanocuprate (II). It is a very stable complex.

The copper atom present inside the coordination sphere does not separate out to

form copper ions and cyanide ions due to strong bond between them.

It does not ionize to give Cu2+ ions and hence on adding H2S,

since there are no copper ions present so no precipitate of copper sulfide is formed.


Question 9.15.

Discuss the nature of bonding in the following coordination entities on the basis of valence bond theory:

(i) [Fe (CN) 6]4– (ii) [FeF6]3– (iii) [Co (C2O4)3]3– (iv) [CoF6]3–


(i) [Fe (CN) 6] 4−

In the above coordination complex, iron exists in the +II oxidation state.

Fe2+: Electronic configuration is 3d6 Orbitals of Fe2+ ion:


As CN is a strong field ligand, it causes the pairing of the unpaired 3d electrons.


Since there are six ligands around the central metal ion, the most feasible hybridization is d2sp3.

d2sp3 hybridized orbitals of Fe2+ are:


6 electron pairs from CN ions occupy the six hybrids d2sp3 orbitals.

Hence, the geometry of the complex is octahedral and the complex is diamagnetic (as there are no unpaired electrons).


(ii) [FeF6] 3−

In this complex, the oxidation state of Fe is +3.

Orbitals of Fe+3 ions:


There are 6 F ions. Thus, it will undergo d2sp3 or sp3d2 hybridization. As F is a weak field ligand,

it does not cause the pairing of the electrons in the 3d orbital. Hence, the most feasible hybridization is sp3d2.

sp3d2 hybridized orbitals of Fe are:


                       6 electrons pairs from F- ions

Hence, the geometry of the complex is found to be octahedral.

(iii) [Co (C2O4)3]3−

Cobalt exists in the +3 oxidation state in the given complex.

Orbitals of Co3+ ion:


Oxalate is a weak field ligand. Therefore, it cannot cause the pairing of the 3d orbital electrons.

As there are 6 ligands, hybridization has to be either sp3d2 or d2sp3 hybridization.


sp3d2 hybridization of Co3+:

The 6 electron pairs from the 3 oxalate ions (oxalate anion is a bidentate ligand) occupy these sp3d2 orbitals.


Hence, the geometry of the complex is found to be octahedral.

(iv) [CoF6]3−

Cobalt exists in the +3 oxidation state.

Orbitals of Co3+ ion:


Again, fluoride ion is a weak field ligand. It cannot cause the pairing of the 3d electrons.

As a result, the Co3+ ion will undergo sp3d2 hybridization. sp3d2 hybridized orbitals of Co3+ ion are:


Hence, the geometry of the complex is octahedral and paramagnetic.


Question 9.16.

Draw figure to show the splitting of d orbitals in an octahedral crystal field?


In octahedral complex the splitting of the d orbital will be such a way that the dx2-y2 and dz2 orbitals which face towards the

axes along the direction of the ligand will experience more repulsion and will be raised in the energy and the

other three orbitals which are directed between the axes are lowered in energy.


 Question 9.17.

What is spectrochemical series? Explain the difference between a weak field ligand and a strong field ligand.


A spectrochemical series is the arrangement of common ligands in the increasing order of their crystal-field splitting energy (CFSE) values.

The ligands present on the R.H.S of the series are strong field ligands while that on the L.H.S are weak field ligands.

Also, strong field ligands cause higher splitting in the d orbitals than weak field ligands.

I < Br < S2− < SCN < Cl< N3 < F < OH < C2O4 2− ~H2O < NCS ~H < CN < NH3< en ~SO32− < NO2− < phenol < CO


Question 9.18.

What is crystal field splitting energy? How does the magnitude of Δo decide the actual configuration of d orbitals in a coordination entity?


The degenerate d-orbitals (in a spherical field environment) split into two levels i.e., eg and t2g in the presence of ligands.

The splitting of the degenerate levels due to the presence of ligands is called the crystal-field splitting while the

energy difference between the two levels (eg and t2g) is called the crystal-field splitting energy. It is denoted by ∆o.

After the orbitals have split, the filling of the electrons takes place. After 1 electron (each) has been filled in the three t2g orbitals,

the filling of the fourth electron takes place in two ways. It can enter the eg orbital (giving rise to t2g3eg1 like electronic configuration)

or the pairing of the electrons can take place in the t2g orbitals (giving rise to t2g4 eg0 like electronic configuration).

If the ∆o value of a ligand is less than the pairing energy (P), then the electrons enter the eg orbital. On the other hand,

if the ∆o value of a ligand is more than the pairing energy (P), then the electrons enter the t2g orbital.


Question 9.19.

[Cr (NH3)6] 3+ is paramagnetic while [Ni (CN) 4] 2– is diamagnetic. Explain why?


Cr is in the +3 oxidation state i.e., d3 configuration.

Also, NH3 is a weak field ligand that does not cause the pairing of the electrons in the 3d orbital. Cr3+.


Therefore, it undergoes d2sp3 hybridization and the electrons in the 3d orbitals remain unpaired.

Hence, it is paramagnetic in nature.

In [Ni (CN) 4]2−, Ni exists in the +2 oxidation state i.e., d8 configuration.



CN is a strong field ligand. It causes the pairing of the 3d orbital electrons.

Then, Ni2+ undergoes dsp2 hybridization.

As there are no unpaired electrons, it is diamagnetic.


Question 9.20.

A solution of [Ni (H2O) 6] 2+ is green but a solution of [Ni (CN) 4] 2– is colourless. Explain?


In [Ni (H2O) 6]2+, H2O: is a weak field ligand. Therefore, there are unpaired electrons in Ni2+.

In this complex, the d electrons from the lower energy level an be excited to the higher energy level i.e.,

the possibility of d−d transition is present. Hence, [Ni (H2O) 6] 2+ is coloured.

In [Ni (CN) 4]2−, the electrons are all paired as CN is a strong field ligand.

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


Question 9.21.

[Fe (CN) 6] 4– and [Fe (H2O) 6] 2+ are of different colours in dilute solutions. Why?


The colour of a particular coordination compound depends on the magnitude of the crystal- field splitting energy, ∆.

This CFSE in turn depends on the nature of the ligand. In case of [Fe (CN) 6] 4− and [Fe (H2O) 6]2+,

the colour differs because there is a difference in the CFSE.

Now, CN is a strong field ligand having a higher CFSE value as compared to the CFSE value of water.

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


Question 9.22.

Discuss the nature of bonding in metal carbonyls.


The metal-carbon bonds in metal carbonyls have both σ and π characters.

A σ bond is formed when the carbonyl carbon donates a lone pair of electrons to the vacant orbital of the metal.

A π bond is formed by the donation of a pair of electrons from the filled metal d orbital into the vacant anti-bonding π* orbital

(also known as back bonding of the carbonyl group).

The σ bond strengthens the π 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.


Question 9.23.

Give the oxidation state, d orbital occupation and coordination number of the central metal ion in the following complexes:

(i) K3 [Co (C2O4)3] (ii) cis-[Cr (en) 2Cl2] Cl (iii) (NH4)2[CoF4]

 (iv) [Mn (H2O) 6] SO.


(i) K3 [Co (C2O4)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 Co3+ is t2g6eg0.

(ii) cis-[Cr (en) 2Cl2] Cl

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

The d orbital occupation for Cr3+ is t2g3.

(iii) (NH4)2[CoF4]

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 Co2+ is eg 4t2g3.

(iv) [Mn (H2O) 6] SO4

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 t2g3 eg 2


Question 9.24.

Write down the IUPAC name for each of the following complexes and indicate the oxidation state,

electronic configuration and coordination number. Also give stereochemistry and magnetic moment of the complex:

(i) K [Cr (H2O) 2(C2O4)2].3H2O (ii) [Co (NH3)5Cl-] Cl2 (iii) CrCl3 (py) 3

 (v) K4 [Mn (CN) 6]

 (iv) Cs [FeCl4].


  • Potassium diaquadioxalatochromate (III) trihydrate.

The complex is an anion with chromium as central atom, 2 water molecules and 2 oxolate ions with -2 negative charges on each.

Balance overall charge as 0, we get oxidation state of Cr as:

X + 2(0) + 2(-2) = -1

X = + 3.

Electronic configuration of Cr: 3d3, t2g3


As both the water molecule and oxolate ion are weak field ligand, so they do not cause pairing up of electron.

μ = √ [n (n + 2)]

μ = √ [3×5]

μ = 4BM.

(ii) [Co (NH3)5Cl] Cl2

IUPAC name: Pentaamminechloridocobalt (III) chloride

Oxidation state of Co = +3

Coordination number = 6 Shape: octahedral.

Electronic configuration: d6: t2g6



2 Isomers

Magnetic Moment = 0

(iii) CrCl3 (py) 3

IUPAC name: Trichloridotripyridinechromium (III)

Oxidation state of chromium = +3

Electronic configuration for d3 = t2g3

Coordination number = 6 Shape: octahedral.



Question 9.25.

What is meant by stability of a coordination compound in solution?

State the factors which govern stability of complexes.


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 ß  --> ML3

Stability constant β= [ML3]/ [M] [L] 3

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: The greater the charge on the central metal ion, the greater is the stability of the complex.

  1. 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.


Question 9.26.

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


When a ligand bonds with the central metal ion in such a way that it forms a ring,

then it is found that the metal - ligand bond is more stable.

In other way a compound which forms chelate ring are more stable than the complexes without chelate rings. T

his effect is known as chelate effect.

For Example:-


Question 9.27.

Discuss briefly giving an example in each case the role of coordination compounds in:

(i) Biological systems

(ii) Medicinal chemistry (iii) analytical chemistry and (iv) extraction/metallurgy of metals.


  • 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.

  • 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.

  • 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) 2].

From this solution, gold is later extracted by the addition of zinc metal.


Question 9.28.

How many ions are produced from the complex Co (NH3)6Cl2 in solution?

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


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

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


Question 9.29.

Amongst the following ions which one has the highest magnetic moment value?

(i) [Cr (H2O) 6] 3+ (ii) [Fe (H2O) 6]2+ (iii) [Zn (H2O) 6]2+


(i) No. of unpaired electrons in [Cr (H2O) 6] 3+ = 3

Then,   μ = √ [n (n + 2)]

=√3 (3+2)


= 4 BM

(ii) No. of unpaired electrons in [Fe (H2O) 6]2+ = 4

Then, μ = √ [n (n + 2)]

 μ = √ [4 (4 + 2)]


= 5 BM

(iii) No. of unpaired electrons in [Zn (H2O) 6]2+ = 0

Hence, [Fe (H2O) 6] 2+ has the highest magnetic moment value.


Question 9.30.

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

(i) +1 (ii) +3 (iii) –1 (iv) –3


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) 4].

Therefore, the oxidation number of Co in the given complex is −1. Hence, option (iii) is correct.


 Question 9.31.

Amongst the following, the most stable complex is

(i) [Fe (H2O) 6] 3+ (ii) [Fe (NH3)6]3+ (iii) [Fe (C2O4)3]3– (iv) [FeCl6]3–


We know that the stability of a complex increases by chelation.

Therefore, the most stable complex is [Fe (C2O4)3]3−.


Question 9.32.

What will be the correct order for the wavelengths of absorption in the visible region for the following?

[Ni (NO2)6] 4–, [Ni (NH3)6]2+, [Ni (H2O) 6]2+


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:

H2O < NH3 < NO2

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

o(H2O) < ∆o(NH3) < ∆o(NO2)

Hence, the wavelengths of absorption in the visible region will be in the order:

[Ni (H2O) 6]2+ > [Ni (NH3)6]2+ > [Ni (NO2)6]4-

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