CBSE

Class–12 Subject Chemistry

NCERT Solutions
Chapter – 09
Coordination Compounds

In-text question

1. Write the formulas for the following coordination compounds:

(i) Tetraamminediaquacobalt(III) chloride

(ii) Potassium tetracyanonickelate(II)

(iii) Tris(ethane-1,2-diamine) chromium(III) chloride

(iv) Amminebromidochloridonitrito-N-platinate(II)

(v) Dichloridobis(ethane-1,2-diamine)platinum(IV) nitrate

(vi) Iron(III) hexacyanoferrate(II)

Ans. (i)

(ii)

(iii)

(vi)

(v)

(vi)

2. Write the IUPAC names of the following coordination compounds:

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(i)

(ii)

(iii)

(iv)

(v)

(vi)

Ans. (i) Hexaamminecobalt(III) chloride

(ii) Pentaamminechloridocobalt(III) chloride

(iii) Potassium hexacyanoferrate(III)

(iv) Potassium trioxalatoferrate(III)

(v) Potassium tetrachloridopalladate(II)

(vi) Diamminechlorido(methanamine)platinum(II) chloride

3. Indicate the types of isomerism exhibited by the following complexes and draw the
structures for these isomers:

(i)

(ii)

(iii)

(iv)

Ans. (i) Both geometrical (cis-, trans-) isomers for can exist. Also,

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optical isomers for cis-isomer exist.

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

(ii) Two optical isomers for exist.

Two optical isomers are possible for this structure.

(iii)

A pair of optical isomers:

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It can also show linkage isomerism.

and

It can also show ionization isomerism.

(iv) Geometrical (cis-, trans-) isomers of can exist.

4. Give evidence that and are ionization

isomers.

Ans. 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.

5. Explain on the basis of valence bond theory that ion with square

planar structure is diamagnetic and the ion with tetrahedral geometry is

paramagnetic.

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Ans. Ni is in the +2 oxidation state i.e., in configuration.

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

It now undergoes hybridization. Since all electrons are paired, it forms diamagnetic

compound.

In case of , Cl- ion is a weak field ligand. Therefore, it does not lead to the pairing

of unpaired 3d electrons. Therefore, it undergoes hybridization.

Since there are 2 unpaired electrons in this case, due to weak field ligand so, it is
paramagnetic in nature.

6. is paramagnetic while is diamagnetic though both are

tetrahedral. Why?

Ans. Though both and are tetrahedral, their magnetic characters

are different. This is due to a difference in the nature of ligands leading to paramagnetism.
is a weak field ligand and it does not cause the pairing of unpaired 3d electrons. Hence,

is paramagnetic.

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In , Ni is in the zero oxidation state i.e., it has a configuration of .

But CO is a strong field 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 hybridization.

Since no unpaired electrons are present in this case, is diamagnetic.

7. is strongly paramagnetic whereas is weakly

paramagnetic. Explain.

Ans. In both and , Fe exists in the +3 oxidation state i.e., in

configuration.

Since is a strong field ligand, it causes the pairing of unpaired electrons. Therefore,
there is only one unpaired electron left in the d-orbital.

Therefore,

=1.732 BM

On the other hand, is a weak field ligand. Therefore, it cannot cause the pairing of

electrons. This means that the number of unpaired electrons is 5.

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Therefore,

= 6 BM

Thus, it is evident that is strongly paramagnetic, while is

weakly paramagnetic.

8. Explain is an inner orbital complex whereas is an

outer orbital complex.

Ans.

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9. Predict the number of unpaired electrons in the square planar ion.

Ans.

In this complex, Pt is in the +2 state. It forms a square planar structure. This means that it
undergoes hybridization. Now, the electronic configuration of Pd(+2) is .

being a strong field ligand causes the pairing of unpaired electrons. Hence, there are

no unpaired electrons in .

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

Ans.

Hence, hexaaquo manganese (II) ion has five unpaired electrons, while hexacyano ion has
only one unpaired electron. So pairing takes place and low spin complexes are formed with
this 1 unpaired electron.

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11. Calculate the overall complex dissociation equilibrium constant for the

ion, given that for this complex is

Ans. =

The overall complex dissociation equilibrium constant is the reciprocal of the overall
stability constant, .

Chapter End Question

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

Ans. 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.

2. solution mixed with solution in 1:1 molar ratio gives the test of

ion but solution mixed with aqueous ammonia in 1:4 molar ratio does not

give the test of ion. Explain why?

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Ans.

Both the compounds i.e., and 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. breaks into

and ions. Hence, it gives a positive test for 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 does not show the test for . The ions

present in the solution of are and .

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

Ans. (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:

= cationic complex

= anionic complex

= neutral complex

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(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, ,

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, , 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 , the coordination number of the central atom

(Ni) is 4.

(vi) 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:

(a)

(b) Tetrahedral

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(v) Homoleptic complexes:

These are those complexes in which the metal ion is bound to only one kind of a donor

group. For eg: 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.:

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

Ans. 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., , Cl - etc.

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

(a) Ethane-1, 2-diamine

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(b) Oxalate ion

(c)Ambidentate ligands:

Ligands that can attach themselves to the central metal atom through two different atoms
are called ambidentate ligands. For example:

(a)

(The donor atom is N)

(The donor atom is oxygen)

(b)

(The donor atom is S)

(The donor atom is N)

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

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(i)

(ii)

(iii)

(iv)

(v)

Ans. (i)

Let the oxidation number of Co be x.

The charge on the complex is +2.

(ii)

Let the oxidation number of Pt be x.

The charge on the complex is –2.

x + 4(–1) = –2

x = +2

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(iv)

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

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

(viii) Hexaammineplatinum(IV)

(ix) Tetrabromidocuprate(II)

(x) Pentaamminenitrito-N-cobalt(III)

Ans. (i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

(ix)

(x)

7. Using IUPAC norms write the systematic names of the following:

(i)

(ii)

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(iii)

(iv)

(v)

(vi)

(vii)

(viii)

(ix)

Ans. (i) Hexaamminecobalt(III) chloride

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

(iii) Hexaaquatitanium(III) ion

(iv) Tetraamminechloridonitrito-N-Cobalt(III) chloride

(v) Hexaaquamanganese(II) ion

(vi) Tetrachloridonickelate(II) ion

(vii) Hexaamminenickel(II) chloride

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

(ix) Tetracarbonylnickel(0)

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

example of each.

Ans.

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(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:

and

Yellow form Red form

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(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.

and

(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)5SO4]Br and

[Co(NH3)5Br ]SO4.

(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.

Violet Blue-green Dark green

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

(i)

(ii)

Ans. (i) For , no geometric isomer is possible as it is a bidentate ligand.

(ii)

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Two geometrical isomers are possible.

10. Draw the structures of optical isomers of:

(i)

(ii)

(iii)

Ans. (i)

(ii)

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(iii)

11. Draw all the isomers (geometrical and optical) of:

(i)

(ii)

(iii)

Ans. (i)

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In total, three isomers are possible.

(ii)

Trans-isomers are optically inactive.

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Cis-isomers are optically active.

(iii)

12. Write all the geometrical isomers of and how many of

these will exhibit optical isomers?

Ans.

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.

13. Aqueous copper sulphate solution (blue in colour) gives:

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(i) a green precipitate with aqueous potassium fluoride, and

(ii) a bright green solution with aqueous potassium chloride

Explain these experimental results.

Ans. Aqueous exists as . It is blue in colour due to the presence

of

ions.

(i) When KF is added:

(ii) When KCl is added:

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

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 is passed through this solution?

Ans.

i.e.,

Thus, the coordination entity formed in the process is . is

a very stable complex, which does not ionize to give ions when added to water. Hence,

ions are not precipitated when is passed through the solution.

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

(i)

(ii)

(iii)

(iv)

Ans. (i) In the above coordination complex, iron exists in the +II oxidation

state.

: Electronic configuration is

Orbitals of ion:

As 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
.

hybridized orbitals of are:

6 electron pairs from ions occupy the six hybrid orbitals.

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Then,

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

(ii)

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

Orbitals of ion:

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

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

hybridized orbitals of Fe are:

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

(iii)

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

Orbitals of 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 or

hybridization.

hybridization of :

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

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

(iv)

Cobalt exists in the +3 oxidation state.

Orbitals of ion:

Again, fluoride ion is a weak field ligand. It cannot cause the pairing of the 3d electrons. As a
result, the ion will undergo hybridization.

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 hybridized orbitals of ion are:

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

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

Ans.

The splitting of the d orbitals in an octahedral field takes palce in such a way that

experience a rise in energy and form the level, while and

experience a fall in energy and form the level.

Chapter End Question

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

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Ans. A spectrochemical series is the arrangement of common ligands in the increasing order
of their field strength and 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.
Strong ligands are able to produce strong field whereas weak ligands produce weak field.

Weak ligands can not cause pairing of electrons and form high spin complexes whereas
strong field ligands can cause pairing of electrons and thus form low spin complexes.

18. What is crystal field splitting energy? How does the magnitude of decide the

actual configuration of d-orbitals in a coordination entity?

Ans. The difference between energies of two sets of d-orbitals is called crystal field splitting
energy i.e. delta note. The degenerate d-orbitals (in a spherical field environment) split into
two levels i.e., and in the presence of ligands. The splitting of these degenerate levels

due to the presence of ligands is called the crystal-field splitting while the energy difference
between the two levels ( and ) is called the crystal-field splitting energy. It is denoted

by .

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

It can enter the orbital (giving rise to like electronic configuration) or the pairing

of the electrons can take place in the orbitals (giving rise to like electronic

configuration). If the value of a ligand is less than the pairing energy (P), then the

electrons enter the orbital. On the other hand, if the value of a ligand is more than the

pairing energy (P), then the electrons enter the orbital.

Although orbital splitting energies are not sufficiently large to force pairing so low spin
configuration are rarely observed.

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19. is paramagnetic while is diamagnetic. Explain why?

Ans. Cr is in the +3 oxidation state i.e., configuration. Also, is a weak field ligand

that does not cause the pairing of the electrons in the 3d orbital.

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

unpaired. Hence, it is paramagnetic in nature.

In , Ni exists in the +2 oxidation state i.e., configuration.

is a strong field ligand. It causes the pairing of the 3d orbital electrons. Then,
undergoes hybridization.

As there are no unpaired electrons, it is diamagnetic with square planar shape.

20. A solution of is green but a solution of is colourless.

Explain.

Ans. In , is a weak field ligand. Therefore, there are unpaired

electrons in . 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,

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 light is absorbed from visible region and complementary green colour is radiated. So,

is green coloured.

In , the electrons are all paired as is a strong field ligand. Therefore, d-

d transition is not possible in . Hence, it is colourless.

21. and are of different colours in dilute solutions. Why?

Ans. 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 and , the colour differs because there is a difference

in the CFSE. Now, 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 from visible region occurs
for the intra d-d transition and complementary colours are radiated. Hence, the transmitted
colour also differs.

22. Discuss the nature of bonding in metal carbonyls.

Ans. The metal-carbon bonds in metal carbonyls have both and characters. bond is
formed when the carbonyl carbon donates a lone pair of electrons to the vacant orbital of the
metal. A bond is formed by the donation of a pair of electrons from the filled metal d
orbital into the vacant anti-bonding orbital (also known as back bonding of the carbonyl
group). The bond strengthens the 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.

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

(i)

(ii) cis-

(iii)

(iv)

Ans. (i)

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 is .

(ii) cis-

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

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The d orbital occupation for is .

(iii)

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 is .

(iv)

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 .

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)

(ii)

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(iii)

(iv)

(v)

Ans. (i) Potassium diaquadioxalatochromate (III) trihydrate.

Oxidation state of chromium = 3

Electronic configuration:

Coordination number = 6

Shape: octahedral

Stereochemistry:

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Magnetic moment,

(ii)

IUPAC name: Pentaamminechloridocobalt(III) chloride

Oxidation state of Co = +3

Coordination number = 6

Shape: octahedral.

Electronic configuration: .

Stereochemistry:

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Magnetic Moment = 0

(iii)

IUPAC name: Trichloridotripyridinechromium (III)

Oxidation state of chromium = +3

Electronic configuration for

Coordination number = 6

Shape: octahedral.

Stereochemistry:

Both isomers are optically active. Therefore, a total of 4 isomers exist.

Magnetic moment,

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(iv)

IUPAC name: Caesium tetrachloroferrate (III)

Oxidation state of Fe = +3

Electronic configuration of

Coordination number = 4

Shape: tetrahedral

Stereochemistry: optically inactive

Magnetic moment:

(v)

Potassium hexacyanomanganate(II)

Oxidation state of manganese = +2

Electronic configuration:

Coordination number = 6

Shape: octahedral.

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Streochemistry: optically inactive

Magnetic moment,

= 1.732

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

Ans. 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.

For this reaction, the greater the value of the stability constant, the greater is the proportion
of 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:

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(a) Charge on the central metal ion: The greater the charge on the central metal ion, the
greater is the stability of the complex.

(b) Basic nature of the ligand: A more basic ligand will form a more stable complex.

(c) Presence of chelate rings: Chelation increases the stability of complexes.

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

Ans. 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:

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

(i) biological system

(ii) medicinal chemistry

(iii) analytical chemistry

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(iv) extraction/metallurgy of metals

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

28. How many ions are produced from the complex in solution?

(i) 6

(ii) 4

(iii) 3

(iv) 2

Ans. (iii) The given complex can be written as .

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Thus, along with two Cl- ions are produced.

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

(i)

(ii)

(iii)

Ans. (i) No. of unpaired electrons in = 3

Then,

(ii) No. of unpaired electrons in = 4

Then,

(iii) No. of unpaired electrons in

Hence, has the highest magnetic moment value.

30. The oxidation number of cobalt in is

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(i) +1

(ii) +3

(iii) -1

(iv) -3

Ans. We know that CO is a neutral ligand and K carries a charge of +1.

Therefore, the complex can be written as . Therefore, the oxidation

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

31. Amongst the following, the most stable complex is

(i)

(ii)

(iii)

(iv)

Ans. We know that the stability of a complex increases by chelation. Therefore, the most

stable complex is .

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32. What will be the correct order for the wavelengths of absorption in the visible

region for the following:

Ans. 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:

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

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

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