NCERT Solutions Class 12th Chemistry. FREE Flip-BOOK by Study Innovations. 472 Pages

Class XII Chapter 9 – Coordination Compounds Chemistry

Trans-isomers are optically inactive.
Cis-isomers are optically active.
(iii) [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?
Answer
[Pt(NH3)(Br)(Cl)(py)

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Class XII Chapter 9 – Coordination Compounds Chemistry

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.
Answer
Aqueous CuSO4 exists as [Cu(H2O)4]SO4. It is blue in colour due to the presence of
[Cu[H2O)4]2+ ions.
(i) When KF is added:

(ii) When KCl is added:

In both these cases, the weak field ligand water is replaced by the F− and Cl− 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?
Answer

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Class XII Chapter 9 – Coordination Compounds Chemistry

i.e.,

Thus, the coordination entity formed in the process is K2[Cu(CN)4]. is a

very stable complex, which does not ionize to give Cu2+ ions when added to water.

Hence, Cu2+ ions are not precipitated when H2S(g) is passed through the solution.

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−
Answer
(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 hybrid d2sp3orbitals.
Then,

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Class XII Chapter 9 – Coordination Compounds Chemistry

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 ion:

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:

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+:

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Class XII Chapter 9 – Coordination Compounds Chemistry

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.
Answer

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Class XII Chapter 9 – Coordination Compounds Chemistry

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 eg level, while dxy, dyzand dzx experience a
fall in energy and form the t2g level.

Question 9.17:
What is spectrochemical series? Explain the difference between a weak field ligand and a
strong field ligand.
Answer
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− < C2O42− ∼ H2O < NCS− ∼ H− < CN− <
NH3 < en ∼ SO32− < NO2− < phen < CO

Question 9.18:
What is crystal field splitting energy? How does the magnitude of ∆o decide the actual
configuration of d-orbitals in a coordination entity?
Answer
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.

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Class XII Chapter 10 – Haloalkanes and Haloarenes Chemistry

Name the following halides according to IUPAC system and classify them as alkyl, allyl,
benzyl (primary, secondary, tertiary), vinyl or aryl halides:

(CH3)2CHCH(Cl)CH3
CH3CH2CH(CH3)CH(C2H5)Cl
CH3CH2C(CH3)2CH2I
(CH3)3CCH2CH(Br)C6H5
CH3CH(CH3)CH(Br)CH3
CH3C(C2H5)2CH2Br

CH3C(Cl)(C2H5)CH2CH3
CH3CH=C(Cl)CH2CH(CH3)2
CH3CH=CHC(Br)(CH3)2
-ClC6H4CH2CH(CH3)2
-ClCH2C6H4CH2C(CH3)3
-Br-C6H4CH(CH3)CH2CH3
Answer

2−Chloro−3−methylbutane
(Secondary alkyl halide)

3−Chloro−4−methyhexane
(Secondary alkyl halide)

1−Iodo−2, 2 −dimethylbutane

Class XII Chapter 10 – Haloalkanes and Haloarenes Chemistry

(Primary alkyl halide)

1−Bromo−3, 3−dimethyl−1−phenylbutane
(Secondary benzyl halide)

2−Bromo−3−methylbutane
(Secondary alkyl halide)

1−Bromo−2−ethyl−2−methylbutane
(Primary alkyl halide)

3−Chloro−3−methylpentane
(Tertiary alkyl halide)

Class XII Chapter 10 – Haloalkanes and Haloarenes Chemistry

3−Chloro−5−methylhex−2−ene
(Vinyl halide)

4−Bromo−4−methylpent−2−ene
(Allyl halide)

1−Chloro−4−(2−methylpropyl) benzene
(Aryl halide)

1−Chloromethyl−3−(2, 2−dimethylpropyl) benzene
(Primary benzyl halide)

1−Bromo−2−(1−methylpropyl) benzene
(Aryl halide)

Class XII Chapter 10 – Haloalkanes and Haloarenes Chemistry

Give the IUPAC names of the following compounds:
CH3CH(Cl)CH(Br)CH3
CHF2CBrClF
ClCH2C≡CCH2Br
(CCl3)3CCl
CH3C( -ClC6H4)2CH(Br)CH3
(CH3)3CCH=CClC6H4I-

Answer

2−Bromo−3−chlorobutane

1−Bromo−1−chloro−1, 2, 2−trifluoroethane
1−Bromo−4−chlorobut−2−yne

2−(Trichloromethyl)−1,1,1,2,3,3,3−heptachloropropane

Class XII Chapter 10 – Haloalkanes and Haloarenes Chemistry

2−Bromo−3, 3−bis(4−chlorophenyl) butane

1−chloro−1−(4−iodophenyl)−3, 3−dimethylbut−1−ene

Write the structures of the following organic halogen compounds.
2-Chloro-3-methylpentane
-Bromochlorobenzene
1-Chloro-4-ethylcyclohexane
2-(2-Chlorophenyl)-1-iodooctane
Perfluorobenzene
4-tert-Butyl-3-iodoheptane
1-Bromo-4-sec-butyl-2-methylbenzene
1,4-Dibromobut-2-ene

Answer

2-Chloro-3-methylpentane

Class XII Chapter 10 – Haloalkanes and Haloarenes Chemistry

-Bromochlorobenzene

1-Chloro-4-ethylcyclohexane
2-(2-Chlorophenyl)-1-iodooctane
Perfluorobenzene

4-Tert-Butyl-3-iodoheptane

Class XII Chapter 10 – Haloalkanes and Haloarenes Chemistry

1-Bromo-4-sec-butyl-2-methylbenzene

1,4-Dibromobut-2-ene

Which one of the following has the highest dipole moment?
CH2Cl2
CHCl3
CCl4

Answer

Dichlormethane (CH2Cl2)
µ = 1.60D

Chloroform (CHCl3)
µ = 1.08D

Class XII Chapter 10 – Haloalkanes and Haloarenes Chemistry

Carbon tetrachloride (CCl4)
µ = 0D
CCl4 is a symmetrical molecule. Therefore, the dipole moments of all four C−Cl bonds
cancel each other. Hence, its resultant dipole moment is zero.
As shown in the above figure, in CHCl3, the resultant of dipole moments of two C−Cl
bonds is opposed by the resultant of dipole moments of one C−H bond and one C−Cl
bond. Since the resultant of one C−H bond and one C−Cl bond dipole moments is
smaller than two C−Cl bonds, the opposition is to a small extent. As a result, CHCl3 has
a small dipole moment of 1.08 D.
On the other hand, in case of CH2Cl2, the resultant of the dipole moments of two C−Cl
bonds is strengthened by the resultant of the dipole moments of two C−H bonds. As a
result, CH2Cl2 has a higher dipole moment of 1.60 D than CHCl3 i.e., CH2Cl2 has the
highest dipole moment.
Hence, the given compounds can be arranged in the increasing order of their dipole
moments as:
CCl4 < CHCl3 < CH2Cl2

A hydrocarbon C5H10 does not react with chlorine in dark but gives a single monochloro
compound C5H9Cl in bright sunlight. Identify the hydrocarbon.
Answer
A hydrocarbon with the molecular formula, C5H10 belongs to the group with a general
molecular formula CnH2n. Therefore, it may either be an alkene or a cycloalkane.
Since hydrocarbon does not react with chlorine in the dark, it cannot be an alkene. Thus,
it should be a cycloalkane.
Further, the hydrocarbon gives a single monochloro compound, C5H9Cl by reacting with
chlorine in bright sunlight. Since a single monochloro compound is formed, the
hydrocarbon must contain H−atoms that are all equivalent. Also, as all H−atoms of a
cycloalkane are equivalent, the hydrocarbon must be a cycloalkane. Hence, the said
compound is cyclopentane.

Cyclopentane (C5H10)

Class XII Chapter 10 – Haloalkanes and Haloarenes Chemistry

The reactions involved in the question are:

Write the isomers of the compound having formula C4H9Br.
Answer
There are four isomers of the compound having the formula C4H9Br. These isomers are
given below.

1−Bromobutane

2−Bromobutane

1−Bromo−2−methylpropane

2−Bromo−2−methylpropane
Write the equations for the preparation of 1−iodobutane from

Class XII Chapter 10 – Haloalkanes and Haloarenes Chemistry

1-butanol
1-chlorobutane
but-1-ene.
Answer

What are ambident nucleophiles? Explain with an example.
Answer
Ambident nucleophiles are nucleophiles having two nucleophilic sites. Thus, ambident
nucleophiles have two sites through which they can attack.
For example, nitrite ion is an ambident nucleophile.

Nitrite ion can attack through oxygen resulting in the formation of alkyl nitrites. Also, it
can attack through nitrogen resulting in the formation of nitroalkanes.

Class XII Chapter 10 – Haloalkanes and Haloarenes Chemistry

Which compound in each of the following pairs will react faster in SN2 reaction with OH−?
CH3Br or CH3I
(CH3)3CCl or CH3Cl

Answer
In the SN2 mechanism, the reactivity of halides for the same alkyl group increases in

the order. This happens because as the size increases, the halide ion becomes a better
leaving group.
R−F << R−Cl < R−Br < R−I
Therefore, CH3I will react faster than CH3Br in SN2 reactions with OH−.

The SN2 mechanism involves the attack of the nucleophile at the atom bearing the
leaving group. But, in case of (CH3)3CCl, the attack of the nucleophile at the carbon atom
is hindered because of the presence of bulky substituents on that carbon atom bearing
the leaving group. On the other hand, there are no bulky substituents on the carbon
atom bearing the leaving group in CH3Cl. Hence, CH3Cl reacts faster than (CH3)3CCl in
SN2 reaction with OH−.

Predict all the alkenes that would be formed by dehydrohalogenation of the following
halides with sodium ethoxide in ethanol and identify the major alkene:

1-Bromo-1-methylcyclohexane
2-Chloro-2-methylbutane
2,2,3-Trimethyl-3-bromopentane.
Answer

Class XII Chapter 10 – Haloalkanes and Haloarenes Chemistry

1−bromo−1−methylcyclohexane
In the given compound, all β-hydrogen atoms are equivalent. Thus, dehydrohalogenation
of this compound gives only one alkene.

In the given compound, there are two different sets of equivalent β-hydrogen atoms
labelled as and . Thus, dehydrohalogenation of the compound yields two alkenes.

Saytzeff’s rule implies that in dehydrohalogenation reactions, the alkene having a
greater number of alkyl groups attached to a doubly bonded carbon atoms is preferably
produced.
Therefore, alkene (I) i.e., 2-methylbut-2-ene is the major product in this reaction.

2,2,3-Trimethyl-3-bromopentane
In the given compound, there are two different sets of equivalent β-hydrogen atoms
labelled as and . Thus, dehydrohalogenation of the compound yields two alkenes.

Class XII Chapter 10 – Haloalkanes and Haloarenes Chemistry

According to Saytzeff’s rule, in dehydrohalogenation reactions, the alkene having a
greater number of alkyl groups attached to the doubly bonded carbon atom is preferably
formed.
Hence, alkene (I) i.e., 3,4,4-trimethylpent-2-ene is the major product in this reaction.

How will you bring about the following conversions?
Ethanol to but-1-yne
Ethane to bromoethene
Propene to 1-nitropropane
Toluene to benzyl alcohol
Propene to propyne
Ethanol to ethyl fluoride
Bromomethane to propanone
But-1-ene to but-2-ene
1-Chlorobutane to n-octane
Benzene to biphenyl.

Answer

Class XII Chapter 10 – Haloalkanes and Haloarenes Chemistry

Class XII Chapter 10 – Haloalkanes and Haloarenes Chemistry

Class XII Chapter 10 – Haloalkanes and Haloarenes Chemistry

Explain why
the dipole moment of chlorobenzene is lower than that of cyclohexyl chloride?
alkyl halides, though polar, are immiscible with water?
Grignard reagents should be prepared under anhydrous conditions?

Answer

In chlorobenzene, the Cl-atom is linked to a 2 hybridized carbon atom. In cyclohexyl
chloride, the Cl-atom is linked to a 3 hybridized carbon atom. Now, 2 hybridized
carbon has more s-character than 3 hybridized carbon atom. Therefore, the former is
more electronegative than the latter. Therefore, the density of electrons of C−Cl bond
near the Cl-atom is less in chlorobenzene than in cydohexyl chloride.
Moreover, the −R effect of the benzene ring of chlorobenzene decreases the electron
density of the C−Cl bond near the Cl-atom. As a result, the polarity of the C−Cl bond in
chlorobenzene decreases. Hence, the dipole moment of chlorobenzene is lower than that
of cyclohexyl chloride.

To be miscible with water, the solute-water force of attraction must be stronger than
the solute-solute and water-water forces of attraction. Alkyl halides are polar molecules
and so held together by dipole-dipole interactions. Similarly, strong H-bonds exist
between the water molecules. The new force of attraction between the alkyl halides and
water molecules is weaker than the alkyl halide-alkyl halide and water-water forces of
attraction. Hence, alkyl halides (though polar) are immiscible with water.

Grignard reagents are very reactive. In the presence of moisture, they react to give
alkanes.

Class XII Chapter 10 – Haloalkanes and Haloarenes Chemistry

Therefore, Grignard reagents should be prepared under anhydrous conditions.

Give the uses of freon 12, DDT, carbon tetrachloride and iodoform.
Answer

Freon-12 (dichlorodifluoromethane, CF2Cl2) is commonly known as CFC. It is used as a
refrigerant in refrigerators and air conditioners. It is also used in aerosol spray
propellants such as body sprays, hair sprays, etc. However, it damages the ozone layer.
Hence, its manufacture was banned in the United States and many other countries in
1994.

DDT ( −dichlorodiphenyltrichloroethane) is one of the best known insecticides. It is
very effective against mosquitoes and lice. But due its harmful effects, it was banned in
the United States in 1973.

It is used for manufacturing refrigerants and propellants for aerosol cans.
It is used as feedstock in the synthesis of chlorofluorocarbons and other chemicals.
It is used as a solvent in the manufacture of pharmaceutical products.
Until the mid 1960’s, carbon tetrachloride was widely used as a cleaning fluid, a
degreasing agent in industries, a spot reamer in homes, and a fire extinguisher.

Iodoform was used earlier as an antiseptic, but now it has been replaced by other
formulations-containing iodine-due to its objectionable smell. The antiseptic property of
iodoform is only due to the liberation of free iodine when it comes in contact with the
skin.

Write the structure of the major organic product in each of the following reactions:

Class XII Chapter 10 – Haloalkanes and Haloarenes Chemistry

Answer

Class XII Chapter 10 – Haloalkanes and Haloarenes Chemistry

Write the mechanism of the following reaction:

Answer
The given reaction is:

The given reaction is an SN2 reaction. In this reaction, CN− acts as the nucleophile and
attacks the carbon atom to which Br is attached. CN− ion is an ambident nucleophile and
can attack through both C and N. In this case, it attacks through the C-atom.

Arrange the compounds of each set in order of reactivity towards SN2 displacement:
2-Bromo-2-methylbutane, 1-Bromopentane, 2-Bromopentane
1-Bromo-3-methylbutane, 2-Bromo-2-methylbutane, 3-Bromo-2- methylbutane
1-Bromobutane, 1-Bromo-2,2-dimethylpropane, 1-Bromo-2-methylbutane, 1-

Bromo-3-methylbutane.
Answer

Class XII Chapter 10 – Haloalkanes and Haloarenes Chemistry

An SN2 reaction involves the approaching of the nucleophile to the carbon atom to which
the leaving group is attached. When the nucleophile is sterically hindered, then the
reactivity towards SN2 displacement decreases. Due to the presence of substituents,
hindrance to the approaching nucleophile increases in the following order.
1-Bromopentane < 2-bromopentane < 2-Bromo-2-methylbutane
Hence, the increasing order of reactivity towards SN2 displacement is:
2-Bromo-2-methylbutane < 2-Bromopentane < 1-Bromopentane

Since steric hindrance in alkyl halides increases in the order of 1° < 2° < 3°, the
increasing order of reactivity towards SN2 displacement is
3° < 2° < 1°.
Hence, the given set of compounds can be arranged in the increasing order of their
reactivity towards SN2 displacement as:
2-Bromo-2-methylbutane < 2-Bromo-3-methylbutane < 1-Bromo-3-methylbutane
[2-Bromo-3-methylbutane is incorrectly given in NCERT]

Class XII Chapter 10 – Haloalkanes and Haloarenes Chemistry

The steric hindrance to the nucleophile in the SN2 mechanism increases with a decrease
in the distance of the substituents from the atom containing the leaving group. Further,
the steric hindrance increases with an increase in the number of substituents. Therefore,
the increasing order of steric hindrances in the given compounds is as below:
1-Bromobutane < 1-Bromo-3-methylbutane < 1-Bromo-2-methylbutane
< 1-Bromo-2, 2-dimethylpropane
Hence, the increasing order of reactivity of the given compounds towards SN2
displacement is:
1-Bromo-2, 2-dimethylpropane < 1-Bromo-2-methylbutane < 1-Bromo-3- methylbutane
< 1-Bromobutane

Out of C6H5CH2Cl and C6H5CHClC6H5, which is more easily hydrolysed by aqueous KOH?
Answer

Hydrolysis by aqueous KOH proceeds through the formation of carbocation. If
carbocation is stable, then the compound is easily hydrolyzed by aqueous KOH. Now,

forms 1°-carbocation, while forms 2°-carbocation, which is

Class XII Chapter 10 – Haloalkanes and Haloarenes Chemistry

more stable than 1°-carbocation. Hence, is hydrolyzed more easily than
by aqueous KOH.

-Dichlorobenzene has higher m.p. and lower solubility than those of - and
-isomers. Discuss.
Answer

-Dichlorobenzene is more symmetrical than -and -isomers. For this reason, it fits
more closely than -and -isomers in the crystal lattice. Therefore, more energy is
required to break the crystal lattice of -dichlorobenzene. As a result, -dichlorobenzene
has a higher melting point and lower solubility than -and -isomers.

How the following conversions can be carried out?
Propene to propan-1-ol
Ethanol to but-1-yne
1-Bromopropane to 2-bromopropane
Toluene to benzyl alcohol
Benzene to 4-bromonitrobenzene
Benzyl alcohol to 2-phenylethanoic acid
Ethanol to propanenitrile
Aniline to chlorobenzene
2-Chlorobutane to 3, 4-dimethylhexane
2-Methyl-1-propene to 2-chloro-2-methylpropane
Ethyl chloride to propanoic acid
But-1-ene to n-butyliodide
2-Chloropropane to 1-propanol

Class XII Chapter 10 – Haloalkanes and Haloarenes Chemistry

Isopropyl alcohol to iodoform
Chlorobenzene to -nitrophenol
2-Bromopropane to 1-bromopropane

Chloroethane to butane
Benzene to diphenyl

-Butyl bromide to isobutyl bromide
Aniline to phenylisocyanide
Answer

Class XII Chapter 10 – Haloalkanes and Haloarenes Chemistry