Page 246 Coordination Compounds Chap 5

CHAPTER 5
Coordination Compounds

SUMMARY Hexadentate ligand has six donar atoms example

EDTA 4- .
Ambidentate Ligand
Unidentate ligands which have more than one
1. DEFINITION OF SOME IMPORTANT TERMS
coordination atoms are called ambidentate ligands
Coordination Compounds or Complex Compounds example NO -2 .
Those compounds in which the central metal atom
is linked to ions or neutral molecules by coordinate
bond are called coordinate compounds or complex e.g.
K 4 [Fe (CN) 6].
Complex Ion
Chelate
Ion formed by coordinate bond between central metal
atom and donor is called complex ion. When a bidentate or a polydentate ligand is attached
There are three types of complexes: by two or more donar atoms to the central atom
forming a ring, the ligand is called chelate ligand
1. Cationic complex e.g., [Co (NH 3) 6] 3+
and the complex is called chelate the effect is called
2. Anionic complex e.g., [Ag (CN) 2] -
chelate effect.
3. Neutral complex e.g., [Ni (CO) 4].
Coordination Number
Double Salts
Total number of coordinate bonds formed with the
These salt are formed by the combination of two simple central metal atom or ion by the ligands.
salts in equimolar ratio. They lose then identity in the
solution e.g., FeSO 4 (NH 4) 2 SO 4 $ 6H 2 O . Coordination Ion
The ionizable group which are written outside the
FeSO 4 (NH 4) 2 SO 4 $ 6H 2 O $ Fe2+ + 2NH +4 + 2SO 24- + 6H 2 O
bracket are called counter ion.
(Mohr's salt)
Complex compound do not lose their identity in the
solution.
K 4 [Fe (CN) 6] $ 4K+ + [Fe (CN) 6] 4-
Ligand
The donor species which donate a pair of electrons to
the metal atom is called ligand.
Denticity Oxidation Number
The number of coordinating groups present in a ligand Oxidation number or oxidation state of the central
is called the denticity of the ligand. metal atom is calculated from the charge on the
Monodentate ligand have one donar atoms example complex ion. [Fe (CN) 6] 4-
NH 3, H 2 O . x + 6 (- 1) = - 4
Bidentate ligand can donate two lone pair of electrons x =+ 2
example : en, ox.
Oxidation state of Fe = + 2
Page 248 Coordination Compounds Chap 5

It is a diamagnetic complex due to absence of

unpaired electron.
8. Formation of [CoF6] 3-

Primary valencies are represented by dotted lines

and secondary valencies by solid lines.
2. Limitations of Werner’s theory:
(i) Why only certain elements form coordination
compounds?
(ii) Why the coordination sphere has a definite
geometry?
It is strongly paramagnetic due to presence of five
(iii) Why do coordination compounds possess
unpaired electrons.
definite magnetic and optical properties?
9. Drawbacks of valence bond theory:
3. Valence Bond Theory:
(i) It cannot explain why some complexes of a
(i) Metal ligand bond is formed by the donation
metal ion in a particular oxidation state are
of pairs of electron by ligand to the metal
inner orbital complexes while some other
atom.
complexes of the same metal ion in the some
(ii) Metal atom must possess vacant orbitals to oxidation state are outer spin orbital complex.
accommodate these electrons. These orbitals
(ii) It could not give any explanation for the
are obtained by hybridisation.
colour of the complex.
(iii) Sometimes the unpaired (n-1) d electrons pair
(iii) It does not distinguish between weak and
up prior to hybridization for making some (n-
strong ligands.
1) d orbitals vacant.
(iv) It does not give a clear explanation of
(iv) Metal-ligand bonds are formed by the overlap
thermodynamic stabilities of coordination
of these orbitals with ligand orbitals, these
compounds.
bonds are of equal strength and directional in
nature. 5. CRYSTAL FIELD THEORY
4. Octahedral complexes are formed by d2 sp3 or
sp3 d2 hybridisation. Square planar complexes The main postulates of crystal field theory are as
are formed by dsp2 hybridisation, tetrahedral follows:
complexes are formed by sp3 hybridisation. (i) Metal ligand bonds are ionic having electrostatic
5. d2 sp3 hybridized complex involve the inner (n-1) interaction, similar to ions in a crystal, therefore
d orbitals these complexes are called inner orbital named as crystal field theory (CFT).
or low spin complexes. (ii) Ligand is treated as a point of negative charge.
Examples : [Co (NH 3) 6] 3+ , [Fe (CN) 6] 3- The arrangement of the ligands around the central
6. sp3 d2 hybridized complex involve the outer (nd) ion is such that the repulsion between them are
orbitals, these complexes are called outer orbital minimum.
or high spin complexes. (iii) In free transition metal ion all the five d-orbitals
Example : [Fe (H 2 O) 6] 3+ , [(CoF6)] 3- have equal energies i.e. they are degenerate. Due
7. Formation of [Co (NH 3) 6] 3+ to ligand the degeneracy is split because those
orbitals which have lobes along the axes towards
ligands fell greater repulsion. The splitting of the
degenerate level due to presence of ligand is called
crystal field splitting.
Page 250 Coordination Compounds Chap 5

5. IUPAC name of H 2 6PtCl 6@ is – The no. of unpaired electrons in 6Ni ^COh4@ = 0 . So,
(a) Hydrogen hexachloro platinate (IV) the magnetic moment = 0 .
The number of unpaired electrons in 7Fe ^CNh6A = 0
-4
(b) Hydrogen hexachloro platinate (II)
(c) Hydrogen hexa chlorido Pt (IV) So, the magnetic moment = 0 .
Therefore, the highest magnetic moment is present for
(d) Hydrogen hexa chlorido Pt (II)
the ion 6Fe ^H 2 Oh6@Cl 2 .
Ans : DELHI 2018
Thus (b) is correct option.
IUPAC name of H 2 6PtCl 6@ is hydrogen hexachloro
platinate (IV). 9. Which of the following compounds has tetrahedral
Thus (a) is correct option. geometry?
(a) 6Ni ^CNh4@2- (b) 6Pd ^CNh4@2-
6. Mercuric chloride reacts with ammonia gas and forms
white precipitate. The molecular formula of white (c) 6PdCl 4@2- (d) 6NiCl 4@2-
precipitate is – Ans : DELHI 2016, SQP 2012
(a) HCl 2 $ 2NH 3 (b) Hg ^NH 3h2 Cl 2 3
Coordinate compounds with Sp hybridization for
(c) Hg ^NH 2h Cl 2 (d) Hg ^NH 2h Cl central metal can show tetrahedral geometry mostly
with weak-ligand like Cl-ligand can form such
Ans : COMP 2018
complexes.
Mercury(II) chloride when reacted with ammonia
Since, They do not go for go for pairing.
gas or ammonium chloride produces a fusible white
In [Nicl 4] 2- , Ni is in Ni2+
precipitate of diammine mercury(II) chloride.
Outer most electronic configuration for Ni2+ is-
HgCl 2 + 2NH 3 ^g h $ Hg ^NH 3h2 Cl 2
fusible white precipitate
Thus (b) is correct option.
7. The EAN of cobalt in the complex ion 6Co ^enh2 Cl 2@5
is :
(a) 27 (b) 36 Chlorine is a weak-ligand, d-electron of Ni2+ do not go
for pairing and [NiCl 4] 2- show sp3 hybridization and
(c) 33 (d) 35 tetrahedral geometry.
Ans : SQP 2017 Thus (d) is correct option.
EAN of cobalt in complex [CO (en) 2 Cl 2] + .
10. What is the coordination number of sodium in sodium
Atomic number = 27 oxide ^Na 2 Oh ?
Ground electronic configuration = 3d 7 4s 2 (a) 6 (b) 4
Co+3 electronic configuration = 3d 6 4s0 (c) 8 (d) 2
EAN = 27 - (+ 3) + (4 # 2) + (2 # 2) = 36 Ans : DELHI 2012, OD 2010
Thus (b) is correct option. The coordination number of Na in Na 2 O is 4. Na+
8. Which complex has maximum paramagnetic moment ion is surrounded by four oxide ions and O 2- ion is
value amongst the following – surrounded by eight Na+ ions, hence the coordination
number of Na in Na 2 O is 4.
(a) 6Cr ^H 2 Oh6@3+ (b) 6Fe ^H 2 Oh6@Cl 2
Thus (b) is correct option.
(c) 6Fe ^CNh6@4- (d) 6Ni ^COh4@
11. The oxidation state of nickel in 6Ni ^COh4@ is :
Ans : SQP 2018 (a) 1 (b) 0
Magnetic moment ^m h = n ^n + 2h (c) 2 (d) 3
where n = no. of unpaired electrons Ans : OD 2009
The no. of unpaired electrons in 6Cr ^H 2 Oh6@ = 3 . So,
3+
Oxidation number is the charge left on the central
the magnetic moment atom when all the bonding pair of electrons are
= 3 (3 + 2) = 15 removed. Oxidation state of metal atom in metal
carbonyl is always zero. Here also oxidation state of
The no. of unpaired electrons in 6Fe ^H 2 Oh6@Cl 2 = 4 . Ni is zero.
So, the magnetic moment
Thus (b) is correct option.
= 2 ^4 + 2h = 12
Page 252 Coordination Compounds Chap 5

20. The complexes 6Co ^NH 3h6@6Cr ^CNh6@ and 24. The coordination number of a central metal atom in a
6Cr ^NH 3h6@6Co ^CNh6@ are the examples of which type complex is determined by
of isomerism? (a) the number of ligands around a metal ionbonded
(a) Linkage isomerism by sigma and pi-bonds both.
(b) Ionisation isomerism (b) the number of ligands around a metal ion bonded
by pi-bonds.
(c) Coordination isomerism
(c) the number of ligands around a metal ion bonded
(d) Geometrical isomerism
by sigma bonds.
Ans : DELHI 2004
(d) the number of only anionic ligands bonded to the
Coordination isomerism occurs when cationic and
metal ion.
anionic complexes of different metal ions are present
in a salt. Interchange of ligand between the complexes Ans : COMP 2017

give isomers e.g. The coordination number of central metal atom in a

6Co ^NH 3h6@6Cr ^CNh6@ is an isomer of 6Co ^CNh6@6Cr ^NH 3h6@ complex is equal to number of monovalent ligands,
Thus (c) is correct option. twice the number of bidentate ligands and so on,
around the metal ion bonded by coordinate bonds.
21. Which of the following pairs represent linkage isomers? Hence coordination number = no. of s bonds formed
(a) 7Pd ^P Ph 3h2 ^NCSh2A and 7Pd ^P Ph 3h2 ^SCNh2A by metals with ligands.
Thus (c) is correct option.
(b) 7Co ^NH 3h5 NO 3A SO 4 and 7Co ^NH 3h5 SO 4A NO 3
Coordination compounds have great importance
(c) 6PtCl 2 ^NH 3h4@ Br2 and 6PtBr2 ^NH 3h4@Cl 2
25.
in biological systems. In this context which of the
(d) 6Cu ^NH 3h4@6PtCl 4@ and 6Pt ^NH 3h4@6CuCl 4@ following statements is incorrect?
Ans : FOREIGN 2007 (a) Cynacobalamin is B 12 and contains cobalt
-
The SCN ion can coordinate through S or N atom (b) Haemoglobin is the red pigment of blood and
giving rise to linkage isomerism contains irons
M ! SCN thiocyanato (c) Chrolophylls are green pigments in plants and
contain calcium
M ! NCS isothiocyanato
(d) Carboxyperptidase - A is an enzyme and contains
Thus (a) is correct option.
zinc.
22. Which of the following is common donor atom in ligands?
Ans : SQP 2000
(a) Arsenic (b) Nitrogen
The chlorophyll molecule plays an important role in
(c) Oxygen (d) Both (b) and (c) photosynthesis, contain porphyrin ring and the metal
Ans : OD 2012, SQP 2007 Mg not Ca.
In the formation of a coordinate bond, the ligands donate Thus (c) is correct option.
a pair of electrons to the metal atom. Further nitrogen and
oxygen has great tendency to donate the pair of electrons 26. Among the following, the compound that is both
in most of the compounds. Therefore both nitrogen and paramagnetic and coloured is
oxygen are common donor atoms in ligands. (a) K 2 Cr2 O 7 (b) ^NH 4h2 ^TiCl 6h
Thus (d) is correct option. (c) CoSO 4 (d) K 3 6Cu ^CNh4@
23. Which of the following may be considered to be an Ans : DELHI 2003
organometallic compound? In ^NH 4h2 ^TiCl 6h , Ti 4+ ^3d0 4s0h has no unpaired
(a) Nickel tetracarbonyl (b) Chlorophyll electrons.
(c) K 3 6Fe ^C 2 O 4h3@ (d) 6Co ^enh3@Cl 3 In K 2 Cr2 O 7 , Cr6+ ^3p6 4d0h has no unpaired electrons.
Ans : SQP 2015
In CoSO 4 , Co2+ ^d7h has three unpaired electrons in d
Organometallic compounds are those compounds -orbitals so it is both paramagnetic and coloured.
in which there is a bond which involve metal. In In K 3 6Cu ^CNh4@ , Cu+ ^3d10h , no unpaired electron.
chlorophyll there is bond involving metal Mg. Thus (c) is correct option.
Thus (b) is correct option.
Page 254 Coordination Compounds Chap 5

33. Hybridisation of Ag in the linear complex 6Ag ^NH 3h2@+ 36. The number of unpaired electrons in the complex ion
is 6CoF6@3- is (Atomic no.: Co = 27)
(a) dsp2 (b) sp (a) zero (b) 2
(c) sp2 (d) sp3 (c) 3 (d) 4
Ans : OD 2017 Ans : DELHI 2015

In 6Ag ^NH 3h2@+ ground state configuration of Co here is in + 3 oxidation state

Ag = 4d10 5s1 excited state configuration of Ag Unpaired electrons = 4 and sp3 d2 hybridisation and
octahedral shape.
= 4d10 5s0 .
Thus (d) is correct option.
sp hybridisation in Ag+ ion.
37. Ammonia forms the complex ion [Cu (NH 3) 4] 2+ with
copper ions in alkaline solutions but not in acidic
solutions. What is the reason for it?
(a) In acidic solutions protons coordinate with
ammonia molecules forming NH +4 ions and NH 3
molecules are not available
(b) In alkaline solutions insoluble Cu (OH) 2 is
The two sp hybrid orbitals in Ag+ overlap with the precipitated which is soluble in excess of any
filled atomic orbitals of N in NH 3 molecules. alkali
(c) Copper hydroxide is an amphoteric substance
(d) In acidic solutions hydration protects copper ions
Ans : FOREIGN 2000
(acid medium) +
p 3 + H+
NH NH 4
Thus (a) is correct option.

Thus (b) is correct option.

38. 6EDTA@4- is a
(a) monodentate ligand
34. Correct name of K 4 6Fe ^CNh6@ is
(b) bidentate ligand
(a) Potassium ferricyanide
(c) quadridentate ligand
(b) Potassium ferrocyanide
(d) hexadentate ligand
(c) Potassium hexacyanoferrate (II)
Ans : OD 2005
(d) Potassium hexacyanoferrate (III)
6EDTA@ is a hexadentate ligand, because it has six
4-

Ans : SQP 2013

donor atoms and donate 6 pair of electrons to central
Oxidation state of iron is + 2 in K 4 6Fe ^CNh6@ metal atom in the complex.
So, its correct name is potassium hexacyanoferrate Thus (d) is correct option.
(II).
39. K 4 6Fe ^CNh6@ is a
Thus (c) is correct option.
(a) double salt
35. CH 3 - Mg - Br is an organometallic compound due (b) complex compound acid
to
(c) acid
(a) Mg - Br bond (b) C - Mg bond
(d) base
(c) C - Br bond (d) C - H bond Ans : SQP 2016
Ans : COMP 2009, OD 2005 Complex compounds do not dissociate into constituent
Compounds that contain at least one carbon metal ions.
bond are known as organometallic compounds. K 4 6Fe ^CNh6@ $ 4K+ 6Fe ^CNh6@4-
In CH 3 - Mg - Br (Grignard’s reagent) a bond is
Hence, It is a complex because no CN- is formed on
present between carbon and Mg (Metal) hence it is an
dissociation.
organometallic compound.
Thus (b) is correct option.
Thus (b) is correct option.
Page 256 Coordination Compounds Chap 5

46. Among the following complexes, optical activity is 50. Which of the following will show paramagnetism
possible in corresponding to 2 unpaired electrons? (Atomic
(a) 6Co ^NH 3h6@3+ numbers: Ni = 28, Fe = 26)
(a) 6FeF6@3- (b) 6NiCl 4@2-
(b) 6Co ^H 2 Oh2 ^NH 3h2 Cl 2@+
(c) 6Fe ^CNh6@3- (d) 6Ni ^CNh4@2-
(c) 6Cr ^H 2 Oh2 Cl 2@
+
Ans : SQP 2000

(d) 6Co ^CNh5 NC@ As in 6NiCl 4@2- Chloride ion being a weak ligand is
Ans : DELHI 2014
not able to paired the electron in d orbital.
It is optically active when two Cl atoms are in cis Thus (b) is correct option.
position. 51. Which of the following complexes will have four
Thus (b) is correct option. different isomers?
47. The total number of possible isomers for the complex (a) 6Co ^enh2 Cl 2@Cl (b) 6Co ^enh^NH 3h Cl 2@Cl
compound 6Cu II ^NH 3h4@6Pt II Cl 4@ (c) 6Co ^PPH 3h2 Cl 2@Cl (d) 6Co ^enh3 Cl 3@
(a) 3 (b) 6
Ans : COMP 2005
(c) 5 (d) 4 Complex 6Co ^enh^NH 3h Cl 2@Cl will have four different
Ans : FOREIGN 2003 isomers.
The total number of isomers for the complex compound (i) Geometrical isomers
6Cu11 ^NH 3h4@6Pt11 Cl 4@ is four.
These four isomers are
6Cu ^NH 3h3 Cl@6Pt ^NH 3h Cl 3@ ,
6Cu ^NH 3h Cl 3@6Pt ^NH 3h3 Cl@ ,
6CuCl 4@6Pt ^NH 3h4@ and 6Cu ^NH 3h4@6PtCl 4@
The isomer 6Cu ^NH 3h2 Cl 2@6Pt ^NH 3h2 Cl 2@ does not
exist due to both parts being neutral.
Thus (d) is correct option.
(ii) Optical isomers
48. IUPAC name of 6Pt ^NH 3h3 ^Brh^NO 2h Cl@Cl is
(a) Triamminechlorobromonitroplatinum (IV) chloride
(b) Triamminebromonitrochloroplatinum (IV) chloride
(c) Triamminebromochloronitroplatinum (IV) chloride
(d) Triamminnenitrochlorobromoplatinum (IV) chloride
Ans : SQP 2016

We know that IUPAC name of 6Pt ^NH 3h3 ^Brh^NO 2h Cl@Cl

is triamminebromochloronitroplatinum (IV) chloride
Thus (b) is correct option.
Thus (c) is correct option.
52. The complex ion 6Co ^NH 3h6@3+ is formed by sp3 d2
49. The hypothetical complex chlorodiaquatriammine
hybridisation. Hence the ion should possess
cobalt (III) chloride can be represented as
(a) Octahedral geometry
(a) 6CoCl ^NH 3h3 ^H 2 Oh2@Cl 2
(b) Tetrahedral geometry
(b) 6Co ^NH 3h3 ^H 2 Oh@Cl 3
(c) Square planar geometry
(c) 6Co ^NH 2h3 ^H 2 Oh2 Cl@
(d) Tetragonal geometry
(d) 6Co ^NH 3h3 ^H 2 Oh3@Cl 3
Ans : DELHI 2006
Ans : OD 2011, FOREIGN 2007
According to VSEPR theory, a molecule with six
Chlorodiaquatriammine cobalt (III) chloride is
bond pairs must be octahedral.
6CoCl ^NH 3h3 ^H 2 Oh2@Cl 2 Thus (a) is correct option.
Thus (a) is correct option.
Page 258 Coordination Compounds Chap 5

60. The number of unpaired electrons in the complex

6Cr ^NH 3h6 Br3@ is (Atomic no. Cr = 24)
= 4 unpaired e- (a) 4 (b) 1
(c) 2 (d) 3
Ans : FOREIGN 2017

= 2 unpaired e - In 6Cr ^NH 3h6 Br3@ , Cr is in + 3 oxidation state etc.

As Ni has minimum no. of unpaired e- thus this is

++

least paramagnetic.
Thus (b) is correct option.
57. A co-ordination complex compound of cobalt has the
molecular formula containing five ammonia molecules,
one nitro-group and two chlorine atoms for one cobalt
atom. One mole of this compound produces three
mole ions in an aqueous solution. On reacting this
solution with excess of AgNO 3 solution, we get two Its ion is octahedral in nature. Due to the presence of
moles of AgCl precipitate. The ionic formula for this three unpaired electrons it is paramagnetic.
complex would be Thus (d) is correct option.
(a) 6Co ^NH 3h4 ^NH 2h Cl@6^NH 3h Cl@ 61. Which statement is incorrect?
(b) 6Co ^NH 3h5 Cl@6Cl ^NO 2h@ (a) Ni ^COh4 - Tetrahedral, paramagnetic
(c) 6Co ^NH 3h5 ^NO 2h@Cl 2 (b) 6Ni ^CNh4@2- - Square planar, diamagnetic
(d) 6Co ^NH 3h5@6^NO 2h2 Cl 2@ (c) Ni ^COh4 - Tetrahedral, diamagnetic
Ans : COMP 2001 (d) 6NiCl 4@2- Tetrahedral, paramagnetic
6Co ^NH 3h5 ^NO 2h@Cl 2 $ 6Co ^NH 3h5 ^NO 2h@ + 2Cl-
++
Ans : OD 2015
-
2Cl + 2AgNO 3 $ 2AgCl + 2NO -
3
(CO) carbonyl group being a strong ligand paired all
electrons present in d -orbital of Ni. Hence from dsp2
Thus (c) is correct option.
hybrid orbitals and hence the shape of molecule is
58. Oxidation number of Ni in 7Ni ^C 2 O 4h3A4- is square planar.
(a) 3 (b) 4 In Ni ^COh4 complex, Ni will have 3d10
configuration. It has tetrahedral geometry but
(c) 2 (d) 6
diamagnetic as there are no unpaired electrons.
Ans : DELHI 2011, SQP 2004
Thus (a) is correct option.
Oxidation number of Ni in 7Ni ^C 2 O 4h3A4-
62. CuSO 4 when reacts with KCN forms CuCN, which is
= x + 3 (- 2) = - 4 insoluble in water. It is soluble in excess of KCN due
x =- 4 + 6 = 2 to formation of the following complex
Thus (c) is correct option. (a) K 2 6Cu ^CNh4@ (b) K 3 6Cu ^CNh4@
59. Which of the following will give maximum number is (c) CuCN 2 (d) Cu 6K Cu ^CNh4@
isomers? Ans : COMP 2000
(a) 6Ni ^C 2 O 4h^enh2@2+ Copper sulphate react with KCN to give white ppt
(b) 6Ni ^enh^NH 3h4@2+ of Cu ^CNh2 and cyanogen gas. The insoluble copper
cynaide dissolve in excess of KCN and give soluble
(c) 6Cr ^SCNh2 ^NH 3h4@+ potassium cuprocyanide
(d) 6Co ^NH 3h4 Cl 2@ CuSo 4 + 2KCN $ K 2 SO 4 + Cu ^CNh2
Ans : SQP 2015
2Cu ^CNh2 $ 2CuCN + CN - CN
6Cr ^SCNh2 ^NH 3h4@ shows likange, gemoetrical and
+
insoluble cyanogen
optical isomerism. Hence produces maximum no. of
CuCN + 3KCN $ K 3 6Cu ^CNh4@
isomers. Soluble
Thus (c) is correct option. Thus (b) is correct option.
Chap 5 Coordination Compounds Page 259

63. According to IUPAC nomenclature sodium Ans : OD 2017

nitroprusside is named as
(a) Sodium pentacyanonitrosyl ferrate (III)
(b) Sodium nitroferrocyanide
(c) Sodium nitroferrocyanide
(d) Aodium pentacyanonitrosyl ferrate (II)
Ans : DELHI 2005

IUPAC nomenclature sodium nitroprusside

Na 2 6Fe ^CNh5 NO@ is sodium pentacyanonitrosyl
ferrate (III) because in it NO is neutral ligand. Hence
2 # O.N. of Na + O.N. of Fe + 5 # O.N. of CN Transform of 6M ^AAh2 a 2@n ! does not shows optical
1 # O.N. of NO = 0 isomerism.
Thus (c) is correct option.
2 # (+ 1) + O.N. of Fe + 5 # (- 1) + 1 # 0 = 0
66. Atomic number of Cr and Fe are respectively 25 and
Thus (a) is correct option.
26, which of the following is paramagnetic?
64. Which of the following coordination compounds would (a) 6Cr ^COh6@ (b) 6Fe ^COh5@
exhibit optical isomerism?
(a) pentamminenitrocobalt (III) iodide (c) 6Fe ^CNh6@-4 (d) 6Cr ^NH 3h6@3+

(b) diamminedichloroplatinum (II) Ans : SQP 2015

3+ 0 3
(c) trans-dicyanobis (ethylenediamine) chromium Cr and 4s 3d electronic configuration with 3
(III) chloride unpaired electrons, hence paramagnetic. In other
cases pairing of d -electrons take place in presence of
(d) tris-(ethylendiamine) cobalt (III) bromide
strong field ligands such as CO or CN- .
Ans : FOREIGN 2009, OD 2003
In Cr ^COh6 molecule 12 electrons are contributed
The optical isomers are pair of molecules which are by CO group and it contains no odd electron.
non super imposable mirror images of each other

The two optically active isomers are collectively called

enantiomers.
Thus (d) is correct option.
65. Which one of the following is expected to exhibit
optical isomerism? (en=ethylenediamine)
Thus (d) is correct option.
(a) cis - 6Pt ^NH 3h2 Cl 2@ 67. Which one of the following is an inner orbital complex
(b) trans - 6Pt ^NH 3h2 Cl 2@ as well as diamagnetic in behaviour? (Atomic no.:
Zn = 30, Cr = 24, Co = 27, Ni = 28)
(c) cis - 6Co ^enh2 Cl 2@
(a) 6Zn ^NH 3h6@2+ (b) 6Cr ^NH 3h6@3+
(d) trans - 6Co ^enh2 Cl 2@ (c) 6Co ^NH 3h6@3+ (d) 6Ni ^NH 3h6@2+
Chap 5 Coordination Compounds Page 261

72. 6Cr ^H 2 Oh6 Cl 3@ (atomic no. of Cr=24) has a magnetic Ans : COMP 2011

moment of 3.83 B.M. The correct distribution of 3d The octahedral coordination compounds of the type
-electrons in the Chromium of the complex is MA 3 B 3 exhibit fac-mer isomerism.
(a) 3dxy , (3dx - y ) 1, 3dyz
1 2 2 (b) 3dxy , 3dyz , 3dxz
1 1 1 1

(c) 3dxy , 3dyz , 3dz

1 1 2 (d) (3dx - y ) 1, 3dz , 3dxz
2 2 2 1

Ans : SQP 2016

m = n (n + 2)
3.83 = n (n + 2)
on solving n = 3
as per magnetic moment, it has three unpaired
electron.
Cr3+ will have configuration as
Cr 1s2 2s2 2p6 3s2 3p6 3d 4 4s2
Cr3+ 1s2 2s2 2p6 3s2 3p6 3d3

So, 3dxy , 3dyz , 3dxz

1 1 1

Thus (a) is correct option.

Thus (b) is correct option.
75. Which of the following complex ion is not expected to
73. Which of the following will give a pair of enantiomorphs? absorb visible light?
(a) 6Cr ^NH 3h6@6Co ^CNh6@ (b) 6Co ^enh2 Cl 2@Cl (a) 6Ni ^CNh4@2- (b) 6Cr ^NH 3h6@3+
(c) 6Pt ^NH 3h4@6PtCl 6@ (d) 6Co ^NH 3h4 Cl 2@NO 2
(c) 6Fe ^H 2 Oh6@2+ (d) 6Ni ^H 2 Oh6@2+
(en = NH 2 CH 2 CH 2 NH 2)
Ans : SQP 2013
Ans : SQP 2015
6Ni ^CNh4@ : Number of unpaired electrons = 0
2-

Non super-imposable mirror images are called optical

isomers and may be described as ‘chiral’. They are 6Cr ^NH 3h6@3+ : Number of unpaired electrons = 3
also called enantiomers and rotate plane polarised
6Fe ^H 2 Oh6@2+ : Number of unpaired electrons = 4
light in opposite directions.
6Ni ^H 2 Oh6@2+ : Number of unpaired electrons = 2
Thus (a) is correct option.
76. Crystal field stablisation energy for high spin d 4
octahedral complex is
(a) - 1.8D 0 (b) - 1.6D 0 + P
(c) - 1.2D 0 (d) - 0.6D 0
Ans : DELHI 2015, OD 2011

d 4 in high spin octahedral complex

Thus (b) is correct option.
74. Which of the following does not show optical
isomerism?
(a) 6Co ^NH 3h3 Cl 3@0 (b) 6Co ^enh Cl 2 ^NH 3h2@+

(c) 6Co ^enh3@3+ (d) 6Co ^enh2 Cl 2@+

CFSE = [0.6 # 1] + [- 0.4 # 3] = - 0.6D 0
(en = ethylenediamine)
Thus (d) is correct option.
Chap 5 Coordination Compounds Page 263

83. The complex, 6Pt ^Pyh^NH 3h BrCl@ will have how many 87. Which of the following complex compounds will
geometrical isomers? exhibit highest paramagnetic behaviour?
(a) 3 (b) 4 (Atom No.: Ti = 22, Cr = 24, Co = 27, Zn = 30)
(c) 0 (d) 2 (a) 6Ti ^NH 3h6@3+ (b) 6Cr ^NH 3h6@3+
Ans :
(c) 6Co ^NH 3h6@3+ (d) 6Zn ^NH 3h6@2+
OD 2017

Complexes of the type M ABCD may exist in three

Ans : FOREIGN 2015
isomeric forms.
(a) 6Ti ^NH 3h6@ : 3d configuration and thus has one
3+ 1

unpaired electron.
(b) 6Cr ^NH 3h6@3+ : In this complex Cr is in + 3
oxidation state.

Similarly, 6Pt ^Pyh^NH 3h BrCl@ may exist in three

isomeric form in which M = PtA = Py, B = NH 3,
C = Br, D = Cl .
Thus (a) is correct option.
84. On mixing concentration NH 4 OH to Cu2+ salt, the
following blue complex is formed Thus, the complex is paramagnetic.
(a) 6Cu ^NH 4h4@ 2+
(b) 6Cu ^NH 3h2@ 2+
(c) 6Co ^NH 3h6@3+ : In this complex are cobalt ion is in
+ 3 oxidation state with 3d6 configuration.
(c) 6Cu ^NH 3h4@ 2+
(d) 6Cu ^NH 4h2@2+
Ans : SQP 2015

On mixing concentration NH 4 OH to a Cu2+ salt, the

NH 4 OH + Cu2+ $ Cu ^OHh2
blue ppt.

6Cu ^NH 3h4@+

NH 4 OH

dark blue solution

Thus (c) is correct option.

85. 6Cr ^H 2 Oh6@Cl 3 (atomic no. of Cr = 24) has a magnetic

moment of 3.83 B.M. The correct distribution of 3d (inner orbital or d2 sp3 hybrid orbital low spin
electrons in the Chromium of the complex is complex) diamagnetic
(a) 3dxy , (3dx - y ) 1, 3dyz
1 2 2 1 (d) In this complex Zn exists as
Zn++ ion
(b) 3dxy , 3dyz 3dxz
1 1 1

Zn++ ion : 3d10 4s0

(c) 3dxy , 3dyz 3dz
1 1 2

(d) (3dx - y ) 1, 3dz , 3dxz

2 2 2 1

Ans :
Zn++ ion in 7Zn ^NH 3h4A2+
COMP 2012, OD 2007

Thus (b) is correct option.

86. A square planar complex is formed by hybridisation of
which atomic orbitals?
(a) s, px , py, dyz (b) s, px , py, dx - y 2 2

(c) s, px , py, dz 2 (d) s, py, pz , dxy

Ans : DELHI 2002 Due to presence of paired electrons complex is
A square planar complex is formed by hybridisation of diamagnetic in nature.
s, px , py and dx - y atomic orbitals.
2 2 Thus (b) is correct option.
Thus (b) is correct option.
Chap 5 Coordination Compounds Page 265

93. Assertion : The [Ni (en)3] Cl2 (en = ethylene-diamine) all the 3d 5 electrons are not paired up. One electron
has lower stability than [Ni (NH3)6]Cl2. remains unpaired. So, it is paramagnetic.
Reason : In [Ni (en)3]Cl2, the geometry of Ni is trigonal Fe2+ in presence of CN- in K 4 (Fe(CN)6)
bipyramidal.
(a) Both Assertion and Reason are correct and
Reason is a correct explanation of the Assertion.
(b) Both Assertion and Reason are correct but Reason
is not the a correct explanation of the Assertion.
(c) Assertion is correct but Reason is incorrect.
Fe3+ in presence of CN- in K 3 [Fe (CN)6]
(d) Both the Assertion and Reason are incorrect.
Ans : SQP 2020

[Ni (en)3] Cl2 is more stable than [Ni (NH 3) 6] Cl 2

because ethylene-diamine is a bidentate ligand, hence
it forms chelating ring with Ni2+ ion.
Thus (d) is correct option. However, the reason is false because crystal field
splitting in ferrocyanide is less than in ferricyanide
94. Assertion : [Co(NO2)3 (NH3)3] does not show optical
ion (higher the oxidation state of the metal, greater
isomerism.
the crystal field splitting).
Reason : It has a plane of symmetry.
Thus (c) is correct option.
(a) Both Assertion and Reason are correct and
Reason is a correct explanation of the Assertion. 96. Assertion : Ethylenediaminetetraacetate ion forms an
octahedral complex with the metal ion.
(b) Both Assertion and Reason are correct but Reason
Reason : It has six donor atoms which coordinate
is not the a correct explanation of the Assertion.
simulatneously to the metal ion.
(c) Assertion is correct but Reason is incorrect. (a) Both Assertion and Reason are correct and
(d) Both the Assertion and Reason are incorrect. Reason is a correct explanation of the Assertion.
Ans : DELHI 2020 (b) Both Assertion and Reason are correct but Reason
Optical isomerism is found in octahedral complexes is not the a correct explanation of the Assertion.
with 1, 2 or 3 sysmmetrical bidentate ligands only. (c) Assertion is correct but Reason is incorrect.
Since given compound is not having any bidentate
(d) Both the Assertion and Reason are incorrect.
ligand, it will not show optical isomerism. It is
because it has plane of symmetry, a plane which is Ans : COMP 2006
perpendicular to equatorial plane. Thus (a) is correct option.
Thus (a) is correct option. 97. Assertion : A chelating ligand must possess two or
95. Assertion : Potassium ferro-cyanide is diamagnetic, more lone pairs at such a distance that it may form
whereas potassium ferricyanide is paramagnetic. suitable strain free rings at the metal ion.
Reason : Crystal field splitting in ferrocyanide ion is Reason : H2N - NH2 is a chelating ligand.
greater than that of ferricyanide ion. (a) Both Assertion and Reason are correct and
(a) Both Assertion and Reason are correct and Reason is a correct explanation of the Assertion.
Reason is a correct explanation of the Assertion. (b) Both Assertion and Reason are correct but Reason
(b) Both Assertion and Reason are correct but Reason is not the a correct explanation of the Assertion.
is not the a correct explanation of the Assertion. (c) Assertion is correct but Reason is incorrect.
(c) Assertion is correct but Reason is incorrect. (d) Both the Assertion and Reason are incorrect.
(d) Both the Assertion and Reason are incorrect. Ans : OD 2018
Ans : FOREIGN 2015 H 2 N - NH 2 does not act as chelating ligand. The
In potassium ferrocyanide, Fe is in the form Fe2+ and coordination by hydrazine leads to a three membered
in potassium ferricyanide, Fe is in the form Fe3+ CN- highly unstable strained ring and thus, it does not act
is a strong leg-and. So, it will pair up all the 3d 6 as chelating agent.
electrons of Fe3+ and make it diamagnetic. In Fe2+, Thus (c) is correct option.
Chap 5 Coordination Compounds Page 267

NH3 is a strong field ligand, it causes pairing of 3d 6

electrons.
Thus,

(i) Coordination number of the complex is 6. Hence,

hybridisation is d 2 sp3 , so geometry is octahedral.
(ii) Since, there is one unpaired electron, so it is
paramagnetic.
104. Draw the structural isomer of 6Co ^NH 3h5 NO 2@Cl 2 and (i) Coordination number of the complex is 6. Hence,
name the type of isomerism. hybridisation is d 2sp3 .
Ans : COMP 2012 (ii) All electrons are paired. So, it is diamagnetic.
The isomerism is linkage isomerism, i.e. more than
106. What do you mean by ligand?
one atom in a monodentate ligand may functions as
Ans : COMP 2008
donor. For this, in NO -2 ion, either nitrogen or an
oxygen atom may functions as donor. The donor species (atom, molecule or ion) which
donate a pair of electrons to the metal atom is called
ligand.

SHORT ANSWER QUESTIONS

107. (a) Write the IUPAC names of the following :

Pentaamminemtrito-O-cobalt (III) ion
(i) [Co (NH 3) 5 (ONO)] 2+
(ii) K 2 [NiCl 4]
or
(b) (i) What is a chelate complex ? Give one example.
(ii) What are heteroleptic complexes ? Give one
example.
Ans : COMP 2023

(a) (i) Pentaamminenitrito-o-cobalt(III) ion

Pentaamminenitro-N-cobalt (III) ion
(ii) Potassium tetrachloridonickelate(11)
105. For the complex ion of 6Co ^NH 3h6@3+ or
(i) state the hybridisation of the complex. (b) (i) A chelate complex is a type of coordination
(ii) state the magnetic nature of the complex. compound where a ligand forms a ring
Ans : OD 2014, SQP 2011 structure with central metal ion, holding it
Oxidation number of Co in the complex 6Co ^NH 3h6@3+ in place. Example : ethylene diamine ligand
is co-ordinate to the copper ion throiigh both
its nitrogen atom, forming a chelate ring
x + 6 # ^0 h = + 3 , x = + 3
structure.
Electronic configuration of Co is 6Ar@3d 7 4s2 . (ii) Heteroleptic complexes are co-ordination
compounds containing different types of
ligands. Example : 8Fe ^CNh5 NOB2- .

108. Discuss the following terms :

(a) Coordination Number
(b) Effective Atomic Number
Chap 5 Coordination Compounds Page 269

115. Define a complex ion and write the types of complexes.

Ans : COMP 2004

Ion formed by coordinate bond between central metal

atom and donars is called complex ion.
Types of complexes:
(i) Cationic complex, e.g. [Co (NH 3) 6] 3+
(ii) Anionic complex, e.g. [Ag (CN) 2] -
(iii) Neutral complex, e.g. [Ni (CO) 4] 120. Define the following:
(i) Coordination number
116. Write and IUPAC name of the following coordination (ii) Coordination sphere
compounds : (iii) Counter ions.
(a) K 4 6Fe ^CNh6@ (b) Ni ^COh4 Ans : DELHI 2012

(c) K 2 6Pt ^Clh6@ (d) Fe 4 6Fe ^CNh6@3 (i) Coordination Number : Total number of
coordinate bonds formed with the central metal
Ans : DELHI 2017
atom or ion by the ligands.
(a) K 4 6Fe ^CNh6@ Potassium hexacyanoferrate (II) Example : The coordination number of cobalt in
(b) Ni ^COh4 tetracarbonylnickel (0) [CoCl (NH 3) 5] Cl 2 is six.
(ii) Coordination Sphere : The central atom and the
(c) K 2 6Pt ^Clh6@ Potassium hexachlorido palatinate ligands are enclosed in square brackets and are
(IV) called coordination sphere.
(d) Fe 4 6Fe ^CNh6@3 iron (III) hexa cyano ferrate (II) (iii) Counter Ions : The ionizable group which are
written outside the bracket are called counter
117. What do you mean by coordination compound? ions.
Ans : COMP 2010, DELHI 2006

Those compounds in which the central metal atom

is linked to ions or neutral molecules by coordinate
bond (donation of lone pairs of electrons) are called
coordinate compounds or complex compounds.

118. What do you mean by chelate? 121. Give oxidation state and co-ordination number of
Ans : OD 2000 central metal ion in (NH 4) 2 [CoF4].
When a bidentate or a polydentate ligand is attached Ans : DELHI 2008

by two or more donor atoms to the central atom Assume that the oxidation state of metal atom is x .
forming a ring, the ligand is called chelating ligand (NH 4) 2 [CoF4]
and the complex is called chelate, the effect is called
chelate effect. 2 (+ 1) + x + 4 (- 1) = 0
x =+ 2
Hence, Oxidation state of Co = + 2
Coordination number of Co = 4
122. What do you mean by denticity?
Ans : FOREIGN 2014

The number of coordinating groups (donor atoms)

present in a ligand is called the denticity of the ligand.
119. What are ambidentate ligands?
(i) Monodentate or Unidentate Ligand : Ligand has
Ans : OD 2006
one donar atom.
Those unidentate ligands which have more than one Example: NH 3 , H 2 O , Cl- .
coordinating atoms are called ambidentate ligands. (ii) Bidentate Ligand : Ligand which can donate two
Example: lone pair of electrons to central metal atom.
Example:
Chap 5 Coordination Compounds Page 271

129. How does the magnitude of T0 decide the actual

configuration of d-orbitals in a coordination entity?
Ans : FOREIGN 2018
th
If T0 < P (pairing energy), the 4 electron enters
one of the eg orbitals giving the configuration t 23g e 1g
thus forming high spin complexes. The ligands for
which T0 < P are called weak field ligands. If T0 > P,
the 4 th electron pairs up in one of the t 2g orbitals
giving the configuration t 24g e 0g thus forming low spin 132. Out of the following two coordination entities, which
complexes. The ligand for which T0 > P are called is chiral (optically active)?
strong field ligands.
(i) cis-[CrCl 2 (ox) 2] 3-
130. [Fe (H 2 O) 6] 3+ is strongly paramagnetic whereas
(ii) trans-[CrCl 2 (ox) 2] 3- .
[Fe (CN) 6] 3- is weakly paramagnetic. Explain.
Ans : FOREIGN 2005
Ans : OD 2001

[Fe (CN 6)] 3- : Iron is in + 3 oxidation state (

3d5 ) hybridisation is d2 sp3 . There is only one
unpaired electron which makes the complex weakly
paramagnetic.
[Fe (H 2 O) 6] 3+ : Iron is in + 3 oxidation state (3d5)
hybridisation is sp3 d2 .

133. Draw figures to show splitting of degenerated orbitals

in square planar crystal field and tetrahedral crystal
field.
[Fe (H 2 O) 6] 3+ is strongly paramagnetic due to presence Ans : FOREIGN 2001
of five unpaired electrons. (i) Splitting of d-orbitals in a square planar crystal
131. Name the following coordination entities and draw field.
the structures of their stereoisomers:
(i) [Cr (C 2 O 4) 3] 3-
(ii) [Co (NH 3) 3 Cl 3].
Ans : FOREIGN 2002
(i) [Cr (C 2 O 4) 3] 3- trioxalatochromate (III) ion.

(ii) Splitting of d-orbitals in a tetrahedral crystal

field.

(ii) [Co (NH 3) 3 Cl 3] triamminetrichloridocobalt (III).

Chap 5 Coordination Compounds Page 273

140. Write the IUPAC names of the following: (i) K [Cr (H 2 O) 2 (C 2 O 4) 2]

(i) [Co (NH 3) 6] Cl 3 (ii) [CO (en) 3] Cl 3
(ii) [Co (NH 3) 5 Cl] Cl 2 (iii) [CO (NH 3) 5 (NO 2)] (NO 3) 2
(iii) K 3 [Fe (CN) 6] (iv) [Pt (NH 3) (H 2 O) Cl 2]
(iv) K 3 [Fe (C 2 O 4) 3] Ans : COMP 2002

(v) K 2 [PdCl 4] (i) Both geometrical (cis, trans) and optical isomer
for cis-form.
(vi) [Pt (NH 3) 2 Cl (NH 2 CH 3)] Cl
Ans : OD 2002
(i) Hexaamine cobalt (III) chloride.
(ii) Pentaamine chloridocobalt (III) chloride.
(iii) Potassium hexacyanoferrate (III).
(iv) Potassium trioxalatoferrate (III).
(v) Potassium tetrachlorido palladate (II).
(vi) Diamminechlorido (methylamine) platinum (II)
chloride.
141. Discuss briefly giving an example in each case the role
of coordination compounds in
(i) Biological system
(ii) Analytical chemistry
(iii) Medicinal chemistry
(iv) Extraction/metallurgy of metals.
Ans : DELHI 2014, OD 2010
(i) Biological System : Biologically important
coordination compounds are: (ii) Two optical isomers can exist.
(a) Chlorophyll is a coordination compound of
Magnesium.
(b) Haemoglobin of blood is a coordination
compound of Iron.
(c) Vitamin B 12 is a coordination compound of
Cobalt.
(ii) Analytical Chemistry :
(a) Presence of Co2+ ion is tested by K 3 [Fe (CN) 6] .
(b) Presence of Fe3+ ion is tested by K 4 [Fe (CN) 6] .
(c) In volumetric analysis the solutions of metal (iii) Linkage isomers [Co (NH 3) 5 ONO] (NO 3) 2 and
ion can be titrated against the solution of the [Co (NH 3) 5 NO 2] (NO 3) 2 .
polydendate ligands like EDTA. (iv) Geometrical isomerism.
(iii) Medicinal Chemistry :
(a) EDTA is used for treatment of lead poisoning.
(b) The platinum complex cis-platin
[Pt (NH 3) 2 Cl 2] is used in cancer therapy.
(iv) Extraction/metallurgy of Metals : The noble
metals like silver and gold are extracted from their
ores through the formation of cyanide complexes
[Ag (CN) 2] - and [Au (CN) 2] - respectively. 143. Explain on the basis of valence bond theory that
142. Indicate the types of isomerism exhibited by the [Ni (CN) 4] 2- ion with square planar structure is
following complexes and draw the structures for these diamagnetic and the [NiCl 4] 2- ion with tetrahedral
isomers: geometry is paramagnetic.
CBSE Chapterswise Question Bank 2025
Includes Solved Exam Papers 20 Years (2024-2005)
Click to Purcahse any NODIA Book From Amzaon

CLASS 12

Also Available for Class 11 for All Subjects

For more details whatsapp at 95301 43210
CBSE Chapterswise Question Bank 2025
Includes Solved Exam Papers 20 Years (2024-2005)
Click to Purcahse any NODIA Book From Amzaon

CLASS 10

Also Available for Class 9 for All Subjects

For more details whatsapp at 95301 43210
Page 274 Coordination Compounds Chap 5

Ans : COMP 2011

In the presence of CO ligand the 4s electrons shift
Nickel in [Ni (CN) 4] 2- is the + 2 oxidation state, i.e. to 3d orbital to pair up the 3d electrons. Thus,
nickel is present as Ni2+ ion. there is no unpaired electron present. Hence it is
diamagnetic.

146. Give evidence that [Co (NH 3) 5 Cl] SO 4 and

[CO (NH 3) 5 SO 4] Cl are ionisation isomers.
Ans : COMP 2018

When they are dissolved in water, they will give

different ions in the solution which can be tested by
adding AgNO 3 solution and BaCl 2 solution. When
Cl- ions are the counter ions, a white precipitate will
be obtained with AgNO 3 solution. If SO 2- 4 ions are
the counter ions, a white precipitate will be obtained
with BaCl 2 solution.
[Co (NH 3) 5 Cl] SO 4 (aq) + BaCl 2 (aq) " BaSO 4 (s) .
ppt

[Co (NH 3) 5 Cl] SO 4 (aq) + AgNO 3 " No reaction

dsp2 - hybrid orbitals
[Co (NH 3) 5 SO 4] Cl (aq) + BaCl 2 (aq) " No reaction
- Diamagnetic
CN- is a strong ligand as it approaches the metal ion, [Co (NH 3) 5 SO 4] Cl (aq) + AgNO 3 (aq) " AgCl (s) .
ppt
so the electrons should get paired up.
In [NiCl 4] 2- , Cl- provides a weak ligand field. It is 147. [Fe (H 2 O) 6] 3+ is strongly paramagnetic whereas
therefore unable to pair up the unpaired electrons of [Fe (CN) 6] 3- is weakly paramagnetic. Explain.
the 3d orbital. Ans : SQP 2008
Hence, the hybridisation is sp3 and it is paramagnetic. In both the complexes, Fe is in + 3 oxidation state
144. A solution of [Ni (H 2 O) 6] 2+ is green but a solution of with the configuration 3d 5 , CN- is a strong field
[Ni (CN) 4] 2- is colourless. Explain. ligand. In its presence, 3d electrons pair up leaving
only one unpaired electron. The hybridisation is
Ans : DELHI 2002
2+ 8 d 2 sp3 forming inner orbital complex. H 2 O is a weak
In [Ni (H 2 O) 6] : Ni is in + 2 state (3d ). It has
ligand. In its presence 3d electrons do not pair up.
two unpaired electrons which do not pair up in the
The hybridisation is sp3 d 2 forming an outer orbital
presence of the weak H 2 O ligand. The d-d transition
complex containing five unpaired electrons, hence it is
absorb red light and the complementary light emitted
strongly paramagnetic.
is green.
In [Ni (CN) 4] 2- : Ni is in + 2 state (3d8) but in presence 148. Explain [Co (NH 3) 6] 3+ is an inner orbital complex
of strong CN- ligand, the two unpaired electron in the whereas [Ni (NH 3) 6] 2+ is an outer orbital complex.
3d orbitals pair up. Thus there is no unpaired electron Ans : DELHI 2004
present therefore it is colourless. In [Co (NH 3) 6] 3+ ,
145. [NiCl 4] 2- is paramagnetic while [Ni (CO) 4] is Oxidation state of Co = + 3
diamagnetic though both are tetrahedral. Why?
Electronic configuration = 3d 6
Ans : COMP 2003
2-
In presence of NH 3 , 3d electrons pair up leaving two
In [NiCl 4] , Ni is in + 2 oxidation state with electronic
d -orbitals empty. Hence, the hybridisation is d 2 sp3
configuration = 3d 8 4s0 .
forming an inner orbital complex.
In [Ni (NH 3) 6] 2+ ,
Oxidation state = + 2
Electronic configuration = 3d 8
-
Cl is a weak ligand. It cannot pair up the electrons in In the presence of NH 3 , 3d electrons do not pair up.
3d orbitals. Hence, it is paramagnetic. In [Ni (CO) 4] The hybridisation involved is sp3 d 2 and it forms an
, Ni is in zero oxidation state and its configuration is outer orbital complex.
- 3d 8 4s2 .
Page 276 Coordination Compounds Chap 5

156. Specify the oxidation numbers of the metals in the 159. Aqueous copper sulphate solution (blue in colour) gives:
following coordination entities: (i) A green precipitate with aqueous potassium fluoride.
(i) [Co (H 2 O) (CN) (en) 2] 2+ (ii) A bright green solution with aqueous potassium
(ii) [CoBr2 (en) 2] + chloride. Explain these experimental results.
Ans : OD 2007
(iii) [PtCl 4] 2-
Aqueous CuSO 4 solution exists as [Cu (H 2 O) 4] SO 4
(iv) [Cr (NH 3) 3 Cl 3] which has blue colour due to [Cu (H 2 O) 4] 2+ ions.
(v) K 3 [Fe (CN) 6] (i) When KF is added, the weak H 2 O ligands are
Ans : COMP 2016
replaced by F- ligands forming [CuF4] 2- ions
which is a green precipitate.
(i) x + (0) + (- 1) + (0) = + 2
[Cu (H 2 O) 4] 2+ + 4F- " [CuF4] - + 4H 2 O
x =+ 3 (Green ppt)

-
(ii) x - 4 =- 2 (ii) When KCl is added, Cl ligands replace the weak
x =+ 2 H 2 O ligands forming [CuCl 4] 2- ion which has
bright green colour.
(iii) x + 0 + 3 (- 1) = 0
[Cu (H 2 O) 4] 2+ + 4Cl- " [CuCl ] 4
2-
+ 4H 2 O
x-3 = 0 Tetrachlorocuprate II
(Green solution)
x =+ 3
160. What is spectrochemical series? Explain the difference
(iv) x + 2 (- 1) + 0 = + 1
between a weak field ligand and a strong field ligand.
x =+ 3 Ans : DELHI 2018
(v) x + 6 (- 1) = - 3 The arrangement of ligands in the order of their
x =+ 3 increasing field strength the increasing crystal
field splitting energy (CFSE) values is called
157. Using IUPAC norms write the systematic names of spectrochemical series.
the following:
The ligands with small value of CFSE (T0 ) are
(i) [Co (NH 3) 6] Cl 3
called weak field ligands.
(ii) [Pt (NH 3) 2 Cl (NH 2 CH 3)] Cl 161. What is crystal field splitting energy? How does the
(iii) [Ti (H 2 O) 6] 3+ magnitude of T0 decide the actual configuration of d
-orbitals in a coordination entity?
(iv) [Co (NH 3) 4 Cl (NO 2)] Cl
Ans : SQP 2016
Ans : OD 2004
When ligands approaches a transition metal ion, the d
(i) Hexaammine cobalt (III) chloride -orbitals splits up into two sets one with lower energy
(ii) Diammine dichlorido (methylamine) platinum and other with higher energy. The difference of energy
(II) chloride between the two set of orbitals is called crystal field
(iii) Hexaaqua titatium (III) ion splitting energy (T0 for octahedral field).
(iv) Tetraammine chloridonitrito-N-Cobalt (III) If T0 < P (pairing energy) then the 4th electron
chloride enters one of the e g orbitals giving the configuration
158. A solution of [Ni (H 2 O) 6] 2+ is green but a solution of t 23g e g1 , thus forming high spin complexes.
[Ni (CN) 4] 2- is colourless. Explain. If T0 > P , the 4th electron gets paired up in
Ans : DELHI 2005 one of the t2g orbitals giving the configuration t 24g e g0
2+
In [Ni (H 2 O) 6] , Ni is again in + 2 oxidation state thereby forming low spin complexes.
with the configuration 3d 8 , i.e. it has two unpaired Such ligands for which T0 > P are called strong
electrons which do not pair up in the presence of weak field ligands.
H 2 O ligand. Hence, it is coloured. The d -d transition 162. Amongst the following the most stable complex is:
absorbs red light and the complementary light (i) [Fe (H 2 O) 6] 3+
emitted is green. In case of [Ni (CN) 4] 2-, Ni is again in
+ 2 oxidation state with the configuration 3d 8 but in (ii) [Fe (NH 3) 6] 3+
the presence of strong CN- ligand, the two unpaired (iii) [Fe (C 2 O 4) 3] 3-
electrons in the 3d orbitals pair up. Thus, there is no
(iv) [FeCl 6] 3-
unpaired electron present. Hence it is colourless.
Page 278 Coordination Compounds Chap 5

Ans : SQP 2002 Ans : OD 2008

The oxidation states of Cr is + 3 , Fe is + 2 and that Coordination Entity

of Zn is also + 2. This entity usually constitutes a central metal atom
Electronic configuration of Cr3+ = , or ion, to which are attached a fixed number of other
no. of unpaired electrons = 3 (inner orbital complex). atoms or ions or groups by coordinate bond. Examples
are [Ni (CO) 4], [COCl 3 (NH 3) 3] etc.
Electronic configuration of Fe2+ = 3d 6 , no. of complex
unpaired electrons = 4 (outer orbital complex). Ligands
Electronic configuration of Zn2+ = 3d 10 , no. of It is ion having at least one lone pair of electrons and
unpaired electrons = 0 (inner orbital complex). are capable of forming a coordinate bond with the
central metal atom/ion in the coordination entity.
m = n (n + 2)
Examples are : Cl- , (OH) - , (CN) -, etc.
Hence, (ii) has the highest magnetic moment. Coordination Number
169. What will be the correct order for the wavelengths of The total number of coordinate bonds with which
absorption in the visible region for the following : central metal atom/ion is linked to ligands in the
coordination entity is called the coordination number
[Ni (NO 2) 6] 4- , [Ni (NH 3) 6] 2+ , [Ni (H 2 O) 6] 2+ ?
of the central metal atom/ion.
Ans : COMP 2015
Coordination Polyhedron
As metal ion is fixed, the increasing field strengths
The spatial arrangement of the ligands which are
(CFSE) values of the ligands form the spectrochemical
directly attached to the central metal atom/ion
series are in order:
defines a coordination polyhedron about the central
H 2 O < NH 3 < NO -2 metal atom/ion.
Thus, the energies absorbed for excitation will be in Examples :
the order: [Co (NH 3) 6] 3+ is octahedral, [Ni (CO) 4] is tetrahedral.
[Ni (H 2 O) 6] 2+ < [Ni (NH 3) 6] 2+ < [Ni (NO 2) 6] 4- Homoleptic and Heteroleptic Complexes
The order of wavelength absorbed will be opposite Complexes in which a metal is bound to only one kind
of it. of donor group are known as homoleptic complexes.
Example : [Co (NH 3) 6] 3+ .
Since, E = hc
l Complexes in which a metal is bound to more than one
kind of donor groups are called hetroleptic complexes.
170. What is meant by stability of a coordination compound
Example : [Co (NH 3) 4 Cl 2] + .
in solution? State the factors which govern stability of
complexes. 172. Give the oxidation state, d -orbital occupation and
Ans : SQP 2011 coordination number of the central metal ion in the
The stability of a complex in solution refers to the following complexes:
degree of association between the two species involved (i) K 3 [Co (C 2 O 4) 3]
in the state of equilibrium. The magnitude of the (ii) cis -[Cr (en) 2 Cl 2] Cl
equilibrium constant for the association, quantitatively (iii) (NH 4) 2 [CoF4]
expresses the stability. (iv) [Mn (H 2 O) 6] SO 4
Factors which govern the stability of a complex are: Ans : FOREIGN 2019, OD 2011
(i) Smaller the size of cation, greater will be the (i) Oxidation state of K 3 [Co (C 2 O 4) 3] 3-
stability of complex, e.g. Fe3+ ion forms more x - 6 =- 3
stable complex than Fe2+ .
(ii) Greater the charge on the central metal ion, more (Coordination no. = - 6 )
stable will be the complex. x =+ 3
(iii) Stronger the ligand, more stable will be the d -orbital occupation of Co 3+
= 3d 6 = t 26g e g0
complex formed, e.g. CN- forms more stable
No. of unpaired electrons = 0
complex than NH 3 .
(ii) cis -[Cr (en) 2 Cl 2] + Cl-
171. Explain with two examples each of the following: Oxidation State, x = + 3 (Coordination no. = - 6 )
Coordination entity, ligands, coordination number,
Cr3+ = 3d 3 = t 23g
coordination polyhedron, homoleptic and heteroleptic
complexes. (No. of unpaired electrons = 3 )
Page 280 Coordination Compounds Chap 5

Ans : OD 2017
(b) Haemoglobin is a coordination compound of
(a) Based on his extensive research on coordination iron.
compound. Alfred warner in 1893 put forward (c) Vitamin B 12 , cyanocobalamin, the
a theory to explain the nature of bonding in
antipernicious anaemia factor is a coordination
coordination or complex compounds. In brief the
compound of cobalt.
warner’s theory may be stated as follows:
1. Central ion in any complex ion/compound 177. How would you account for the fact that [Fe (CN 6)] 3-
exhibits two types of valencies : Primary is weakly paramagnetic while [Fe (CN) 6] 4- is
valency and Secondary valency. diamagnetic?
2. The primary valency is ionisable and Ans : SQP 2004

corresponds to the oxidation state of the [Fe (CN) 6] 3-

metal forming the central ion. The secondary Iron is in + 3 oxidation state = 3d 5
valency is non-ionisable.
3. Every central ion has a fixed number of
secondary valencies. This number is called
the coordination number of the central ion.
4. The primary valency of the metal ion is
always satisfied by negative ion.
5. The secondary valencies are satisfied by
negative ions or neutral molecules.
6. Every element must satisfy all its primary
and secondary valencies. In order to meet this
requirement the negative ion may have a dual
role, i.e., in satisfying both the primary and The presence of one unpaired electron makes the
secondary valencies. complex weakly paramagnetic.
(b) Monodentate ligands have only one atom capable [Fe (CN) 6] 4-
of binding to a central atom or ion. H 2 O is a Iron is in + 2 oxidation state = 3d 6
ligand, oxygen is the donor atom binding to the
metal.
176. Briefly describe the importance of coordination
compounds in:
(i) Qualitative analysis
(ii) Extraction of metals
(iii) Biological systems
Ans : FOREIGN 2003
(i) Qualitative Analysis
(a) EDTA is used for the estimation of Ca2+ and
Mg2+ ions in hard water.
There is no unpaired electron therefore it is
(b) Ni2+ ion is tested and estimated by DMG diamagnetic.
(dimethyl glyoxime).
(ii) Extraction of Metals 178. Explain bonding in coordination compounds with the
help of crystal field theory.
(a) Silver and gold are extracted by treating zinc
with their cyanide complexes. Ans : COMP 2014, OD 2006

(b) Bauxite is purified by forming complexes with The main points of crystal field theory are:
NaOH. 1. Metal ligand bonds are ionic having electrostatic
interaction similar to ions in a crystal, therefore
(c) Impure nickel is converted to [Ni (CO) 4],
named as crystal field theory (CFT).
which is decomposed to yield pure nickel.
2. Ligand is treated as a point of negative charge.
(iii) Biological Systems
The arrangement of the ligands around the central
(a) The pigment responsible for photosynthesis, ion is such that the repulsion between them are
i.e. chlorophyll is a coordination compound of minimum.
Mg.
Page 282 Coordination Compounds Chap 5

6. Ending the Name of Metal Ion or Atom : When 20. K2 [Pd Cl 4] potassium
the complex ion is anionic, the name of central tetrachloridopalladate (II)
metal atom ends in - ate and in cationic or
21. [Ti (H2 O) 6] 3+ hexaaquatitanium (III) ion
neutral complexes, the name of metal is written
as such. 22. [Co (en) 3] 3+ tris (ethane-1, 2-diamine)
7. Oxidation State of Central Metal Ion : Oxidation cobalt (III) ion
state is indicated by a Roman numeral and 23. [Co (NH3) 4 Cl (NO2)] Cl tetraamminechloridonitrito-
written after metal in brackets. N-cobalt (III) chloride
Example: 24. [NiCl 4] 2- tetrachloridonickelate (II) ion
Complex Name 25. [Ni (NH3) 6] Cl2 hexaamminenickel (II) chloride
Compound
26. [Mn (H2 O) 6] 2+
hexaaqua manganese (II) ion
1. K 4 [Fe (CN) 6] potassium hexacyanoferrate
(II)
181. Discuss the nature of bonding in the following
2. K3 [Fe (CN) 6] potassium hexacyanoferrate coordination entities on the basis of valence bond theory.
(III)
(i) [Fe (CN) 6] 4-
3. [Co (NH3) 6] Cl3 hexaammine cobalt (III) (ii) [FeF6] 3-
chloride
(iii) [Co (C 2 O 4) 3] 3-
4. K3 [Co (CN) 5 (NO)] potassium pentacyanonitrosyl (iv) [CoF6] 3-
cobaltate (II)
Ans : FOREIGN 2008
5. K [Pt (NH3) Cl3] potassium 4-
(i) [Fe (CN) 6] : Iron is in + 2 oxidation state
amminetrichloridoplatinate (II)
(3d 6), Hybridisation is d 2 sp3 , octahedral and
6. [Pt (NH3) 2 Cl (NO2)] diamminechloridonitrito-N- diamagnetic.
platinum (II)
(ii) [FeF6] 3- : Iron is in + 3 oxidation state (3d 5),
7. [Ni (CO) 4] tetracarbonyl nickel (O) hybridisation sp3 d 2 , octahedral.
8. K3 [Cr (C2 O 4) 3] potassium
trioxalatochromate (III)
9. [Co (NH3) 4 (H2 O) (Cl)] Cl2 tetraamineaquachlorido cobalt
(III) chloride
10. K2 [Zn (OH) 4] potassium tetrahydroxozincate
(II)
11. [Ag (NH3) 2] [Ag (CN) 2] diamminesilver (I)
dicyanoargentate (I)
12. K3 [Al (C2 O 4) 3] potassium trioxalato
aluminate (III)
[FeF6] 3- is strongly paramagnetic due to presence
13. Hg [Co (SCN) 4] mercury of five unpaired electrons.
tetrathiocyanatocobaltate (III)
(iii) [Co (C 2 O 4) 3] 3- : Cobalt is in + 3 oxidation state,
14. [Co (NH3) 5 (CO3)] Cl pentaammine carbonatocobalt (3d 6) hybridisation is d 2 sp3 , octahedral.
(III) chloride
15. K3 [Fe (C2 O 4) 3] potassium trioxalatoferrate
(III)
16. [Pt (NH3) 2 Cl (NH2 CH3)] Cl diamminechloridomethanamine
platinum (II) chloride
17. [Cr (NH3) 3 (H2 O) 3] Cl3 triammine triaquachromium
(III) chloride
18. [Co (en) 3] 2 (SO 4) 3 tris (ethylenediammine) cobalt
(III) sulphate
19. [CoCl2 (en) 2] Cl dichloridobis (ethane-1,
2-diamine) cobalt (III) chloride
[Co (C 2 O 4) 3] 3- is diamagnetic due to absence of
unpaired electron.
Page 284 Coordination Compounds Chap 5

(d) Octahedral Complex (Co. No. = 6 ) :

[CoCl 2 (en) 2] + . trans [CoCl 2 (en) 2] +

(e) Octahedral Complex (Co. No. = 6 ) :

[Pt (NH 3) (Br) (Cl) (I) (NO 2) (py)] forms 15 Square planar complexes do not show optical
different geometrical isomers. because they contain a plane of symmetry.
(ii) Optical Isomerism : Optical isomers rotate the 183. (i) Draw all the possible isomers having the formula
plane polarised light in opposite direction i.e.,
Cr [(NH 3) 4 Cl 2] + .
right and left are called dextro (d) and laevo (l)
(ii) Illustrate the following with an example:
isomers.
The d and l isomers of a compound are called (a) Linkage isomerism
enantiomers. (b) Coordination isomerism.
The two isomers are non-superimposable mirror (iii) Why is [NiCl 4] 2- is paramagnetic (Ni = 28 )?
image of each other are called Chiral and this Ans : DELHI 2001
property is called Chirality. (i)
Examples:
(a) [Co (en) 3] 3+

(ii) (a) Linkage Isomerism : The isomerism in which

a ligand can form linkage with metal through
different atoms, e.g., nitro group (- NO 2) can
link to metal either through nitrogen atom or
(b) [CoCl 2 (en) 2] + : Its isomers is optically active through oxygen atom, e.g.
but trans isomer is optically inactive. [Co (NH 3) 5 ONO] Cl 2
dichlorobis (ehtane-1, 2-diamine) IUPAC Name : Pentaamminenitrito-o-cobalt
cobalt (III) ion. (III) chloride.
cis [CoCl 2 (en) 2] + [Co (NH 3) 5 NO 2] Cl 2
IUPAC Name : Pentaamminenitrito-N-cobalt
(III) chloride.
Page 286 Coordination Compounds Chap 5

Overall dissociation constant is the reciprocal of

overall stability constant.
Hence, overall dissociation constant.
K = 1
b4
= 1
2.1 # 1013
= 4.7 # 10-14

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

CASE BASED QUESTIONS
Nickel is in + 2 oxidation state = 3d 8 .

188. Hard water does not form leathers with soap.

Neelam uses a washing powder containing sodium
metapolyphosphate and ethylene diamine tetracetate
(EDTA) while Manju is using ordinary washing
powder.
(i) Which washing powder is more effective for
washing clothes in hard water ?
(ii) Which type of ligand is EDTA ?
(iii) Give an example of monodentate ligand.
(iv) Draw the structure of ethylene diamine (en).
[Ni (CN) 4] 2- ion is diamagnetic since it contains Ans :
no unpaired electron. (i) Washing powder containing EDTA.
(iii) [Ni (CO) 4] (ii) It is a hexadentate ligand.
Nickel is in 0 oxidation state = 3d 8 4s2 . (iii) NH 3, H 2 O (any one).
(iv)

Ethylene diamine (en)

189. Anjali went to an exhibition, where she found now-a-
days younger generation girls are interested in wearing
ornaments made up of white metal platinum.
(i) Are you interested in wearing ornaments made of
[Ni (CO) 4] is diamagnetic since it contains no platinum or will you save platinum by not wearing
unpaired electron.
it ? Give reason.
(ii) Name the life saving drug prepared from platinum.

NUMERICAL QUESTIONS (iii) Which disease can be cured by that complex ?

(iv) Write down the structure of cis platin.
187. Calculate the overall complex dissociation equilibrium Ans :
constant for Cu (NH 3) 2+
4 ion, give that b 4 for this (i) No, I will not wear ornaments made up of platinum
complex is 2.1 # 1013 . but I will save it for medicinal use.
Ans : OD 2012 (ii) Cis platin.
Overall stability constant, (iii) It is used to cure cancer.
b 4 = 2.1 # 1013 (iv) cis 6Pt ^NH 3h2 C1 2@.
Page 288 Coordination Compounds Chap 5

192. There is growing interest in the use of chelate therapy

in medicinal chemistry. An example is the treatment
of problems caused by the presence of metals in toxic
proportions in plan/animal systems. Thus, excess
of copper and iron are removed by the chelating
ligands D -penicillamine and desferrioxime B via the
formation of coordination compounds. EDTA is used
in the treatment of lead poisoning. Some coordination
compounds of platinum effectively inhibit the growth
of tumours. Examples are : cis - platin and related
compounds.
(i) What are chelates ?
(ii) Write the chemical name of EDTA.
Ans :
(i) When a di-or polydentate ligand uses its two or
more donoratoms to bind a single metal ion, it is
said to be a chelate ligand.
(ii) Ethylenediaminetetraacetic acid.

***********
Chap 6 Haloalkanes and Haloarenes Page 289

CHAPTER 6
Haloalkanes and Haloarenes

SUMMARY 2. Halogen derivatives of arenes are called haloarenes.

E.g. X - C 6 H 5 .

1. ALKYL HALIDES OR HALOALKANES

Organic compounds in which one or more hydrogen atom
of an alkane have been substituted by a halogen atom.
For example, chloromethane, CH 3 Cl , dibromoethane,
CH 2 BrCH 2 Br , etc. Haloalkanes can be formed by direct
reaction between alkanes and halogens using ultraviolet
radiation. They are usually prepared by the reaction of (i) Vicinal Dihalides : Halogen derivatives where
an alcohol with a halogen carrier. the two halogen atoms are attached on the
adjacent carbon atom.
2. REACTIONS OF HALOALKANES INVOLVE CH 2 Cl
1. Nucleophilic substitution E.g., y
2. Elimination reaction CH 2 Cl
3. Reaction with metals (ii) Geminal Dihalide : Where the two halogen
Nucleophilic substitution reactions are categorised as atoms are attached to the same carbon atom.
S N1 and S N2 on the basis of kinetic properties. For e.g., CH 3 CHBr2 .
A S N2 reaction proceeds with stereo-chemical
Nomenclature
inversion while a S N1 reaction proceeds with
racemisation. Structure Common Name IUPAC Name

Optically Active Molecule CH3 CH2 CH (Cl) CH3 Sec-Butyl 2-Chlorobutane

chloride
It is a molecule that cannot be superimposed on its
mirror image. It is also called chiral molecule. (CH3) 3 CCH2 Br Neo-Pentyl 1-Bromo-2,
bromide 2-dimethyl
Asymmetric Carbon propane
It is the carbon atom that is bonded to four different
(CH3) 3 CBr Tert-Butyl 2-Bromo
groups. It is also called chiral carbon. bromide -2-methyl propane
Enantiomers
CH2 - CHCl Vinyl chloride Chloroethane
These are optical isomers that are mirror images of
each other. CH2 - CHCH2 Br Allyl bromide 3-Bromo propene
Diastereoisomers
o-Chlorotoluene 1-chloro-2-
These are optical isomers that are not mirror images methylbenzene
of each other.

3. FORMATION OF HALOALKANES
Benzyl chloride a -Chlorotoluene/
RH + X 2 $ R - X + HX Chloromethyl
1. Halogen derivatives of alkanes are called benzene
haloalkanes. For e.g., X - CH 3 .
Chap 6 Haloalkanes and Haloarenes Page 291

6.5 Mechanism of Nucleophilic Substitution Reactions 3. Higher level of methylene chloride in air causes
dizziness, nausea, tingling and numbness in
S N1 S N2 fingers and toes.
4. In humans, direct skin contact with methylene
1. Nucleophilic substitution Nucleophilic substitution chloride causes intense burning and mild redness
is uni-molecular. is bimolecular. of the skin, direct contact with eyes can also burn
2. It takes place in two It takes place in one the cornea.
steps. step.
7.2 Trichloromethane (Chloroform)
3. They are first order They are mostly
reactions. second order reactions. Applications
1. Chloroform is employed as a solvent for fats,
4. It leads to It leads to inversion of
alkaloids and other substances.
racemisation. configuration.
2. In the production of refrigerant, freon R-22.
Step:1
Harmful Effects of Chloroform
(CH 3) 3 CBr (Reactant)
1. Breathing of even about 900 parts of chloroform
per million parts of air for a short period of time
can cause dizziness, fatigue and headache.
2. Chronic chloroform exposure may cause damage
to the liver and kidneys.
3. Chloroform is slowly oxidised by air in the presence
Step:2 of light to an extremely poisonous gas named
carbonyl chloride, also known as phosgene. It is
therefore stored in closed dark coloured bottles
completely filled so that air can be kept out.

7.3 Tri-iodomethane (Iodoform)

6.6 Wurtz-Fitting Reaction
A yellow volatile solid sweet smelling haloform, made
by the haloform reaction.
Application
It was earlier used as an antiseptic. Due to its
objectionable smell, it has been replaced by other
formulations containing iodine.
7. POLYHALOGEN COMPOUNDS 7.4 Tetrachloromethane (Carbon tetrachloride), CCl 4
Carbon compounds containing more than one A colourless volatile liquid with a characteristic colour,
halogen atom are usually referred to as polyhalogen virtually soluble in water but it is miscible with many
compounds. organic liquids, such as ethanol and benzene.
7.1 Dichloromethane Applications
1. Used in the manufacture of refrigerants and
Applications
propellants, for example aerosol.
1. Widely used as a solvent as a paint remover and 2. In the synthesis of chlorofluorocarbons and other
as a processing solvent in the manufacture of chemicals.
drugs.
3. It is widely used as a cleaning fluid.
2. It is also used in metal cleaning and as a finishing
4. It is also used as a fire extinguisher.
solvent.
Harmful Effects of CCl 4
Harmful Effects of Dichloromethane
1. Most common effects are dizziness, slight
1. Methylene chloride can severely harm the human
headache, nausea and vomiting, which an even
central nervous system.
cause permanent damage to the nerve cells.
2. Exposure to lower level of methylene chloride
2. Exposure to CCl 4 can make the heart beat
in air can lead to slightly impaired hearing and
irregular or even stops it.
vision.
Chap 6 Haloalkanes and Haloarenes Page 293

of carbocation. Here 1c alkyl group has only

one alkyl group. Hence it is hardly form 1c
carbocation by SN1 reaction. Instead it will prefer
SN2 reaction and avoid the formation of unstable
carbocation.
Thus (a) is correct option.
6. Alkyl halide is converted into an alcohol by :
(a) Addition reaction
(b) Substitution reaction
(c) Elimination reaction
(d) Dehydrohalogenation reaction
Ans : SQP 2009, DELHI 2007
Thus (b) is correct option.
Alkyl halides are converted into alcohols by
substitution reaction. When alkyle halides are treated 9. Ethyl acetate reacts with CH 3 MgBr to form :
with aqueous KOH, alcohols are formed. This is a (a) Secondary Alcohol (b) Tertiary Alcohol
nucleophilic substitution reaction. Equation for the (c) Primary Alcohol (d) Carboxylic Acid
reaction is as follows : Ans : COMP 2017
aq. KOH
R-X R - OH
Thus (b) is correct option.
7. Which of the following reactions is an example of
nucleophilic substitution reaction ?
(a) 2RX + 2NA " R - R + 2NaX
(b) RX + H 2 " RH + HX
(c) RX + Mg " RH + HX
(d) RX + KOH " ROH + KX
Ans : FOREIGN 2000

In nucleophilic substitution, a nucleophile provides an

electron pair to the substrate and the leaving group
departs with an electron pair.

These are usually written as S N (S stands for

substitution and N for nucleophilic) and are common
in aliphatic compounds especially in alkyl halides and
acyl halides. Thus (b) is correct option.
Thus (d) is correct option. 10. Which of the following possesses highest melting
8. Chlorobenzene give DDT when it reacts with : point?
(a) charcoal (b) chloral (a) Chlorobenzene (b) m -dichlorobenzene

(c) naphthalene (d) benzenoid (c) o -dichlorobenzene (d) p -dichlorobenzene

Ans : DELHI 2009

Ans : FOREIGN 2005

DDT is synthesised by heating a mixture of chloral Para-di-chlorobenzene has most symmetrical structure
(1 mol) with chlorobenzene (2 mol) in the presence of than others. It is found as crystalline lattice form,
concentrated H 2 SO 4 therefore, it has highest melting point (52cC) due to
symmetrical structure.
Page 294 Haloalkanes and Haloarenes Chap 6

13. obtained by chlorination of

n- butane, will be
(a) l -form (b) d-form
(c) Meso form (d) Racemic mixture
Ans : SQP 2017
Thus (d) is correct option. Chlorination of n- butane taken place via free radical
formation i.e.,
11. Which one is most reactive towards S N 1 reaction?
(a) C 6 H 5 CH (C 6 H 5) Br
(b) C 6 H 5 CH (CH 3) Br
(c) C 6 H 5 C (CH 3) (C 6 H 5) Br
(d) C 6 H 5 CH 2 Br
Ans : OD 2008

S N 1 reactions involve the formation of carbocations,

hence higher the stability of carbocation, more will
Cl $ may attack on either side and give a racemic
be reactivity of the parent alkyl halide. Thus tertiary
mixture of 2 chloro butane which contain 50% d-form
carbocation formed from (c) is stabilized by two phenyl
and 50% l -form.
groups and one methyl group, hence most stable.
Thus (d) is correct option.
Thus (c) is correct option.
14. A Grignard reagent may be made by reacting
12. In a S N 2 substitution reaction of the type
magnesium with
R - Br + Cl- $ R - Cl + Br- (a) Methyl amine (b) Diethyl ether
which one of the following has the highest relative (c) Ethyl iodide (d) Ethyl alcohal
rate?
Ans : COMP 2011
(a) CH 3 - CH 2 - CH 2 Br
Dry
CH 3 CH 2 l + Mg Ether
CH 3 CH 2 Mgl
(b) Ethyl magnesium
iodide

Thus (c) is correct option.

15. Benzene reacts with n- propyl chloride in the presence

(c) of anhydrous AlCl 3 to give
(a) 3-Propyl - 1 - chlorobenzene
(d) CH 3 CH 2 Br (b) n-Propylbenzene
Ans : SQP 2015, OD 2011 (c) No reaction
2
The rate of S N substitution reaction is maximum in (d) Isopropylbenzene
case of CH 3 CH 2 Br because S 2N mechanism is followed
Ans : DELHI 2007
in case of primary and secondary halides i.e., S 2N
reaction is favoured by small groups on the carbon Anhyd.
atom attached to halogens so C 6 H 6 + CH 3 CH 2 CH 2 Cl AlCl 3

Thus (d) is correct option.

16. Which chloro derivative of benzene among the

following would undergo hydrolysis most readily
with aqueous sodium hydroxide to furnish the
Thus (d) is correct option. corresponding hydroxy derivative?
Page 296 Haloalkanes and Haloarenes Chap 6

23. Oxidation of primary alcohols with chlorine yields 26. Chloropicrin is obtained by the reaction of
(a) Acyl chloride (b) Alkyl chloride (a) steam on carbon tetrachloride
(c) Aldehyde (d) Ketone (b) nitric acid on chlorobenzene
Ans : COMP 2009 (c) chlorine on picric acid
Primary alcohol on oxidation produce an aldehyde. (d) nitric acid on chloroform
X2 {O}
ROH RX -2HX
RCHO Ans : SQP 2017
Pri.Chloride Aldehyde
Chloropicrin is nitrochloroform. It is obtained by the
Thus (c) is correct option. nitration of chloroform with HNO 3 .
24. The replacement of chlorine of chlorobenzene to give HNO 3
phenol requires drastic conditions, but the chlorine of HCCl 3 O 2 NCCl 3
Chloroform Chloropicrin
2,4-dinitrochlorobenzene is readily replaced since,
Chloropicrin is a liquids, poisonous and used as an
(a) nitro groups make the aromatic ring electron rich insecticide and a war gas.
at ortho/para positions.
Thus (d) is correct option.
(b) nitro groups withdraw electrons from the meta
27. Bromine gives substitution reaction products with
positions of the aromatic ring.
benzene readily in presence of
(c) nitro groups donate electrons at meta position. (a) Sunlight
(d) nitro groups withdraw electrons from ortho/para (b) Platinum
positions of the aromatic ring.
(c) FeBr3 at 35c C
Ans : SQP 2014
(d) FeBr3 at high temperature
- NO 2 group is electron attractive group, so it is able
to deactivate the benzene ring. Ans : COMP 2001

Thus (c) is correct option.

28. Haloforms are trihalogen derivatives of
Hence withdraw of electrons from ortho & para (a) Ethane (b) Methane
position cause easy removal of - Cl atom due to (c) Propane (d) Benzene
development of + ve charge on o - and p positions. Ans : DELHI 2003
Thus (d) is correct option. Haloform compounds with the formula CHX 3 , where
25. Which of the following is responsible for depletion of X is a halogen atom.
the ozone layer in the upper strata of the atmosphere? Haloforms are trihalogen derivatives of methane.
(a) Polyhalogens (b) Ferrocene Example : Chloroform CHCl 3 .
(c) Fullerens (d) Freons Thus (b) is correct option.
Ans : OD 2017, DELHI 2011 29. Which of the following compounds is used as a
Chloroflurocarbons, e.g. CF2 Cl 2 , CHF2 Cl 2, refrigerant?
HCF2 CHCl 2 . These are non-inflammable colour (a) Acetone (b) CCl 4
less and stable upto 550cC . These are emitted as (c) CF4 (d) CCl 4 F2
propellants in aerosol spray, cans refrigerators, fire
fighting reagents etc. They are chemically inert and Ans : FOREIGN 2008

hence do not react with any substance with which Under ordinary conditions freon is a gas. Its boiling
they come in contract and therefore float through the point is - 29.8cC . It can easily be liquefied. It is
atmosphere. There they absorb UV-rays and due to chemically inert. It is used in air-conditioning and
this they produce free atomic chlorine which cause in domestic refrigerators for cooling purposes (as
depletion of ozone layer. refrigerant)
Thus (d) is correct option. Thus (d) is correct option.
Page 298 Haloalkanes and Haloarenes Chap 6

Hence B - I being a weakest bond break most easily.

Hence R - I is most reactive.
Thus (b) is correct option.
41. When CH 3 CH 2 CHCl 2 is treated with NaNH 2 , the
product formed is
(a) CH 3 - CH = CH 2
(b) CH 3 - C / CH
(c) CH 3 - C / CH 2
(d) CH 3 - CH / CH
Ans : OD 2016
NaNH 2
CH 3 - CH 2 - CHCl 2 O
NaNH 2
CH 3 - CH = CHCl CH 3 - C / CH
Thus (c) is correct option. O
Final Product
Thus (b) is correct option.
39. An organic compound A(C 4 H 9 Cl) on reaction with
Na/diethyl ether gives a hydrocarbon which on 42. Identify X , Y , Z in the following series
monochlorination gives only one chloro derivative, alc $ KOH Br2 KCN
C2H5I X Y Z
then A is
(a) tert-butyl chloride (b) sec-butyl chloride (a) BrCH 2 CH 2 Br

(c) isobutyl chloride (d) n-butyl chloride (b) BrCH 2 CH 2 CN

Ans : FOREIGN 2000

(c) CN - CH 2 - CH 2 - CN
(d) Br - CH - CH - CN
Ans : SQP 2000

Thus (b) is correct option.

43. Industrial preparation of chloroform employs acetone
and
(a) Phosgene
Thus (a) is correct option.
(b) Calcium hypochlorite
40. Reactivity order of halides for dehydrohalogenation is
(a) R - F > R - Cl > R - Br > R - I (c) Chlorine gas

(b) R - I > R - Br > R - Cl > R - F (d) Sodium chloride

(c) R - I > R - Cl > R - Br > R - F Ans : COMP 2006, OD 2014

By distilling ethanol or acetone with a paste of

(d) R - F > R - I > R - Br > R - Cl
bleaching powder (laboratory and commercial
Ans : SQP 2006 method).
The order of atomic size of halogens decrease in the CaOCl 2 + H 2 O $ Ca (OH) 2 + Cl 2
order I > Br > Cl > F
Cl 2 , so obtained acts as a mild oxidising as well as
i.e. on moving down a group atomic size increases.
chlorinating agent
Further the bond length of C - X
(a) CH 3 COCH 3 + 3Cl 2 $
bond decreases in the order
C - I > C - Br > C - Cl > C - F CCl 3 $ CO $ CH 3 + 3HCl (Chlorination)
and hence the bond dissociation energy decreases in (b) 2CCl 3 $ CO $ CH 3 + Ca (OH) 2 $
the order 2CHCl 3 + (CH 3 COO) 2 Ca (Hydrolysis)
R - F > R - Cl > R - Br > R - I Thus (b) is correct option.
Page 300 Haloalkanes and Haloarenes Chap 6

50. Which of the following will have the maximum dipole 54. The number of structural and configurational isomers
moment? of a bromo compound C 5 H 9 Br , formed by the addition
(a) CH 3 F (b) CH 3 Cl of HBr to 2-pentyne respectively
(c) CH 3 Br (d) CH 3 I (a) 1 and 2 (b) 2 and 4
(c) 4 and 2 (d) 2 and 1
Ans : OD 2015

CH 3 Cl has higher dipole moment than CH 3 F due Ans : FOREIGN 2002

to much longer C - Cl bond length than the C - F Addition of HBr of 2-pentyne gives two structural
bond. The much longer bond length of the C-C isomers (I) and (II)
bond outweighs the effect produced by lower electro- CH 3 - C = C - CH 2 CH 3
HBr
CH 3 C (Br) = CHCH 2 CH 3
negativity of Cl than that of F. (I)

+ CH 3 CH = C (Br) CH 2 CH 3
Thus (b) is correct option. (II)

51. The reaction conditions leading to the best yields of Each one of these will exist as a pair of geometrical
C 2 H 5 Cl are isomers. Thus, there are two structural and four
uv light configurational isomers.
(a) C 2 H 6 (excess) + Cl 2
Thus (b) is correct option.
dark
(b) C 2 H 6 + Cl 2 room temperature 55. Zerevitinov’s determination of active hydrogen in a
(c) C 2 H 6 + Cl 2 (excess)
uv light compound is based
(a) Na (b) CH 3 Mgl
uv light
(d) C 2 H 6 + Cl 2 (c) Zn (d) Al
Ans : SQP 2002
Ans : OD 2017, FOREIGN 2016
Chlorination beyond monochlorination during the Number of active hydrogen in a compound corresponds
preparation of alkyl halides in presence of UV light to the number of moles of CH 4 evolved per mole of
can be suppressed by taking alkane in excess. the compound.
Thus (a) is correct option. - OH , - NH 2 , - SH , - OH or
52. The reaction of toluene with chlorine in presence of CH 3 MgI
- C = CH CH 4 - (2CH 4 from - NH 2)
ferric chloride gives predominantly
Thus (b) is correct option.
(a) benzoyl chloride
56. 2-Bromopentane is heated with potassium ethoxide in
(b) m -chlorotoluene
ethanol. The major product obtained is
(c) benzyl chloride
(a) 2-ethoxypentane (b) pentene-1
(d) o - and p - chlorotoluene (c) trans-2-pentene (d) cis-pentene-2
Ans : COMP 2006 Ans : SQP 2011
The given reaction is an example of electrophilic Potassium ethoxide is a strong base and
substitution. Further, CH 3 group in toluene is o , p 2-bromopentance is a 2c bromide, so elimination
-directing reaction predominates
Thus (d) is correct option. CH 3 CH (Br) CH 2 CH 2 CH 3
53. A salt solution is treated with chloroform drops. Then OC 2 H -5
CH 3 CH = CHCH 2 CH 3 + CH 2 = CHCH 2 CH 2 CH 3
it is shaken with chlorine water. Chloroform layer Pentene - 2 (major) trans Pentene - 1 (min or) cis

becomes violet. Solution contains. Since trans - alkene is more stable than cis. Thus
(a) NO -2 ion (b) NO -3 ion trans - pentene-2 is the main product.
Thus (c) is correct option.
(c) Br- ion (d) I- ion
Ans : DELHI 2015
57. C - X bond is strongest in
Cl 2 water being aqueous and CHCl 3 being organic (a) CH 3 Cl (b) CH 3 Br
form two layers. Cl 2 water oxidizes I- " I 2 which (c) CH 3 F (d) CH 3 I
shows its colour (violet) in organic layer (CHCl 3). Ans : COMP 2001
This is layer test which is used for detection of I- and
Because of the small size of F, the C-F bond is
Br- in a solution. Br+ gives brown layer.
strongest in CH 3 F .
Thus (d) is correct option.
Thus (c) is correct option.
Page 302 Haloalkanes and Haloarenes Chap 6

Ans : OD 2008

Thus (a) is correct option.

In this reaction both hydrogen and halogen atom has
72. The above structural formula refers to
been removed so it is known as dehydro halogenation
reaction.
Thus (d) is correct option.
68. Which of the following alkyl halides is used as a
methylating agent
(a) C 2 H 5 Br (b) C 6 H 5 Cl
(c) CH 3 I (d) C 2 H 5 Cl (a) BHC (b) DNA

Ans : SQP 2017

(c) DDT (d) RNA
C 2 H 5 Br and C 6 H 5 Cl are ethyl letting agents, while Ans : OD 2016

C 6 H 5 Cl is inert. Thus (c) is correct option.

Thus (c) is correct option. 73. The reaction of toluene with Cl 2 in presence of FeCl 3
69. Which of the following is not used in Friedel-Crafts gives X and reaction in presence of light gives Y .
reaction? Thus, X and Y are
(a) N-Phenyl acetanilide (b) Bromobenzene (a) X = Benzal chloride,
(c) Benzene (d) Chlorobenzene Y = o -chlorotoluene
Ans : COMP 2009 (b) X = m - Chlorotoluene
N-Phenyl acetanilide, C 6 H 5 N (C 6 H 5) COCH 3 , Y = p -Chorotoluene
precipitates out to a complex with anhydrous AlCl 3 .
(c) X = o -and p -Chlorotoluene
Thus (a) is correct option.
Y = Trichloromethyl-benzene
70. A set of compounds in which the reactivity of halogen
(d) X = Benzyl chloride
atom in the ascending order is
(a) chlorobenzene,vinyl chloride, chloroethane Y = m -Chlorotoluene
(b) chloroethane, chlorobenzene, vinyl chloride Ans : SQP 2010

(c) vinyl chloride, chlorobenzene, chloroethane

(d) vinyl chloride, chloroethane, chlorobenzene
Ans : DELHI 2015, SQP 2014

On the same basis as above

CH 3 CH 2 Cl > CH 2 = CHCl > C 6 H 5 Cl
Thus (a) is correct option.
71. Mg reacts with RBr best in
(a) C 2 H 5 OC 2 H 5
(b) C 6 H 5 OCH 3
Thus (c) is correct option.
(c) C 6 H 5 N (CH 3) 2
74. Which one of the following is most reactive towards
(d) Equally in all the three nucleophilic substitution reaction?
Ans : FOREIGN 2000 (a) CH 2 = CH - Cl (b) C 6 H 5 Cl
Although all the three compounds can be used (c) CH 3 CH = CH - Cl (d) ClCH 2 - CH = CH 2
for preparing Grignard reagents, diethyl ether is
considered as the best because it provides electron Ans : COMP 2013

pairs to Mg of the reagent fully for coordination, in More the stability of the carbecation, higher will be
case of C 6 H 5 OCH 3 and C 6 H 5 N (CH 3) 2 electron pair the reactivity of the parent chloride. Allylchloride>
on O and N are partialy delocalised over the benzene Vinyl chloride > Chlorobenze
and hence are less available for coordination with Mg. Thus (d) is correct option.
Page 304 Haloalkanes and Haloarenes Chap 6

81. Assertion : CHCl3 is stored in dark bottles. 84. Assertion : Reimer-Tiemann reaction of phenol with
Reason : CHCl3 is oxidised in dark. CCl4 in NaOH at 340 K gives salicylic acid as the
(a) Both Assertion and Reason are correct and major product.
Reason is a correct explanation of the Assertion. Reason : The reaction occurs through intermediate
formation of dichlorocarbene.
(b) Both Assertion and Reason are correct but Reason
is not the a correct explanation of the Assertion. (a) Both Assertion and Reason are correct and
Reason is a correct explanation of the Assertion.
(c) Assertion is correct but Reason is incorrect.
(b) Both Assertion and Reason are correct but Reason
(d) Both the Assertion and Reason are incorrect. is not the a correct explanation of the Assertion.
Ans : SQP 2015
(c) Assertion is correct but Reason is incorrect.
CHCl3 is stored in dark bottles to prevent oxidation of
(d) Both the Assertion and Reason are incorrect.
CHCl3 in-presence of sunlight.
Thus (c) is correct option. Ans : OD 2006

Dichlorocarbene (CCl2) attacks on the orthodox-

82. Assertion : SN2 reaction of an optically active aryl
position of the phenol-ate ion to form an intermediate
halide with an aqueous solution of KOH always gives
which on hydrolysis gives salicylic acid.
an alcohol with opposite sign of rotation.
Thus (c) is correct option.
Reason : SN2 reactions always proceed with retention
of configuration. 85. Assertion : Alkyl iodide can be prepared by treating
(a) Both Assertion and Reason are correct and alkyl chloride/bromide with NaI in acetone.
Reason is a correct explanation of the Assertion. Reason : NaCl/NaBr are soluble in acetone while NaI
is not.
(b) Both Assertion and Reason are correct but Reason
is not the a correct explanation of the Assertion. (a) Both Assertion and Reason are correct and
Reason is a correct explanation of the Assertion.
(c) Assertion is correct but Reason is incorrect.
(b) Both Assertion and Reason are correct but Reason
(d) Both the Assertion and Reason are incorrect. is not the a correct explanation of the Assertion.
Ans : OD 2020
(c) Assertion is correct but Reason is incorrect.
Assertion if false, because aryl halides do not undergo
(d) Both the Assertion and Reason are incorrect.
nucleophilic substitution under ordinary conditions.
This is due to resonance, because of which the carbon Ans : DELHI 2009

chlorine bond acquires partial double bond character, Alkyl halides on treatment with NaI in presence of
hence it becomes shorter and stronger and thus acetone forms alkyl iodide. This is called Finkelstein
cannot be replaced by nucleophiles. Also, Reason is reaction.
false because SN2 reactions proceeds with inversion R - X + NaI
acetone
R - I + NaX
of configuration.
Here NaI is soluble in acetone but NaBr/NaCl are not
Thus (d) is correct option.
soluble. Hence due to precipitation of slat, equilibrium
83. Assertion : Alkyl-benzene is not prepared by Friedel- is shifted to forward direction. Hence assertion is true
Crafts alkylation of benzene. but reason is false.
Reason : Alkyl halides are less reactive than acyl Thus (c) is correct option.
halides.
86. Assertion : SN2 reactions always proceed with
(a) Both Assertion and Reason are correct and inversion of configuration.
Reason is a correct explanation of the Assertion.
Reason : SN2 reaction of an optically active aryl
(b) Both Assertion and Reason are correct but Reason halide with an aqueous solution of KOH always gives
is not the a correct explanation of the Assertion. an alcohol with opposite sign of rotation.
(c) Assertion is correct but Reason is incorrect. (a) Both Assertion and Reason are correct and
(d) Both the Assertion and Reason are incorrect. Reason is a correct explanation of the Assertion.
Ans : DELHI 2013 (b) Both Assertion and Reason are correct but Reason
Alkyl halides give polyalkylation products. is not the a correct explanation of the Assertion.
Thus (c) is correct option. (c) Assertion is correct but Reason is incorrect.
(d) Both the Assertion and Reason are incorrect.
Page 306 Haloalkanes and Haloarenes Chap 6

93. Draw the structure of 2-bromopentane. 98. Write the IUPAC name of the following compound.
Ans : SQP 2015

Ans : OD 2010
94. Write the IUPAC name of ^CH 3h2 CH .CH ^Clh CH 3 .
Ans : DELHI 2019

99. Write the IUPAC name of the following compound.

95. Write the IUPAC name of the following compound

CH 2 = CHCH 2 Br .
Ans : OD 2013, FOREIGN 2010

Ans : DELHI 2018

96. Write the IUPAC name of

CH 3 - CH - CH 2 - CH = CH 2
y Numbering will be done in alphabetically manner.
Cl
100. A hydrocarbon C 5 H 12 gives only one monochlorination
Ans : DELHI 2002
product. Identify the hydrocarbon.
Ans : COMP 2003

Hydrocarbon which gives only one monochlorination

product is

(As double bond is given preference over halogen)

97. Write the IUPAC name of

101. Draw the structure of the following compound:

3-(4-chloropheny1)-2-methylpropane
Ans : FOREIGN 2014

Ans : FOREIGN 2006

Page 308 Haloalkanes and Haloarenes Chap 6

Ans : COMP 2003

114. Complete the following chemical equation, Peroxide

Ans : SQP 2001

Peroxide
CH 3 CH 2 CH = CH 2 + HBr
CH 3 - CH 2 - CH 2 - CH 2 Br
1 - bromobu tane (Major)
111. Give the IUPAC name of the following compound.
+ CH 3 - CH 2 - CH ^Brh CH 3
^Minorh

115. Write the IUPAC name of ClCH 2 C / CCH 2 Br .

Ans : OD 2004

Ans : DELHI 2005

116. Draw the structure of the following compound

4-bromo-3 -methylpent-2-ene.
Ans : COMP 2014

112. Write the IUPAC name of the following compound:

117. Draw the structure of the compound, 1-chloro-4-

ethylcyclohexane.
Ans : OD 2012
Ans : OD 2018, DELHI 2007

118. Write the balanced chemical equation for ethyl

113. Write a chemical reaction in which the iodide ion chloride treated with alcoholic potassium hydroxide.
replaces the diazonium group in a diazonium salt. Ans : OD 2019

Ans : COMP 2008 When ethyl chloride is treated with alcoholic KOH, it
undergoes elimination reaction to form ethene.
C 2 H 5 Cl + alc. KOH $ C 2 H 4 + KCl + H 2 O

119. What will be the product formed when chlorobenzene

is heated with sodium metal in the presence of dry
ether?
Ans : COMP 2018

Chlorobenzene is heated with sodium metal in the

presence of dry ether gives diphenyl compound.
Chap 6 Haloalkanes and Haloarenes Page 309

Ans : OD 2010

When iodo-form is heated with silver powder,

acetylene (ethyne) is formed.
CH 3 I + 6Ag + CH 3 I $ CH / CH + 5AgI
Acetylene

126. Out of ethyl bromide and ethyl chloride which has

higher boiling point and why?
120. Give the balanced equation for the following name Ans : OD 2000
reaction.
The boiling point of ethyl bromide is higher due to
Wurtz-Fittig reaction
greater magnitude of the van der Waals’ forces which
Ans : DELHI 2016
depends upon molecular size of the halogen atom.
127. What are polyhalogen compounds? Give examples.
Ans : COMP 2012

Carbon compounds containing more than one halogen

121. How can the following conversions be brought about ? atom are called polyhalogen compounds, example is
Chlorobenzene to phenol chloroform, Iodoform.
Ans : COMP 2015 128. Which isomer of C 5 H 10 gives a single monochloro
compound C 3 H 9 Cl in bright sunlight ?
Ans : COMP 2023

The isomer of C 5 H 10 gives a single monochloro

compound C 5 H 9 Cl in bright sunlight is cyclopentane.
This reaction is an example of photo chlorination
where chlorine react with cyclopentane in presence of
sunlight to form a single monochloro compound.
122. Write the chemical equation for the following reaction
129. Arrange the following compounds in increasing order
Acetone with sodium hypoiodite.
of reactivity towards SN 2 reaction :
Ans : OD 2005
2-Bromopentane, 1-Bromopentahe, 2-Bromo-2-
In the preparation of iodoform, sodium hypoiodite can
methylbutane.
also be used in place of sodium hydroxide with acetone.
Ans : COMP 2015, SQP 2013
CH 3 COCH 3 + 3NaOI $ CH 3 COCl 3 + 3NaOH
The order of reactivity towards S N 2 reaction is as
CH 3 COCI 3 + NaOH $ CH 3 COONa + CHI 3 follows :
1-Bromopentane < 2-Bromopentanc < 2-Bromo-2-
123. Draw the structure of the following organic compound: methylbutane
2-chloro-3-methylpentane.
This is because the reactivity order of S N 2 reaction is
Ans : FOREIGN 2012
1° 2 2° 2 3°
130. Why p -dichlorobenzene has higher melting point than
those of ortho- and meta-isomers ?
Ans : COMP 2023

This is because p -isomer is more symmetrical i.e.

more closely packed arrangement of molecules in the
crystal lattice.
124. Out of ethyl bromide and ethyl chloride, which has
higher boiling point and why? 131. Identify A and B in the following :
Ans : COMP 2001

For the same alkyl group, boiling point increases with

the size of the halogen atom. Thus, boiling point of
ethyl bromide is higher than ethyl chloride.
125. What happens when iodo-form is heated with silver
powder? Write the chemical equation.
Chap 6 Haloalkanes and Haloarenes Page 311

137. Explain as to why haloarenes are much less reactive (ii) Diastereoisomers : Those pairs of stereoisomers
than haloalkanes towards nucleophilic substitution which are not mirror images of each other and
reactions. differ in optical rotation.
Ans : FOREIGN 2005 Enantiomers : They are non-superimposable
Haloarenes are much less reactive than haloalkanes mirror images of molecules with each other. They
towards the nucleophilic substitution reactions due to have optical rotation equal in magnitude but
the following reasons: opposite in sign.
(i) Resonance Effect : In haloarenes the electron pair 140. Does the presence of two chiral carbon atoms always
on the halogen atom is in conjugation with the p make the molecule optically active? Explain giving an
-electrons of the ring and the following resonating example.
structures are possible. Ans : DELHI 2010

No the presence of two chiral carbon atoms not always

make the molecule optically active. For example, meso-
tartaric acid is optically inactive because if upper half of
molecule rotates the plane polarised light towards the
left, then lower half rotate it towards the right, such that
net optical rotation is zero due to internal compensation.
C - Cl bond acquires a partial double bond
character due to resonance. As a result, the bond
cleavage in haloarene is difficult than in case of
haloalkanes and therefore, they are less reactive
towards nucleophilic substitution reaction.
(ii) The C - Cl bond length in haloalkane is 177 pm
while in haloarene is 169 pm. Since it is difficult to
break shorter bond than a longer bond. Therefore,
haloarenes are less reactive than haloalkanes
towards nucleophilic substitution reaction.
(iii) Phenyl cation is less stable than the alkyl 141. Account for the following:
carbocation. The treatment of an alkyl chloride with aqueous KOH
leads to the formation of alcoholic whereas in the
138. Complete the following reaction equations : presence of alcoholic KOH, alkene is the major product.
(i) C 6 H 5 N 2 Cl + KI $ Ans : DELHI 2001

Aqueous KOH gives OH- ions which can replace

(ii) .................. Cl- ions and carry out the nucleophilic substitution
Ans : SQP 2012
reaction, whereas alc. KOH gives C 2 H 5 O- which is
a stronger nucleophile and abstracts H+ from the b
(i) C 6 H 5 N 2 Cl- + KI $ C 6 H 5 I + KCl + N 2
carbon atom and carry out the elimination reaction,
thus leading to the formation of an alkene.
(ii)
142. Write the structure of the main product:
(i) Chlorination of benzene in the presence of UV
139. Point out the difference between: light.
(i) Chirality and chiral centre. (ii) Propene is treated with HBr is presence of benzoyl
(ii) Diastereoisomers and Enantiomers. peroxide.
Ans : FOREIGN 2018, OD 2011 (iii) Chlorobenzene is treated with NaOH at 623 K
(i) Chirality : The phenomena of a molecule containing and high pressure.
a carbon atom attached to four different atoms Ans : SQP 2006
or groups of atom and thus making the mirror
image of the molecule non-superimposable on the
molecule is called chirality. (i)
Chiral Centre : The carbon atom which is
attached to four different atoms or groups of
atoms is called the chiral centre.
Chap 6 Haloalkanes and Haloarenes Page 313

Ans : OD 2011
(ii)
(i)

(iii)

(ii)
(iv)

155. Which one of the following has the highest dipole

moment?
(i) CH 2 Cl 2
152. Give the reason of dipole in C - X bond.
(ii) CHCl 3
Ans : SQP 2019

Since halogen atoms are more electronegative than (iii) CCl 4

carbon, the carbon halogen bond is polarized; the Ans : COMP 2008

carbon atom bears a positive charge whereas the

halogen atom bears a partial negative charge.

153. Which would undergo S N 1 reaction faster in the

following pair and why?
156. How Alkyl halides are prepared from alcohols?
Ans : DELHI 2015, COMP 2005

The hydroxyl group of an alcohol is replaced by

halogen on reaction with concentrated halogen acids,
phosphorous halides or thionyl chloride.
Ans : FOREIGN 2011 ZnCl 2
R - OH + HCl R - Cl + H 2 O
3° carbocation > 2° carbocation > 1° carbocation
ZnCl 2
> CH +3 . Therefore, the reactivity of alkyl halides C 2 H 5 OH + HCl C 2 H 5 Cl + H 2 O
towards S N 1 reactions decreases in the same order.
3CH 3 OH + PCl 3 $ 3CH 3 Cl + H 3 PO 3
Among the given compounds bromoethane is primary
alkyl halide which forms a 1° carbocation intermediate C 2 H 5 OH + PCl 5 $ C 2 H 5 Cl + POCl 3 + HCl
in the S N 1 reaction. The other compound is 2-chloro red P/X
2-methylpropane which is a tertiary alkyl halide which C 2 H 5 OH X 2 = Br2, I 2
R-X
forms a tertiary carbocation intermediate in the S N 1 C 2 H 5 OH + SOCl $ C 2 H 5 Cl + SO 2 - + HCl -
2
reaction. Ethanol Thionyl chloride Chloroethane

Hence, 2-bromo-2-methylpropane undergoes an S N 1 Preparation with thionyl chloride is preferred because

reaction faster than bromoethane. the other two products SO 2 and HCl are escapable
154. Write the isomers of the compound having formula gases.
C 4 H 9 Br . 157. Complete the following reactions:
Ans : FOREIGN 2002 Cl 2 /UV light
(i) CH 3 CH 2 CH 2 CH 3
C 4 H 9 Br follows the formula C n H 2n + 1 Br (Saturated
Compound). The compound has four isomers,
(i) (ii)
Chap 6 Haloalkanes and Haloarenes Page 315

163. Of the two bromo derivatives C 6 H 5 CH (CH 3) Br and

C 6 H 5 CH (C 6 H 5) Br , which one is more reactive in S N 1
substitution reaction and why?
Ans : SQP 2002

In S N 1 reactions, reactivity depends upon the stability

167. Write the structural formula of the organic compounds
of the intermediate carbocation.
A, B, C and D in the following sequence of reaction:

Ans : DELHI 2014

The carbocation (b) is stabilised by + R effect of the

two C 6 H 5 -group but the carbocation (a) is stabilized
by + R effect of one C 6 H 5 group and + I effect of the
- CH 3 group. Since + R effect is much stronger than
+ 1 effect, therefore carbocation (b) is more stable
than carbocation (a). Thus C 6 H 5 CH (C 6 H 5) Br is more
reactive than C 6 H 5 CH (CH 3) Br in S N 1 substitution
reaction.
164. What is a racemic mixture? Give one example.
Ans : SQP 2007, OD 2005

A 50 : 50 mixture of two enantiomers of any optically

active compound is called a racemic mixture. It is
optically inactive. Since rotation caused by the molecules
of one enantiomer is exactly cancelled by equal and 168. A hydrocarbon C 5 H 10 does not react with chlorine
opposite rotation caused by the same molecules of the but gives a single monochloro compound C 5 H 9 Cl in
other enantiomers. For example an equimolar mixture bright sunlight. Identify the hydrocarbon.
of (+)-2-bromobutane and (-)-2- bromobutane. Ans : SQP 2004

165. Which compound in each of the following pairs will Hydrocarbon C 5 H 10 follow C n H 2n , it may be an alkene
react faster in S N 2 reaction with OH- ? or cycloalkane.
(i) CH 3 Br or CH 3 I Since the hydrocarbon does not react with Cl 2 in the
absence of bright sunlight, it cannot be an alkene but
(ii) (CH 3) 3 CCl or CH 3 Cl must be a cycloalkane.
Ans : DELHI 2013 Because C 5 H 9 Cl is a single monochloro compound,
(i) Since I ion is a better leaving group than Br-
-
therefore, all the ten hydrogen atoms of the
ion, therefore CH 3 I reacts faster than CH 3 Br in cycloalkane must be equivalent, hence the cycloalkane
S N 2 reaction with OH- ion. is cyclopentane.
(ii) 1° alkyl halides are more reactive than tert-
alkyl halides in S N 2 reaction on steric ground.
Therefore CH 3 Cl will react at a faster rate than
(CH 3) 3 CCl in a S N 2 reaction with OH- ion.
166. A small amount of ethanol is usually added to
chloroform bottles. Why?
Ans : DELHI 2009 169. The treatment of alkyl chloride with aqueous KOH
0.6-1% ethanol is added to chloroform bottle because leads to the formation of alcohols but in presence
ethanol combines with the poisonous phosgene gas of alcoholic KOH, alkenes are the major products.
and convert it into non-toxic diethyl carbonate. Explain why?
Chap 6 Haloalkanes and Haloarenes Page 317

176. How will you convert CH 3 - CH 2 - Br into following? LONG ANSWER QUESTIONS
(i) CH 3 - CH 3
(ii) C 2 H 5 - O - C 2 H 5 179. What are haloalkanes and haloarenes? Give their
(iii) CH 3 - CH 2 - CN classification.
Ans : OD 2017 Ans : FOREIGN 2015
(i) Reduction : When one or more hydrogen atoms of aliphatic or
Ni
CH 3 CH 2 Br + H 2 T
CH 3 CH 3 + HBr . aromatic hydrocarbon are replaced by halogen
(ii) By Williamson’s Synthesis Method : atom (s), alkyl halide (haloalkane) and aryl halide
(haloarene) are formed respectively.
+
T
CH 3 CH 2 Br + C 2 H 5 O- Na C 2 H 5 OC 2 H 5 + NaBr Classification
(iii) By the use of KCN : Haloalkanes and haloarenes may be classified as
CH 3 CH 2 Br + KCN $ CH 3 CH 2 CN + KBr follows:
(i) On the Basis of Number of Halogen Atoms :
177. Chloroform is stored in dark coloured bottles Depending upon number of halogen atoms (F,
completely filled so that air is kept out. Explain. Cl, Br, I) they are classified as mono, di or poly
Ans : SQP 2001 halogen compounds. For example,
Chloroform is stored in dark coloured bottles because
it is oxidised by air in presence of flight to form
an extremely posionous gas carbonyl chloride or
phosgene.

178. The presence of electron withdrawing group at

ortho and para positions increases the reactivity of
haloarenes toward nucleophilic substitution reaction.
Explain.
Ans : COMP 2019

The presence of electron with drawing group such

as, - NO 2 , - CN at O- and p -position withdraws
electrons from the benzene ring thus facilitates d, e, f are types of arylhalide or haloarene in
the attract of the nucleophile on haloarene. The which-X is attach to an aromatic ring.
carbocations thus formed is stabilized by resonance (ii) Compounds Containing sp3 C - X Bond
as shown below : (X = F, Cl, Br, I) :
(a) Alkyl Halides or Haloalkanes (R - X ) :
They are classified as primary, secondary and
tertiary according to the nature of carbon to
which halogen is attached.

(b) Allylic Halides : In these compounds halogen

atom is bonded to an sp3 hybridised carbon
atom next to carbon-carbon double bond
(C = C) i.e. to an allylic carbon.
Page 318 Haloalkanes and Haloarenes Chap 6

(iii) Kharasch effect

(iv) Swarts reaction
(v) Finkelstein reaction
(vi) Hundsdiecker reaction
(vii) Sandmeyer reaction
(viii) Preparation of Iodobenzene
(c) Benzylic Halides : In these compounds the
(ix) Balz-Schiemann reaction
halogen atom is bonded to an sp3 -hybridised
carbon atom next to an aromatic ring. (x) Gattermann reaction.
Ans : COMP 2007
(i) Allylic Halogenation : Reactions in which
halogenation occurs at the allylic position of an
alkene are called allylic halogenation reaction. It
occurs in alkene other then ethene.
Example :
CH - CH = CH 2 + Cl 2 $
3
Propene
(iii) Compounds Containing sp2 C - X Bond :
Cl - CH 2 - CH = CH 2 + HCl
Vinylic Halides : In these compounds halogen 3 - Chloroprop - 1 - ene

atom is bonded to an sp2 hybridised carbon atom (ii) Markovnikov’s Rule : The addition of
of a carbon-carbon double bond (C = C). unsymmetrical reagents like HX, H 2 O etc. to
unsymmetrical alkenes occurs in such a way that
the negative part of adding molecule goes to that
carbon atom of the double bond which carries
lesser number of hydrogen atoms.
Example :
d+d-
180. Write IUPAC names of the following : CH 3 CH = CH 2 + HI $
CH 3 CH 2 CH 2 I + CH 2 CHICH 3
(Minor) (Major)
(i) (ii)
(iii) Kharasch Effect : In presence of peroxide like
benzoyl peroxide (C 6 H 6 CO - O - O - COC 6 H 5),
the addition of HBr to unsymmetrical alkenes
takes place opposite to Markovnikov’s rule. This
is known as Peroxide effect or Kharasch effect.
(iii) (iv)
Example :
(C 6 H 5 COO) 2
CH 3 CH = CH 2 + HBr CH 3 CH 2 - CH 2 Br
(Propene) 1 - Bromopropane

(iv) Swarts Reaction : Synthesis of alkyl fluoride

(v) (vi) (fluoroalkane) by heating an alkyl chloride/
bromide in the presence of metallic fluoride like
AgF, Hg 2 F2 or SbF3 is called Swarts reaction.
Ans : FOREIGN 2010, DELHI 2008 H 3 C - Br + AgF $ H 3 C - F + AgBr
(i) 4-Bromopent-2-ene (v) Finkelstein Reaction : Preparation of alkyl iodide
(ii) 3-Bromo-2-methylbut-1-ene by the reaction of alkyl chlorides/bromides with
(iii) 4-Bromo-3-methylpent-2-ene NaI in dry acetone is known as finkelstein reaction.
(iv) 1-Bromo-2-methylbut-2-ene 182. Write the following:
(v) 1-Bromobut-2-ene (i) Hundsdiecker reaction
(vi) 3-Bromo-2-methylpropene (ii) Sandmeyer reaction
181. Write the following: (iii) Preparation of Iodobenzene
(i) Allylic halogenation (iv) Balz-Schiemann reaction
(ii) Markovnikov’s rule (v) Gattermann reaction.
Chap 6 Haloalkanes and Haloarenes Page 319

Ans : COMP 2007

(i) Hundsdiecker Reaction : (ii)

CCl 4, Reflux
CH 3 CH 2 COOAg + Br2
Silver propanoate

CH 3 CH 2 - Br + CO 2 + AgBr
Bromoethane

(ii) Sandmeyer Reaction : (iii)

(iv)

Ans : SQP 2016, OD 2014

(i)

(iii) Preparation of Iodobenzene :

(ii)

(iv) Balz-schiemann Reaction :

(iii)

(iv)

(v) Gattermann Reaction :

184. What do you mean by nucleophilic substitution

reactions? Give examples.
Ans : COMP 2018

When a nucleophile reacts with haloalkane (the

substrate) having a partial positive charge on the
carbon atom bonded to halogen. A substitution
reaction takes place and halogen atom called leaving
group departs as halide ion. Since the substitution
reaction is initiated by a nucleophile, it is called
183. Draw the structures of major monohalo products in nucleophilic substitution reaction.
each of the following:

(i)
Page 320 Haloalkanes and Haloarenes Chap 6

Example : (a) CH 3 C / N + H 2 O
(i) conc. HCl
CH 3 CONH 2
(ii) H 2 O
Ethanenitrile Ethanamide
Reagent Nucleophile Substitution Class Dil. acid
(b) CH 3 - C / N + 2H 2 O CH 3 COOH + NH 3
(Nu - ) Product of Main Ethanenitrile Ethanoic acid

R-Nu Product Na/C 2 H 5 OH

(c) CH 3 - C / N + 4 [H] CH 3 CH 2 - NH 2
Ethanenitrile Ethanamine
NaOH or OH -
ROH Alcohol
KOH 186. Explain the mechanism of nucleophilic substitution
H2 O H2 O ROH Alcohol reactions.
Ans : DELHI 2008
NaOR' RO - ROR' Ether
There are two types of nucleophilic substitution reactions:
NaI I- R-I Alkyl iodide
(i) S N 2 (Substitution Nucleophilic Bimolecular)
NH3 NH3 R - NH3 Primary : The recation between methyl chloride and
amine hydroxide ion to form methanol and Cl- ion
R'NH2 R'NH2 RNHR’ Secondary follows second order kinetic.
amine Rate = K [CH 3 Cl] [OH-]
R'R''NH R'R''NH RNR'R'' Tert. amine When incoming nucleophile (OH- ) approaches the
KCN - C / N: RCN Nitrile alkyl halide C - Cl bond starts breaking and new
(Cyanide) C - OH bond starts forming. These two process
take place simultaneously in a single step and no
AgCN AgCN : RNC Isonitrile
(isocyanide) intermediate is formed, therefore S N 2 reactions
are concerted reactions i.e., take place in one step.
KNO2 O = N-O R-O-N = O Alkyl nitrite
AgNO2 Ag - O - N = O R - NO2 Nitroalkane
R'COOAg R'COO - R'COOR Ester
LiAlH 4 H- RH Hydrocarbon
R - M+ R'- RR’ Alkane The attack of nucleophile (OH- ) occurs from
the back side and the leaving group (Cl- ) leaves
185. Write the following: from the front side. As a result S N 2 reactions are
(i) Willamson synthesis always accompanied by inversion of configuration.
(ii) Reaction of alkyl halide with dry silver oxide This inversion is called Walden inversion. The
(iii) Uses of alkyl cyanides. reactivity in S N 2 depends upon steric hindrance;
more the steric hindrance slower the reaction.
Ans : FOREIGN 2003
(ii) S N 1 (Substitution Nucleophilic Unimolecular) :
(i) Willamson Synthesis :
S N 1 reactions are generally carried out in polar
protic solvents (like water, alcohol acetic acid
etc.). The reaction between tert-butyl bromide
and hydroxide ion (OH-) to form tert-butyl
alcohol and Br- ion follows I st order kinetics.
(ii) Reaction of Alkyl Halide with Dry Silver Oxide :

(iii) Use of Alkyl Cyanides : Alkyl cyanides R - CN It occurs in two Steps.

are very useful compounds since they can be Step I
converted into (i) amines on partial hydrolysis Tert-butyl bromide undergoes ionization to produce
with conc. HCl or conc. H 2 SO 4 followed by tert-butyl carbocation and bromide ion.
treatment with H 2 O (ii) Carboxylic acid on
complete hydrolysis with dilute mineral acids or
alkalies and (iii) primary amines by reduction
with sodium and alcohol or catalytically with H 2
/Ni or with LiAlH 4 .
Chap 6 Haloalkanes and Haloarenes Page 321

This step is slow and reversible and hence it is the Ans : FOREIGN 2009

rate determining step. (i) on ionization gives 3° carbocation

Step II
Carbocation formed in step I is very reactive, it is
immediately attacked by the nucleophile i.e., OH- ion
to give tert-butyl alcohol.
on ionization gives 2° carbocation

Since 3° carbocations are more stable than 2°

This step is fast and hence does not affect the rate of carbocation, therefore will react fast in
reaction. S N 1 reaction.
We can sum up the order of reactivity of alkyl halides
towards S N 1 and S N 2 reactions as follows: (ii)

Since 2° carbocation are more stable than 1°

carbocation therefore will react faster in
S N 1 reaction.
188. Arrange the compounds of each set in order of
reactivity towards S N 2 displacement:
3° alkyl halide undergo S N 1 reaction very fast (i) (a) 2-Bromo-2-methylbutane
because of the high stability of 3° carbocation for (b) 1-Bromopentane
the same reasons, allylic and benzylic halides show (c) 2-Bromopentane
high reactivity towards the S N 1 reaction, because (ii) (a) 1-Bromobutane-3-methylbutane
the carbocation thus formed gets stabilised through (b) 2-Bromo-2-methylbutane
resonance as shown below:
(c) 2-Bromo-3-methylbutane
(iii) (a) Bromobutane
(b) 1-Bromo-2, 2-dimethylpropane
(c) 1-Bromo-2-methylbutane
(d) 1-Bromo-3-methylbutane.
Ans : SQP 2010, DELHI 2007

(i) (a)
For a given alkyl group, the reactivity of the halide,
R - X follows the same order in both the mechanism
R - I > R - Br > R - Cl > R - F .
(b)
187. In the following pairs of halogen compounds which
compound undergo faster S N 1 reaction?
(c)

(i) and
The reactivity in S N 2 depends upon steric hindrance
hence the order of reactivity in S N 2 reaction is
(ii) and
1° > 2° > 3° therefore order of reactivity is:
Page 322 Haloalkanes and Haloarenes Chap 6

1-Bromopentane>2-Bromopentane>2-Bromo-2- (vii) CH 3 CH 2 CH = CH 2 + HBr

Peroxide

methylbutane.
(viii) CH 3 CH = C (CH 3) 2 + HBr $
Ans : COMP 2004
(ii) (a)
(i) CH 3 CH 2 CH 2 I

(ii)
(b)
(iii)

(iv) CH 3 CH 2 - CN
(v) C 6 H 5 - O - C 6 H 5
(c)
(vi) CH 3 CH 2 CH 2 Cl
(vii) CH 3 CH 2 CH 2 CH 2 Br
The order of reactivity towards S N 2 displacement
is: (viii)
1-Bromo-2-methyl butane (1°)
>2-Bromo-3-methylbutane (2°)
190. Predict the alkenes that would be formed by
>2-Bromo-2-methylbutane (3°) dehydrohalogenation of the following halides. With
sodium ethoxide in ethanol and identity the major
(iii) (a)
alkene.
(i) 1-Bromo-1-methyl cyclohexane
(ii) 2-Chloro-2-methylbutane
(b) (iii) 3-Bromo-2, 2, 3-trimethylpentane
Ans : COMP 2001
(i) In 1-bromo-1-methylcyclohexane, the b
-hydrogens on either side of the Br atom are
(c) equivalent therefore only 1-alkene is formed.

(d)

The order of steric hindrance is b > c > d > a a

hence the order of reactivity is: (ii) 2-chloro-2-methyl butane has two different sets of
1-Bromobutane>1-Bromo-3-methylbutane>1- equivalent b -hydrogen.
Bromo-2-methylbutane>1-Bromo-2,
2-dimethylpropane.
189. Write the structure of the major organic product in
each of the following reaction:
acetone, heat
(i) CH 3 CH 2 CH 2 Cl + Nal
ethanol heat
(ii) (CH 3) 3 CBr + KOH
Water
(iii) CH 3 CH (Br) CH 2 CH 3 + NaOH
aq ethanol
(iv) CH 3 CH 2 Br + KCN
According to Saytzeff’s rule, more highly
(v) C 6 H 5 ONa + C 2 H 5 Cl $
substituted alkene (b) being more stable, is the
(vi) CH 3 CH 2 CH 2 OH + SOCl 2 $ major product.
Chap 6 Haloalkanes and Haloarenes Page 323

(iii) 192. Give the preparation of chlorobenzene from benzene

diazonium chloride and give its reaction with:
(i) Na
(ii) CH 3 Cl is the presence of anhydrous AlCl 3
(iii) H 2 SO 4
(iv) HNO 3 in the presence of cone. H 2 SO 4 .
Ans : DELHI 2006

Preparation

191. Write one use of each of the following:

(i) Chloroform (i) Fitting Reaction :
(ii) Iodoform
(iii) Freons
(iv) DDT
(v) Carbon-tetrachloride
Ans : COMP 2011, SQP 2008
(i) Chloroform or Trichloromethane, CHCl 3 :
(a) It is used as a solvent for rubbers, resins, oil
and fats, alkaloids and other substances.
(b) It is used in the production of the freon
refrigerant. (ii) Friedel-Crafts Alkylation :
(ii) Idoform or Triodomethane, CHI 3 : It was earlier
used as an antiseptic for dressing wounds, now a
days it has been replaced by other formulations
containing iodine because of its unpleasant smell.
(iii) Freons :
(a) It is used as a refrigerant (cooling agent) in
refrigerators and air conditioners.
(b) It is used as a propellant in aerosols and foams
(iii) Sulphonation :
(i.e., hair sprays, deodrants, shaving cream etc.).
(iv) DDT
(a) It is used as insecticide for sugarcane and
fodder crops.
(b) It is used to kill mosquitoes and other insects,
particularly against Anopheles mosquitoes
which spread malaria.
(v) Carbon-tetrachloride or Tetrachloromethane,
CCl 4 : (iv) Nitration :
(a) It is used in the manufacture of refrigerants
and propellants for aerosols cans.
(b) It is used in the synthesis of freons.
(c) It is used as a solvent in the manufacture of
pharmaceuticals.
(d) It is used as a fire extinguisher under the
name pyrene.
Page 324 Haloalkanes and Haloarenes Chap 6

193. Write the following reaction: 194. How will you bring about the following conversions?
(i) Friedel-Crafts acylation (i) Ethanol to but-1-yne
(ii) Wurtz-Fitting reaction (ii) Ethane to bromoethene
(iii) Dow’s process (iii) Propene to 1-nitropropane
(iv) Ullmann biaryl synthesis (iv) Ethanol to ethyl fluoride
Ans : SQP 2003 (v) Toluene be benzyl alcohol
(i) Freidel-Crafts Acylation : (vi) Propene to propyne
(vii) Bromomethane to propanone
(viii) But-1-ene to but-2-ene
Ans : DELHI 2000
(i) Ethanol to but-1-yne :
SOCl 2
CH 3 CH 2 OH -SO 2, - HCl
CH 3 CH 2 - Cl
HC / C- Na+
Sod. acetylide
-NaCl
CH 3 CH 2 - C / CH

:HC HC / C-1 Na+ + NH 3D

Liq. NH 3
/ CH + NaNH 2
Acetylene Sodium acetylide

(ii) Ethane to Bromoethene :

hv
CH 3 - CH 3 + Br2 -HBr
CH 3 CH 2 - Br
Ethane Bromoethane

(ii) Wurtz-Fitting Reaction : KOH (alc) Br2 /CCl 4

-HBr
CH 2 / CH 2 BrCH 2 - CH 2 - Br
Ethene 1, 2 - Dibromoethane

KOH (alc)
-HBr
CH 2 = CHBr
Bromoethene

(iii) Propene to 1-nitropropane :

HBr, ROOR
CH 3 - CH = CH 2 Peroxide effect
CH 3 - CH 2 - Br
Propene Bromoethane

AgNO 2 (alc)
-AgBr
CH 3 - CH 2 - CH 2 - NO 2
1 - Nitropropane

(iv) Ethanol to Ethyl Fluoride :

(iii) Dow’s Process : SOCl 2
CH 3 CH 2 - OH -SO 2, - HCl
CH 3 CH 2 - Cl
Ethanol Chloroethane

AgF
CH 3 - CH 2 - F
Ethyl fluoride

(v) Toluene be Benzyl Alcohol :

(iv) Ullmann Biaryl Synthesis : (Reaction with Cu

powder of Iodobenzene) When an iodobenzene is
heated with Cu powder in a sealed tube, diaryl is
formed.
(vi) Propene to Propyne :
Br2 /CCl 4
CH 3 - CH = CH 2 CH 3 - CHBr - CH 2 Br
Propene 1, 2 - Dibromopropane

2KOH (alc)
-2KBr, - 2H 2 O
CH 3 - C / CH
Propyne

(vii) Bromomethane to Propanone :

KCN (alc.)
CH 3 - Br -KBr
CH 3 - C / N
Bromomethane Acetonitrile
Page 326 Haloalkanes and Haloarenes Chap 6

197. How will you bring about the following conversions? 198. What are haloarenes? How are they classified? Give
(i) 1-Chlorobutane to n-Octane one method each for the preparation of nuclear and
(ii) Benzene to biphenyl side chain substituted haloarenes.
(iii) Ethene to dibromoethane Ans : COMP 2004

(iv) Aniline to iodobenzene Haloarenes

(v) Propene to propan-1-ol The replacement of hydrogen atom (s ) in an aromatic
(vi) 1-Bromopropane to 2-bromopropane hydrocarbon chain by the halogen atom (s ) which
Ans : OD 2010, COMP 2005
results in the formation of aryl halide (haloarene).
Haloarenes contain halogen atom (s ) attached to the
(i) 1-Chlorobutane to n-Octane :
sp2 hybridised carbon atom (s ) of aryl group.
2CH 3 - CH 2 - CH 2 - CH 2 - Cl + 2Na
(i) Nuclear Halogen Derivatives : Halogen derivatives
Dry ether
CH 3 - CH 2 - CH 2 - CH 2 - CH 2 of aromatic hydrocarbons in which the halogen
atom (F, Cl, Br, or I) is directly attached to
- CH 2 - CH 2 - CH 3
an aromatic ring are called aryl halides. Some
(ii) Benzene to Biphenyl : examples of aryl halids are:

(iii) Ethene to Dibromoethane :

Preparation : They are prepared by direct
halogenation of aromatic hydrocarbons.

(iv) Aniline to Iodobenzene :

(ii) Side Chain Halogen Derivatives : Halogen

derivatives of aromatic hydrocarbons in which the
halogen atom is linked to one of the carbon atoms
of the side chain carrying the aryl group are called
aryl halides. For example:
(v) Propene to Propan-1-ol :
CH 3 - CH = CH 2
Propene
HBr/Peroxide
(Anti Mark. Addition)
CH 3 - CH 2 - CH 2 Br
KOH (aq)
-KBr
CH 3 CH 2 CH 2 OH
Propan 1 - ol

(vi) 1-Bromopropane to 2-bromopropane :

CH 3 - CH 2 - CH 2 - Br Preparation : By the direct halogenation of a
1 - Bromopropane
suitable arene.
Page 328 Haloalkanes and Haloarenes Chap 6

(iii) When methl chloride is treated with AgCN, occur, the polarity of the carbon halogen bond is
methyl cyanide is formed: responsible for these substitution reactions. The rate
CH 3 - Cl + AgCN $ CH 3 - C / N + AgCl of S N 1 reactions are governed by the stability of
Methyl chloride Methyl cyanide
carbocation whereas for S N 2 reactions steric factor
is the deciding factor. If the starting material is a
chiral compound, we may end up with an inverted
CASE BASED QUESTIONS product or racemic mixture depending upon the
type of mechanism followed by alkyl halide. Cleavage
of ethers with HI is also governed by steric factor
203. Read the passage given below and answer the following
and stability of carbocation, which indicates that in
questions :
organic chemistry, these two major factors help us in
A farmer cultivating his land near the village pond
deciding the kind of product formed.
was also drawing water from it for irrigation. He used
Answer The following questions :
insecticide excessively to protect his crops and improve
the harvest, over a period of time his agricultural growth (a) Out of chlorobenzene and benzyl chloride, which
improved vastly. But the pond lost his aquatic life. one gets easily hydrolysed by aqueous NaOH and
why?
(i) Give the IUPAC name of compound used as an
insecticides. (b) Predict the stereochemistry of the product formed
if an optically active alkyl halide undergoes
(ii) Contrast the activity in plants and aquatic life
substitution reaction by S N 1 mechanism.
with insecticides.
(c) Following compounds are given to you :
(iii) Is the above used insecticides is bio-degradable ?
2-Bromopentane, 2-Bromo-2-methylbutane,
(iv) Give any alternative which helps farmers.
1-Bromopentane
Ans :
(i) Write the compound which is most reactive
(i) 2, 2-Bis (4-chloropheny1)-1, 1, 1-trichloro-ethane. towards S N 2 reaction.
(ii) Plant growth improved because it is a very (ii) Write the compound which is optically active.
effective insecticides but aquatic life got destroyed
(d) What are the points of similarities between S N 1
as it is highly toxic to fish.
and S N 2 reactions?
(iii) No, above used insecticide i.e., DDT is non-
Ans :
biodegradable.
(a) Benzyl chloride gets easily hydrolysed by aq. NaOH
(iv) Instead of DDT, farmers can use neem powder
due to formation of stable benzyl carbocation.
which will not harm the soil as well as aquatic life.
But due to partial double bond character of C–Cl
204. Raj is a student of class VI, he fell down during bond in chlorobenzene, it does not hydrolyse.
playing. His friends immediately took him to the (b) Stereochemical aspects of nucleophilic substitution
doctor and the doctor dressed the would with the reaction in S N 1 proceeds with racemisation.
halogen compound. (c) (i) 1-Bromopentane
(i) What is the name and formula of the compound ? (ii) 2-Bromopentane
(ii) Write any one method of preparation. (d) Both S N 1 and S N 2 reactions undergo substitution
(iii) The above used halogen compound shows which of a nucleophile by other nucleophile. In both
property ? the reactions configuration of product changes,
(iv) Give an example of halogen compound, used as partially in S N 1 and completely in S N 2 .
an anaesthetic.
206. When haloalkanes with (b -hydrogen atom are
Ans : boiled with alcoholic solution of KOH, they undergo
(i) Iodoform, CHI 3 . elimination of hydrogen halide resulting in the
(ii) CH 3 CH 2 OH + NaOI " CHI 3 + HCOOH formation of alkenes. These reactions are called b
(iii) The above halogen compound has antiseptic -elimination reactions or dehydrohalogenation
property. reactions. These reactions follow Saytzeff’s rule.
(iv) Chloroform. Substitution and elimination reactions often compete
with each other. Mostly bases behave as nucleophiles
205. The substitution reaction of alkyl halide mainly occurs and therefore can engage in substitution or elimination
by S N 1 or S N 2 mechanism. Whatever mechanism reactions depending upon the alkyl halide and the
alkyl halides follow for the substitution reaction to reaction conditions.
Page 330 Haloalkanes and Haloarenes Chap 6

haloalkane and therefore, are less reactive

towards nucleophilic substitution reaction.
(ii) In halorenes, halogen is attached to sp2
-hybridised carbon while in haloalkanes,
halogen is attached to sp3 -hvbridised carbon.
(b)

Read the above passage carefully and answer the (c) The presence of nitro group at ortho and para
following questions : positions withdraws the electron density from the
benzene ring and thus facilitates the attack of the
(a) Chlorobenzene is extremely less reactive towards
nucleophile on aryl halide. The carbanion thus
a nucleophilic substitution reaction. Give two
formed is stabilised by resonance. The negative
reasons for the same.
charge appeared at ortho- and parc-positions with
(b) Write the product formed when p
respect to the halogen substituent is stabilised by
-nitrochlorobenzene is heated with aqueous
- NO 2 , group while in case of meta-nitrobenzene
NaOH at 443 K followed by acidification.
none of the resonating structures bear the
(c) Why NO 2 , group shows its effect only at ortho negative on carbon atom bearing the - NO 2 ,
and para-positions and not at meta-position? group. Therefore, the presence of - NO 2 , group
(d) Aryl halides are extremely less reactive towards at meta position does not stabilise the negative
nucleophilic substitution. Predict and explain the charge and hence no effect on reactivity of - NO 2
order of reactivity of the following compounds , group at meta position.
towards nucleophilic substitution : (d) III > II > I. Presence of an electron withdrawing
group (like - NO 2 ) at ortho and para positions
increases the reactivity of haloarenes towards
nucleophilic substitution.

Ans :
(a) Haloarenes are much less reactive than haloalkanes
towards nucleophilic substitution reactions due to
the following reasons.
(i) Resonance effect : In haloarenes the electron
pairs on halogen atom are in conjugation
with p -electrons of the ring and the following 208. Chemically, chloroform is employed as a solvent for
resonating structures are possible. fats, alkaloids, iodine and other substances. The major
use of chloroform today is in the production of the
freon refrigerant R-22. It was once used as a general
anaesthetic in surgery but has been replaced by less
toxic, safer anaesthetics, such as ether. As might be
expected from its use as an anaesthetic, inhaling
chloroform vapours depresses the central nervous
C-Cl bond acquires a partial double bond system. Breathing about 900 parts of chloroform
character due to resonance. As a result, the per million parts of air (900 parts per million) for a
bond cleavage in haloarene is difficult than short time can cause dizziness, fatigue, and headache.
Chap 6 Haloalkanes and Haloarenes Page 331

Chronic chloroform exposure may cause damage to the

liver (where chloroform is metabolised to phosgene)
and to the kidneys, and some people develop sores
when the skin is immersed in chloroform. Chloroform
is slowly oxidised by air in the presence of light to
an extremely poisonous gas, carbonyl chloride, also
known as phosgene. It is therefore stored in closed
dark coloured bottles completely filled so that air is
kept out.
(i) Which gas is formed when chloroform is oxidised
in air ?
(ii) Name the freon refrigerant used now a days ?
(iii) What might happen if chloroform is inhaled for
longer time ?
Ans :
(i) Phosgene,
(ii) R-22
(iii) Chronic chloroform exposure may cause damage
to the liver (where chloroform is metabolised to
phosgene) and to the kidneys, and some people
develop sores when the skin is immersed in
chloroform.
209. Alkyl halides are best prepared from alcohols, which
are easily accessible. The hydroyl group of an alcohol
Is replaced by halogen reaction on with concentrated
halogen acids, phosphorus halides or thionyl chloride.
Thionyl chloride is preferred because the other two
products are escapable gases. Hence the reaction
gives pure alkyl halides. Phosphorus tribromide and
triiodide are usually generated in situ (produced in the
reaction mixture) by the reaction of red phosphorus
with bromine and iodine respectively. The preparation
of alkyl chloride is carried out either by passing dry
hydrogen chloride gas through a solution of alcohol
or by heating a solution of alcohol in concentrated
aqueous acid.
(i) What is the condition to obtain good yield of
R-I ?
(ii) Is the above method suitable for preparation of
aryl halides. Why and why not ?
Ans :
(i) Good yields of R—I may be obtained by heating
alcohols with sodium or potassium iodide in 95%
phosphoric acid.
(ii) Because the carbon-oxygen bond in phenols has
a partial double bond character and is difficult to
break being stronger than a single bond.

***********