Download P-BLOCK ELEMENTS

UNIT
5
P-BLOCK ELEMENTS
Elements in which last electron enter into p-sub orbit are known as p-block elements
(i) General electronic configuration  ns2 np1–6
(ii) Position in the periodic table
They belong to group 13 to 18 of periodic table.
Most of the p-block elements are non metals.
(iii) Valency and oxidation state:
Elements of group – 13 and 14 show valency equal to valence electron whereas
elements of group 15 to 18 show valency. (8–number of valence electrons)
(iv) Inert pair effect:
The reluctance of a pair of s-electron to take part in bonding is called inert pair
effect.
From top to bottom in group due to poor shielding of d, and f-orbitals, effective
nuclear charge increases, it holds nS electron tightly. This is responsible for inert pair
effect.
It is observed in heavier elements of group 13, 14 and 15.
Ex. Group 13-Galium Indium and Thalium
Group 14-Germanium, tin and lead
Group 15-Arsenic, Antimony and Bismuth
* Ga, In, Ge, Sn, As and Sb show variable valency due to inert pair effect.
221
D:\BOOKS\JANUARY\26-A NEW APPROACH TO INORGANIC CHEMISTRY VOL. 1\UNIT-5\IIIRD PROOF DT. 27/3/12
222 A NEW APPROACH TO INORGANIC CHEMISTRY
* * Thalium, Lead and Bismuth show valency two unit less than their group valency due to
inert pair effect.
(v) P-block elements have high ionization enthalpy, high electron gain enthalpy and high
electronegativity hence, they have strong tendency to form anions
(vi) Metallic character increases down the group.
(vii) Anomalous behaviour: Properties of First member of each group of p-block differs
from the remaining members of its group.
Cause of anomalous behaviour
(i) small size
(ii) high electronegativity
(iii) absence of d-orbital
(iv) ability to form p-p multiple bond with itself and with other elements
eg. C = C, C  C, N  N, C  O, C  N, N = O etc
D:\BOOKS\JANUARY\26-A NEW APPROACH TO INORGANIC CHEMISTRY VOL. 1\UNIT-5\IIIRD PROOF DT. 27/3/12
P-BLOCK ELEMENTS 223
GROUP-13
BORON FAMILY
Members – B, Al, Ga, In and Tl
Table 5.1: Physical Properties of Group 13 Elements
Property
Element
Boron
B
Atomic number
Aluminium
Al
Gallium
Ga
Indium
In
Thallium
Tl
5
13
31
49
81
Atomic mass (g mol–1)
10.81 (11)
26.98 (27)
69.72 (70)
114.82 (115)
204.38 (204)
Electronic configuration
[He] 2s2 2p1
[Ne] 3s2 3p1
[Ar]3d104s24p1
[Kr]4d105s25p1
[Xe]4f145d106s26p1
Atomic radius/pm
(85)
143
135
167
170
Ionic radius M3+/pm
(27)
53.5
62.0
80.0
88.5
Ionic radius M+/pm
–
–
120
140
150
Ionization
tH1
801
577
579
558
589
enthalpy
tH2
2427
1816
1979
1820
1971
(kJ mol–1)
tH3
3659
2744
2962
2704
2877
Electronegativity2.0
1.5
1.6
1.7
1.8
Density / g cm–3 at 289 K
2.35
2.70
5.90
7.31
11.85
Melting point / K
2453
933
303
430
576
Boiling point / K
3923
2740
2676
2353
1730
Ee / V for (M3+/ M)
–
–1.66
–0.56
–0.34
+1.26
Ee / V for (M+ / M)
–
+0.55
–0.79 (acid)
–1.39 (alkali)
–0.18
–0.34
5.1 GENERAL CHARACTERISTICS OF BORON FAMILY (GROUP - 13)
(i) Electronic Configuration – ns2, np1
Nature of elements: Boron is a typical non-metal, Aluminium is a metal but shows
many chemical properties similar to boron.
Gallium, indium and thalium are exclusively metallic.
(ii) Atomic radii: On moving down the group, atomic and ionic radii increases because
one extra shell of electrons is added.
D:\BOOKS\JANUARY\26-A NEW APPROACH TO INORGANIC CHEMISTRY VOL. 1\UNIT-5\IIIRD PROOF DT. 27/3/12
224 A NEW APPROACH TO INORGANIC CHEMISTRY
Atomic radii of gallium (135 p.m.) is less than that of Al (143 p.m.)
Cause: The presence of additional 10 d electrons in gallium offer very poor screening effect
for the valence electrons.
As a result effective nuclear charge increases and electron cloud shrinks.
There is very small increase in atomic radius from Indium to thalium due to very poor shielding
of f-electrons in thalium which increases effective nuclear attraction.
(iii) Ionization Enthalpy: The ionization enthalpy value decreases down the group.
But, decrease is less than that expected, this is because, the d and f electrons
which have very poor screening effect is unable to compensate the increase in
nuclear charge.
The ionisation enthalpy decreases from B to Al due to increase in atomic size. But
ionisation enthalpy from Al to Tl are almost same because the presence of d and f
electrons shield the nuclear charge very imperfectly and compensate the effect in
increasing atomic size.
The sum of the first three ionisation enthalpy iH1, iH2 and iH3 for these elements
are very high. This is why these elements generally do not form trivalent (M 3 )
cations.
(iv) Electronegativity: Electronegativity of group-13 elements decreases from Boron to
Aluminium and then increases slightly.
(iv) Physical Properties: Boron is non-metallic, extremely hard and black coloured solid.
It exists in many allotropic forms. Boron has unusually high melting point due to
strong crystalline lattice.
Rest of the members are soft metals with low melting point. Gallium is low melting
solid. M.P. = 30°C but very high boiling point 2676 K. Therefore it is useful material
for measuring high temperature.
(v) Oxidation State: All the elements of boron family show oxidation state +3. Stability
of +3 oxidation state decreases down the group of periodic table due to inert pair
effect.
However stability of +1 oxidation state increases from B to Tl. B and Al show +3
oxidation state only Ga and In show oxidation state +3 and +1 both whereas Tl forms
compounds in +1 oxidation state only.
Boron has very small size, sum of first three ionization enthalpies is very high. Therefore
Boron does not form B3+ ion. Boron forms only co-valent bonds.
D:\BOOKS\JANUARY\26-A NEW APPROACH TO INORGANIC CHEMISTRY VOL. 1\UNIT-5\IIIRD PROOF DT. 27/3/12
P-BLOCK ELEMENTS 225
Note: Anhydrous AlCl3 is covalent but hydrated AlCl3 is ionic.
Cause: Sum of three ionization enthalpy of aluminium is high but hydration enthalpy of
Al3+ ion more than compensates the sum of three ionization enthalpies. This is why, in water Al
forms ionic compound.
Compounds in trivalent state are electron deficient and have tendency to accept a pair of
electrons to achieve stable electronic configuration. These compounds are known as lewis acid.
Chemical Properties of Boron Family (Group-13)
1. Action of air: Boron is unreactive with air in crystalline form.
Amorphous boron on heating in air forms oxide.
4B + 3O2  2B2O3
Aluminium forms a protective oxide layer on the surface.
Aluminium metal on heating in air form Al2O3.
At high temperature they form nitride also,
2Al + N2  2AlN
2B + N2  2BN
Boron oxide is acidic, aluminium and gallium oxides are amphoteric and oxides of other
elements are basic.
2. Action of acid: Boron is non metal, therefore does not react with acids and alkalis as well.
Aluminium is amphoteric and dissolves in acids and alkalis both.
2Al(s) + 6HCl(aq.)  2AlCl3(aq.) + 3H2(g)
2Al(s) + 2Na OH(aq.) + 6H2O(l)  2Na[Al(OH)4](aq.) + 3H2(g)
Note:
Fe, Al, Cr, Co, Ni become passive in conc. HNO3 due to formation of a protective
oxide layer on the surface.
3. Action of halogen: All the elements react with halogen and form trihalide except Tl.
2E(s) + 3X2(g)  2EX3(s)
Lewis acid strength of boron trihalides are
BF3 < BCl3 < BBr3 < BI3 [where X = F, Cl, Br, I]
T1 I3 does not exist due to very strong oxidising nature of Tl3+.
D:\BOOKS\JANUARY\26-A NEW APPROACH TO INORGANIC CHEMISTRY VOL. 1\UNIT-5\IIIRD PROOF DT. 27/3/12
226 A NEW APPROACH TO INORGANIC CHEMISTRY
5.1.2 Chemistry of Boron
(i) Anomalous properties of Boron
Properties of boron differs from other members of its family
Cause of anomalous behaviour of boron
– Small Size
– Absence of d orbital
– High electronegativity
Difference in properties of Boron and other group-13 elements.
– Boron is non-metal whereas all other elements are metals
– Boron is hard
– Boron oxides are acidic
– Maximum co-valency of boron is 4 while other elements show valency upto 6.
– Boron halide is monomeric where as halides of other elements are dimeric.
(i)
Cl
Cl
(ii)
Cl
Cl
Cl
B
Al
Al
Cl
Cl
Cl
Cl
(ii) Sources of Boron: In combined state important sources of boron are:
–
–
–
–
Orthoboric acid (H3BO3)
Borax (Na2B4O7·10H2O)
Kernite (Na2 B4O7·4H2O)
Colemanite (Ca2B6O11)
(iii) Isotopes of Boron:
Boron exists in two isotopic forms.
B-10 (19%) and B-11 (81%)
(iv) Sources of Aluminium
Aluminium is third most abundant element in the earth crust. Important sources of Al
are
Bauxite (Al2O3·2H2O)
Cryolite (Na3AlF6)
D:\BOOKS\JANUARY\26-A NEW APPROACH TO INORGANIC CHEMISTRY VOL. 1\UNIT-5\IIIRD PROOF DT. 27/3/12
P-BLOCK ELEMENTS 227
5.1.3 Compounds of Boron
1. Borax (Na2B4O7.10H2O)
Its correct formula is Na2[B4O5(OH)4].8H2O
Preparation of Borax
It occurs naturally as tincal (Suhaga) which contains about 50% borax in certain land, lakes.
It is also obtained from the mineral colemanite by boiling it with a solution of Na2CO3 .
Ca2B6O11 + 2Na2CO3  Na2B4O7 + 2CaCO3 + 2NaBO2
colemanite
Properties of borax
Borax
(a) It is white crystalline solid. soluble in water.
(b) Its aqueous solution is alkaline.
Na2B4O7 + 7H2O  2NaOH + 4H3BO3
(c) Action of heat: On heating, borax first looses water molecules and swells up. On
further heating, it melts into a transparent liquid. The liquid solidifies into a glass like
solid known as borax bead.
Na2B4O7·10H2O  Na2B4O7  2Na BO2 + B2O3
This bead when heated with trasition metal salt becomes coloured. This colour of bead gives
clue for the identification of metal.
(II) Ortho Boric Acid [H3BO3]
Preparation
(a) It is prepared by the action of acid on aqueous solution of borax.
Na2B4O7 + 2HCl + 5H2O  2NaCl + 4H3BO3
(b) It can also be prepared by the hydrolysis of boron halides or hydrides.
BCl3 + 3H2O  H3BO3 + 3HCl
Properties of boric acid: (ii) It is a white crystalline solid with soapy touch, sparingly soluble
in water.
It is a weak monobasic acid. H3BO3 is not a protonic acid i.e. It will not donate proton. It acts
as lewis acid.
H3BO3 + 2H2O  B(OH)–4 + H3O+
D:\BOOKS\JANUARY\26-A NEW APPROACH TO INORGANIC CHEMISTRY VOL. 1\UNIT-5\IIIRD PROOF DT. 27/3/12
228 A NEW APPROACH TO INORGANIC CHEMISTRY
Action of heat:
On heating above 370K it gives metaboric acid which on further heating gives boric oxide.
370K

H 3BO 3 
 HBO 2 
 B 2O 3
H
O
H
B
O
O
H
H
O
O
Structure of boric acid:
B
H
O
H
Boric acid exists as a cyclic dimer.
H3BO3 is a planar molecule linked together through hydrogen bonds.
(III) Hydrides of Boron
Boron forms a large number of compounds with hydrogen known as boranes
Boranes has general formula
(i) BnHn+4 and
(ii) BnHn+6
Ex. Diborane (B2H6)
Ex. B4H10.
(i) Preparation of diborane
(a) It is prepared by heating boron trifluoride with lithium aluminium hydride.
4BF3 + 3LiAlH4  2B2H6 + 3LiF + 3AlF3
(i) In laboratory diborane is prepared by the oxidation of sodium borohydride with iodine.
2NaBH4 + I2  B2H6 + 2NaI + H2
(ii) On large scale diborane is produced by the reaction of BF3 with sodium hydride
2BF3 + 6NaH  B2H6 + 6NaF
D:\BOOKS\JANUARY\26-A NEW APPROACH TO INORGANIC CHEMISTRY VOL. 1\UNIT-5\IIIRD PROOF DT. 27/3/12
P-BLOCK ELEMENTS 229
Properties of diborane
(i) It is a colourless, highly toxic gas.
(ii) B.P. – 180K
(iii) It catches fire in air spontaneously.
B2H6 + 3O2  B2O3 + 3H2O : H = – 1976 kJ/mol
(iv) It is hydrolysed by water
B2H6 + 6H2O  2H3BO3 + 6H2
(v) It combines with Lewis base to give adduct (i.e. addition product)
B2H6 + 2CO  2BH3·CO
B2H6 + 2N(Me)3  2BH3·NMe3
(vi) With Ammonia
Diborane react with ammonia to give a diamine aduct, which on further heating gives
borazene known as inorganic benzene.

3B2H6 + 6NH3  3[BH2(NH3)2]+ + 3[BH4]– 
 2B3N3H6 + 12H2
H
|
N
H
B
|
N
H
H
B
|
N
B
|
H
H
Structure of Inorganic benzene or borazine
(vii) Structure of diborane
Two boron and four hydrogen atoms lie in the same plane whereas two hydrogen atom
H
lie above and below the plane known as hydrogen bridge.
B
H
D:\BOOKS\JANUARY\26-A NEW APPROACH TO INORGANIC CHEMISTRY VOL. 1\UNIT-5\IIIRD PROOF DT. 27/3/12
H
H
B
H
H
230 A NEW APPROACH TO INORGANIC CHEMISTRY
Explanation:
B(Ground State) – 1S2 2S2 2p1

2S 2
B(excited state) – 1S2 2S1 2px1 2py1

2 p1

2S 1

2 p1x

2 p1y 2 p
z
Boron atom has four orbitals in valence shell but only three electrons. Two half filled orbital of
each boron atom overlaps with 1s orbital of hydrogen atom to form B–H covalent bonds. Half
filled orbtial of one boron atom, empty orbital of other boron atom and 1s orbital of hydrogen
atom overlap to form 2-electron-3 centre (2e = 3c) bond. This is also called multicentred bond or
banana bond.
H
B
B
Covalent bond
H
2e– – 3C bond
(IV) Sodium Borohydride (NaBH4)
Sodium tetra-hydridoborate is also known as sodium borohydride.
It is prepared by reaction of sodium hydride with diborane in diethylether
2NaH + B2H6  2Na[BH4]
It is used as reducing agent in many organic synthesis.
5.1.4 Uses of Boron and Aluminium
(a) Uses of Boron
(i) Boron fibres are used in making bullet proof vests and air craft parts.
(ii) B-10 is used in nuclear industry as control rods.
D:\BOOKS\JANUARY\26-A NEW APPROACH TO INORGANIC CHEMISTRY VOL. 1\UNIT-5\IIIRD PROOF DT. 27/3/12
P-BLOCK ELEMENTS 231
(iii) Borax and boric acid is used in the manufacture of heat resistant glasses, glass
wool and fiber glass (pyrex).
(iv) Borax is used as flux for soldering metals, medicinal soap.
(v) Boric acid is used as mild antiseptic and for glazing earthen ware.
(b) Uses of Aluminium
Aluminium is used for making pipes, tubes, rods, wires, files etc. It is used for
packing, electrical wires, making utensils.
D:\BOOKS\JANUARY\26-A NEW APPROACH TO INORGANIC CHEMISTRY VOL. 1\UNIT-5\IIIRD PROOF DT. 27/3/12
232 A NEW APPROACH TO INORGANIC CHEMISTRY
GROUP-14
CARBON FAMILY [C, Si, Ge, Sn, and Pb]
Table 5.2: Atomic and Physical Properties of Group 14 Elements
Property
Element
Carbon
C
Atomic number
Silicon
Si
Germanium
Ge
Tin
Sn
Lead
Pb
6
14
32
50
82
Atomic mass (g mol–1)
12.01 (12)
28.09 (28)
72.60 (72)
118.71 (119)
207.2 (207)
Electronic configuration
[He] 2s2 2p2
[Ne]3s2 3p2
[Ar]3d104s24p2
[Kr]4d105s25p2
[Xe]4f145d6s26p2
Covalent radius/pm
77
118
122
140
146
Ionic radius M4+/pm
–
40
53
69
78
–
–
73
118
119
2+
Ionic radius M /pm
Ionization tH1
1086
786
761
708
715
enthalpy
2352
1577
1737
1411
1450
4620
3228
3300
2942
3081
6220
4354
4409
3929
4082
2.5
1.8
1.8
1.8
1.9
(kJ mol–1)
tH2
tH3
tH4
Electronegativity
d
–3
f
Density / g cm
3.51
2.34
5.32
7.26
11.34
Melting point / K
4373
1693
1218
505
600
3550
3123
2896
2024
Boiling point / K
Electrical resistivity/
ohm cm (293 K)
–
14
10
16
– 10
50
50
–5
10
2 × 10–5
5.2.1 General Characteristics of Carbon Family (Group-14)
(i) Electronic Configuration : ns2 np2
(ii) Atomic Radius
Atomic radius increases from carbon to lead but after silicon small increase in radius is
observed. This is due to poor shielding of nuclear charge by d, and f-electrons which increases
nuclear attraction on last electron.
(iii) Ionisation Enthalpy
Ionisation enthalpy of group–14 members are higher than those of group-13 elements
It decreases down the group but from silicon to lead, decrease is small due to poor shielding
of nuclear charge by d and f electrons.
D:\BOOKS\JANUARY\26-A NEW APPROACH TO INORGANIC CHEMISTRY VOL. 1\UNIT-5\IIIRD PROOF DT. 27/3/12
P-BLOCK ELEMENTS 233
(iv) Electronegativity
Members of group 14 are more electronegative than those of group 13.
Electronegativity decreases from carbon to silicon and from Si to lead electronegativity is
almost same.
(v) Nature of Element
C and Si are non-metals, Ge is metalloid whereas Sn and Pb are metals.
(vi) Oxidation State
The common oxidation state of group–14 elements are +4. In heavier members, stability of +4
oxidation state decreases in the sequence due to inert pair effect.
Carbon shows valency 4 only, because it can not expand its octet, other members can expand
their octet using d-orbitals and hence show higher valency.
5.2.2 Chemical Properties of Group-14 Elements
(a) Action of oxygen: All members when heated in oxygen form oxides.
CO2, SiO2 and GeO2 are acidic whereas SnO2 and PbO2 are amphoteric.
(b) Action of water: Only tin decomposes steam to form stannic dioxide and dihydrogen.
Sn + 2H2O  SnO2 + 2H2
(c) Action of halogen: All the elements form halide with halogen.
SnF4 and PbF4 are ionic.
PbI4 does not exist because Pb-I bond is very large and thus, unstable.
Heavier members form dihalide. stability of dihalide increases down the group due to inert
pair effect.
CCl4 does not undergo hydrolysis where as SiCl4 undergoes hydrolysis, why?
Because the central atom Si can accept a lone pair of electron from O atom of H2O in d-orbital
SiCl4 + 4H2O  Si(OH)4 + 4HCl
Carbon atom has no vacant orbital to accept ione pair of H2O
5.2.3 Anomalous behaviour of Carbon
Properties of carbon differ from most of the members of its group.
This is called anomalous behaviour of carbon.
D:\BOOKS\JANUARY\26-A NEW APPROACH TO INORGANIC CHEMISTRY VOL. 1\UNIT-5\IIIRD PROOF DT. 27/3/12
234 A NEW APPROACH TO INORGANIC CHEMISTRY
Cause of Anomalous behaviour of Carbon
(a) small size
(b) large electronegativity
(c) ability to form p–p bond with itself and with other atoms of small size and high
electronegativity.
(d) Absence of d-orbital
(e) Catenation property: The catenation property exist due to strong C-C bond.
(f ) Higher ionisation enthalpy
Heavier elements do not form p-p bond because their atomic orbitals are very large and
diffuse to have effective overlapping.
5.2.4 Allotropes of Carbon
Carbon exists in many allotropic forms.
Carbon
Crystalline
Diamond
Graphite
Amorphous
Buckminster
fullerene
Coal
coke
charcoal gascarbon
Plant charcoal
Peat
lignite
Animal charcoal
Bituminus Anthracite
(i) Diamond
In diamond carbon atom has sp3 hybridisation. Each carbon atom is linked with 4 carbon atoms
along the corners of a tetrahedron. It has a giant, three dimensional structure. It is insulator. It is
hardest naturally occuring substance. It is used as abrasive (for drilling rocks), cutting glass, in
making rocket windows and jwellery.
D:\BOOKS\JANUARY\26-A NEW APPROACH TO INORGANIC CHEMISTRY VOL. 1\UNIT-5\IIIRD PROOF DT. 27/3/12
P-BLOCK ELEMENTS 235
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
Fig. 5.1 Structure of diamond
(ii) Graphite
In graphite, each carbon is sp2 hybridised. Each carbon atom is linked with three carbon atoms,
forming a hexagonal plane. Each carbon atom has a pure p-orbital containing unshared electrons.
These p-orbitals overlap together and form a layer of mobile electrons above and below the
carbon plane. Therefore, carbon layers are flanked by electrons and their layers are stacked over
each other. Layers are held by weak vander waal’s force. This is why graphite is soft. It is conductor
of electricity due to the presence of mobile layer of electrons.
Fig. 5.2 Structure of graphite
Uses of graphite: Graphite is used as electrode, high temperature lubricant, lead pencil etc.
(iii) Buckminster fullerene
It is an allotrope of carbon, made by heating graphite in an electric arc in the presence of inert gas.
It mainly contains molecules like C-60, C-70 and some other molecules containing upto 350
carbon atoms. Each carbon form 3 sigma bonds. The most common is C-60. This ball shaped
molecule has 60 vertices. It contains both single (143.5) and double (138.3) bonds.
It has cage like structure like a soccer ball having hexagonal and pentagonal rings of carbon.
It has 20-six membered ring and 12 five membered rings. Carbon atom is sp2 hydridised. Remaining
electron is delocalised over whole molecule. This is why, it has aromatic character. It is also
known as buckey ball.
D:\BOOKS\JANUARY\26-A NEW APPROACH TO INORGANIC CHEMISTRY VOL. 1\UNIT-5\IIIRD PROOF DT. 27/3/12
236 A NEW APPROACH TO INORGANIC CHEMISTRY
Fig. 5.3 Structure of buckminster fullerene
5.2.5 Uses of Carbon
Graphite fibers are used in tennis rackette, fishing rod, air craft etc.
– Charcoal is used in adsorbing poisonous gases:, in water filters etc.
– Coke is used as reducing agent in metallurgy
– Carbon black is used as a filler in automobile tyres.
Coal is used as fuel in houses and industries. It is used in production of electricity in thermal
power plant.
5.2.6 Compounds of Carbon
Some important inorganic compounds of carbon are as follows :
(i) Carbon monoxide [CO]
Preparation:
(a) Carbon burns in limited air to give carbon monoxide.
2C + O2  2CO
(b) On large scale it is prepared by passing steam over red hot coke.
C + H2O  CO + H2
This mixture is known as water gas or syn gas.
D:\BOOKS\JANUARY\26-A NEW APPROACH TO INORGANIC CHEMISTRY VOL. 1\UNIT-5\IIIRD PROOF DT. 27/3/12
P-BLOCK ELEMENTS 237
(c) It may be prepared by passing air on hot coke.
2C + O2 + 4N2  2CO + 4N2
This mixture is known as producer gas.
(d) On small scale, CO is prepared by heating formic acid with conc. H2SO4.
H 2SO4
HCOOH 
 CO + H 2 O

Properties of carbon monoxide
(i) Carbon monoxide is a Colourless, odourless gas
(ii) It is insoluble in water
(iii) Action of metals: Carbon monoxide combines with many metals to form metal
carbonyls.
Ni + 4CO  Ni(CO)4
This reaction is used for purification of nickel by Mond’s process.
(iv) Reducing property: It reduces oxides of transition metals.
e.g.
Fe2O3(s) + 3CO(g)  2Fe(s) + 3CO2 (g)
ZnO(s) + CO(g)  Zn(s) + CO2(g)
This property is used in extraction of metals from their oxide ores.
(v) Toxic nature: CO has great affinity for iron. It combines with iron of haemoglobin
and form a complex carboxy haemoglobin which is about 300 times more stable than
the oxy-haemoglobin. This prevents oxygen carrier ability of haemoglobin in R.B.C.
and causes death of animals. This is why CO is highly poisonous.
(vi) Structure: CO has 1 and 2 bonds between C and O. Both atom has lone pair of
electron. It has vacant antibonding *) molecular orbitals. This is why CO is  donor
and  acceptor.
: C  O:
This is also called  acid ligand
(ii) Carbon dioxide (CO2)
Preparation:
(i) It is prepared by complete oxidation of carbon in excess of air.
C + O2  CO2
D:\BOOKS\JANUARY\26-A NEW APPROACH TO INORGANIC CHEMISTRY VOL. 1\UNIT-5\IIIRD PROOF DT. 27/3/12
238 A NEW APPROACH TO INORGANIC CHEMISTRY
(ii) Burning of fossil fuel gives CO2.
2C4 H10 + 13O2  8CO2 + 10H2O
(iii) Metal carbonates (Except alkali metal carbonate) on heating decomposes to give CO2.

 CaO + CO2
CaCO3 
(iv) In laboratory, CO2 is prepared by the action of dilute HCl on marble
CaCO3 + 2HCl  CaCl2 + H2O + CO2
Properties of Carbon dioxide
(i) It is colourless, odourless gas.
(ii) It is slightly soluble in water.
H2O + CO2  H2CO3
(iii) In aqueous solution it is a weak dibasic acid.
H2CO3  H+ + HCO3–
HCO3–  H+ + CO32–
H2CO3/HCO3– buffer is present in blood which maintains pH of blood from 7·26 to
7·42
(iv) It combines with alkalis to form metal carbonates.
CO2 gas when passed through lime water gives milky white precipitate of CaCO3 and
milkiness disapperars if CO2 gas is passed for long time.
Ca(OH)2 + CO2  CaCO3 + H2O
CaCO3 + H2O + CO2  Ca (HCO3)2
(iv) CO2 is absorbed by plant and converted to carbohydrate.
6CO2 +12H2O  C6H12O6 + 6O2 + 6H2O
This process is used for making food for plants.
Increase in concentration of CO2 in the atmosphere causes greenhouse effect and it enhances
the temperature of the atmosphere known as global warming which has serious consequences.
Uses of carbon dioxide
(i) Solid CO2 is used as refrigerant for Ice cream and frozen food.
(ii) Gaseous CO2 is used in making soft drinks.
D:\BOOKS\JANUARY\26-A NEW APPROACH TO INORGANIC CHEMISTRY VOL. 1\UNIT-5\IIIRD PROOF DT. 27/3/12
P-BLOCK ELEMENTS 239
(iii) It is also used as fire extinguisher in cinema halls.
(iv) It is also used in making urea.
Structure of carbon dioxide
In CO2 molecule C atom has sp hybridisation it has linear structure.
O=C=O
5.2.7 Compounds of Silicon
(I) Silicon dioxide
95% of earth crust is made of silica and silicate. Silica occurs in many crystallographic forms.
e.g., Quartz, cristobalite, tridimite.
Silicon dioxide has 3-D network structure in which each silicon atom is covalently bonded to
4 oxygen atoms and each oxygen atom is covalently bonded to 2 silicon atoms.
O
|
Si
O | O
O
O
O
Si
Si
|
| O
O Si
O
O
O | O
O
Fig.5.4 Structure of SiO2
Silica is inert however it is attacked by HF
SiO2 + 4HF  SiF4 + 2H2O
SiF4 + 2HF  H2SiF6
Silica dissolves in alkali to form silicate
SiO2 + 2NaOH  Na2SiO3 + H2O
Uses of silica
(i) Quartz is used as piezeolectric materials and also used in clock, radio, TV, mobile.
(ii) Silica gel is used as a drying agent and also used in chromatography.
D:\BOOKS\JANUARY\26-A NEW APPROACH TO INORGANIC CHEMISTRY VOL. 1\UNIT-5\IIIRD PROOF DT. 27/3/12
240 A NEW APPROACH TO INORGANIC CHEMISTRY
(iii) It is used as catalyst
(iv) Kieselguhr is a form of silica used in dynamite
(v) Amorphous silica is used in filtration plants.
(II) Silicones
Organo silicon polymer having R2SiO repeating unit are called silicones.
Preparation of Silicones
It is prepared by hydrolysis of SiCl4, R Si Cl3, or R2 Si Cl2
e.g., (CH3)2 SiCl2 + 2H2O  (CH3)2 Si(OH)2
Condensation polymerisation of this product gives straight chain silicones.
Silicones are surrounded by nonpolar alkyl groups hence are water repelling in nature. They
have high thermal stability. They have high di-electric strength. They are resistant to oxidation.
They are inert to chemicals.
Uses of Silicones:
Silicones are used as sealant, greazes, electrical insulators, making water proof fabrics. It is
bio compatible hence are used in surgical and cosmetic implants.
(iii) Silicates (SiO44–)
Compounds containing SiO44– units are called silicates. Silicates are of six types.
O
|
Si
–
O– | – O
O
1. Ortho silicate which has only one SiO4–
4 unit
O
O
|
|
Si
Si
–
O– | – O | – O
O
O
O
O
|
|
Si
Si
O | O | O
O
O
O
|
Si
|
O
O
2.
Pyro silicate which has two SiO4–
4 units joined by a common
corner
3.
Linear silicate or chain silicate is a polymer of SiO44– with
two common corners forming a striaght chain.
D:\BOOKS\JANUARY\26-A NEW APPROACH TO INORGANIC CHEMISTRY VOL. 1\UNIT-5\IIIRD PROOF DT. 27/3/12
P-BLOCK ELEMENTS 241
6.
4.
Cyclic silicate having two common corner forming a cyclic
ring.
5.
Sheet silicate is also a polymer of SiO4–
4 unit having three
common corner.
3-dimensional Silicate: It is a polymer of SiO4–
unit having all the four common corners.
4
IV Zeolite
Silicates in which some silicon atoms are replaced by Al, atom in 3-D network is called zeolite, it
is negatively charged.
Cations like Na, K, Ca etc. balances the –ve charge.
Ex. Sodium aluminium silicate is known as zeolite.
(i) They are used as ion exchanger in softening of hard water.
(ii) They are also used as catalyst in petro chemical industries for cracking and isomerisation
of hydrocarbons.
Ex. ZSM-5 is used to convert alcohols directly into gasoline.
NCERT TEXT BOOK EXERCISE WITH SOLUTIONS
Q. 1. Discuss the pattern of variation in the oxidation states of (i) B to Tl (iii) C to Pb.
Ans. (i) B andAl show an oxidation state of +3 only due to the presence of two electrons
in the s-and one electron in the p-orbital of the valence shell. All other elements from
Ga to Tl show oxidation states of +1 and +3. The inert pair effect becomes more and
more pronounced, as we move down the group from Ga to Tl. Thus the stability of
+1 oxidation state increases (i.e., Ga < In < Tl) while that of +3 oxidation state
decreases (i.e. Ga > ln> Tl). Thus, +1 oxidation state of Tl is more stable than its
+3 oxidation state.
(ii) Carbon and silicon show an oxidation state of +4 due to the presence of two
D:\BOOKS\JANUARY\26-A NEW APPROACH TO INORGANIC CHEMISTRY VOL. 1\UNIT-5\IIIRD PROOF DT. 27/3/12
242 A NEW APPROACH TO INORGANIC CHEMISTRY
electrons in the s- and two electrons in the p-orbital of the valence shell. All other
elements from Ge to Pb show two oxidation states of +2 and +4. The inert pair effect
becomes more and more pronounced as we move down the group from Ge to Pb.
Thus, the stability of +2 oxidation state increases (i.e., Ge < Sn < Pb) while that of
+4 oxidation state decreases (i.e. Ge > Sn > Pb). Thus, +2 oxidation state of Pb is
more stable than its +4 oxidation state.
Q. 2. How can you explain higher stability of BCl3 as compared to TlCl3?
Ans. Due to the poor shielding of the s-electrons of the valence shell (6s2) by the 3d-, 4d5d an 4f-electrons, inert pair effect is maximum in Tl. Therefore, TlCl3 is unstable. All
the three valence electrons of boron (i.e., two 2s- and one 2p-) take part in bond
formation and hence B forms BCl3. Thus BCl3 is more stable than TlCl3.
Q. 3. Why does boron trifluoride behave as a Lewis acid?
Ans. The B atom in BF3 has only 6 electrons in the valence shell and thus needs two more
electrons to complete its octet. Therefore, it easily accepts a pair of electrons and thus
behaves as Lewis acid.
Q. 4. Consider the compounds BCl3 and CCl4. How will they behave with water?
Justify.
Ans. The B atom in BCl3 has only six electrons in the valence shell. It is an electrondeficient molecule. It easily accepts a pair of electrons donated by water and hence
BCl3 undergoes hydrolysis to form boric acid (H3BO3) and HCl.
BCl3 + 3H2O  H3BO3 + 3 HCl
In contrast, C atom in CCl4 has 8 electrons in the valence shell. It is an electronprecise molecule. It does not accept a pair of electrons from H2O molecule and hence
CCl4 does not undergo hydrolysis in water.
Q. 5. Is boric acid a protic acid? Explain.
Ans. It is not a protic acid since it does not ionize in H2O to give a proton:
H3BO3 + H2O 
 H2BO3– + H3O+
Instead because of the small size of boron atom and presence of only six electrons in
its valence shell, B(OH)3 accepts a lone pair of electrons from the oxygen atom of the
H2O molecule to form a hydrated species. Facilitating the release of a proton.
–
+
··
(HO)3B + : O H2  (HO)3B — O
H
–
 B(OH)4 + H +
H
As a result, B(OH)3 acts as a weak monobasic Lewis acid.
D:\BOOKS\JANUARY\26-A NEW APPROACH TO INORGANIC CHEMISTRY VOL. 1\UNIT-5\IIIRD PROOF DT. 27/3/12
P-BLOCK ELEMENTS 243
Q. 6. Explain what happens when boric acid is heated.
Ans. Boric acid, on heating, loses water molecules successively in three different stages at
different temperatures ultimately giving boron trioxide or boric anhydride.
370 K
H3BO3 
 HBO 2  H 2O
Metaboric acid
Boric acid
4HBO 2
Metaboric acid
419 K


–H 2 O
Red heat
H 2 B4O7 
 2B2 O3 +H 2O
Tetraboric acid
Boron trioxide
Q. 7. Describe the shapes of BF3 and [BH4]–. Assign the hybridization of boron in
these species.
Ans. In BF3, boron is sp2-hybridized and, it is a planar molecule. On the other hand, in
[BH4]– boron is sp3-hybridized and hence [BH4]– is a tetrahedral species.
Empty p-orbital
F
B
F
F
sp2–Hybridized
–
H
|
B
H
H
H
sp3–Hybridized
Q. 8. Write reaction to justify amphoteric nature of aluminium.
Ans. It dissolves both in acids and alkalies evolving dihydrogen.
2Al(s)+ 3H2SO4 (aq)  Al2 (SO4)3(aq) + 3H2 (g)
2Al(s) + 2NaOH (aq) + 6H2O(l)  2Na + [Al(OH)4 ] (aq) + 3H2(g)
Sod. Tetrahydroxoaluminate (III)
Q. 9. What are electron deficient compounds? Are BCl3 and SiCl4 electron deficient
species? Explain.
Ans. Species in which the central atom does not have eight electrons in the valence shell are
called electron deficient molecules. For example,
(i) In BCl3, the central boron atom has only six electrons. Therefore, it is an electron
deficient compound.
(ii) In SiCl4, the central Si atom has 8 electrons. Therefore, SiCl4 is not an electron-deficient
molecule.
Q. 10. Write the resonance structure of CO23– and HCO–3.
Ans. Resonance structures of CO23– ion:
D:\BOOKS\JANUARY\26-A NEW APPROACH TO INORGANIC CHEMISTRY VOL. 1\UNIT-5\IIIRD PROOF DT. 27/3/12
244 A NEW APPROACH TO INORGANIC CHEMISTRY


–
–




–
O—C












–
–
O—C


O


–
O
O 




O


O C




O 
O 




OH
O 
O—C





OH






OC
–


O






Resonance structures of HCO3– ion:
Q.11. What is the state of hybridization of carbon in (a) CO2–3 (b) diamond (c) graphite?
Ans. (a) sp2 (b) sp3 (c) sp2
Q. 12. Explain the difference in properties of diamond and graphite on the basis of
their structures.
Ans.
See text page No. 234-235
Q.13. Rationalise the given statements and give chemical reactions:
(i) lead (II) chloride reacts with Cl2 to give PbCl4
(ii) lead (IV) chloride is highly unstable towards heat
(iii) lead is known not to form an iodide, PbI4
Ans. (i) Due to inert effect, Pb +2 is more stable than Pb +4 oxidation state. Therefore,
lead (II) chloride is more stable than lead (IV) chloride and hence lead (II) chloride
react with Cl2 to form lead IV chloride which is less stable.
PbCl2 ( s )  Cl2 ( g ) 
 PbCl 4 (l )
(ii) Due to inert pair effect PbCl2 is more stable than lead (IV) chloride.Hence PbCl4
on heating decomposes to give lead (II) chloride and Cl2

PbCl4 (l ) 
 PbCl2 ( s )  Cl 2 (g )
(iii) Pb4+ is oxidising agent and I– is a reducing agent hence PbI4 does not exist.
This is due to inert pair effect.
Q.14. Suggest reasons why the B·F bond lengths in BF3 (130 pm) and BF4– (143 pm)
differ.
Ans. BF3 is planar molecule in which B is sp2-hybridized. It has an empty 2p-orbital. F-atom
has three lone pairs of electrons in the 2p-orbitals. Because of similar sizes, p – p 
back bonding occurs in which a lone pair is transferred from F to B as shown below:
D:\BOOKS\JANUARY\26-A NEW APPROACH TO INORGANIC CHEMISTRY VOL. 1\UNIT-5\IIIRD PROOF DT. 27/3/12
P-BLOCK ELEMENTS 245

F
B
F
F
As a result of this back bonding, B–F bonding acquires some double bond character.
In [BF 4]– ion, B is sp3-hybridized and B—F is a purely single bond. Therefore, the
B—F bond length in BF3 is shorter (130 pm) than B—F bond length (143 pm) in
[BF 4]–.
Q.15. If B–Cl bond has a dipole moment, explain why BCl3 molecule has zero dipole
moment.
Ans. Due to electronegativity difference between B (E.N. = 2·0) and Cl (E.N. = 3·0), the
B—Cl bond is polar and hence has a finite dipole moment. Now BCl3 is a planar
molecule in which the three B—Cl bonds are inclined at an angle of 120°. Therefore,
the resultant of three B—Cl bonds is cancelled:
Cl
B
Cl
120°
Resultant
Cl
u=0
As a result, overall dipole moment of BCl3 is zero.
Q.16. Aluminium trifluoride is insoluble in anhydrous HF but dissolves on addition of
NaF. Aluminium trifluoride precipitates out of the resulting solution when gaseous
BF3 is bubbled through. Give reasons.
Ans. (i) Anhydrous HF is a covalent compound and is strongly H–bonded. Therefore, it
does not give F– ions and hence AlF3 does not dissolve in HF. In contrast NaF being
an ionic compound contains F– ions and hence combines with AlF3 to form the soluble
complex.
3NaF +AlF3 

Na 3 [AlF6 ]
Sod. hexafluoroaluminate (III)
( soluble complex )
(ii) Because of smaller size and higher electronegativity, B has much higher tendency
to form complexes than Al, therefore, when BF3 is added to the above solution, AlF3
gets precipitated.
Na 3[AlF6 ]  3BF3 

3Na[BF4 ]
 AlF3 ( s)
Sod. tetrafluoroborate (III)
( soluble complex )
D:\BOOKS\JANUARY\26-A NEW APPROACH TO INORGANIC CHEMISTRY VOL. 1\UNIT-5\IIIRD PROOF DT. 27/3/12
246 A NEW APPROACH TO INORGANIC CHEMISTRY
Q. 17. Suggest a reason as to why CO is poisonous.
Ans. CO combines with iron of haemoglobin irreversibly to form carboxyhaemoglobin
which is much more stable than the oxyhaemoglobin.
Haemoglobin + CO  Carboxyhaemoglobin
As a result, the oxygen carrying capacity of haemoglobin is destroyed and the man
dies of suffocation.
Q. 18. How is excessive content of CO2 responsible for global warming?
Ans. If the concentration of CO2 increases beyond a certain level. This excess CO2 absorbs
heat radiated by the earth. Some of it is dissipated into the atmosphere while the
remaining part is radiated back to the earth and other bodies present on the earth. As
a result, temperature of the earth and other bodies on the earth increases. This is called
global warming and greenhouse effect and CO2 is called a green house gas.
Q. 19. Explain structures of diborane and boric acid.
Ans. See text page no. 228-230.
Q.20. What happens when
(a) Borax is heated strongly, (b) Boric acid is added to water,
(c) Aluminium is treated with dilute NaOH, (d) BF3 is reacted with ammonia?
Ans. (a) When borax is heated strongly, a transparent glassy bead is formed which consists
of sodium metaborate (NaBO2) and boric anhydride.
Heat
Na 2 B4O 7 10H 2O 
 Na 2 B4O 7  10H 2O
Borax
Heat
Na 2 B4O7 
 2NaBO2  B2O3
Sod. metaborate Boric anhydride


Transparent glassy bead
(b) Boric acid acts as a weak Lewis acid and accepts a hydroxide ion of water and
releases a proton into the solution.
–
H—OH + B (OH)3  [B(OH)4] + H
+
(c) Dihydrogen is evolved.
2Al (s) + 2NaOH (aq) + 6H2O(l)  2Na+ [Al(OH)4]– (aq) + 3H2(g)
D:\BOOKS\JANUARY\26-A NEW APPROACH TO INORGANIC CHEMISTRY VOL. 1\UNIT-5\IIIRD PROOF DT. 27/3/12
P-BLOCK ELEMENTS 247
(d) BF3 being a Lewis acid accepts a pair of elecrons from NH3 to form the corresponding
complex.
F3B + : NH3 = F3B NH3
Lewis acid
Lewis base
Complex
Q.21. Explain the following reactions
(a) Silicon is heated with methyl chloride at high temperature in the presence of
copper
(b) Silicon dioxide is treated with hydrogen fluoride
(c) CO is heated with ZnO
(d) Hydrated alumina is treated with aqueous NaOH solutions.
Ans. (a) A mixture of mono-, di- and trimethylchlorosilanes along with a small amount of
tetramethylsilane is formed.
Cu powder
CH3Cl  Si 

 CH3SiCl3  (CH3 )2 SiCl 2  (CH3 )3SiCl + (CH3 ) 4 Si
373 K
Methyl chloride
(b) SiO2 dissolves in HF to from hydrofluorosilicic acid
SiO 2  4HF 
 SiF4  2H 2O
SiF4  2HF 
 H 2SiF6
(c) ZnO is reduced to zinc metal.
ZnO  CO 
 Zn  CO 2
(d) Alumina dissolves to form sodium meta-aluminate
Heat

Al 2 O3  2H 2 O(l ) + 2NaOH( aq )  H 2O(l ) 
 2Na[Al(OH) 4 ]( aq ) 
 2NaAlO 2  3H 2 O
Q. 22. Give reasons:
(i) Conc. HNO3 can be transported in aluminium container.
(ii) A mixture of dilute NaOH and aluminium pieces is used to open drain.
(iii) Graphite is used as lubricant
(iv) Diamond is used as an abrasive
(v) Aluminium alloys are used to make aircraft body
D:\BOOKS\JANUARY\26-A NEW APPROACH TO INORGANIC CHEMISTRY VOL. 1\UNIT-5\IIIRD PROOF DT. 27/3/12
248 A NEW APPROACH TO INORGANIC CHEMISTRY
(vi) Aluminium utensils should not be kept in water overnight.
(vii) Aluminium wire is used to make transmission cables.
Ans. (i) Al reacts with conc. HNO3 to form a very thin film of aluminium oxide on its
surface which protects it from further action.
 Al2 O3 ( s )  6NO 2 ( g )  3H 2 (l )
2Al (s) + 6 HNO3 (conc.) 
Alumina
Thus, Al becomes passive and hence aluminium containers can be used to tansport
conc. HNO3
(ii) NaOH reacts with Al to evolve dihydrogen whose pressure can be used to open
clogged drains.
2Al (s) + 2 NaOH (aq) + 2H2O(l)  2NaAlO2(aq) + 3H2(g)
(iii) Graphite has layered structure in which the different layers are held together by
weak vander walls forces and hence they slip over one another. Therefore, graphite
acts as a lubricant.
(iv) Diamond is very hard and hence can be used as an abrasive.
(v) Aluminium alloys such as duralumin is light, tough and resistant to corrosion and
hence is used to make aircraft body.
(vi) Al reacts with H2O and dissoved O2 to form a thin film of aluminium oxide and
2Al(s) + O2(g) + H2O(l)  Al2O3(s) + H2(g)
Since Al3+ ions are injurious to health, therefore, water should not be kept in aluminium
utensils overnight.
(vii) Aluminium is light and cheaper conductor than Cu. Therefore, it is used in
transmission cables.
Q. 23 Explain why is there a phenomenal decrease in ionization enthalpy from carbon
to silicon?
Ans. Due to increase in atomic size and screening effect, the force of attraction of the
nucleus for the valence electron decreases considerably in Si as compared to C. As a
result, there is a phenomenal decrease in ionization enthalpy from carbon to silicon.
Q. 24. How would you explain the lower atomic radius of Ga as compared to Al.
Ans.
Due to poor shielding of the nuclear charge of Ga by the inner 3d-electrons, the
effective nuclear charge of Ga is geater in magnitude than that of Al. As a result, the
electrons in gallium experience greater force of attraction by the nucleus than in Al and
hence atomic size of Ga (135 pm) is slightly less than that of Al (143 pm).
D:\BOOKS\JANUARY\26-A NEW APPROACH TO INORGANIC CHEMISTRY VOL. 1\UNIT-5\IIIRD PROOF DT. 27/3/12
P-BLOCK ELEMENTS 249
Q.25 What are allotropes? Sketch the structure of two allotropes of carbon namely
diamond and graphite. What is the impact of structure on physical properties
of two allotropes?
Ans. Different physical forms of same element are called allotropes.
See text page no. 234-235
Q.26. (a) Classify following oxides as neutral acidic, basic or amphoteric
CO, B2O3, SiO2, CO2, Al2O3, PhO2, Tl2O3
(b) Write suitable chemical equations to show their nature.
Ans. (a) Neutral oxides: CO; Acidic oxides: B2O3, SiO2, CO2
Amphoteric oxides : Al2O3, PbO2 Basic oxide: Tl2O3
(b) (i) Being acidic B2O3, SiO2 and CO2 react with alkalis to form salts.
B2O 3  2NaOH 
 2NaBO 2
Boric
Sod. metaborate
anhydride
 H 2 O;

SiO 2  2NaOH 
 Na 2SiO3  H 2O
Silica
CO 2 
Carbon dioxide
Sod. silicate
2NaOH 
 Na 2CO3  H 2 O
Sod. carbonate
(ii) Being amphoteric, Al2O3 and PbO2 react with both acids and bases.
Fuse
Al 2O3  2NaOH 

Alumina
2NaAlO 2  H 2O;
Sod. meta aluminate
Al 2O3 + 3H 2SO 4 
 Al2 (SO 4 )3 + 3H 2O
PbO 2  2NaOH 
 Na 2 PbO 2  H 2O;
Lead deioxide
Sod. plumbate
2PbO 2  2H 2SO 4 
 2PbSO 4  2H 2 O  O 2
(iii) Being basic, Tl2O3 dissolves in acids.
Tl2O3 + 6HCl  2TlCl3 + 3H2O
Q.27. In some of the reactions, thallium resembles aluminium whereas in others it
resembles with group I metals. Support this statement by giving some evidences.
Ans. Like aluminium,Tl also shows +3 oxidation state in some of its compounds like TlCl3,
Tl2O3, etc. Like group 1 metals, due to inert pair effect Tl also shows +1 oxidation
state in some of its compounds such as Tl2O, TlCl, TlClO4, etc.
Like group 1 oxides, Tl2O is strongly basic.
D:\BOOKS\JANUARY\26-A NEW APPROACH TO INORGANIC CHEMISTRY VOL. 1\UNIT-5\IIIRD PROOF DT. 27/3/12
250 A NEW APPROACH TO INORGANIC CHEMISTRY
Q.28. When metal X is treated with sodium hydroxide, a white precipitate (A) is
obtained, which is soluble in excess of NaOH to give soluble complex (B).
Compound (A) is soluble in dilute HCl to form compound (C). The compound
(A) when heated strongly gives (D), which is used to extract metal. Identify (X),
(A), (B), (C) and (D). Write suitable equations to support their identities.
Ans. Since metal X reacts with NaOH to give a white ppt. (A) which dissolves in excess
of NaOH to give a soluble complex (B), therefore, metal (X) must be Al; ppt (A) must
be Al(OH)3 and complex (B) must be sodium tetrahydroxoaluminate (III)
2Al+3NaOH 

(X)
Al(OH)3 
 3Na +
Aluminium hydroxide (ppt.)
Al(OH)3 + NaOH 

(A)
Na + [Al(OH) 4 ]
Sod. tetrahydroxoaluminate (III)
Since (A) is amphoteric in nature, it reacts with dil. HCl to from (C) which must be
AlCl3
Al(OH)3 + 3HCl 
 AlCl3  3H 2 O
(A)
(C)
Since (A) on heating gives (D) which is used to extract metal therefore, (D) must be
alumina (Al2O3)

2Al(OH)3 
 Al2 O3  3H 2O
(A)
(D)
Q.29. What do you understand by (a) inert pair effect (b) allotropy (c) catenation?
Ans. See text (a) page No.221, (b) P-291, Self linking prperty of elements to form long
chains is called catenation. Carbon atom has maximum tendency of catenation due to
strong C — C bonds Si, P and S also show catenation property.
Q.30. A certain salt X, gives the following results.
(i) Its aqueous solution is alkaline to litmus.
(ii) It swells up to a glassy material Y on strong heating.
(iii) When conc. H2SO4 is added to a hot solution of X, white crystal of an acid Z
separates out
Write equations for all the above reactions and identify X, Y and Z.
Ans. (i) Since the aqueous solution of salt (X) is alkaline to litmus, it must be the salt of
a strong base and a weak acid.
(ii) Since the salt (X) swells up to a glassy material (Y) on strong heating, therefore,
D:\BOOKS\JANUARY\26-A NEW APPROACH TO INORGANIC CHEMISTRY VOL. 1\UNIT-5\IIIRD PROOF DT. 27/3/12
P-BLOCK ELEMENTS 251
(X) must be borax and (Y) must be a mixture of sodium metaborate and boric
anhydride.
(iii) When conc. H2SO4 is added to a hot solution of X (borax) white crystals of an
acid (Z) separate out, therefore, (Z) must be orthoboric acid. The equations for the
reactions involved in the question are:
Water
(i) Na 2 B4O 7 10H 2O  2NaOH  H 2 B4O 7  8H 2O
(Strong alkali)
Borax (X)
(Weak acid)
Heat
 Na 2 B4 O7  10H 2 O
(ii) Na 2 B4O7  10H 2 O 
(X)
Heat
Na 2 B4O 7 
 2NaBO 2  B2 O3

Glassy material (Y)
(iii) Na 2 B4O7 10H 2O  H 2SO4 
 4H3BO3  Na 2SO4  5H2 O
Boric acid (Z)
Q.31. Write balanced equations for:
(i) BF3 + LiH 


(iv) H3BO3 
(ii) B2H6 + H2O
(iii) NaH + B2H6
(v) Al + NaOH 
(vi) B2H6 + NH3
 B2 H 6  6LiF
Ans. (i) 2BF3 + 6 LiH 
Diborane
 2H 3 BO 3  6H 2
(ii) B 2 H 6  6H 2 O 
Diborane
Orthoboric acid
(iii) 2NaH + B2 H6 
 2Na + [BH 4 ]
Sod. borohydride
(iv)

H 3 BO3 
 HBO 2  H 2 O
Orthoboic acid
Metaboric acid


4HBO 2 
 H 2 B4O 7 
 2B2 O3  H 2O
H O
Metaboric acid
2
Tetraboric acid
Boron trioxide
(v) 2Al+2NaOH+6H2 O 
 2Na + [Al(OH)4 ]– +3H 2
Sod. tetrahydroxoaluminate (III)
(vi) B2 H 6 +2NH3 

2BH3  NH3
Borane-ammonia complex


 B3 N3H6
Borazine
D:\BOOKS\JANUARY\26-A NEW APPROACH TO INORGANIC CHEMISTRY VOL. 1\UNIT-5\IIIRD PROOF DT. 27/3/12
252 A NEW APPROACH TO INORGANIC CHEMISTRY
Q.32. Give one method for industrial preparation and one for laboratory preparation
of CO and CO2 each.
Ans. Carbon monoxide (CO)
Carbon dioxide (CO2)
Limited
Excess
Industrial: 2C(s )  O 2 (g ) 
 2CO(g ) ; C(s )  O 2 (g ) 
CO 2 (g )
air
air
H 2SO 4
Laboratory: HCOOH 
CO+H 2O
Formic acid
CaCO3 ( s)  2HCl (aq) 

CaCl2 ( aq)  CO2 ( g )  H 2O(l )
MULTIPLE CHOICE QUESTIONS
Q.33. An aqueous solution of borax is
(a) neutral
(b) amphoeteric
(c) basic
(d) acidic
Ans. (c) Borax is a salt of a strong base (NaOH) and a weak acid (H3BO3), therefore its
aqueous solution is basic in nature.
Q. 34 Boric acid is polymeric due to
(a) Its acidic nature
(b) the presence of hydrogen bonds
(c) its monobasic nature
(d) its geometry
Ans (b) Boric acid is polymeric due to the presence of hydrogen bonds.
Q. 35 The type of hybridisation of Boron in diborane is
(a) sp
(b) sp2
(c) sp3
(d) dsp2
Ans.(c)In B2H6, B is sp3–hybridized.
Q. 36. Thermodynamically the most stable form of carbon is
(a) diamond
(b) graphite
(c) fullerenes
(d) coal
Ans. (c) Thermodynamically the most stable form of carbon is graphite.
Q. 37. Elements of group 14.
(a) exhibit oxidation state of +4 only
2–
(c) form M and M
4+
ion
(b) exhibit oxidation state of +2 and +4
(d) form M2+ and M4+ ions
D:\BOOKS\JANUARY\26-A NEW APPROACH TO INORGANIC CHEMISTRY VOL. 1\UNIT-5\IIIRD PROOF DT. 27/3/12
P-BLOCK ELEMENTS 253
Ans.(b) Due to inert pair effect, elements of group 14 exhibit oxidation states of +2 and +4.
Q. 38. If the starting material for the manufacture of silicones is RSiCl3, write the structure
of the product formed.
Ans. Hydrolysis of alkyltrichlorosilanes gives cross-linked silicones.
R
R
|
|
Cl— Si — Cl + 
HO—
Si—OH
3 HCl
|
|
Cl
OH

O
R
R
R
|
|
|
|
Polymerization
…O—Si—O—Si—O—Si—R
n HO— Si — OH 
|
|
|
 (n–1)H 2O
|
O
O
O
OH
|
|
|
…O—Si—O—Si—O—Si—R
|
|
|
R
R
O

Cross-linked silicone
QUESTIONS FROM PREVIOUS IIT JEE PAPERS WITH SOLUTIONS
Type 1: Subjective Questions
Q.1. Carbon acts as an abrassive and also as a lubricant, explain.
(1981, 1M)
Ans. The two common allotropes of carbon are diamond and graphite. Diamond is the
hardest, natural substance used as an abrasive while graphite is soft, used as a lubricant.
Q.2 Give reason for the following in one two sentences:
“Solid carbon dioxide is known as dry ice.”
(1993, 1M)
Ans. Solid carbon dioxide sublime, without passing through liquid state hence, known as
dry ice.
Q.3. Give reasons for the following in one or two sentences:
“Graphite is used as a solid lubricant.”
(1985, 1M)
Ans. Graphite has layered structure and the adjacent layers are weakly associated giving
slippery nature hence it is used as solid lubricant.
D:\BOOKS\JANUARY\26-A NEW APPROACH TO INORGANIC CHEMISTRY VOL. 1\UNIT-5\IIIRD PROOF DT. 27/3/12
254 A NEW APPROACH TO INORGANIC CHEMISTRY
Q.4. Each entry in column X is in some way related to the entries in column Y and Z.
Match the appropriate entries.
X
Y
Yeast
Mica
Superphosphate
Carbon fibres
Rock salt
Carborundum
Fermentation
Graphite
Crystalline cubic
Layer structure
Diamond structure
Bone ash
Z
Ethanol
Abrasive
Insulator
Fertilizer
Reinforced plastics
Preservative
(1989, 3M)
Ans.
X
Y
Yeast
Mica
Superphosphate
Carbon fibres
Rock salt
Carborundum
Z
Fermentation
Layer structure
Bone ash
Graphite
Crystalline cubic
Diamond structure
Ethanol
Insulator
Fertilizer
Reinforced plastics
Preservative
Abrasive
Q.5. Write the balanced equation for the preparation of crystalline silicon from SiCl4.
(1990)

Ans. 3SiCl 4  4Al 
 4AlCl3  3Si
vapour
molten
volatilizes
crystalline
Q.6. Anhydrous AlCl3 is covalent. From the data given below, predict whether it
would remain covalent or become ionic in aquaous solution.
(Ionization energy for Al = 5137 kJ mol–1)
Hhydration for Al3+ = –4665 kJ mol–1
Hhydration for Cl– = –381 kJ mol–1
(1992, 2M)
Ans. The total hydration enthalpy of AlCl3
= Hydration enthalpy of Al3+ + 3 × Hydration enthalpy of Cl–
= –4665 + 3 (–381) kJ/mol
= –5808 kJ/mol
Thus hydration enthalpy is more than the energy required for ionization of AlCl3 into
Al3+ and Cl– this is why, AlCl3 becomes ionic in aqueous solution. In aqueous solution,
D:\BOOKS\JANUARY\26-A NEW APPROACH TO INORGANIC CHEMISTRY VOL. 1\UNIT-5\IIIRD PROOF DT. 27/3/12
P-BLOCK ELEMENTS 255
it is ionized completely as
AlCl3 + 6H2O  [Al(H2O)6]3+ + 3Cl–
Q.7. Aluminium sulphide gives a foul odour when it becomes damp. Write a balanced
chemical equation for the reaction.
(1997)
Ans. Al2S3 + 6H2O  2Al(OH)3   3H 2S(g ) 
foul odour
Foul odour on damping of Al2S3 is due to formation of H2S
Q.8. Draw the structure of a cyclic silicate, (Si3O9)6– with proper labelling.
(1998)
O–
O–
–O
Si3O6–9
Ans.
–O
O–
O
Q.9. Compound X on reduction with LiAlH4 gives a hydride Y containing 21.72%
hydrogen alongwith other products. The compound Y reacts with air explosively
resulting in boron trioxide. Identify X and Y. Give balanced reactions involved
in the formation of Y and its reacion with air. Draw the structure of Y.
(2001)
LiAlH
4
 ‘Y ’ a hydride + other compound. Hydride ‘Y’ contain
Ans. Compound ‘X’ 
21.72% hydrogen.

Y  O 2 
 B 2 O3  H 2 O
Therefore, Y is a hydride of boron and it is obtained by reduction of X with LiAlH4.
So, X is either BCl3 or BF3.
4BCl3  LiAlH 4 
 B2 H 6  3AlCl3  3LiCl

X
Y
Other products
Molar mass of B2H6 = 2 × 11 + 6 = 28
D:\BOOKS\JANUARY\26-A NEW APPROACH TO INORGANIC CHEMISTRY VOL. 1\UNIT-5\IIIRD PROOF DT. 27/3/12
256 A NEW APPROACH TO INORGANIC CHEMISTRY
6
% of H in B2H6 =
 100  21.5  21.72
28
Structure of Y(B2H6)
H
3Å
1.3
97°
H
9Å
1.1
122°
B2 H 6  3O 2 
 B2 O3  3H 2 O + heat
Y
H
H
1.77 Å
Q.10. Starting from SiCl4, prepare the following
in steps not exceeding the number given in parenthesis (give reacions only)
(i) Silicon
(ii) Linear silicone containing methyl group only
(iii) Na2SiO3
(2001)

Ans. (i) 3SiCl4  4Al 
 3Si  4AlCl3 ;
(ii) SiCl4 + 2CH3MgCl  (CH3)2 SiCl2 +2MgCl2
OH
|
–HCl
(CH 3 ) 2 SiCl2  H 2 O 
 CH 3 — Si—CH 3
|
OH
CH 3
CH 3
CH 3
|
|
|
Condensation
—O — Si —O—Si—O —Si—O—
|
|
|
CH 3
CH 3
CH 3
 Si(OH) 4  4HCl
(iii) SiCl4  4H 2O 
unstable
heat
Si(OH)4 
 SiO 2  2H 2 O

SiO2  Na 2 CO3 
 Na 2SiO3 +CO2
Q.11. (i) How is boron obtained form borax? Give chemical equations with reaction
conditions. (ii) Write the strucurte of B2H6 and its reaction with HCl.
Ans. (i) Na2 B4O7 + HCl  NaCl + H3 BO3

H 3 BO3  3HCl 
 BCl3 + 3H 2 O

BCl3 +Al 
 B  AlCl3
D:\BOOKS\JANUARY\26-A NEW APPROACH TO INORGANIC CHEMISTRY VOL. 1\UNIT-5\IIIRD PROOF DT. 27/3/12
P-BLOCK ELEMENTS 257
(ii) Structure of B2H6 See page No. 256 Q. No-9
Reaction: B2H6 + HCl  B2H5Cl + H2
Type-II Objective Questions [Only one correct option]
Q.1. Moderate electrical conductivity is shown by
(a) silica
(b) graphite
(c) diamond
(d) None of these
(1982)
Q.2. Which of the following halides is least stable and has doubtful existence?
(1996)
(a) CCl4
(b) GeI4
(c) SnI4
(d) PbI4
Q.3. Which one of the following oxides is neutral?
(a) CO
(b) SnO2
(b) ZnO
(d) SiO2
(1996)
Q.4. In compounds of type ECl3, where E = B, P, As or Bi, the angles Cl—E—Cl for
different E are in the order
(1992)
(a) B >P = As = Bi
(b) B > P > As > Bi
(c) B < P = As = Bi
(d) B < P < As < Bi
Q.5. Identify the correct order of acidic strength of CO2, CuO, CaO, H2O.
(2002)
(a) CaO < CuO < H2O < CO2
(b) H2O < CuO < CaO < CO2
(c) CaO < H2O < CuO < CO2
(d) H2O < CO2 < CaO < CuO
Q.6. H3BO3 is
(a) monobasic acid and weak Lewis acid
(b) monobasic and weak Bronsted acid
(c) monobasic and strong Lewis acid
(d) tribasic and weak Bronsted acid
D:\BOOKS\JANUARY\26-A NEW APPROACH TO INORGANIC CHEMISTRY VOL. 1\UNIT-5\IIIRD PROOF DT. 27/3/12
(2003)
258 A NEW APPROACH TO INORGANIC CHEMISTRY
Q.7. (Me)2 SiCl2 on hydrolysis will produce
(2003)
(a) (Me)2 Si(OH)
(b) (Me)2 Si = O
(c) [—O—(Me)2 Si —O—]n
(d) Me2SiCl(OH)
Q.8. Name the structure of silicates in which three oxygen atoms of [SiO4]4– are
shared.
(2005)
(a) pyrosilicate
(b) sheet silicate
(c) linear chain silicate
(d) three dimensional silicate
Q.9. B(OH)3 + NaOH  NaBO2 + Na[B(OH)4] + H2O
How can this reaction is made to proceed in forward direction?
(a) Addition of cis 1, 2 diol
(b) Addition of borax
(c) Additioin of trans 1, 2 diol
(d) Addition of Na2HPO4
(2005, 3M)
Type-III Objective Questions [One or more than one correct option]
1. In the reaction,
2X + B 6H6  [BH2(X)2]+[BH4]–
the amine(s) is(are)
(2009)
(a) NH3
(b) CH3 NH2
(c) (CH3)2 NH
(d) (CH3)3N
Type-IV Objective Questions
Assertion and Reason Type Question
Read the following questions and answer as per the direction given below:
(a) Statement I is true; Statement II is true; Statement II is the correct explanation of
Statement I.
(b) Statement I is true; Statement II is true, Statement II is not the correct explanation of
Statement I.
(c) Statement I is true; Statement II is false
(d) Statement I is false; Statement II is true
D:\BOOKS\JANUARY\26-A NEW APPROACH TO INORGANIC CHEMISTRY VOL. 1\UNIT-5\IIIRD PROOF DT. 27/3/12
P-BLOCK ELEMENTS 259
1. Statement I: Al (OH)3 is amphoteric in nature
Statement II: Al—O and O—H bonds can be broken with equal ease in Al (OH)3.
(1998)
2. Statement I: Between SiCl4 and CCl4, only SiCl4 reacts with water.
Statement II: SiCl4 is ionic and CCl4 is covalant.
(2001, S)
3. Statement I: Boron always forms covalent bond.
Statement II: Ionisation enthalpy of boron is very high.
(2007)
4. Statement I: In water, orthoboric acid behaves as a weak monobasic acid.
Statement II: In water, orthoboric acid acts as a proton donor.
(2007)
5. Statement I: Pb4+ compounds are stronger oxidising agents than Sn2+ compounds.
Statement II: The higher oxidation states for the group 14 elements are more stable
for the heavier members of the group due to ‘inert pair effect’.
(2008, 3M)
Type-V Objective Questions
Matrix Match Type Questions
1. Match the following:
Column I
(A) Bi3+  (BiO)+
–
Column II
(p) Heat
(B) [AlO2]  Al(OH)3
(q) Hydrolysis
(C) SiO44–  Si2O76–
(r) Acidification
(D) (B4O2–
7 )  [B(OH)3]
(s) Dilution by water
(2006, 6M)
Type-VI Fill in the Blanks
1. The hydrolysis of alkyl substituted chlrososilanes gives ....................
(1991, 1M)
2. The hydrolysis of trialkylchlorosilane R3 SiCl, yields ....................
(1994, 1M)
3. The two types of bonds present in B2H6 are covalent and .................... (1994, 1M)
4. The recently discovered allotrope of carbon (e.g. C 60) is commonly known as
....................
(1994, 1M)
5. A liquid which is permanently super cooled is frequently called ....................
(1997, 1M)
D:\BOOKS\JANUARY\26-A NEW APPROACH TO INORGANIC CHEMISTRY VOL. 1\UNIT-5\IIIRD PROOF DT. 27/3/12
260 A NEW APPROACH TO INORGANIC CHEMISTRY
Type-VII
 True/False Type Questions
1. Carbon tetrachloride burns in air and light to give phosgene gas.
(1983, 1M)
2. Graphite is a better lubricant on the moon than on the earth.
(1987, 1M)
3. All the Al—Cl bonds in Al2Cl6 are equivalent.
(1989, 1M)
4. Diamond is harder than graphite.
(1993, 1M)
5. The basic nature of hydroxide of group 13 (Gr. IIIA) decreases progressively down
the group.
(1993, 1M)
6. The tendency for catenation is much higher for C than for Si.
(1993, 1M)
Type-VIII
 Integer Type Questions
1. The value of n in the molecular formula Ben Al2 Si6 O18 is
(2010)
Objective Questions I
1. (b)
2. (d)
8. (b)
9. (a)
3. (a)
4. (b)
5. (a)
3. (a)
4. (c)
5. (c)
6. (a)
7. (c)
Objective Questions II
1. (a, b, c)
Assertion and Reason
1. (a)
2. (c)
Match the Columns
1. A  q;
B  r;
C  p,
D  q,r;
Fill in the Blanks
1. silicones
2. (R3 SiO)2
4. Buckminister fullerene
3. three centre two electron bond or banana bond
5. amorphous solid eg. glass
D:\BOOKS\JANUARY\26-A NEW APPROACH TO INORGANIC CHEMISTRY VOL. 1\UNIT-5\IIIRD PROOF DT. 27/3/12
P-BLOCK ELEMENTS 261
True/False Type
1. F
2. T
3. F
4. T
5. F
6. T
Integer Type Question
1. (3)
HINTS & SOLUTIONS
Type-II Objective Questions I
Q.1.(b)Graphite has layered structure and conduct electricity moderately.
Q.2.(d)PbI4 is least stable, has doubtful existence due to inert pair effect, the stable oxidation
state of lead is +2
Q.3.(a) Carbon monoxide is a neutral oxide, all others are amphoteric:
Q.4.(b)In BCl3 (sp2 hybridisation) bond angle = 120°.
In PCl3, AsCl3 and BiCl3, central atom is sp3 hybridized.
Bond angle decreases down the group with increasing size of atom. Hence overall
order of bond angle is:
B > P > As > Bi
Q.5.(a) CO2 is acidic oxide, H2O is neutral, CaO is strongly basic and CuO is weakly basic.
Therefore, order of acid strength is:
CaO < CuO < H2O < CO2
Q.6.(a) Orthoboric acid is a weak, monobasic, Lewis acid.
OH
has deficiency of a lone-pair
|
(Lewis acid)
HO—B
|
OH
Q.7.(c) Me2SiCl2 on hydrolysis yields a linear chain silicone as:
D:\BOOKS\JANUARY\26-A NEW APPROACH TO INORGANIC CHEMISTRY VOL. 1\UNIT-5\IIIRD PROOF DT. 27/3/12
262 A NEW APPROACH TO INORGANIC CHEMISTRY
CH 3
CH3
|
|
Cl—Si—Cl + 2H 2O 
 HO— Si—OH  2HCl
|
|
CH
CH 3
3
CH3
CH 3
|
|
Polymerization
nHO—Si—OH 
[—O— Si—O—]n
|
|
CH 3
CH 3
Q.8.(b)In sheet silicates, three out of four oxygen of SiO44– unit are shared as shown below:
Three oxygens of
every tetrahedra are
shared with others
Q.9.(a) Orthoboric acid is a very weak acid, direct neutralization does not complete. However,
addition of cis-diol allow the reaction to go to completion by forming a stable complex
with [B(OH)4] as :
HO
HO
–
B
HO
HO
+
CH2—OH
|
CH2—OH
H2C—O
|
CH2—O
–
B
O—H2C
+ 2H2O
|
O—CH2
Objective
Q.1.(a, b, c) Diborane (B2H6) undergo unsymmetric cleavage with NH 3, primary and
secondary amine while tertiary amine brings about symmetrical cleavage of B2H6 as:
H
H
B
H
H
B
unsymmetric
cleavage
H
H
+
NH3 or 1° amine
or 2° amine
X
D:\BOOKS\JANUARY\26-A NEW APPROACH TO INORGANIC CHEMISTRY VOL. 1\UNIT-5\IIIRD PROOF DT. 27/3/12
[BH2 (X )2 ]+ [BH4]–
P-BLOCK ELEMENTS 263
Type-IV Assertion and Reason Type Question
Q.1.(a) Due to small size and high charge on Al in Al (OH)3 the Al—O and O—H bond
strengths are comparable and compound can give both H+ and HO– under appropriate
conditions as
Al(OH)3 + 3HCl  AlCl3 + 3H2O
base
Al(OH)3 + NaOH  Na[Al(OH)4]
acid
Both statements are correct and statement II is a correct explanation of statement I.
Q.2.(c) SiCl4 reacts with H2O due to vacant d-orbitals available with Si
··
H2O··
SiCl4
No such vacant d- are available with carbon, hence CCl4 does not react with water.
Otherwise, both SiCI4 and CCl4 are covalent.
Statement I is true but statement II is false.
Q.3.(a) Small size and high charge on B3+ makes it highly polarising. Sum of three IE of
Boron is very high. Therefore, in most of its compounds, boron forms covalent bonds.
Hence, both statement I and statement II are true and statement II is a correct explanation
of statement I.
Q.4.(b)Orthoboric acid is a weak, monobasic, Lewis acid and the poor acidic character is due
to p – p back bondings

p p

  Back bonding decreases electron

 deficiency at boron, decreases
HO—B—OH
 its Lewis acid strength.
|

OH

Q.5.(c) In group 14, the higher oxidation state becomes less tenable due to inert pair effect.
Therefore, lead show +2 stable oxidation state. Hence, Pb4+ act as a strong oxidising
agent, itself reduced to Pb2+ very easily.
Type-V Match the Columns
Q.1. (A) Bi3+ hydrolysis to (BiO)+ = q
D:\BOOKS\JANUARY\26-A NEW APPROACH TO INORGANIC CHEMISTRY VOL. 1\UNIT-5\IIIRD PROOF DT. 27/3/12
264 A NEW APPROACH TO INORGANIC CHEMISTRY
(B) [AlO2]– exist in basic medium, on acidification gives Al(OH)3 = r
(C) Orthosilicate (SiO44–) on heating changes into pyrosilicate Si2O76– = p.
(D) Tetraborate ion [B4O2–7] on treatment with dil. acid hydrolyses gradually to orthoboric
acid = q, r.
Type-VI Fill in the Blanks
1. Silicones
2. Dimeric Silicone
R
R
R
|
|
|
–HCl
R3SiCl + H 2 O 
 R — Si —OH 
 R — Si —O — Si —R
|
|
|
R
R
R
dimeric silicone
3. Three centred two electron bonds.
H
H
B
H
B
H
H
Covalent bonds
H
three centred two electron
bridged B—H—B bonds
4. Buckminster fullerene is the name of recently discovered allotrope of carbon.
5. Amorphous solids are commonly known as super cooled liquid eg. glass.
Type-V True/False
1. False Phosgene gas is obtained by treatment of CCl4 with superheated steam:
CCl 4  H 2 O (vapour) 
 COCl2  2HCl
2. True Graphite is better lubricant on moon than on earth because of absence of gravitional
pull on the moon.
3. False In Al2Cl6 Al—Cl bonds are not equivalent:
Cl
Cl
Al
Cl
Cl
Al
Cl
Cl
The bridged Al—Cl bonds
are different from terminal
Al—Cl bonds
D:\BOOKS\JANUARY\26-A NEW APPROACH TO INORGANIC CHEMISTRY VOL. 1\UNIT-5\IIIRD PROOF DT. 27/3/12
P-BLOCK ELEMENTS 265
4. True Graphite has a layered structure of hexagonal carbon rings stacked one over other
and held by weak vanderwaals force. On the other hand, in diamond, each carbon is
tetrahedrally bond to other four carbons extended in three dimensional space, giving
a giant, network structure. Due to this reason, diamond is harder than graphite.
5. True The basic nature of hydroxide of G-13 increases from top to bottom due to increase in
electropositive character.
6. True Due to small size of carbon than silicon, C—C bond is stronger than Si—Si bond,
hence, former is more likely to show catenation than later.
Integer Type Questions
1. BenAl2Si6O18,
2n + 6 + 24 – 36 = 0  n = 3
SECTION B: AIEEE QUESTIONS
Q.1. Graphite is a soft lubricant extremely difficult to melt. The reason for this anomalous
behaviour is that graphite.
(a) is a non-crystalline substance
(b) is an allotropic form of diamond
(c) has molecules of variable molecular masses like polymers
(d) has carbon atoms arranged in large plates of ring of strongly bound carbon atoms with
weak interpolar bonds.
Ans. (d): Graphite has a two-dimensional sheet like structure and each carbon atom makes a
use of sp2 hybridisation.
The above layer structure of graphite is less compact. Further, since the bonding
between the layer involving only weak van der waal’s force these layer can slide
over each other. This gives softness, greasiness and lubricating character of graphite.
Q. 2. The state of hybridization of boron and oxygen atoms in boric acid (H3BO3) are
respectively
(a) sp2 and sp2
(c) sp2 and sp2
(b) sp3 and sp2
(d) sp3 and sp3
D:\BOOKS\JANUARY\26-A NEW APPROACH TO INORGANIC CHEMISTRY VOL. 1\UNIT-5\IIIRD PROOF DT. 27/3/12
266 A NEW APPROACH TO INORGANIC CHEMISTRY
H
··
:
sp3 O
Ans. (b)
B
H
··
O
sp3 ··
sp2
··
H
O
·· sp3
Q.3. In silicon dioxide
(a) each silicon atom is surrounded by four oxygen atoms and each silicon atom is bonded
to two silicon atom
(b) each silicon atom is surrounded by two oxygen atoms and each oxygen atom is bonded
to two silicon atoms
|
(c) silicon atom is bonded to two oxygen atoms
(d) there are double bonds between silicon and oxygen atoms
Ans. (a) :Silicon dioxide exhibits polymorphism. It is a network solid
in which each Si atom is surrounded tetrahedrally by four
oxygen atoms. And each oxygen atom is surround by two Si
atoms.
Q.4. The stability of dihalides of Si, Ge, Sn and Pb increases
steadily in the sequence
O
|
Si
O | O
O
Si —
Si
|
|
Si
O | O
O
|
(a) PbX2 << SnX2 <<GeX2 << SiX2
(b) GeX2 << SiX2 << SnX2 << PbX2
(c) SiX2 << GeX2 << PbX2 << SnX2
(d) SiX2 << GeX2 << SnX2 << PbX2
Ans. (d) :Due to the inert pair effect the stability of dihalides ( of group IV elements ) increases
as we go down the group.
Q.5. Among the following substituted silanes the one which will give rise to cross linked
silicone polymer on hydrolysis is
(a) R3SiCl
(b) R4 Si
(c) RSiCl3
(d) R2SiCl2
Ans. (c) : RSiCl3 on hydrolysis gives a cross linked silicone. The formation can be explained as
D:\BOOKS\JANUARY\26-A NEW APPROACH TO INORGANIC CHEMISTRY VOL. 1\UNIT-5\IIIRD PROOF DT. 27/3/12
P-BLOCK ELEMENTS 267
R
R
R
|
|
|
—O—Si—O—Si—O—Si—O
Cl
OH
|
|
|
|
|
O
O
O
3H 2O
|
|
|
R— Si — Cl 
 R— Si— OH 
—3HCl
—O—Si—O—Si—O—Si—O
|
|
|
|
|
Cl
OH
R
R
R
Cross-linked silicone
Q.6. Alum help in purifying water by
(a) forming Si complex with clay particles
(b) sulpate part which combines with the dirt and removes it
(c) coagulating the mud particles
(d) making mud water soluble.
Ans.(c) : The negatively charged colloidal particles of impurities get coagulated by the AI3+ ions
and settle down and pure water can be decanted off.
Q.7. Aluminium chloride exists as dimmer, Al2Cl6 in solid state as well as in solution of
non-polar solvents such as benzene. When dissolved in water, it gives
(a) Al3 + 3CI–
(c) [Al(OH)6]3– + 3HCI
(b) [Al(H2O)6]3+ + 3CI–
(d) Al2O3 + 6HCI.
Ans.(b) : Al2Cl6 + 12H2O  2[Al(H2O6)] + 6Cl–
3+
Q.8. Among the properties (A) reducing (B) oxidizing (C)complexing, the set of properties
shown by CN– ion towards metal species is
(2004)
(a) A, B
(b) B, C
(c) C, A
(d) A, B, C
Ans.(c) : CN– ions act both as reducing agent as well as good complexing agent.
Q.9. Heating an aqueous solution of aluminium chloride to dryness will give
(a) AlCI3
(b) Al2Cl6
(c) Al2O3
(d) Al(OH)Cl2
Ans.(b) :
D:\BOOKS\JANUARY\26-A NEW APPROACH TO INORGANIC CHEMISTRY VOL. 1\UNIT-5\IIIRD PROOF DT. 27/3/12
268 A NEW APPROACH TO INORGANIC CHEMISTRY
Q.10. A metal, M forms chlorides in +2 and +4 oxidation state. Which of the following
statements about these chlorides is correct?
(a) MCl2 is more volatile than MCl4
(b) MCl2 is more soluble in anhydrous ethanol than MCl4
(c) MCl2 is more ionic than MCl4
(d) MCl2is more easily hydrolysed than MCl4
Ans. (c) :The compounds showing an oxidation state of +4 are covalent compound whereas
those which show +2 oxidation state are ionic in nature.
(c) MCl2 is more ionic than MCl4
Q.6. In curing cement plasters water is sprinkled from time to time. This helps in
(a) Keeping it cool
(b) developing interlocking needle-like crystals of hydrated silicates
(c) hydrating sand and gravel mixed with cement
(d) converting sand into silicic acid.
(2003)
Ans.(b): Water develops interlocking needle-like crystals of hydrated silicates. The reactions
involved are exothermic, water is sprinkled to shift the reaction towards right.
Q.7. Among Al2O3, SiO2, P2O3 and SO2 the correct order of acid strength is
(a) SO2 < P2O3 < SiO2 < Al2O3
(b) SiO2 < SiO2 < AlO3 < P2O3
(c) Al2O3 < SiO2 < SO2 < P2O3
(d) Al2O3 < SiO2 < P2O3 < SO2.
(2004)
Ans. (d): Acidity of the oxides of non-metals increases with increasing the electronegativity
and oxidation number of the element.
Al2 O3 < SiO2 < P2O3 < SO2
Q. 9. Which of the following oxides is amphoteric in character?
(a) CaO
(b) CO2
(c) SiO2
(d) SnO2
Ans.(d): CaO-basic, CO2 and SiO2-acidic, SnO2-amphoteric, as it reacts both with acids and
bases.
D:\BOOKS\JANUARY\26-A NEW APPROACH TO INORGANIC CHEMISTRY VOL. 1\UNIT-5\IIIRD PROOF DT. 27/3/12
P-BLOCK ELEMENTS 269
SnO2 + 4HCl  SnCl4 + 2H2O
SnO2 + 2NaOH  Na2SnO3 + H2O
Q.13. Identify the wrong statement in the following:
(a) Acid rain is mostly because of oxide of nitrogen and sulphur
(b) Chlorofluorocarbons are responsible for ozone layer depletion
(c) Greenhouse effect is responsible for global warming
(d) Ozone layer does not permit infrared from the sun to reach the earth
Ans.(d): The thick layer of ozone called ozone blanket which is effective in absorbing harmful
ultraviolet rays given out by the sun acts as a protective shield. It does not permit the
ulta violet rays from sun to reach the earth.
Q.14. Which one of the following is the correct statement
(a) B2H6·2NH3 is known as inorganic benzene
(b) Boric acid is a protonic acid
(c) Beryllium exhibits coordination number six
(d) Chlorides of both beryllium and aluminium have bridged chloride structure in solid.
Ans. (a) Wrong – Inorganic benzene is B3N3H6
(b) Wrong: Boric acid is weak monobasic acid but (Ka = 1.0 ×10–9). it does not act as a
protonic acid (i.e., proton donor) but behaves as a Lewis acid by accepting a pair of
electrons from OH– ions.
B(OH)3 + 2H2O  [B(OH)4]– + H3O+
(c) Wrong: Beryllium exhibits max coordination number of four as it has only four available
orbitals in its valency shell.
(d) Correct
Cl
Cl—Be
Cl
Be—Cl
Cl
Polymeric structure of BeCl2
Cl
Al
Cl
Cl
Al
Cl
Cl
Dimeric structure of Al2Cl6
D:\BOOKS\JANUARY\26-A NEW APPROACH TO INORGANIC CHEMISTRY VOL. 1\UNIT-5\IIIRD PROOF DT. 27/3/12
270 A NEW APPROACH TO INORGANIC CHEMISTRY
B. MULTIPLE CHOICE QUESTIONS FOR PRACTICE
(ONLY ONE CORRECT OPTION)
1.
2.
3.
4.
5.
6.
Carborundum is obtained when silica is heated at high temperature with
(a) carbon
(b) carbon monoxide
(c) carbon dioxide
(d) calcium carbonate
In the aluminothermic process, Al acts as
(a) flux
(b) oxidising agent
(c) reducing agent
(d) solder
Which of the following is a component of ruby?
(a) CaCO3
(b) MgCO3
(c) Al2O3
(d) Al (OH)3
Read lead is
(a) PbO
(b) Pb3O4
(c) PbO2
(d) Pb4O3
The structure and hybridisation of Si(CH3)4 is
(a) bent, sp
(b) trigonal, sp2
(c) octahedral, d2sp3
(d) tetrahedral, sp3
The shape of gaseous SnCl2 is
(a) tetrahedral
(c) angular
7.
8.
(b) linear
(d) T-shaped
Mark the oxide which is amphoteric in character
(a) CO2
(b) SiO2
(c) SnO2
(d) CaO
Which of the following is most stable ?
(a) Sn2+
(c) Si2+
(b) Ge2+
(d) Pb2+
D:\BOOKS\JANUARY\26-A NEW APPROACH TO INORGANIC CHEMISTRY VOL. 1\UNIT-5\IIIRD PROOF DT. 27/3/12
P-BLOCK ELEMENTS 271
9.
Of the following compounds, the most acidic is
(a) As2O3
(c) Sb2O3
(b) P2O5
(d) Bi2O3
10. Which one of the following is an amphoteric oxide ?
(a) Na2O
(c) B2O3
(b) SO2
(d) ZnO
11. Which one of the following compounds has the smallest bond angle in its molecule ?
(a) OH2
(c) NH3
(b) SH2
(d) SO2
12. Which is not a mineral of aluminium ?
(a) Anhydride
(c) Corundum
(b) Bauxite
(d) Diaspore
13. The metal used for making radiations shield is
(a) Pb
(c) Zn
(b) Fe
(d) Si
14. For making good quality mirrors, plates of flint glass are used. These are obtained by
floating molten glass over a liquid metal which does not solidify before glass. The metal
used can be
(a) tin
(b) sodium
(c) magnesium
(d) mercury
15. (Me)2SiCl2 on hydrolysis will produce
(a) (Me)2 Si(OH)2
(b) (Me)2Si==O
(c) [—O—(Me)2Si—O—]n
(d) Me2SiCl(OH)
16. Alum is widely used to purify water since
(a) it forms complex with clay particles
(b) it coagulates the mud particles
(c) it exchanges Ca2+ and Mg2+ ions present in hard water
(d) its sulphate ion is water purifier
D:\BOOKS\JANUARY\26-A NEW APPROACH TO INORGANIC CHEMISTRY VOL. 1\UNIT-5\IIIRD PROOF DT. 27/3/12
272 A NEW APPROACH TO INORGANIC CHEMISTRY
ANSWERS
1. (a)
2. (c)
3. (c)
4. (b)
5. (d)
6. (c)
7. (c)
10. (d)
11. (b)
12. (a)
13. (a)
14. (a)
15. (a)
16. (b)
8. (d)
9. (b)
PROBLEMS FOR PRACTICE
A. Subjective Questions
Q.1. Give chemical reaction in support of each of the following statements.
(i) The + 1 oxidation state gets stabilised progressively from Ga to Tl in group-13
(ii) Tin (II) is a reducing agent whereas Pb(II) is not
(iii) Ga(I) undergoes disproportionation reaction
(iv) PbO2 can act as an oxidising agent
Q.2. What happens when
(i) Tin dissolves in hot alkaline solution.
Q.3. Draw structural formula of the following
(i) Al2Cl6
Q.4. Assign appropriate reason for each of the following
(i) Aluminum is highly electropositive yet it is used as a structural material
(ii) Anhydrous AlCl3 acts as Catalyst
(iii) No form of elemental silicon is conductor
(iv) The tendency to exhibit +2 oxidation state increases with increase in atomic number
in group – 14
(v) Bismuth is a strong oxidizing agent in pentavalent state
(vi) Unlike In+, Tl+ is stable w.r.t. disproportionation reaction
(vii) PbO2 is stronger oxidizing agent than SnO2 Q.5. Present a comparative account of the following
(i) Structure of beryllium chloride and aluminum chloride
(ii) Action of water on CCl4 and SlCl4
D:\BOOKS\JANUARY\26-A NEW APPROACH TO INORGANIC CHEMISTRY VOL. 1\UNIT-5\IIIRD PROOF DT. 27/3/12
P-BLOCK ELEMENTS 273
Q.6. Give appropriate reason for the following:
(i) SiF62– is known but SiCl62– is not known
(ii) Molten aluminum bromide is poor conductor of electricity.
o————o
D:\BOOKS\JANUARY\26-A NEW APPROACH TO INORGANIC CHEMISTRY VOL. 1\UNIT-5\IIIRD PROOF DT. 27/3/12