p-block

Table of Contents

 Theory .................................................................................................................................. 2

 Exercise - 1 : Basic Objective Questions ............................................................................ 28

 Exercise - 2 : Previous Year Competition Questions ........................................................ 38

 Exercise - 3 : Advanced Objective Questions ................................................................... 42

 Exercise - 4 : Previous Year Other Competition Questions ............................................. 48

 Answer Key ......................................................................................................................... 51

 2 p-block (GROUP 13 & 14) MEDICAL

THEORY
Introduction :
Group 13 to 18 of the periodic table of elements constitute the p–block. The p–block contains
metals, metalloids as well as non–metals.
The p–block elements have general valence shell electronic configuration ns2 np1–6.
The first member of each group from 13–17 of the p–block elements differ in many respects from the other
members of their respective groups because of small size, high electronegativity and absence of d–orbitals.
The first member of a gorup also has greater ability to form p–p multiple bonds to itself (e.g. C = C, C 
C, N  N) and to element of second row (e.g C = O, C = N, C  N, N = O) compared to the other members of
the same group.
The highest oxidation of p–block element is equal to the group number minus 10. Moving down the group,
the oxidation state two less than the highest group oxidation state becomes more stable in groups 13 to 16 due to
inert pair effect (reluctance of s-subshell electrons to participate in chemical bonding)
TRENDS IN PROPERTIES OF p-BLOCK ELEMENTS.
Electronegativity,
ionization enthalpy,
oxidizing power.

B C N O F Ne

Covalent radius, Al Si P S Cl Ar Electronegativity,

van der Waals' radius, enthalpy of atomization
metallic character Ga Ge As Se Br Kr (except for N2, O2, F2),
ionization enthalpy,
In Sn Sb Te I Xe oxidizing power.

Tl Pb Bi Po At Rn

Covalent radius,
van der Waals' radius,
enthalpy of atomization
(upto group 14),
metallic character

(A) GROUP 13 ELEMENTS : THE BORON FAMILY

Boron is a typical non-metal, aluminium is a metal but shows many chemical similarities to boron , and gallium,
indium and thallium are almost exclusively metallic in character,
Electronic Configuration:
The outer electronic configuration of these elements is ns2 np1.
Atomic Radii :
On moving down the group, for each successive member one extra shell of electrons is added and therefore, atomic
radius is expected to increases. Atomic radius of Ga is less than of Al. The presence of additional 10 d-electrons
offer only poor screening effect for the outer electrons from the increased nuclear charge in gallium. Consequentlym,
the atomic redius of gallium (135 pm) is less than that of aluminium (143 pm).
Ionization Enthalpy:
The ionisation enthalpy values as expected from the general trends do not decrease smoothly down the group . The
decreases from B to Al is associated with increases in size. The observed discontinuity in the ionisation enthalpy
values between Al and Ga, and between In and TI are due to inability of d- and f electrons, which have low screening
effect, to compensate the increase in nuclear charge. The sum of the first three ionisation enthalpies for each of the
elements is very high .
 3 p-block (GROUP 13 & 14) MEDICAL

Electronegativity:
Down the group, electronegativity first decreases from B to Al and then increases marginally. This is because of the
discrepancies in atomic size of the elements.
Physical Properties
Boron is non-metallic in nature . It is extremely hard and black coloured solid. It exists in many allotropic forms.
Due to very strong crystalline lattice, boron has unusually high melting point. Rest of the member are soft metals
with low melting point and high electrical conductivity. Gallium with low melting point (303 K), could exist in liquid
state during summer. Its high boiling point (2676 K) makes it a useful material for measuring high temperatures.
Density of the elements increases down the group from boron to thallium.
Atomic & physical properties :

Element B Al Ga In Tl
Atomic Number 5 13 31 49 81
Atomic Mass 10.81 26.98 69.72 114.82 204.38
2 1 2 1 10 2 1 10 2 1 14
Electronic configuration [He] 2s 2p [Ne] 3s 3p [Ar] 3d 4s 4p [Kr] 4d 5s 5p [Xe] 4f 5d10 6s26p1
Atomic Radius / pm 85 143 135 167 170
3+
Ionic Radius M / pm – 53.5 62 80 88.5
 800 577 578 558 590
Ionization enthalpy
 2427 1816 1979 1820 1971
/ (kJ mol–1)
 3659 2744 2962 2704 2877
Electronegativity 2.0 1.5 1.6 1.7 1.8
–3
Density/[g cm (293 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

Chemical Properties :
Oxidation state and trends in chemical reactivity :
Due to small size of boron, the sum of its first three ionization enthalpies is very high. This prevents it to form + 3
ions and force it to form only covalent compounds. But as we move from B to Al, the sum of the first three ionisation
enthalpies of Al considerably decreases, and is therefore able to form Al+3 ions. However, down the group, due to
poor shielding, effective nuclear charge holds ns electrons tightly (responsible for inert pair effect) and thereby,
restricting their participation in bonding. As a result of this only p-orbital electron may be involved in bonding. In fact
in Ga, In and Tl, both + 1 and + 3 oxidations states are observed. The relative stability of + 1 oxidations state
progressively increases for heavier elements: Al < Ga < In < Tl. In thallium +1 oxidation state is predominant and
+ 3 oxidation state highly oxidising in character. The compound in +1 oxidation state , as expected from energy
considerations, are more ionic than those in + 3 oxidations state.
In trivalent state, the number of electrons around the central atom in a molecule of the compounds of these
elements (e.g., boron in BF3) will be only six.
Such electron deficient molecules have tendency to accept a pair of electrons to achieve stable electronic
configuration and thus, behave as Lewis acids. The tendency to behave as Lewis acid decreases with the increases
in the size down the group. BCl3 easily accepts a lone pair of electrons from ammonia to form BCl3. NH3. In trivalent
state most of the compounds being covalent are hydrolysed in water. The trichloride on hydrolysis in water form
tetrahedral [ M (OH)4]– species; Aluminium chloride in acidified aqueous solution form octahedral [ Al(H2O)6]3+ ion.
(i) Reactivity towards air
Boron is unreactive in crystalline form. Aluminium forms a very thin oxide layer on the surface which protects the
metal from further attack. Amorphous boron and aluminium metal on heating in air form B2O3 and Al2O3 respectively.
With dinitrogen at high temperature they form nitrides.
 
2E(s) + 3 O2 (g)  2 E2O3(s) ; 2E(s) + N2(g)  2 EN (s).
 4 p-block (GROUP 13 & 14) MEDICAL

The nature of these oxides varies down the group. Boron trioxide is acidic and reacts with basic (metallic) oxides
forming metal borates. Aluminium and gallium oxides are amphoteric and those of indium and thallium are basic in
their properties.

(ii) Reactivity towards acids and alkalies

Boron does not react with acids and alkalies even at moderate temperature; but aluminium dissolves in mineral
acids and aqueous alkalies and thus shows amphoteric character.
Aluminimum dissolved in dilute HCl and liberates dihydrogen. However, concentrated nitric acid renders aluminium
passive by forming protective oxide layer on the surface . Aluminium also reacts with aqueous alkali and liberates
dihydrogen .
2 Al(s) + 6 HCl(aq)  2 Al3+ (aq) + 6 Cl–(aq) + 3 H2(g)
2Al(s) + 2NaOH (aq) + 6H2O (1)  2Na+ [Al(OH)4]– (aq) + 3H2(g)
Sodium tetrahydroxoaluminate (III)
(iii) Reactivity towards halogens
These elements react with halogen to form trihalides (except Tl I3 )..
2E(s) + 3X2 (g)  2EX3 (s) (X = F, Cl Br, I)
IMPORTANT TRENDS AND ANOMALOUS PROPERTIES OF BORON
The tri-chlordes, bromides and iodies of all these elements being covalent in nature are hydrolysed in water.
Species like tetrahedral [M(OH)4]– and octahedral [M(H2O)6]3+ ,except in boron, exist in aqueous medium.
It is due to the absence of d orbitals that the maximum covalence of boron is 4. Since the d-orbitals are available
with Al and other elements, the maximum covalence can be expected beyond 4.

 BORON (B):
 OCCURRENCE :
Boron occurs in nature in the form of the following minerals:
(i) Borax (Na+)2B4O72- .10H2O.(Boron is part of an anionic complex), (ii) Boric acid H3BO3,
(iii) Kernite Na2B4O7 . 4H2O & (iv) Colemanite Ca2 B6O11. 5H2O
 EXTRACTION OF BORON :
(i) By the reduction of B2O3 with magnesium, sodium or potassium in the absence of air :
Na2B4O7 + 2HCl + 5H2O  4H3BO3 + 2NaCl

2H3BO3  B2O3 + 3H2O ; B2O3 + 3Mg  2B + 3MgO
The product thus obtained is boiled with HCl and filtered when K2O or MgO dissolves leaving
behind elemental boron. It is thoroughly washed to remove HCl and then dried finally.
Brown amorphous powder of B is obtained in this way.
(ii) From potassium fluoroborate (KBF4) by heating it with potassium metal.

KBF4 + 3K  4KF + B.
It is then treated with dilute HCl to remove KF and B is then washed and dried.
(iii) In small quantities in pure form (crystalline boron) by the
(i) Reduction of BBr3 with H2 on a heated titanium metal filament at 1275-1475 K
The vapours of Br2 are absorbed in Cu and the residual vapours of boron are condensed.
(ii) Decomposition of BI3 vapours by means of high tension arc (80 kV) through tungsten
electrodes.
2BI3  2B + 3 I2 (Van Arkel mehod).
 5 p-block (GROUP 13 & 14) MEDICAL

 PROPERTIES :
(i) It exists in five forms, four of which are crystalline and one is amorphous. All crystalline forms are very
hard made up of clusters of B12 units. All crystalline forms are black in appearance and chemically inert. Melting
points are around 2300°C. But amorphous form is brown and chemically active.
(ii) Reaction with air :
4B + 3O2   2B2O3
700C

High temperatur e, pressure
2B + N2   2BN ;
700C
 BN + 3H2O       H3BO3 + NH3
(iii) Action of alkalies and acids :
2B + 2NaOH + 2H2O  2NaBO2 + 3H2
oxidation
2B + 3H2SO4    2H3BO3 + 3SO2
oxidation
2B + 6HNO3    2H3BO3 + 6NO2
(iv) Reaction with Mg and Ca :
3Mg + 2B  Mg3B2
3Ca + 2B  Ca3B2
Mg3B2 on consequent hydrolysis gives diborane.
hydrolysis
Mg3B2 + 6HCl     3MgCl2 + B2H6 ; B2H6 + 6H2O  2H3BO3 + 6H2
(v) Reducing properties :
3SiO2 + 4B  2B2O3 + 3Si
3CO2 + 4B  2B2O3 + 3C
(vi) It decomposes steam liberating hydrogen gas.
2B + 3H2O(steam)  B2O3 + 3H2
 USES :
Boron is used in the construction of high impact-resistant steel and, since it absorbs neutrons, in
reactor rods for controlling atomic reactions.
 COMPOUNDS OF BORON :
 BORON TRIOXIDE (B 2 O 3 ) :
 PREPARATION :

 PROPERTIES :
It is a weakly acidic oxide and reacts with alkalies or bases to form borates.
3Na2O + B2O3  2Na3BO3 (sodium orthoborate).
It reacts with water slowly to form orthoboric acid. When heated with transition metal salts, it forms
coloured compounds.
H2O + B2O3  2HBO2 ; HBO2 + H2O H3BO3

3B2O3 + Cr2(SO4)3  3SO3  + 2Cr(BO2)3(green)

2B2O3 + 2Cu(NO3)2  4NO2  + O2  + 2Cu(BO2)2 (blue)
B2O3 + P2O5 2BPO4
 ORTHOBORIC ACID (H 3 BO 3 ) :
Among the oxyacids of boron are
 6 p-block (GROUP 13 & 14) MEDICAL

 PREPARATION :
(i) It is precipitated by treating a concentrated solution of borax with sulphuric acid.
Na2B4O7 + H2SO4 + 5H2O  Na2SO4 + 4H3BO3 
(ii) From Colemanite: Powdered colemanite is suspended in water and excess SO2 is passed through it.
On filtering and cooling the filtrate, white crystals of H3BO3 are obtained.
Ca2B6O11 + 2SO2 + 11H2O  2Ca(HSO3)2 + 6H3 BO3
 PROPER TIES:
(i) It is a weak monobasic acid and in aqueous solution the boron atom completes its octet by
removing OH- from water molecules:
B(OH)3(aq) + 2H2O()  B(OH)4– (aq) + H3O+(aq).
It, therefore, functions as a Lewis acid and not as a proton donor.
It behaves as strong acid when a polyhydroxy compound such as glycol or glycerol is added to its
aqueous solution. The acidity is due to the high stability of the conjugate bone chelate complex.

Ethanol does not form similar complex but catechol, salicylic acids form similar complexes.

When heated it first forms metaboric acid (HBO2) and then boron trioxide.

Orthoboric acid is greasy to touch less soluble in cold water but more soluble in hot water. It has a layered
structure in which planar BO3 units are joined by hydrogen bonds.

H B H H

H B H H B

H H H H

H B H H B H

H B H
H
H
 7 p-block (GROUP 13 & 14) MEDICAL

 TEST FOR BORATE RADICAL :

When boric acid is heated with ethyl alcohol, the evolved gas is burned forming a green edged flame.
H3BO3 + 3C2H5OH  B(OC2H5)3 + 3H2O
ethyl borate (volatile)
 USES :
It is an antiseptic and its water solution is used as an eyewash. It is also used in glass, enamel and pottery
industry.
 BORAX (Na 2 B 4 O 7 .10H 2 O) :
 PREPARATION: It is found in nature but can also be prepared by the following methods.
(i) From Colemanite.
When colemanite powder is heated with Na2CO3 solution, the following reaction occurs with the
precipitation of CaCO3.
Ca2B6O11 + 2Na2CO3  2CaCO3  + Na2B4O7 + 2NaBO2
The filterate is cooled when white crystals of borax are precipitated. The mother liquor on treatment
with CO2 converts NaBO2 to Na2B4O7 which precipitates out on crystallization.
4NaBO2 + CO2  Na2B4O7 + Na2CO3
(ii) From orthoboric acid.
Borax is obtained by the action of Na2CO3 on orthoboric acid.
4H3BO3 + Na2CO3  Na2B4O7 + 6H2O + CO2 
 PROPER TIES:
(i) Borax is a white powder, less soluble in cold water, more soluble in hot water.

(ii) Its aqueous solution is alkaline because of its hydrolysis to weak acid H3BO3 and strong alkali
NaOH.
Na2B4O7 + 7H2O  4H3BO3 + 2NaOH
(iii) Action of heat.
When borax powder is heated, it first swells due to loss of water in the form of steam but at 740oC
it becomes converted into colourless transparent borax bead.

Na2B4O7.10H2O +  Na2B4O7 + 10 H2O
Na2B4O7   2NaBO2 + B2O3 (borax bead)
740 º C

(iv) Action of acids :
Na2B4O7 + 2HCl + 5H2O  2NaCl + 4H3BO3 (boric acid)
On cooking, the white flakes of boric acid are obtained
NaOH H2 O 2
(v) Na2B4O7   NaBO2   Na2 [(OH)2 B(O–O)2 B (OH)2] 6H2O
 Correct formula of borax is Na2[B4O5(OH)4] . 8H2O
 BORAX-BEAD TEST : Boric anhydride reacts with certain metal salts such as, Ni2+, Co2+, Cr3+, Cu2+, Mn2+
etc. to form coloured metaborates. The colour of the metaborates can be used to identify the metallic ions (cations)
in salts.
 8 p-block (GROUP 13 & 14) MEDICAL


Na2B4O7. 10H2O  
10H2O
Na2B4O7 
740 º C
 2NaBO 2  B 2O 3 ; CuO + B2O3
 
 Cu(BO2)2 (blue bead)
glassy mass

 USES :
It is used
(i) in borax bead test, (ii) in purifying gold, (iii) as flux during welding of metals and (iv) in production of glass.
 DIBORANE (B2H6) : Binary compounds of B with H are called boron hydrides or boranes. These compounds
form following two types of series :
BnHn+4 - B2H6, B5H9, B6H10, B10H14
BnHn+6 - B4H10, B5H11, B6H12, B9H15
The chemistry of diborane has aroused considerable interest because of its usefulness in many
synthetic reactions and also because the elucidation of its structure helped to clarify the basic concepts about
the structure of electron deficient compounds.
 PREPARATION :
(i) 4BF3 + 3LiAlH4   2B2H6 + 3LiF + 3AlF3
ether

silent electric
(ii) 2BCl3 + 6H2 (excess) disch
  B H + 6HCl
arg e 2 6

ether
(iii) 8BF3 + 6LiH   B2H6 + 6LiBF4
ether
(iv) 2NaBH4 + 2   B2H6 + 2Na + H2
ether
(v) 3NaBH4 + 4BF3   3NaBF + 2B H
450 K 4 2 6

(vi) It can also be prepared by treating NaBH4 with concentrated H2SO4 or H3PO4.
2NaBH4 + H2SO4  B2H6 + 2H2 + Na2SO4
2NaBH4 + 2H3PO4  B2H6 + 2H2 + 2NaH2PO4
(vii) 2BF3 + 6NaH   B2H6 + 6NaF (Industrial method)
450 K


 PROPERTIES :
(i) Diborane is a colourless gas (boiling point 183 K).
(ii) It is rapidly decomposed by water with the formation of H3BO3 & H2:
B2H6 + 6H2O  2H3BO3 + 6H2
(iii) Mixtures of diborane with air or oxygen inflame spontaneously producing large amount of heat.
Diborane has a higher heat of combustion per unit weight of fuel than most other fuels. It is therefore
used as a rocket fuel.
B2H6 + 3O2  B2O3 + 3H2O H = – 1976 kJ mol–1
(iv) Pyrolysis of B2H6 in sealed vessels at temperatures above 375 K is an exceedingly complex
process producing a mixture of various boranes, eg, B4H10, B5H9, B6H12, and B10H14.
(v) Diborane undergoes a facile addition reaction with alkenes and alkynes in ether solvents at room
temperature to form organoboranes. This reaction is known as hydroboration reaction
(vi) B2H6 + HCl  B2H5Cl + H2 (vii) B2H6 + 6MeOH  2B(OMe)3 + 6H2
 9 p-block (GROUP 13 & 14) MEDICAL

(vii) Cleavage reactions

(a) B2H6 + 2Me3N  2Me3NBH3 (b) B2H6 + 2Me3P  2Me3PBH3
(c) B2H6 + 2CO     2BH3CO (borane carbonyl)
200 º C, 20 atm

(viii) B2H6 + 2Et2S  2Et2SBH3

(ix) 3B2H6 + 6NH3  + –
   B2H6 . 2NH3 or [BH2(NH3)2] BH4 
low temprature
 B3N3H6 (borazole) + 12H2
200 º C


(x) B2H6 + 2KOH + 2H2O  2KBO2 + 6H2 (xv) B2H6 + 6Cl2  2BCl3 + 6HCl
(xi) B2H6 + 2LiH  2LiBH4

 ALUMINIUM (Al) :

 EXTRACTION (HALL-HEROULT PROCESS):

The aluminium is extracted from ore bauxite (Al2O3.2H2O). The ore is first purified by Bayere’s process.
The anhydrous Al2O3 is mixed with Na3AlF6 & CaF2 & then fused. The fused mixture is subjected to electrolytic
reduction when aluminium is obtained at cathode.
 Aluminium is purified by Hoope’s process

 PROPERTI ES :
(i) It is a silvery metal with a density of 2.7 g/cc, having a melting point of 660oC, and is a good
conductor of heat and electricity. It is malleable and ductile.
(ii) Action of air: Dry air has no action on aluminium. But moist air forms a thin layer of Al2O3 on its
surface and it loses its luster. At very high temperatures it burns to form Al2O3 and AlN.
(iii) Reaction with halogens: When gaseous halogens are passed over aluminium, its halide are formed in
an anhydrous form. 2Al + 3Cl2  2AlCl3
(iv) Action of alkalies: When warmed with concentrated NaOH, it liberates H2 gas and a colourless
solution of sodium aluminate is formed.
2Al + 2NaOH + 2H2O  2NaAlO2 + 3H2
(v) Action of acids: Aluminium reacts with dilute H2SO4 and dilute HCl but concentrated HNO3 does
not react with aluminium because aluminium becomes passive by the action of concentrated HNO3
forming a protective oxide layer on the surface.
2Al + 3H2SO4  Al2(SO4)3 + 3H2 2Al + 6HCl  2AlCl3 + 3H2
(vi) Reaction with N2: When N2 gas is passed over heated aluminium, aluminium nitride is formed. Hot
aluminium thus acts as an absorbing agent for N2.
2Al + N2  2AlN
AlN reacts with hot water to form Al(OH)3 and NH3
(vii) Reaction with water: Aluminium does not react with cold water. It is very slowly attacked by boiling
water or steam.
2Al + 3H2O  Al(OH)3 + 3H2
(viii) Action of HgCl2 solution: When aluminium is added to HgCl2 solution mercury is liberated.
3HgCl2 + 2Al  2AlCl3 + 3Hg
 10 p-block (GROUP 13 & 14) MEDICAL

(ix) Reduction of oxides of metals: When oxides of less reactive metal than aluminium is heated with
aluminium, the other metal is liberated.

3MnO2 + 4Al +  2Al2O3 + 3Mn; 
Cr2O3 + 2Al +  Al2O3 + 2Cr
 USES :
It is extensively used
(i) for manufacture of cooking and household utencils.
(ii) as aluminium plating for tanks, pipes, iron bars and other steel objects to prevent corrosion.
(iii) for manufacture of aluminium cables.
(iv) for making precision instruments, surgical apparatus, aircraft bodies, rail coaches, motorboats, car
 COMPOUNDS OF ALUMINIUM :
 ALUMINIUM OXIDE (Al 2 O 3 ) :
It is also called alumina. It occurs in nature in the form of bauxite and corundum. It is also found in the form
of gems. Some important aluminium oxide gems are:
(A) Topaz-yellow, (B) Sapphire-blue, (C) Ruby-red, (D) Amethyst-violet, (E) Emerald-green
 PREPARATION:
Pure Al2O3 is obtained by igniting Al2(SO4)3, Al(OH)3 or ammonium alum.

Al2(SO4)3 +  Al2O3 + 3SO3 ; 
2Al(OH)3 +  Al2O3 + 3H2O

(NH4)2SO4.Al2(SO4)3.24H2O  2NH3 + Al2O3 + 4SO2  + 25H2O
 PROPERTIES :
It is a white amorphous powder insoluble in water but soluble in acids (forming eg., AlCl3) as well as
alkalies (forming NaAlO2) , Thus amphoteric in nature. It is a polar covalent compound.
 USES :
It is used
(i) for the extraction of aluminium.
(ii) for making artificial gems.
(iii) for the preparation of compounds of aluminium.
(iv) in making furnace linings. It is a refractory material.
(v) as a catalyst in organic reactions.
 ALUMINIUM CHLORIDE (AlCl 3 .6H 2 O) :
It is a colourless crystalline solid, soluble in water. It is covalent. Anhydrous AlCl3 is a deliquescent white
solid.
 PREPARATION :
(i) By dissolving aluminium, Al2O3, or Al(OH)3 in dilute HCl :
2Al + 6HCl  2AlCl3 + 3H2 Al2O3 + 6HCl  2AlCl3 + 3H2O; Al(OH)3 +
3HCl  AlCl3 + 3H2O
The solution obtained is filtered and crystallized when the crystals of AlCl3.6H2O are obtained.
(ii) Anhydrous AlCl3 is obtained by the action of Cl2 on heated aluminium.
(iii) By heating a mixture of Al2O3 and coke and passing chlorine over it.
Al2O3 + 3C + 3Cl2  2AlCl3 (anhydrous) + 3CO
 11 p-block (GROUP 13 & 14) MEDICAL

 PROPERTIES :
(i) Action of heat:
Hydrated salt when heated strongly is converted to Al2O3.
2AlCl3.6H2O   Al2O3 + 6HCl + 9H2O
(ii) Action of moisture on anhydrous AlCl3:
When exposed to air, anhydrous AlCl3 produces white fumes of HCl
AlCl3 + 3H2O  Al(OH)3 + 3HCl
(iii) Action of NH3:
Anhydrous AlCl3 absorbs NH3 since the latter is a Lewis acid.
AlCl3 + 6NH3  AlCl3.6NH3 (white solid)
(iv) Action of NaOH solution:
When NaOH solution is added dropwise to an aqueous AlCl3 solution, a gelatinous precipitate of
Al(OH)3 is first formed which dissolves in excess of NaOH solution to give a colourless solution of
sodium aluminate.
AlCl3 + 3NaOH  Al(OH)3 + 3NaCl ; Al(OH)3 + NaOH  NaAlO2 + 2H2O
This reaction is important as a test to distinguish between an aluminium salt from salts of Mg, Ca, Sr,
and Ba. (When NaOH solution is added to their salt solutions, a white precipitate of hydroxide
forms which does not dissolve in excess of NaOH).
(v) Action of NH4OH solution:
When NH4OH solution is added to a solution of AlCl3, a white precipitate of Al(OH)3 is formed
which does not dissolve in excess of NH4OH.
AlCl3 + 3NH4OH  Al(OH)3 (white gelatinous) + 3NH4Cl
This reaction is important as a test to distinguish an Al salt from a Zn salt. (With a Zn salt a white
precipitate of Zn(OH)2 is formed which dissolves in excess of NH4OH solution).
(vi) Hydrolysis with water:
When AlCl3 is dissolved in water, it undergoes hydrolysis rapidly to produce Al(OH)3 which is a
weak base and HCl which is a strong acid. Hence the solution is acidic to litmus.
[Al(H2O)6]3+ [Al(H2O)5OH]+2 + H+
The complex cation has a high tendency to get dimerised.
2[Al(H2O)5OH]2+  [(H2O)4 (H2O)4 ]+4 + 2H2O
(vii) 4LiH + AlCl3  LiAlH4 + 3LiCl
 USES :
It is used :
(i) as catalyst for cracking of petroleum.
(ii) as catalyst in Friedel-Crafts reactions.
(iii) for preparing aluminium compounds.
 ALUMS ; M 2 SO 4 . M 2 (SO 4 ) 3 . 24H 2 O OR MM (SO 4 ) 2 . 12H 2 O
Alums are transparent crystalline solids having the above general formula where M is a univalent metal or
positive radical and M is a trivalent metal. Some important alums are:
(i) Potash alum K2SO4 . Al2(SO4)3 . 24H2O (ii) Chrome alum K2SO4 . Cr2(SO4)3 . 24H2O
(iii) Ferric alum K2SO4 . Fe2(SO4)3. 24H2O (iv) Ammonium alum (NH4)2SO4 . Al2(SO4)3 . 24H2O
Alums are double salts which when dissolved in water produce metal ions(or ammonium ions) and the
sulphate ions.
 12 p-block (GROUP 13 & 14) MEDICAL

 PREPARATION :
Alums can be prepared by fusing M2SO4 & M’2(SO4)3 in 1 : 1 molar ratio & the resulting mass is
dissolved into water. From the solution thus obtained, alums are crystallised.
 USES :
It is used
(i) as a mordant in dye industry
(ii) as a germicide for water purification
(iii) as a coagulating agent for precipitating colloidal impurities from water.

(B) GROUP 14 ELEMENTS : THE CARBON FAMILY

Carbon (C), silicon (Si), germanium (Ge), tin (Sn) and lead (Pb) are the members of group 14. Naturally occurring
carbon contains two stable isotopes:12C and 13C. In addition to these third isotopes, 14C is also presents , it is a
radioactive isotope with half-life 5770 years and used for radiocarbon dating. Silicon is a very important component
of ceramices, glass and cement. Germanium exists only in traces. Tin occurs mainly as cassiterite, SnO2 and
lead as galena, PbS. Ultrapure form of germanium and silicon are used to make transistors and semiconductor
devices.

Electronic Configuration :
The valence shell electronic configuration of these elements is ns2 np2.

Covalent Radius :
There is a considerable increase in covalent radius from C to Si, thereafter from Si to Pb a small increase in radius
is observed. This is due to the presence of completey filled d and f orbitals in heavier members.

Ionization Enthalpy :
The first ionization enthalpy of group 14 members is higher than the corresponding members of group 13. The
influence of inner core electron is visible here also. In general the ionisation enthalpy decreases down the group.Small
decreases in iH from Si to Ge to Sn and slight increase in iH from Sn to Pb is the consequence of poor shielding
effects of intervening d and f–orbitals and increases in size of the atom.

Electronegativity :
Due to small size, the elements of this group are slightly more electronegative than group 13 elements. The
electronegativity value for elements from Si to Pb are almost the same.

Physical Properties :
All group 14 members are solids. Carbon and silicon are non-metals, germanium is metalloid whereas tin and lead
are soft metals with low melting points. Melting points and boiling points of group 14 elements are much higher than
those of corresponding elements of group 13.
 13 p-block (GROUP 13 & 14) MEDICAL

ATOMIC & PHYSICAL PROPERTIES

Element C Si Ge Sn Pb

Atomic Number 6 14 32 50 82

Atomic Mass 12.01 28.09 72.60 118.71 207.2

2 2 2 2 10 2 2 10 2 2 14 10 2 2
Electronic configuration [He] 2s 2p [Ne] 3s 3p [Ar] 3d 4s 4p [Kr] 4d 5s 5p [Xe] 4f 5d 6s 6p

Atomic Radius / pm 77 118 122 140 146

+4
Ionic Radius M / pm – 40 53 69 78

 1086 786 761 708 715

Ionization enthalpy
 2352 1577 1537 11411 1450
/ (kJ mol–1)
 4620 3228 3300 2942 3081

Electronegativity 2.5 1.8 1.8 1.8 1.9

Melting point / K 4373 1693 1218 505 600

Boiling point / K – 3550 3123 2896 2024

Chemical Properties :
Oxidation states and trends in chemical reactivity :
The group 14 elements have four electrons in outermost shell. The common oxidation states exhibited by these
elements are + 4 and + 2. Carbon also exhibits negative oxidation states.Since the sum of the first four ionization
enthalpies is very high, compound in +4 oxidation state are generally covalent in nature. In heavier members the
tendency to show +2 oxidation state increases in the sequence Ge < Sn < Pb. It is due to the inability of ns2
electrons of valence shell to participate in bonding. The relative stabilities of these two oxidation states vary down
the group. Carbon cannot exceed its covalence more than 4. Other elements of the group can do so. It is because
of the presence of d orbital in them. Due to this, their halides undergo hydrolysis and have tendency to form
complexes by accepting electron pairs from donor species. For example, the species like ,SiF62– . [GeCl6]2–
,[Sn(OH)6]2– exist .
(i) Reactivity towards oxygen :
All members when heated in oxygen form oxides. There are mainly two types of oxides, i.e. monoxide and dioxide
of formula MO and MO2 respectively. SiO only exists at high temperature. Oxides in higher oxidation states of
elements are generally more acidic than those in lower oxidation state. The dioxides – CO2, SiO2 and GeO2 are
acidic, whereas SnO2 and PbO2 are amphoteric in nature. Among monoxides, CO is neutral, GeO is distinctly
acidic whereas SnO and PbO are amphoteric .
(ii) Reactivity towards water :
Carbon , silicon and germanium are not affected by water . Tin decomposes steam to form dioxide and dihydrogen
gas. Lead is unaffected by water, probably becauses of a protective oxide film formation.
(iii) Reactivity towards halogen :
These elements can form halides of formula MX2 and MX4 (where X = F, Cl Br, I). Except carbon all other members
react directly with halogen under suitable condition to make halides. Most of the MX4 are covalent in nature.
Exceptions are SnF4 and PbF4, which are ionic in nature . PbI4 does not exist because Pb–I bond initially formed
during the reaction does not release enough energy to unpair 6s2 electrons and excite one of them to higher orbital
to have four unpaired electrons around lead atom. Heavier members Ge to Pb are able to make halides of formula
MX2 . Stability of dihalides increases down the group. Except CCl4 other tetrachlorides are easily hydrolysed by
water because the central atom can accommodate the lone pair of electrons from oxygen atom of water molecules
in d orbital.
 14 p-block (GROUP 13 & 14) MEDICAL

IMPORTANT TRENDS AND ANOMALOUS BEHAVIOUR OF CARBON

Like first member of other groups, carbon also differs from rest of the members of its group. It is due to its smaller
size, higher electronegativity, higher ionisation enthalpy and unavailability of d orbitals. Accommodate only four
pairs of electrons around it. This would limit the maximum covalence to four whereas other members can expand
their covalence due to the presence of d orbitals, Carbon also has unique ability to form p-p multiple bonds with
itself and with other atoms of small size and high electronegativity. Few example of multiple bonding are C = C, C
C, C = O C = S and C N. Heavier elements do not form p-p bonds because their atomic orbital are too large
and diffuse to have effective overlapping .
Carbon atoms have the tendency to link with one another through covalent bonds to form chains and rings. This
property is called catenation. This is becauses C–C bonds are very strong. Down the group the size increases
tendency to show catenation decreases. This can be clearly seen from bond enthalpies values. The order of
catenation is C > > Si > Ge  Sn. Lead does not show catenation. Due to the property of catenation and p-p bonds
formation, carbon is able to show allotropic forms.

ALLOTROPES OF CARBON
Carbon exhibits many allotropic forms; both crystallic as well as amorphous. Diamond and graphite are two well-
known crystalline forms of carbon. In 1985 third form of carbon known as fullerenes was discovered by H.W. Kroto,
E Smalley and R.F.Curl.
Diamond :
It has a crystalline lattice. In diamond each carbon atom undergoes sp3 hybridisation and linked to four other carbon
atoms by using hybridised orbitals in tetrahedral fashion. The C–C bond length is 154 pm. The structure extends in
space and produces a rigid three dimensional network of carbon atoms .In this structure directional covalent bonds
are presents throughout the lattice.
It is very difficult to break extended covalent bonding and therefore, diamond is a hardest substance on the earth. It
is used as an abrasive for sharpening hard tools in making dyes and in the manufacture of tungsten filament for
electric light bulbs.
Graphite :
Graphite has layered structure. Layers are held by van der Waal’s forces and distance between two layers is 340
pm. Each layer is composed of planar hexagonal rings of carbon atoms. C – C bond length within the layer is 141.5
pm Each carbon atom in hexagonal ring undergoes sp2 hybridisation and make three sigma bonds with three
neighbouring carbon atoms. Fourth electron forms a  bond. The electrons are delocalised over the whole sheet.
Electrons are mobile and , therefore graphite conducts electricity along the sheet. Graphite cleaves easily between
the layers and therefore, it is very soft and slippery. For this reason graphite is used as a dry lubricant in machines
running at high temperature, where oil cannot be used as a lubricant.

Fullerenes :
Fullerenes are made by the heating of graphite in an electrical arc in the presence of inert gases such as helium or
argon. Fullerences are the only pure form of carbon becauses they have smooth structure without having ‘dangling’
bonds. Fulleren are cage like molecules. C60 molecule has a shape like soccer ball and called Buckminsterfullerene.
It contains twenty six -membered rings and twelve five membered rings. A six membered ring is fused with six or five
membered rings but a five membered ring can only fuse with six membered rings. All the carbon atoms are equal
and they undergo sp2 hybridisation. Each carbon atom forms three sigma bonds with other three carbon atoms. The
remaining electron at each carbon atom is delocalised in molecular orbitals, which in turn give aromatic character
to molecule. This ball shaped molecule has 60 vertices and each one is occupied by one carbon atom and it also
contains both single and double bonds with C – C distance of 143.5 pm and 138.3 pm respectively. Spherical
fullerenes are also called bucky balls in short. It is very important to know that graphite is thermodynamically most
stable allotrope of carbon and, therefore, fH(–) values of diamond and fullerene, C60 are 1.90 and 38.1 kJ mol–1,
respectively. Carbon black is obtained by burning hydrocarbons in a limited supply of air.
 15 p-block (GROUP 13 & 14) MEDICAL

Uses of carbon :
Graphite fibres embedded in plastic material form high strength, lightweight composites. The composites are used
in products such as tennis rackets, fishing rods, aircrafts and canoes. Being good conductor, graphite is used for
electrodes in batteries and industrial electrolysis. Crucibles made from graphite are inert to dilute acids and alka-
lies. Being highly porous, activated charcoal is used in adsorbing poisonous gases; also used in water filters to
remove organic contaminators and in air conditioning system to control odour. Carbon black is used as black
pigment in black ink and as filler in automobile tyres. Coke is used as a fuel and largely as a reducing agent in
metallurgy. Diamond is a precious stone and used in jewellery. It is measured in carats (1 carat = 200 mg.).
 PROPERTIES OF CARBON :
(i) Carbon in any form will react with oxygen at a sufficiently high temperature to give carbon dioxide;
in a deficiency of oxygen, carbon monoxide is formed as well.
(ii) C(s) + 2S(s)  CS2(l) (iii) Ca(s) + 2C(s)  CaC2(s)
(iii) C(s) + 2F2(g)  CF4(g)
(iv) It will reduce steam, forming water gas, and many oxides of metals; these reductions are of industrial
importance.

C + H2O(g)  CO + H2 ; Fe2O3 + 3C  2Fe + 3CO
(v) It is not attacked by dilute acids, but concentrated nitric acid and sulphuric acid are reduced if
warmed with carbon according to the equations:
C(s) + 4HNO3(aq)  2H2O(l) + 4NO2(g) + CO2(g) ; C(s) + 2H2SO4(l)  2H2O(l) + 2SO2(g)
+ CO2(g)
 OXIDES OF CARBON :
 CARBON DIOXIDE (CO 2 ) :
 PREPARATION :
(i) In the laboratory it can be conveniently made by the action of dilute hydrochloric acid on marble
chips:
CO32-(aq) + 2H+(aq)  CO2(g) + H2O(l)
(ii) Industrially it is produced as a by-product during the manufacture of quicklime and in fermentation
processes:
CaCO3(s)  CaO(s) + CO2(g) ; C6H12O6(aq){glucose} 2C2H5OH(aq) + 2CO2(g)
 PROPER TIES:
(i) It is a colourless, odourless and heavy gas which dissolves in its own volume of water at ordinary
temperature and pressure. Like all gases, it dissolves much more readily in water when the pressure
is increased and this principle is used in the manufacture of soda water and fizzy drinks.
(ii) CO2 is easily liquefied (critical temperature = 31.1oC) and a cylinder of the gas under pressure is a
convenient fire extinguisher. When the highly compressed gas is allowed to expand rapidly solid
carbon dioxide (‘dry ice’) is formed. Solid carbon dioxide sublimes at –78oC and, since no massy
liquid is produced, it is a convenient means of producing low temperatures.
 16 p-block (GROUP 13 & 14) MEDICAL

(iii) Carbon dioxide is the acid anhydride of carbonic acid, which is a weak dibasic acid and ionises in to
steps as follows :
H2CO3(aq) + H2O (l) (reversible) HCO3– (aq) + H3O+ (aq)
HCO3– (aq) + H2O (l) (reversible) CO32– (aq) + H3O+ (aq)
H2CO3 / HCO3– buffer system helps to maintain pH of blood between 7.26 to 7.42.
A solution of carbonic acid in water will slowly turn blue litmus red and when the solution is boiled,
all the CO2 is evolved.
(iv) Carbon dioxide readily reacts with alkalies forming the carbonate and, if CO2 is in excess, the
hydrogen carbonate. This is the basis of the lime-water test for CO2 gas.
Ca(OH)2(aq) + CO2(g)  CaCO3(s) + H2O(liq) ;
CaCO3(s) + H2O(liq) + CO2(g)  Ca(HCO3)2(aq)
The above reaction accounts for the formation of temporarily hard water.
(v) Carbon dioxide, which is normally present to the extent of ~ 0.03% by volume in the atmosphere, is
removed from it by the process known as photosynthesis. It is the process by which green plants
convert atmospheric CO2 into carbohydrates such as glucose. The overall chemical change can be
expressed as :
hv
6 CO2 + 12 H2O  C6H12O6 + 6 O2 + 6 H2O
Chlorphyll

By this process plants make food for themselves as well as for animals and human beings.But the
increase in combustion of fossil fuels and decomposition of limestone for cement manufacture in
recent years seem to increase the CO2 content of the atmosphere. This may lead to increase in
green house effect and thus, raise the temperature of the atmosphere which might have serious
consequences.
(vi) Gaseous CO2 is extensively used to carbonate soft drinks. Being heavy and non–supporter of
combustion it is used as fire extinguisher. A substantial amount of CO2 is used to manufacture urea.
 CARBON MONOXIDE (CO) :
 PREPARATION:
(i) It forms together with CO2, when carbon or carbonaceous matter is oxidized by air or oxygen. It is
also produced when CO2 is reduced by red- hot carbon; this reaction is of importance in metal
extractions.
C(s) + CO2(g)  2CO(g)
(ii) In the laboratory it can be prepared by dehydrating methanoic acid with concentrated sulphuric acid:
373 K
HCOOH (liq)   CO(g) + H2O
conc .H SO
2 4

(iii) If oxalic acid is dehydrated in the same way, CO2 is formed as well.
conc . H2SO 4 , 
H2C2O4  –H O
 CO + CO
2
2

(iv) On commercial scale it is prepared by the passage of steam over hot coke. The mixture of CO and
H2 thus produced is known as water gas or synthesis gas.sss
17 p-block (GROUP 13 & 14) MEDICAL

C (s) + H2O (g)    CO (g) + H2(g) (water gas).

473 1273 K

When air is used instead of steam, a mixture of CO and N2 is produced, which is called producer
gas.
1273 K
2 C (s) + O2 (g) + 4 N2 (g)   2 CO (g) + 4 N2 (g) (Producer gas).
Water gas and producer gas are very important industrial fuels. Carbon monoxide in water gas or
producer gas can undergo further combustion forming carbon dioxide with the liberation of heat.
(v) Zn + CO2  ZnO + CO
(vi) K4Fe(CN)6 + 6H2SO4 (conc.) + 6H2O 
 2K2SO4 + FeSO4 + 3(NH4)2SO4 + 6CO
(vii) HCN + 2H2O  HCOOH + 2NH3 (absorbed by H2SO4)

HCOOH  H2O + CO
 PROPERTIES :
(i) Carbon monoxide is a colourless, odourless gas which burns in air with a blue flame, forming CO2.
It is exceedingly poisonous, combining with the haemoglobin in the blood more readily than oxygen,
so that normal respiration is impeded very quickly. Ordinary gas masks are no protection against the
gas, since it is not readily adsorbed on active charcoal. In the presence of air, a mixture of
manganese (IV) oxide and copper(II) oxide catalytically oxidizes it to CO2, and this mixed catalyst
is used in the breathing apparatus worn by rescue teams in mine disasters.
(ii) Carbon monoxide is a powerful reducing agent, being employed industrially in the extraction of iron
and nickel:
Fe2O3(s) + 3CO(g)  2Fe(s) + 2CO2(g) ; NiO(s) + CO(g)  Ni(s) + CO2(g)
(iii) It reacts with many transition metals, forming volatile carbonyls; the formation of nickel carbonyl
followed by its decomposition is the basis of the Mond’s process for obtaining very pure nickel:
90 º C 180 º C
Ni(s) + 4CO(g)   Ni(CO)4(liq)   Ni(s) + 4CO(g)
(iv) In addition to reacting with oxygen, carbon monoxide combines with sulphur to give carbonyl
sulphide and with chlorine in the presence of light to give carbonyl chloride (phosgene), used in the
production of polyurethane foam plastics. Phosgene is an exceedingly poisonous gas.
CO(g) + S(s)  COS(s) (carbonyl sulphide) ; CO(g) + Cl2(g)  COCl2(g) (carbonyl
chloride)
(v) Although carbon monoxide is not a true acid anhydride since it does not react with water to produce
an acid, it reacts under pressure with fused sodium hydroxide to give sodium methanoate:
dil. HCl
NaOH(liq) + CO(g)  HCOONa(s)   HCOOH(aq)
(vi) With hydrogen under pressure and in the presence of zinc oxide or chromium (III) oxide catalyst it
reacts to give methanol; this reaction is of industrial importance.
CO(g) + 2H2(g)  CH3OH(liq)
(vii) CO is readily absorbed by an ammoniacal solution of copper (I) chloride to give CuCl.CO.2H2O. It
reduces an ammonical solution of silver nitrate to silver (black) and, in the absence of other gaseous
reducing agents, this serves as a test for the gas. It can be estimated by reaction with iodine
pentoxide, the iodine which is produced quantitatively being titrated with standard sodium
thiosulphate solution.
5CO(g) + 2O5(s)  2(s) + 5CO2(g)
 18 p-block (GROUP 13 & 14) MEDICAL

 CARBON SUBOXIDE (C 3 O 2 ) :
This is an evil-smelling gas and can be made by dehydrating propanedioic acid (malonic acid), of which it is
the anhydride, with phosphorus pentoxide:
3 CH2(COOH)2 + P4O10  3C3O2 + 4H3PO4
When heated to about 200oC, it decomposes into CO2 and C:
C3O2(g)  CO2(g) + 2C(s)
The molecule is thought to have a linear structure: O=C=C=C=O.
 CARBONATES (CO 3 2– ) AND BICARBONATES (HCO 3 – )
Carbonic acid is a dibasic acids giving rise to two series of salts, carbonates (normal salts) and bicarbonates
(acid salts) due to successive removal of the replaceable hydrogens from H2CO3.
H2CO3 + NaOH  NaHCO3 + H2O ; NaHCO3 + NaOH  Na2CO3 + H2O
 PREPARATION :
(i) With NaOH :
2NaOH + CO2  Na2CO3 ; Na2CO3 + H2O + CO2  2NaHCO3
(ii) By precipitation :
BaCl2 + Na2CO3  BaCO3  + 2NaCl
 CARBIDES :
The binary compounds of carbon with other elements (less electronegative or of similar electronegativity)
are called carbides. They are classified into following 3 categories :
(i) Ionic (ii) Covalent (iii) Interstitial (or metallic)
(i) Ionic carbides (or salt like carbides) : Generally formed by elements of I, II & III group (Boron is
exception). Based on the product obtained on hydrolysis, they are further sub-classified into three types.
(a) Methanides
These give CH4 on reaction with H2O.
Al4C3 + 12H2O  4Al (OH)3 + 3CH4 ; Be2C + 4H2O  2Be (OH)2 + CH4
These carbides contain C4– ions in their constitution.
(b) Acetylides
These give C2H2 on reaction with H2O.
CaC2 + 2H2O  Ca (OH)2 + C2H2 ; Al2 (C2)3 + 6H2O  2Al (OH)3 + 3C2H2
SrC2 + 2H2O  Sr (OH)2 + C2H2
Such compounds contain C22– [: C  C :]2 ions.
(c) Allylides
These give 1-propyne on reaction with H2O.
Mg2C3 + 4H2O  2Mg (OH)2 + CH3– C  CH
..
Such compounds contain C34– [: C – C  C :] 4 ions.
..

(ii) Covalent carbides

Compounds like CH4, CO2, CS2 can be considered to be covalent carbides. Besides these, some giant molecules
like SiC are also examples of covalent carbides.
 19 p-block (GROUP 13 & 14) MEDICAL

(iii) Interstitial or metallic carbides

Such carbides are formed by transition metals in which carbon atoms occupy interstitials in the crystal
structure of metals.

 CARBORUNDUM (SiC) :
 PREPARATION:
elect.
SiO2 + 3C furnace
 SiC + 2CO
2000 º C

 PROPERTIES :
(i) It is a very hard substance (Hardness = 9.5 Moh)
(ii) On heating it does not melt rather decomposes into elements.
(iii) Not attacked by acids. However, it gives the following two reactions at high temperature.
 
SiC + 2NaOH + 2O2  Na2SiO3 + CO2 + H2O ; SiC + 4Cl2  SiCl4 + CCl4
 It has a diamond like structure in which each atom is sp3 hybridized. Therefore each atom is tetrahedrally
surrounded by 4 atoms of other type.
 SILICON :
Silicon is the second most abundant element occurring in the earth’s crust (about 28 per cent by weight) as
the oxide, silica, in a variety of forms, e.g., sand, quartz and flint, and as silicates in rocks and clays.
 PREPARATION :
(i) The element is obtained from silica by reduction with carbon in an electric furnace:
SiO2(s) + 2C(s)  Si(s) + 2CO(g)
Extremely pure silicon is obtained from ‘chemically’ pure silicon by the method of zone refining.

(ii) SiO2 + 2Mg  2MgO + Si
 PROPERTIES :
Silicon is a very high melting-point solid with the same structure as diamond. The non-existence of an
allotrope with the graphite structure clearly shows the inability of silicon atoms to multiple bond with themselves. In
the massive form, silicon is chemically rather unreactive but powdered silicon is attacked by the halogens and
alkalies:
(i) Si(powdered) + 2Cl2(g)  SiCl4(liq)
(ii) Si(powdered) + 2OH-(aq) + H2O(liq)  SiO32-(aq) + 2H2(g)
(iii) It is not attacked by acids except hydrofluoric acid, with which it forms hexafluorosilicic acid:
Si(s) + 6HF(g)  H2SiF6(aq) + 2H2(g)

(iv) Si + 2KOH + H2O  K2SiO3 + 2H2 (v) Na2CO3 + Si  Na2SiO3 + C

(vi) 2Mg + Si  Mg2Si (Magnesium silicide)
 COMPOUNDS OF SILICON:
Silicon Dioxide SiO2
Silicon dioxide, commonly known as silica, occurs in several crystallographic forms. Quartz, cristobalite and
tridymite are some of the crystalline forms of silica, and they are interconvertable at suitable temperature. Silicon
dioxide is a covalent, three-dimensional network solid in which each silicon atom is covalently bonded in a tetrahe-
dral manner to four oxygen atoms. Each oxygen atom in turn covalently bonded to another silicon atoms. Each
corner is shared with another tetrahedron. The entire crystal may be considered as giant molecule in which eight
 20 p-block (GROUP 13 & 14) MEDICAL

membered rings are formed with alternates silicon and oxygen atoms. Silica in its normal form is almost non-
reactive because of very high Si – O bond enthalpy. It resists the attack by halogens, dihydrogen and most of the
acids and metals even at elevated temperatures. Howevers it is attacked by HF and NaOH.
SiO2 + 2 NaOH  Na2SiO3 + H2O
SiO2 + 4 HF  SiF4 + 2 H2O
Quartz is extensively used as a piezoelectric material ; it has made possible to develop extremely accurate clocks,
modern radio and television broadcasting and mobile radio communications. Silica gel used as a drying agent and
as a support for chromatographic materials and catalysts. Kieselghur, an amorphous form of silica is used in
filtration plants.
 SILICATES :
Binary compouds of silicon with oxygen are called silicates but they contain other metals also in their structures.
(i) Since the negativity difference b/w O & Si is about 1.7, so Si–O bond can be considered 50% ionic &
50% covalent.
rSi  4
(ii) If we caluclate the radius ratio = 0.29
rO 2 
It suggests that the co-ordination no. of silicon must be 4 and from VBT point of view we can say
that Si is sp3 hybridized. Therefore silicate structures must be based upon SiO4–4 tetrahedral units
(iii) SiO4–4 tetrahedral units may exist as discrete units or may polymerise into larger units by sharing
corners.
 CLASSIFICATION OF SILICATES :
(A) Orthosilicates :
These contain discrete [SiO4]4– units i.e., there is no sharing of corners with one another as shown is figure.

e.g. Zircon (ZrSiO4), Forsterite of Olivine (Mg2SiO4), Willemite (Zn2SiO4)

(B) Pyrosilicate :
In these silicates two tetrahedral units are joined by sharing oxygen at one corner thereby giving [Si2O7]6– units.

 (–) charge will be present on the oxygen atoms which is bonded with one Si atom.
e.g. Thorteveitite (Sc2Si2O7), Hemimorphite (Zn3(Si2O7) Zn(OH)2H2O)
(C) Cyclic silicates :
If two oxygen atoms per tetrahedron are shared to form closed rings such that the structure with general
formula (SiO32–)n or (SiO3)n2n– is obtained, the silicates containing these anions are called cyclic silicates. Si3O96– and
Si6O1812– anions are the typical examples of cyclic silicates.
 21 p-block (GROUP 13 & 14) MEDICAL

–
O

–
O
– – –
O O O
O O O
O– O –

–
O
O O O O

– –
O O O– –
O
– –
O O O –
O O O O
–

O –

6–
–
O 12–
Si3O9 Si6O18

(D) Chain silicates :

Chain silicates may be further classified into simple chain & double chain compounds.
In case of simple chains two corners of each tetrahedron are shared & they form a long chain of
tetrahedron. Their general formula is also same as the cyclic silicates i.e. (SiO3)n2n–

Similarly, double chain silicates can be drawn in which two simple chains are joined together by shared
oxygen. Such compounds are also known as amphiboles. The asbestos mineral is a well known example of double
chain silicates. The anions of double chain silicates have general formula (Si4O11)n6n– .
– –
O O

– –
O O
O O
O O O O O
O– O– O–

O O O

O– O
–
O
–

O O O O O O
–
O– O–

– –
O O

e.g., Synthetic silicates (Li2SiO3, Na2SiO3), Spondumene (LiAl(SiO3)2),

Enstatite (MgSiO3), Diopside (CaMg(SiO3)2 ), Tremolite (Ca2Mg5(Si4O11)2 (OH)2 ), etc.
(E) Two dimensional sheet silicates :
In such silicates, three oxygen atoms of each tetrahedral are shared with adjacent SiO44– tetrahedrals. Such
sharing forms two dimension sheet structure with general formula (Si2O5)n2n–
e.g. Talc (Mg(Si2O5)2 Mg(OH)2 , Kaolin Al2(OH)4 (Si2O5)
(F) Three dimenstional sheet silicates :
These silicates involve all four oxygen atom in sharing with adjacent SiO44– tetrahedral units.
e.g. Quartz, Tridymite, Crystobalite, Feldspar, Zeolite and Ultramarines.
 22 p-block (GROUP 13 & 14) MEDICAL

 SILICONES :
Silicones are synthetic organosilicon compounds having repeated R2SiO units held by Si – O – Si linkages.
These compounds have the general formula (R2SiO)n where R = alkyl or aryl group.
The silicones are formed by the hydrolysis of alkyl or aryl substituted chlorosilanes and their subsequent
polymerisation. The alkyl or aryl substituted chlorosilanes are prepared by the following reactions.
Cu
(a) RCl + Si 300
 R SiCl + R SiCl + RSiCl
ºC 3 2 2 3

(b) RMgCl + SiCl4  RSiCl3 + MgCl2

2RMgCl + SiCl4  R2SiCl2 + 2MgCl2 ; 3RMgCl + SiCl4  R3SiCl + 3MgCl2
After fractional distillation, the silane derivatives are hydrolysed and the ‘hydroxides’ immediately condense
by intermolecular elimination of water. The final product depends upon the number of hydroxyl groups originally
bonded to the silicon atom:

In this manner several molcules may combine to form a long chain polymer whose both the ends will be
occupied by –OH groups. Such compounds are generally represented from the following formula.

The polymer chain depicted above is terminated by incorporating a small quantity of the monochlorosilane
derivative into the hydrolysis mixture.
 Silicones can be prepared from the following types of compounds only.
(i) R3SiCl (ii) R2SiCl2 (iii) RSiCl3
 Silicones from the hydrolysis of (CH3)3 SiCl
(CH3)3 SiCl  (CH3)3 Si (OH)
H2 O

 Silicones from the hydrolysis of a mixture of (CH3)3 SiCl & (CH3)2 SiCl2
 23 p-block (GROUP 13 & 14) MEDICAL

The dichloro derivative will form a long chain polymer as usual. But the growth of this polymer can be
blocked at any stage by the hydrolysis product of mono-chloro derivative.

 Silicones from the hydrolysis of trichloro derivative

When a compound like CH3SiCl3 undergoes hydrolysis, a complex cross-linked polymer is
obtained.
 The hydrocarbon layer along the silicon-oxygen chain makes silicones water-repellent.
Products having the physical properties of oils, rubbers, and resins can be produced using silicones. Silicone
fluids (say as hydraulic systems of planes) are thermally stable and their viscosity alters very little with temperature,
and silicone rubbers retain their elasticity at much lower temperatures than ordinary rubber. Silicone varnishes are
such excellent insulators and so heat-resistant that insulating wiring with them enabled motors to work over-loads
that would have set fire to the insulation formerly used. A whole new field of chemistry and technology, civilian as
well as military, has been opened up by the development of silicones.

TIN AND LEAD :

 COMPOUNDS OF TIN :
 STANNOUS OXIDE (SnO) :
 PREPARATION:
(i) By heating stannous hydroxide, Sn(OH)2, in absence of air.
Sn(OH)2  SnO + H2O
(ii) By heating stannous oxalate, SnC2O4 in absence of air.
SnC2O4  SnO + CO + CO2 
 PROPER TIES:
(i) SnO is an amphoteric dark grey or black solid oxide, insolulbe in water. It dissolves in
acids to form stannous salts.
SnO + 2HCl  SnCl2 + H2O ; SnO + H2SO4  SnSO4 + H2O
(ii) SnO dissolves in hot NaOH solution to form (soluble) sodium stannite and water.
SnO + 2NaOH  Na2SnO2 + H2O
stannites are only known in aqueous solutions. Stannites absorb oxygen from air and are oxidised to
stannate which are stable in nature.
2 Na2SnO2 + O2  2 Na2SnO3
 USES :
For the preparation of stannous chloride and stannous sulphate.
 STANNOUS CHLORIDE (SnCl 2 ·2H 2 O) :
It is a colourless solid soluble in water. Its solution becomes milky on standing due to its hydrolysis
to Sn(OH)2 and HCl. It aqueous solution is acidic to litmus. It is a strong reducing agent. It is soluble
in alcohol and ether also.
 24 p-block (GROUP 13 & 14) MEDICAL

 PREPARATION :
(i) Sn + 2HCl(concentrated)  SnCl2(aq) + H2
(ii) SnO + 2HCl  SnCl2(aq) + H2O
The solution on crystallization gives colourless crystals of SnCl2·2H2O.
 PROPERTIES :
(i) Reaction with Hg2Cl2 solution: When SnCl2 solution is added to an aqueous solution of mercuric
chloride, a
silky white precipitate of mercurous chloride, Hg2Cl2 is formed which turns black due to further
reduction of Hg2Cl2 to black mercury.
2HgCl2 + SnCl2  Hg2Cl2 + SnCl4 ; Hg2Cl2 + SnCl2  2Hg + SnCl4
(ii) It reduces ferric chloride, FeCl3 to ferrous chloride, FeCl2.
2FeCl3(brown solution) + SnCl2  2FeCl2 (colourless solution) + SnCl4
(iii) It is hydrolysed with water to produce white precipitate of Sn(OH)2
SnCl2 + 2H2O  Sn(OH)2 (white) + 2HCl
As it produces a weak base and strong acid its aqueous solution is acidic. Its hydrolysis can be
prevented by adding concentrated HCl to it during the process of its preparation.
 USES :
(i) In dye industry as a reducing agent.
(ii) For the test of mercuric salt.
(iii) For the preparation of other stannous compounds.
 STANNIC OXIDE (SnO 2 ) :
 PREPARATION:
(i) By burning Sn in air
Sn + O2  SnO2
(ii) By heating Sn with concentrated HNO3

Sn + 4HNO3  H2SnO3 + 4NO2 + H2O ;H2SnO3  H2O + SnO2
 PROPERTIES :
(i) It is a white solid insoluble in water. It is weakly acidic
(ii) It dissolves in conc. H2SO4 to form stannic sulphate.
SnO2 + 2H2SO4  Sn(SO4)2 + 2H2O
(iii) It also dissolves in conc. Alkalies to form alkali metal stannate solution.
SnO2 + 2NaOH  Na2SnO3 (sodium stannate) + H2O
 STANNIC CHLORIDE (SnCl 4 ) :
 PREPARATION:
(i) By the action of Cl2 gas on heated Sn
Sn + 2Cl2  SnCl4
(ii) By the action of Cl2 on stannous chloride
SnCl2 + Cl2  SnCl4
 25 p-block (GROUP 13 & 14) MEDICAL

 PROPERTIES :
(i) It is a colourless fuming liquid, Bp = 114oC. It is covalent.
(ii) Action of moisture: It absorbs moisture and becomes converted into hydrated stannic chlorides,
SnCl4·3H2O, SnCl4·5H2O, SnCl4·6H2O, and SnCl4·8H2O. SnCl4 . 5 H2O is known as “butter of
tin” or “oxymercurate of tin”.
(iii) Hydrolysis with water: It easily gets hydrolysed in water and produces strong acid HCl. Hence its
aqueous solution is acidic to litmus. It hydrolyses more rapidly than SnCl2
SnCl4 + 4H2O  H4SnO4(colloidal white precipitate) + 4HCl
(iv) SnCl4 is a Lewis acid. Hence it has a tendency to accept lone pair of electrons from NH3, PH3 etc
and form adducts such as SnCl4·4NH3
(v) It dissolves in concentrated HCl forming H2SnCl6 and in presence of ammonium chloride, it forms
ammonium salts of this acid.
SnCl4 + 2 HCl  H2SnCl6
SnCl4 + 2 NH4Cl  (NH4)2SnCl6
 USES : For the preparation of stannic compounds.
 COMPOUNDS OF LEAD :
 LITHARGE (PbO) :

PbO is prepared by heating Pb at 180oC. It is a volatile yellow compound. 2Pb + O2  2PbO
It is an amphoteric oxide and dissolves in acids as well as in alkalis
 
PbO + 2HNO3  Pb(NO3)2 + H2O ; PbO + 2NaOH  Na2PbO2 (sodium plumbate) + H2O
It is used in rubber industry and in the manufacture of flint glasses, enamels, and storage batteries.
 LEAD DIOXIDE (PbO 2 ) :
The fact that it does not liberate H2O2 by the action of dilute HCl suggest the above formula (It is a dioxide
not a peroxide)
 PREPARATION :

(i) PbO + NaOCl  PbO2 (insoluble) + NaCl
(ii) Pb3O4 + 4HNO3 (dilute  2Pb(NO3)2 + PbO2 + 2H2O

 PROPERTIES :
(i) It is a chocolate coloured insoluble powder. On heating at 440oC it gives the monoxide:
440 º C
2PbO2   2PbO + O2

(ii) It oxidizes HCl to Cl2:

PbO2 + 4HCl  PbCl2 + 2H2O + Cl2
(iii) It dissolves in conc. NaOH solution:
PbO2 + 2NaOH  Na2PbO3(sodium plumbate) + H2O
(iv) It oxidises Mn salt to permanganic acid:
2MnSO4 + 5PbO2 + 6HNO3  2HMnO4 + 2PbSO4 + 3Pb(NO3)2 + 2H2O
 26 p-block (GROUP 13 & 14) MEDICAL

(v) It reacts with SO2 at red heat to form lead sulphate:


PbO2 + SO2  PbSO4
(vi) It reacts with conc. HNO3 to evolve oxygen gas.
PbO2 + 2HNO3  Pb(NO3)2 + 1/2O2 + H2O
 USES :
It is used in match industry for making ignition surface of match boxes and in the preparation of KMnO4.
 RED LEAD (Pb 3 O 4 ) :
 PREPARATION:
It is prepared by heating PbO at 450oC for a long time.
450 º C
6PbO + O2   2Pb3O4

 PROPERTIES :
(i) It is a red powder insoluble in water but when heated with conc. HNO3 it gives a red precipitate of
PbO2
Pb3O4 + 4HNO3  2Pb(NO3)2 + PbO2 + 6H2O
(ii) When heated above 550oC, it decomposes into PbO

Pb3O4  6PbO + O2
(iii) It oxidizes conc. HCl to chlorine
Pb3O4 + 8HCl  3PbCl2 + 4H2O + Cl2
(iv) When heated with conc. H2SO4 it evolves oxygen
2Pb3O4 + 6H2SO4  6PbSO4 + 6H2O + O2
 USES :
It is used as an oxidizing agent, for making red paint, for making special lead cement and for making
flint glass.
 LEAD CHLORIDE (PbCl 2 ) :
 PREPARATION:
(i) Pb(NO3)2 + 2HCl  PbCl2 + 2HNO3
(ii) Pb(NO3)2 + 2NaCl  PbCl2 + 2NaNO3
(iii) Pb(CH3COO)2 + 2HCl  PbCl2 + 2CH3COOH
(iv) PbO + 2HCl  PbCl2 + H2O
(v) Pb(OH)2 + 2HCl  PbCl2 + 2H2O
(vi) Pb(OH)2·PbCO3 (basic lead carbonate) + 4HCl  2PbCl2 + CO2 + 3H2O
 PROPERTIES :
It is a white crystalline solid, insoluble in cold water but soluble in boiling water.It dissolves in
concentrated HCl forming a complex ion.
2 HCl + PbCl2 (reversible) H2PbCl4 (chloroplumbous acid)
 USES :
It is used for making pigments for paints.
 27 p-block (GROUP 13 & 14) MEDICAL

 LEAD TETRACHLORIDE (PbCl 4 ) :

 PREPARATION:
It is prepared by the following methods:
(i) By dissolving PbO2 in cold conc. HCl
PbO2 + 4HCl  PbCl4 + 2H2O
PbCl4 dissolves in excess of HCl to form a stable solution of H2PbCl6.
PbCl4 + 2HCl  H2PbCl6
 When NH4Cl is added to a solution of chloroplumbic acid, a yellow precipitate of ammonium
chloroplumbate is formed.
H2PbCl6 + 2NH4Cl  (NH4)2PbCl6 + 2HCl
 When crystals of ammonium chloroplumbate is added to ice cold conc. H2SO4, lead tetrachloride is
formed and separates as a yellow oily liquid.
(NH4)2PbCl6 + H2SO4  PbCl4 + (NH4)2SO4 + 2HCl
(ii) By the action of Cl2 on a solution of PbCl2 in conc. HCl
PbCl2 + Cl2  PbCl4
 PROPERTIES :
(i) It is a yellow oily liquid which solidifies at –10oC and is soluble in organic solvents like ethanol and
benzene.
(ii) Rapid hydrolysis with water forms PbO2 precipitate
PbCl4 + 2H2O  PbO2 + 4HCl
 USES :
It is used for making stannic compounds.
 28 p-block (GROUP 13 & 14) MEDICAL

EXERCISE - 1 : BASIC OBJECTIVE QUESTIONS

Group - 13 8. Borax bead test is responded by:

1. Boric acid H3BO3 and BF3 have the same number of elec- (a) divalent metals (b) heavy metals
trons. The former is a solid and the latter is a gas. The (c) light metals
reason is:
(d) metals which form coloured metaborates
(a) BF3 is a Lewis acid while B(OH)3 is not
9. Borazole, B3N3H6 is related to benzene as:
(b) they have different geometrics
(a) isoelectronic (b) isostructural
(c) In BF3, F– is smaller in size than OH– in B(OH)3
(c) both (a) and (b) (d) none of these
(d) No molecular association is possible in BF3 while it
is possible in B(OH)3 due to hydrogen bonding 10. Reactivity of borazole is higher than that of benzene
because:
2. Ionisation enthalphy ( 1H1 in kJ mol-1 ) for the elements (a) borazole is a polar compound
of group 13 follows the order:
(b) borazole is a non-polar compound
(a) B > Al > Ga > In > Tl (b) B < Al < Ga < In < Tl
(c) borazole is electron deficient compound
(c) B < Al > Ga < In > Tl (d) B > Al  Ga > In < Tl
(d) of localised electrons in it
3. On the addition of mineral acid to an aqueous solution
[Note: In benzene, -electrons are delocalised over the
of borax, the following compound is formed:
ring which give stability to benzene ring.]
(a) boron hydride (b) orthoboric acid 11. In diborane:
(c) metaboric acid (d) pyroboric acid (a) 4 bridged hydrogens and two terminal hydrogens
are present
4. Which of the following compounds is formed when bo-
ron trichloride is treated with water? (b) 2 bridged hydrogens and four terminal hydrogens
are present
(a) H3BO3 + HCl (b) B2H6 + HCl
(c) 3 bridged hydrogens and three terminal hydrogens
(c) B2O3 + HCl (d) None of these are present
5. The hardest compound of boron is: (d) none of the above
(a) boric acid (b) boron nitride 12. The bonds present in borazole are:
(c) boron carbide (d) boron hydride (a) 9, 9 (b) 6, 6

6. When strongly heated, orthoboric acid leaves a residue (c) 9, 6 (d) 12, 3
of: 13. BCl3 does not exist as a dimer but BH3 exists as B2H6
(a) metaboric acid (b) tetraboric acid because
(a) Cl2 is more electronegative than hydrogen
(c) boric anhydride (d) boron
(b) large size of chlorine atom does not fit between small
7. Boric acid is prepared from borax by the action of : sized boron atoms, while small-sized hydrogen
(a) Hydrochloric acid (b) sodium hydroxide atoms occupy the space between boron atoms

(c) carbon dioxide (d) sodium carbonate (c) there is p-dback bonding in BCl3
(d) both (b) and (c)
 29 p-block (GROUP 13 & 14) MEDICAL

14. The main factor responsible for weak acidic nature of 22. Specify the coordination geometry around and hybrid-
B—F bonds in BF3 is: ization of N and B atoms in 1:1 complex of BF3 and NH3:
(a) high electronegativity of F (a) N: tetrahedral, sp3 ; B:tetrahedral, sp3
(b) three centred two electron bonds in BF3 (b) N: pyramidal, sp3 ; B:pyramidal , sp3
(c) p—d back bonding (c) N:pyramidal, sp3 ; B:planar, sp3
(d) p—p back bonding (d) N:pyramidal, sp3 ; B:tetrahedral, sp3
15. Which one of the following statements regarding BF3 is 23. H3BO3 is:
not correct? (a) monobasic and weak Lewis acid
(a) It is a Lewis acid (b) monobasic and weak Bronsted acid
(b) It is an ionic compound (c) monobasic acid and strong Lewis acid
(c) It is an electron deficient compound (d) tribasic acid and weak Bronsted acid
(d) It forms adducts 24. Which of the following compounds is known as inor-
16. The number of OH units directly linked to boron atoms ganic benzene?
in Na2B4O7.10H2O is: (a) B6H6 (b) C5H5B
(a) 2 (b) 3 (c) C3N3H3 (d) B3N3H6
(c) 4 (d) 10 25. A mixture of boron trichloride and hydrogen is subjected
to silent electric discharge to form ‘A’ and HCl. ‘A’ is
17. From B2H6, all the following can be prepared except:
mixed with NH3 and heated to 200oC to form ‘B’. The
(a) B2O3 (b) H3BO3 formula of ‘B’ is:
(c) B2(CH3)6 (d) NaBH4 (a) H3BO3 (b) B2O3

18. The power of halides of boron to act as Lewis acids (c) B2H6 (d) B3N3H6
decreases in the order: 26. Diborane reacts with ammonia under different conditions
(a) BF3 > BCl3 > BBr3 (b) BBr3 > BCl3 > BF3 to give a variety of products. Which one among the
following is not fomed in these conditions?
(c) BCl3 > BF3 > BBr3 (d) BCl3 > BBr3 > BF3
(a) B2H6.2NH3 (b) B12H12
19. Which of the following is a false statement ?
(c) B3N3H6 (d) (BN)n
(a) BH3 is not a stable compound
27. Boron cannot form which one of the following anions?
(b) Boron hydrides are formed when dil. HCl reacts with
Mg3B2 (a) BF6 (b) BH 4
(c) All the B—H bond distances in B2H6 are equal
(c) [B(OH)4]– (d) BO 2
(d) The boron hydrides are readily hydrolysed
20. In B2H6: [Hint: Due to non-availability of d-orbitals, boron is un-
able to expand its octet. The maximum covalency of bo-
(a) there is direct boron-boron bond
ron is 4.]
(b) the boron atoms are linked through hydrogen bridges 28. Which one is the wrong statement ?
(c) the structure is similar to C2H6 (a) Aluminium is the most abundant metal in the earth’s
(d) all the atoms are in one plane crust
21. BF3 is used as catalyst in several industrial processes (b) Aluminium is not affected by strong alkalies
due to its: (c) Aluminium becomes passive with conc. HNO3
(a) strong reducing nature
(d) Aluminium when heated in the atmosphere of
(b) weak reducing action nitrogen forms a nitride of the formula AIN.
(c) strong Lewis acid nature
[Hint: Al is affected by strong alkalies with evolution of
(d) weak Lewis acid character hydrogen. ]
 30 p-block (GROUP 13 & 14) MEDICAL

29. Alumina is: Al 2 (SO 4 )3 + 6NaOH  2Al(OH)3 + 3Na 2SO 4

(a) acidic (b) basic
Al(OH)3 + NaOH  NaAlO 2 + 2H 2 O
(c) neutral (d) amphoteric Soluble

30. The element which exists in liquid state for a wide range The other sulphates K2SO4 and Na2SO4 are also soluble
of temperature and can be used for measuring high tem- in water. Thus, a clear solution is obtained.]
perature is: 38. Aluminium vessels should not be washed with materials
containing washing soda since :
(a) B (b) Al
(a) washing soda reacts with aluminium to form soluble
(c) In (d) Ga aluminate
31. When alumina is heated with carbon in nitrogen atmo- (b) washing soda is expensive
sphere, the products are: (c) washing soda is easily decomposed
(a) Al + CO (b) Al + CO2 (d) washing soda reacts with aluminium to form insoluble
aluminium oxide
(c) Al + CO + CO2 (d) AlN + CO
39. Aluminium is more reactive than iron. But aluminium is
32. Aluminium does not react with: less easily corroded than iron because:
(a) NaOH (b) HCl (a) oxygen forms a protective oxide layer on aluminium
(c) N2 (d) HNO3 (b) aluminium is a noble metal
33. Alumina may be converted into anhydrous aluminium (c) iron undergoes reaction easily with water
chloride by:
(d) iron forms mono and divalent ions
(a) heating it with conc. HCl
40. Al2O3 formation from aluminium and oxygen involves
(b) heating in a current of dry chlorine
evolution of a large quantity of heat, which makes alu-
(c) heating it with rock salt minium used in:
(d) mixing it with carbon and heating the mixture in a
(a) deoxidiser (b) confectionary
current of dry chlorine
(c) indoor photography (d) thermite welding
34. Which of the following statements about anhydrous alu-
minium chloride is correct? 41. The main factor responsible for weak acidic nature of
(a) It exists as AlCl3 molecule B—F bonds in BF3 is:
(b) It is a strong Lewis base (a) high electronegativity of F
(c) It sublimes at 100oC under vaccum (b) three centred two electron bonds in BF3
(d) It is not easily hydrolysed (c) p—d back bonding
35. When a solution of sodium hydroxide is added in excess
(d) p—p back bonding
to the solution of potash alum, we obtain:
(a) a white precipitate (b) bluish white precipitate 42. When aluminium is heated with conc. H2SO4
(c) a clear solution (d) a crystalline mass (a) aluminium becomes passive
36. Which one of the statements is incorrect for aluminium (b) hydrogen is liberated
metals?
(c) oxygen is liberated
(a) It is a good conductor of heat and electricity
(d) sulphur dioxide is liberated
(b) It is malleable and ductile
(c) It can be welded and cast but difficult to solder 43. When aluminium hydroxide dissolves in NaOH solution,
(d) Its alloys are heavy the product is:
37. When excess of NaOH solution is added in potash alum, (a) [Al(H2O)3(OH)3] (b) [Al(H2O)4(OH)2]+
the product is:
(c) [Al(H2O)2(OH)4]– (d) [Al(H2O)5(OH)]2+
(a) a bluish white precipitate
(b) clear solution 44. AlCl3 on hydrolysis gives:
(c) a white precipitate (d) a greenish precipitate (a) Al2O3. H2O (b) Al(OH)3
[Hint: Alum consists Al2 (SO4)3 reacts with NaOH. (c) Al2O3 (d) AlCl3.6H2O
 31 p-block (GROUP 13 & 14) MEDICAL

45. The stability of +1 oxidation state increases in the se- (d) Due to large size of X, B – X distance decreases.
quence: 51. Aluminium exhibits +3 oxidation state. As we move down
(a) Tl < In < Ga < Al (b) In < Tl < Ga < Al the group, +1 oxidation state gets more stable. This is a
(c) Ga < In < Al < Tl (d) Al < Ga < In < Tl consequence of

46. Group 13 elements show +1 and +3 oxidation states. (a) increasing size of the atom
Relative stability of +3 oxidation state may be given as (b) inert pair effect
(a) Tl3  In 3  Ga 3  Al3  B3 (c) electron deficient nature
(d) p - p bonding.
(b) B3  Al3  Ga 3  In 3  Tl3
52. The shapes and hybridisation of BF 3 and BH 4
(c) Al3  Ga 3  Tl3  In 3  B3 respectively are
(a) BF3 - Trigonal, sp2 hybridisation ;
(d) Al3  B3  Ga 3  Tl3  In 3
47. Which of the following is not true regarding the nature BH 4 - Square planar, sp3 hybridisation
of halides of boron ? (b) BF3 - Triangular, sp3 hybridisation ;
(a) Boron trihalides are covalent.
BH 4 - Hexagonal, sp3d hybridisation
(b) Boron trihalides are planar triangular with sp2
hybridisation (c) BF3 - Trigonal, sp2 hybridisation ;
(c) Boron trihalides act as Lewis acids.
BH 4 - Tetrahedral, sp3 hybridisation
(d) Boron trihalides cannot be hydrolysed easily.
48. Which of the following does not show similarity (d) BF3 - Tetrahedral, sp3 hybridisation ;
between boron and aluminium ?
BH 4 - Tetrahedral, sp3 hybridisation.
(a) Both form oxides of type M2O3 when heated with
53. Electropositive charcter for the elements of group 13
oxygen at high temperature.
follows the order
(b) Both dissolve in alkalies and evolve hydrogen.
(a) B > Al > Ga > In > Tl (b) B < Al < Ga < In < Tl
(c) Hydroxides of both the elements are basic in nature.
(c) B < Al > Ga < In > Tl (d) B < Al > Ga > In > Tl
(d) Both form nitrides of MN type when heated with N2.
54. The reason behind the lower atomic radius of Ga as
49. The decreasing order of power of boron halides to act as compared to Al is
Lewis acids is
(a) poor screening effect of d-electrons for the outer
(a) BF3  BCl3  BBr3 (b) BBr3  BCl3  BF3 electrons from increased nuclear charge
(b) increased force of attraction of increased nuclear
(c) BCl3  BF3  BBr3 (d) BCl3  BBr3  BF3 charge on electrons
50. In BX3, B – X distance is shorter than what is expected (c) increased ionisation enthalpy of Ga as compared to
theoretically because (X = F, Cl, Br, I) Al
(d) anomalous behaviour of Ga.
55. Which of the following hydroxides is acidic ?
(a) Al(OH)3 (b) Ga (OH)3
(a) sp3 hybridisation of B is responsible for shorter (c) Tl (OH)3 (d) B(OH)3
B – X distance.
56. Thermite is a mixture of iron oxide and
(b) B – X has a double bond character due to back-
bonding. (a) aluminium powder (b) zinc powder

(c) Dimerisation takes place in BX3 which is responsible (c) iron turnings (d) copper turnings.
for shorter B – X distance.
 32 p-block (GROUP 13 & 14) MEDICAL

57. A metal M reacts with sodium hydroxide to give a white (d) A red coloured Co(BO2)2 bead is formed.
precipitate X which is soluble in excess of NaOH to give
64. In diborane,
Y. Compound X is soluble in HCl to form a compound Z.
Identify M, X, Y and Z. (a) four bridged hydrogen atoms and two terminal
hydrogen atoms are present
M X Y Z
(b) two bridged hydrogen atoms and four terminal
(a) Si SiO2 Na2SiO3 SiCl4 hydrogen atoms are present
(b) Al Al(OH)3 NaAlO2 AlCl3 (c) three bridged hydrogen atoms and three terminal
(c) Mg Mg(OH)3 NaMgO3 MgCl2 hydrogen atoms are present
(d) there are no bridged hydrogen atoms in diborane,
(d) Ca Ca(OH)2 Na2CO3 NaHCO3
only hydrogen bonds are present.
58. Boron is unable to form BF63 ions due to 65. Match the column I with column II and mark the
appropriate choice.
(a) non-availability of d-orbitals
Column I Column II
(b) small size of boron atom
(A) Borax (i) Na3AlF6
(c) non-metallic nature (B) Inorganic benzene (ii) Na2B4O7.10H2O
(d) it is a strong Lewis acid. (C) Cryolite (iii) Al2O3.2H2O
59.   (D) Bauxite (iv) B3N3H6
Na 2 B4 O7 .10H 2 O  X 
 YZ
(a) (A)  (ii), (B)  (iv), (C)  (i), ((D))  (iii)
X, Y and Z in the reaction are
(b) (A)  (i), (B)  (ii), (C)  (iii), (D)  (iv)
(a) X  Na 2 B4 O 7 , Y  NaBO2 , Z  H 3 BO3 (c) (A)  (ii), (B)  (iii), (C)  (i), (D)  (iv)
(b) X  Na 2 B4 O7 , Y  B2 O3 , Z  H 3 BO3 (d) (A)  (iii), (B)  (i), (C)  (ii), (D)  (iv)
66. The type of hybridization of boron in bidorane is
(c) X  B2 O3 , Y  NaBO 2 , Z  B(OH)3 (a) sp-hybridization (b) sp2-hybridization
(d) X  NaBO 2 , Y  B2 O3 , Z  B(OH)3 (c) sp3-hybridization (d) sp3d2-hybridication.
67. An aqueous solution of boric acid is found to be weakly
60. Na 2 B4 O7  X  H3 BO3 . What is X in the reaction ? acidic in nature. This acidic character arises due to the
(a) Aqueous solution of NaOH following reasons.
(a) It is a protic acid which donates protons in aqueous
(b) Dilute nitric acid
solution.
(c) Conc. H2SO4 or HCl
(b) It is a Lewis acid which abstracts OH– from water
(d) Water and leaves H+ to make the solution acidic.
61. Which of the following compounds is formed in borax (c) It gives metaboric acid when dissolved in water.
bead test ? (d) It is prepared by reaction of borax with sulphuric
(a) Metaborate (b) Tetraborate acid hence it behaves as an acid.
(c) Triborate (d) Orthoborate 68. Boron nitride can be represented by the given structure.
62. Boric acid has a polymeric layer structure in which
planar BO3 units are joined by
(a) covalent bonds
(b) two centre - two electron bonds
(c) coordinate bonds
(d) hydrogen bonds.
63. What happens when a mixture of cobalt oxide and borax
is heated in a flame on a loop of platinum wire ? The structure of BN is similar to
(a) A transparent white bead is formed. (a) graphite (b) diamond
(b) A bright pink coloured NaBO2 bead is formed. (c) benzene (d) pyridine.
(c) A blue coloured Co(BO2)2 bead is formed.
 33 p-block (GROUP 13 & 14) MEDICAL

69. NaBH4 + I2  X + Y + Z (b) in making enamel and pottergy glazes

450 K (c) as a flux in soldering

BF3 + NaH  X+P
(d) in making optical glasses.
BF3 + LiAlH4  X + Q + R
77. The element which exists in liquid state for a wide range
X, Y, Z, P, Q and R in the reactions are of temperature and can be used for measuring high
X Y Z P Q R temperature is

(a) Na4B4O7 NaI HI HF LiF AlF3 (a) B (b) Al

(c) Ga (d) In
(b) B2H6 NaI H2 NaF LiF AlF3
78. Which of the following is a Lewis acid ?
(c) B2H6 BH3 NaI B3N3H6 Al2F6 AlF3
(a) AlCl3 (b) MgCl2
(d) BH3 B2H6 H2 B3N3H6 LiF AlF3
(c) CaCl2 (d) BaCl2
70. Which of the following compounds are formed when
BCl3 is treated with water ? 79. The geometry of a complex species can be understood
from the knowledge of type of hybridisation of orbitals
(a) H3BO3 (b) B2H6 of central atom. The hybridisation of orbitals of central
(c) B2O3 (d) HBO2 atom in [B(OH)4]– and the geometry of the complex are
respectively
71. Which is the hardest compound of boron ?
(a) sp3, tetrahedral (b) sp3, square planar
(a) B2O3 (b) BN
(c) B4C (d) B2H6 (c) sp3d2, octahedral (d) dsp2, square planar.

72. What are X and Y in the reaction ? 80. Which of the following oxides is acidic in nature ?

heat
3B2 H 6  6X  3 [BH 2 (X) 2 ] [BH 4 ]   Y  12H 2 (a) B2O3 (b) Al2O3

(c) Ga2O3 (d) In2O3

(a) X = NH3, Y = B3N3H6 (b) X = CO, Y = BH3CO
(c) X = NaH, Y = NaF (d) X = NF3, Y = B3N3 81. The exhibition of highest co-ordination number depends
on the availability of vacant orbitals in the central atom.
73. On hydroloysis, diborane produces Which of the following elements is not likely to act as
(a) H3BO2 + H2O2 (b) H3BO3 + H2 central atom in MF63 ?
(c) B2O3 + O2 (d) H3BO3 + H2O2 (a) B (b) Al
74. What happens when diborane reacts with Lewis bases ? (c) Ga (d) In
(a) It forms boron trihydride (BH3) due to cleavage.
82. Ionisation enthalpy ( i H1 kJ mol1 ) for the elements
(b) It undergoes cleaveage to give borane adduct BH3L of Group 13 follows the order
(where, L = Lewis base). (a) B > Al > Ga > In > Tl (b) B < Al < Ga < In < Tl
(c) It oxidises to give B2O3. (c) B < Al > Ga > In > Tl (d) B > Al < Ga > In < Tl
(d) It does not react with Lewis bases. 83. In the structure of diborane
75. Which is not the use of orthoboric acid ? (a) all hydrogen atoms lie in one plane and boron atoms
(a) As an antiseptic and eye wash. lie in a plane perpendicular to this plane

(b) In glass industry. (b) 2 boron atoms and 4 terminal hydrogen atoms lie in
the same plane and 2 bridging hydrogen atoms lie in
(c) In glazes for pottery. the perpendicular plane
(d) In borax - bead test.
(c) 4 bridging hydrogen atoms and boron atoms lie in
76. Borax is not used one plane and two terminal hydrogen atoms lie in a
(a) as a styptic to stop bleeding plane perpendicular to this plane
(d) all the atoms are in the same plane.
 34 p-block (GROUP 13 & 14) MEDICAL

84. A compound X, of boron reacts with NH3 on heating to (b) In nitrogen, lone pair of electrons is present at
give another compound Y which is called inorganic nitrogen hence it has pyramidal shape.
benzene. The compound X can be prepared by treating
BF3 with lithium aluminium hydride. The compounds X (c) Both (a) and (b) are correct.
and Y are represented by the formulas (d) Both (a) and (b) are wrong.
(a) B2H6, B3N3H6 (b) B2O3, B3N3H6 91. Identify X in the reaction :
(c) BF3, B3N3H6 (d) B3N3H6, B2H6

X  2H 2 O   XO2  2H 2
85. Which of the following ions is the most stable ? (steam)

(a) Sn2+ (b) Ge2+ (a) C (b) Si

(c) Si2+ (d) Pb2+ (c) Ge (d) Sn

Group-14 92. All members of group 14 when heated in oxygen form

oxides. Which of the following is the correct trend of
86. The tendency of group 14 elements to show +2 oxidation oxides ?
state increases in the order of
(a) Dioxides CO2, SiO2 and GeO2 are acidic while SnO2
(a) C < Si < Sn < Pb < Ge (b) C < Si < Ge < Sn < Pb and PbO2 are amphoteric.
(c) Ge < Sn < Pb < C < Si (d) Pb < Sn < Ge < C < Si (b) CO, GeO, SnO and PbO are amphoteric.
87. An element of group 14 forms two oxides one of which is (c) Monoxides react with haemoglobin to form toxic
highly poisonous and neutral. Other oxide can be easily compounds.
liquefied and compressed to give a solid which is used
as a refrigerant under the name of drikold. The element (d) All oxides burn with blue flame.
and the oxides are 93. In which of the following the inert pair effect is most
(a) Si, SiO, SiO2 (b) Pb, PbO, PbO2 prominent ?
(a) C (b) Ge
(c) C, CO, CO2 (d) Sn, SnO, SnO2
(c) Si (d) Pb
88. The members of group 14 form tetrahalides of the type
MX4. Which of the following halides cannot be readily 94. Identify the wrong example from the following for the
group 14 elements.
hydrolysed by water ?
(a) Element which forms most acidic dioxide-Carbon
(a) CX4 (b) SiX4
(b) Element which is affected by water - Lead
(c) GeX4 (d) SnX4 (c) Commonly found in +2 oxidation state - Lead
89. Which of the following hydrides is least stable to (d) Element used as semiconductor - Silicon
hydrolysis ?
95. Maximum ability of catenation is shown by
(a) CH4 (b) SiH4 (a) silicon (b) lead
(c) SnH4 (d) PbH4 (c) germanium (d) carbon.
96. Which one of the following is not the characteristic
90. Trimethylammonia is pyramidal whereas trisilylammonia
property of carbon ?
is a planar molecule. Why ?
(a) It exhibits catenation.
(b) It forms compounds with multiple bonds.
(c) Its melting point and boiling point are exceptionally
high.
(d) It shows semi-metallic character.
(a) In silicon atom, the electron pair of nitrogen is involved
in p - d bonding.
 35 p-block (GROUP 13 & 14) MEDICAL

97. There is a large number of carbon compounds due to (b) Fullerenes are pure forms of carbon.
(a) tetravalency of carbon (c) Fullerenes have open cage structure like ice.
(b) strong catenation property of carbon (d) C60 is called Buckminsterfullerene.
(c) allotropic property of carbon
105. The most stable form of carbon at high temperature is X.
(d) non-metallic character of carbon. The C – C bond length in diamond is Y while C – C bond
98. Carbon shows a maximum covalency of four whereas length in graphite is Z.
other members can expand their covalence due to What are X, Y and Z respectively ?
(a) absence of d-orbitals in carbon (a) Graphite, 1.42 Å, 1.54 Å
(b) ability of carbon to form p - p multiple bonds
(b) Coke, 1.54 Å, 1.84 Å
(c) small size of carbon
(c) Diamond, 1.54 Å, 1.42 Å
(d) catenation of carbon.
(d) Fullerene, 1.54 Å, 1.54 Å
99. Which of the following is not true about structure of
diamond and graphite ? 106. Which of the following is not a use of graphite ?

(a) In diamond, each carbon in sp3 hybridised while in (a) For electrodes in batteries.
graphite each carbon is sp2 hybridised. (b) Crucibles made from graphite are used for its
(b) In diamond, carbon atoms are closely packed in inertness to dilute acids and alkalies.
crystal lattice while graphite has layer structure. (c) For adsorbing poisonous gases.
(c) Diamond is a hard substance while graphite is a soft (d) Lubricant at high temperature.
substance.
107. Silicon is an important constituent of
(d) Graphite is thermodynamically very less stable as
compared to diamond and is amorphous form of (a) sand (b) atmosphere
carbon. (c) plants (d) water bodies
100. Buckminsterfullerence is 108. Match the column I with column II and mark the
(a) graphite (b) diamond appropriate choice.
(c) C-60 (d) quartz. Column I Column II
101. In graphite, C atom is in .......... state. (A) Coal gas (i) CO + H2
3
(a) sp (b) sp (B) Synthesis gas (ii) CH4
2
(c) sp (d) None of these (C) Producer gas (iii) H2 + CH4 + CO
102. In graphite, the layers of carbon atoms are held by (D) Natural gas (iv) CO + N2
(a) covalent bonds (b) coordinate bonds (a) (A)  (i), (B)  (ii), (C)  (iii), (D)  (iv)
(c) van der Waals forces (d) ionic bonds. (b) (A)  (iii), (B)  (i), (C)  (iv), (D)  (ii)
103. Identify the incorrect statement.
(c) (A)  (iv), (B)  (iii), (C)  (ii), (D)  (i)
(a) Graphite is thermodynamically most stable allotrope
(d) (A)  (i), (B)  (iii), (C)  (ii), (D)  (iv)
of carbon.
(b) Other forms of elemental carbon like coke, carbon 109. In SiO 44  , the tetrahedral molecule, two oxygen atoms
black, charcoal are impure forms of graphite. are shared in
(c) All allotropes of carbon have thermodynamically (a) sheet silicates
same stability.
(b) double-chain silicates
(d) Charcoal and coke are obtained by heating wood in
absence of air. (c) chain silicates

104. Which of the following does not depict properties of (d) three-dimensional silicates.
fullerenes ?
(a) Fullerenes are made by heating graphite.
 36 p-block (GROUP 13 & 14) MEDICAL

110. Complete the following reactions : 115. Which of the following bonds is shown in silicones ?
(i) SiO2 + 2NaOH  X + H2O
(ii) SiO2 + 4HF  Y + 2H2O
(a) (b)
Cu powder
(iii) Si  2CH3 Cl 
570 K  Z
X Y Z

(a) Na2SiO3 SiF4 (CH3)2SiCl2

(c) (d)
(b) H2SiO3 SiF2 CH3SiCl3

(c) Na2SiO3 H2SiO3 (CH3)3SiCl 116. Which of the following is not true about structure of
carbon dioxide ?
(d) Na2SiO3 H2SiF4 (CH3)2SiCl2
(a) In CO2, carbon is sp - hybridised.
111. When excesss of carbon dioxide is passed through lime
water, the milkiness first formed disappears due to (b) C forms two sigma bonds one with each oxygen atom
(a) the reversible reaction taking place and two p  p bonds.

(b) formation of water soluble calcium bicarbonate (c) CO2 is a linear covalent compound
(c) huge amount of heat evolved during the reaction (d) It is a polar molecule.
(b) formation of water soluble complex of calcium. 117. Carbon monoxide acts as a donor and reacts with certain
112. Which of the following is the correct statement about metals to give metal carbonyls. This is due to
silicones ? (a) presence of one sigma and two pi bonds between C
and
(a) They are made up of SiO44 units.
(b) presence of a lone pair on carbon atom in CO
(b) They are polymers made up of R2SiO units. molecule

(c) They are water soluble compounds. (c) presence of lone pair on oxygen atom in CO molecule

(d) They are hydrophillic in nature. (d) poisonous nature of CO.

113. A type of zeolite used to convert alcohols directly into 118. Mark the example which is not correct.
gasoline is (a) Non-combustible heavy liquid used as fire
extinguisher - CCl4
(a) zeolite A
(b) Blocks used to shield radioactive materials - Lead
(b) zeolite L
(c) Element which has property of leaving mark on
(c) zeolite Beta paper-Graphite
(d) ZSM-5 (d) A gas in solid form used as a refrigerant - Carbon
monoxide.
114. Which of the following properties correctly explain SiO2 ? 119. Which of the following oxides can act as a reducing
(a) Linear, basic agent ?
(a) CO (b) CO2
(b) Tetrahedral, acidic
(c) SnO2 (d) PbO2
(c) Tetrahedral, basic
120. Glass and cement are two important examples of
(d) Linear, acidic (a) man-made silicates (b) silicones
(c) zeolites (d) organic polymers.
 37 p-block (GROUP 13 & 14) MEDICAL

121. Match the column I with column II and mark the 126. Which is not a method of preparing carbon monoxide on
appropriate choice. a commercial scale ?
Column I Column II 4731273 K
(a) C(s)  H 2 O(g)   CO(g)  H 2(g)
(A) Used as lubricant (i) Carbon dioxide
(B) Oxide with three- dimensional (ii) Graphite 1273 K
(b) 2C (s)  O 2(g)  4N 2(g)  2CO(g)  4N 2(g)
structure

(C) Used in solar cells (iii) Silica (c) 2C(s)  O2(g)   2CO(g)
(D) Anhydride of carbonic acid (iv) Silicone
373 K
(d) HCOOH 
conc. H2SO4  H 2 O  CO
(a) (A)  (iv), (B)  (iii), (C)  (ii), (D)  (i)
(b) (A)  (iv), (B)  (i), (C)  (iii), (D)  (ii) 127. What happens when silicon is heated with methyl
chloride in presence of copper as a catalyst at 573 K ?
(c) (A)  (iii), (B)  (ii), (C)  (i), (D)  (iv)
(a) Methyl substituted chlorosilanes are formed.
(d) (A)  (ii), (B)  (iii), (C)  (iv), (D)  (i)
(b) Only Me4Si is formed.
122. Which of the following acids cannot be stored in glass ?
(c) Polymerised chains of (CH3)3SiCl are formed.
(a) HF (b) HCl
(d) Silicones are formed.
(c) H2SO4 (d) HI
128. Catenation i.e., linking of similar atoms depends on size
123. Which property of CO 2 makes it of biological and and electronic configuration of atoms. The tendency of
geochemical importance ? catenation in Group 14 elements follows the order
(a) Its acidic nature. (a) C > Si > Ge > Sn (b) C > > Si > Ge  Sn
(b) Its colourless and odourless nature. (c) Si > C > Sn > Ge (d) Ge > Sn > Si > C
(c) Its low solubility in water. 129. Silicon has a strong tendency to form polymers like
(d) Its high compressibility. silicones. The chain length of silicone polymer can be
controlled by adding
124. CO2 is not a poisonous gas but there is increase in
(a) MeSiCl3 (b) Me2SiCl2
concentration of CO2 in the atmosphere due to burning
of fossil fuels and decompostion of limestone. The (c) Me3SiCl (d) Me4Si
increase in concentration of CO2 may lead to
130. The most commonly used reducing agent is
(a) increase in photosynthesis in plants
(a) AlCl3 (b) PbCl2
(b) higher concentration of CO2 in water
(c) SnCl4 (d) SnCl2
(c) increase in greenhouse effect, thus raising the
temperature 131. Dry ice is

(d) increase in formation of metal carbonates. (a) solid NH3 (b) solid SO2
125. An oxide X in its normal form is almost non-reactive due (c) solid CO2 (d) solid N2.
to very high X – O bond enthalpy. It resists the attack by
halogens, hydrogen and most of acids and metals even 132. Cement, the important building material is a mixture of
at elevated temperatures. It is only attacked by HF and oxides of several elements. Besides calcium, iron and
NaOH. The oxide X is sulphur, oxides of elements of which of the group(s) are
present in the mixture ?
(a) SiO2 (b) CO2
(a) Group 2 (b) Groups 2, 13 and 14
(c) SnO2 (d) PbO2
(c) Groups 2 and 13 (d) Groups 2 and 14
 38 p-block (GROUP 13 & 14) MEDICAL

EXERCISE - 2 : PREVIOUS YEAR COMPETITION QUESTIONS

Group - 13
9. Which amont the following is not a borane ?(AMU 2009)
1. The tendency of BF3, BCl3 and BBr3 to behave as Lewis
(a) B2H6 (b) B3H6
acid decreases in the sequence (AIPMT 2010)
(c) B4H10 (d) None of these
(a) BF3 > BCl3 > BBr3 (b) BCl3 > BF3 > BBr3
10. Darking of surface painted with white lead is due to
(c) BBr3 > BCl3 > BF3 (d) BBr3 > BF3 > BCl3 (CPMT 2009)
2. The strongest Lewis acid among boron halides is (a) H2S (b) CO2
(RPMT 2010) (c) Cu (d) O2
(a) BBr3 (b) BCl3 11. Product of the following reaction is
(c) BI3 (d) BF3 Al 4 C3  D 2 O 
 (PMT 2008)
3. BF3 is (RPMT 2010)
(a) Al(OD)3 + CD4 (b) Al(OD)2 + CD4
(a) electron-deficient compound
(c) Al(OD)4 + CD4 (d) Al(OD)3 + CD
(b) Lewis base
12. Value of x in potash alum, K2SO4.Alx (SO4)3 . 24H2O is
(c) used as rocket fuel
(CPMT 2007)
(d) ionic compound (a) 4 (b) 1
4. Borax is used as a cleansing agent because on dissolving (c) 2 (d) None of these
in water, it gives (RPMT 2010)
13. In borax bead test which compound is formed ?
(a) alkaline solution (b) acidic solution (Punjab PMET 2007)
(c) bleaching solution (d) amphoteric solution (a) Ortho borate (b) Meta borate
5. Inorganic benzene is (PMT 2010) (c) Double oxide (d) Tetra borate
(a) B3H3N3 (b) BH3NH3 14. For which one of the following minerals, the composition
given is incorrect ? (Kerala CEE 2007)
(c) B3H6N3 (d) H3B3N6
(a) Glauber’s salt – Na2SO4 . 10H2O
6. The stability of +1 oxidation state increases in the sequence
(b) Borax – Na2B4O7 . 7H2O
(AIPMT 2009)
(c) Cornallite – KCl . MgCl2 . 6H2O
(a) Al < Ga < In < Tl (b) Tl < In < Ga < Al
(d) Soda ash – Na2CO3
(c) In < Tl < Ga < Al (d) Ga < In < Al < Tl
15. The colour imparted by Co(II) compounds to glass is
7. The lead of the lead pencils melts at (AIIMS 2009)
(AIIMS 2006)
(a) 2000ºC (b) 350ºC
(a) green (b) deep-blue
(c) 3170ºC (d) 75ºC (c) yellow (d) red
8. Assertion : Coloured cations can be identified by borax 16. Assertion : Silicones are hydrophobic in nature.
bead test.
Reason : Si–O–Si linkages are moisture sensitive.
Reason : Transparent bead (NaBO2 + B2O3) forms coloured (AIIMS 2006)
bead with coloured cation. (AIIMS 2009)
(a) Both Assertion and Reason are true and Reason is the
(a) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.
correct explanation of Assertion. (b) Both Assertion and Reason are true but Reason is not
(b) Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
the correct explanation of Assertion. (c) Assertion is true but Reason is false.
(c) Assertion is true but Reason is false. (d) Both Assertion and Reason are false.
(d) Both Assertion and Reason are false.
 39 p-block (GROUP 13 & 14) MEDICAL

17. The hardest substance among the following is 26. Which of the following is not attacked by hot sodium
(Kerala CEE 2004) hydroxide solution ? (Manipal 2010)
(a) Be2C (b) graphite (a) Silicon (b) Carbon
(c) titanium (d) B4C (c) Tin (d) Lead

18. Which of the following is not a protomic acid ? (2003) 27. Which type of silicate is shown in the given figure
(Guj. CET 2010)
(a) B(OH)3 (b) PO(OH)3
(a) Orthosilicate (b) Pyrosilicate
(c) SO(OH)2 (d) SO2(OH)2
(c) Meta silicate (d) None of these
19. Which of the following does not exist in free state ?(2004) 28. Which of the following is not hydrolysed easily ?
(a) BF3 (b) BCl3 (OJEE 2010)
(c) BBr3 (d) BH3 (a) CCl4 (b) SiCl4
20. In diborane, the two H – B – H angles are nearly : (2005) (c) GeCl4 (d) SnCl4

(a) 60º, 120º (b) 95º, 120º 29. The main reason that SiCl4 is easily hydrolyzed as compared
to CCl4 is that (Kerala CEE 2009)
(c) 95º, 150º (d) 120º, 180º
(a) Si—Si bond is weaker.
21. Which of the following is the electron deficient molecule ? (b) SiCl4 can form hydrogen bonds.
(2005) (c) SiCl4 is covalent.
(a) B2H6 (b) C2H6 (d) Si can extend its coordination number beyond four.

(c) PH3 (d) SiH4 30. Silica is soluble in (Haryana PMT 2009)
(a) HCl (b) HNO3
22. The number of isomers possible for disubstituted borazine,
B3N3H3X2 is : (2006) (c) H2SO4 (d) HF
(a) 3 (b) 4 31. Which of the following is used for making optical
instruments ? (AFMC 2008)
(c) 6 (d) 2
(a) SiO2 (b) Si
Group-14
(c) SiH4 (d) SiC
23. Name the type of the structure of silicate in which one
oxygen atom of [SiO4]4– is shared ? (AIPMT 2011) 32. Which one of the following anions is present in the chain
structure silicates ? (CBSE AIPMT)
(a) Three dimensional (b) Linear chain silicate
(a) Si 2 O 76 (b) (Si 2 O52 ) n
(c) Sheet silicate (d) Pyrosilicate
24. Chemical formula of phosgene is (AFMC 2010) (c) (SiO32 )n (d) SiO 44
(a) COCl2 (b) CaOCl2 33. Litharge is chemically (AIIMS 2007)
(c) CaCO3 (d) COCl (a) PbO (b) PbO2
25. Graphite is a good conductor of electricity. Its electrical (c) Pb3O4 (d) Pb(CH3COO)2
conductivity is due to the fact that (CPMT 2010) 34. Supercritical CO2 is used as (AIIMS 2007)
(a) it is an allotrope of carbon. (a) dry ice.
(b) it has C-atoms arranged in large plates of rings of (b) fire fighting.
strongly bound C-atoms. (c) a solvent for extraction of organic compounds from
2
(c) in it C-atoms are sp hybridised. natural sources.
(d) it is a non-crystalline substance. (d) a highly inert medium for carrying out various reactions.
 40 p-block (GROUP 13 & 14) MEDICAL

35. White lead is (CPMT 2007) (d) graphite is covalent and diamond is ionic.
(a) Pb3O4 (b) PbO 42. The hardest substance is (BHU 2003)
(c) 2PbCO3 . Pb(OH)2 (d) Pb(CH3COO)2 . Pb(OH)2 (a) iron (b) steel
36. Which glass will not crack when temperature changes ? (c) diamond (d) graphite
(DUMET 2006) 43. What is the number of free electrons present on each carbon
(a) Pyrex (b) Boron silicate atom in graphite ?

(c) Calcium silicate (d) Flint (a) Zero (b) 3

(c) 2 (d) 1
37. Assertion : SiF62 is known but SiCl 62 is not.
44. Which one of the following statements about the zeolites
Reason : Size of fluorine is small and its lone pair of is false ? (2004)
electrons interacts with d-orbitals of Si strongly.
(AIIMS 2005) (a) Zeolites are aluminosilicates having three dimensional
network
(a) Both Assertion and Reason are true and Reason is the
correct explanation of Assertion. (b) Some of the SiO44– units are replaced by AlO45– and
(b) Both Assertion and Reason are true but Reason is not AlO69– ions in zeolites
the correct explanation of Assertion.
(c) They are used as cation exchangers
(c) Assertion is true but Reason is false.
(d) They have open structure which enables them to take
(d) Both Assertion and Reason are false.
up small molecules
38. Sindoor is represented by (RPMT 2005)
45. Which of the following cuts ultraviolet rays ? (2004)
(a) Pb(NO3)2 (b) PbCO3Pb (OH)2
(a) Soda glass (b) Crooke’s glass
(c) Pb(OH)24PbCO3 (d) Pb3O4
(c) Pyrax (d) None of these
39. Assertion : PbI4 is a stable compound.
Mixed
Reason : Iodide stabilizes higher oxidation state.
(AIIMS 2003) 46. Which of the following structure is similar to graphite ?

(a) Both Assertion and Reason are true and Reason is the (NEET 2013)
correct explanation of Assertion. (a) BN (b) B
(b) Both Assertion and Reason are true but Reason is not
(c) B4C (d) B2H6
the correct explanation of Assertion.
(c) Assertion is true but Reason is false. 47. The basic structural unit of silicates is (NEET 2013)
(d) Both Assertion and Reason are false.
(a) SiO– (b) SiO 44
40. The purest form of coal is (CPMT 2003)
(a) peat (b) anthracite (c) SiO 32 (d) SiO 24

(c) bituminous (d) lignite 48. Boric acid is an acid because its molecule (NEET 2016)
41. Graphite is soft while diamond is hard because (a) gives up a proton
(BHU 2003)
(b) accepts OH– from water releasing proton.
(a) graphite is in powder form.
(c) combines with proton from water molecule.
(b) diamond has sp2-hybridisation but graphite has sp3-
hybridisation. (d) contains replacable H+ ion.
(c) graphite is in planar form while diamond is in tetrahedral
form.
 41 p-block (GROUP 13 & 14) MEDICAL

49. AIF3 is soluble in HF only in presence of KF. It is due to 52. Which one of the following elements is unable to form
the formation of
MF63 ion? (NEET 2018)
(a) K3[AlF6] (b) AlH3
(a) Al (b) B
(c) K[AlF3H] (d) K3[AlF3H3] (c) Ga (d) In
50. The element Z = 114 has been discovered recently. It will 53. The correct order of atomic radii in group 13 elements is
belong to which of the following family/group and (NEET 2018)
electronic configuration ? (NEET 2017) (a) B < Al <Ga< In <Tl
(a) Carbon family, [Rn]5f146d107s27p2 (b) B <Ga< Al <Tl< In
(c) B < Al < In <Ga<Tl
(b) Oxygen family, [Rn]5f146d107s27p4
(d) B <Ga< Al < In <Tl
(c) Nitrogen family, [Rn]5f146d107s27p6

(d) Halogen family, [Rn]5f146d107s27p5

51. It is because of inability of ns2 electrons of the valence

shell to participate in bonding that (NEET 2017)

(a) Sn2+ is oxidizing while Pb4+ is reducing.

(b) Sn2+ and Pb2+ are both oxidizing and reducing.

(c) Sn4+ is reducing while Pb4– is oxidizing.

(d) Sn2+ is reducing while Pb4– is oxidizing.

 42 p-block (GROUP 13 & 14) MEDICAL

EXERCISE - 3 : ADVANCED OBJECTIVE QUESTIONS

1. All questions marked “S” are single choice questions
2. All questions marked “C” are comprehension based questions
3. All questions marked “A” are assertion–reason type questions
(A) If both assertion and reason are correct and reason is the correct explanation of assertion.
(B) If both assertion and reason are true but reason is not the correct explanation of assertion.
(C) If assertion is true but reason is false.
(D) If reason is true but assertion is false.

o o
100 C 160 C Re d hot
1. (S) Which of the following is correct? 6. (S) H 3 BO3   X   Y  B2 O 3 ;
(a) The members of BnHn+6 are less stable than BnHn+4 X and Y respectively are:
series
(a) X = Metaboric acid ; Y = Tetraboric acid
(b) Diborane is coloured and unstable at room tempera
ture (b) X = Borax; Y = Metaboric acid

(c) The reaction of diborane with oxygen is endothermic (c) X = Tetraboric acid ; Y = Metaboric acid

(d) All of the above (d) X = Tetraboric acid ; Y = Borax

2. (S) In which of the following, a salt of the type KMO2 is 7. (S) A gas other than HCl is obtained in :
obtained? Silent electric
(a) BCl3 + H 2 
discharge 
(a) B2 H 6  KOH(aq.) 
(b) LiAlH 4 + BF3 
(b) Al  KOH(aq.) 
(c) Both (d) None of these
(c) Both 8. (S) Hydride of boron occurs as B2H6 but B2Cl6 does not
(d) None of these exist. This is because :
3. (S) Which of the following compounds is formed in borax (a) p-d back bonding is possible in B2H6 but not in
bead test? B2Cl6
(a) Metaborate (b) tetraborate (b) boron and hydrogen have almost equal values of
(c) Double oxide (d) Orthoborate electronegativity

4. (S) Borax is used as a cleansing agent because on dissolv- (c) boron and chlorine have almost equal atomic sizes
ing in water, it gives: (d) small hydrogen atoms can easily fit in between
(a) alkaline solution (b) acidic solution boron atoms but large chlorine atoms do not

(c) bleaching solution (d) neutral solution 9. (S) When an inorganic compound (X) having 3C—2e as
well as 2e—2e bonds reacts with ammonia gas at a cer-
5. (S) Boric acid is used in carrom boards for smooth gliding of tain temperature, gives a compound (Y), isostructural
pawns because: with benzene. Compound (X) with ammonia at a high
(a) H3BO3 molecules are loosely chemically bonded and temperature produces a substance (Z). Which of the
hence soft following is not correct?
(b) its low density makes it fluffy (a) (X) is B2H6
(c) it is chemically inert with the plywood (b) (Z) is known as inorganic graphite
(d) H-bonding in H3BO3 gives it a layered structure (c) (Y) is B3N3H6 (d) (Z) is soft like graphite
 43 p-block (GROUP 13 & 14) MEDICAL

10. (S) Boron does not form B3+ cation easily. It is due to: 18. (S) In the purification of bauxite by Hall’s process:
3+
(a) energy required to form B ion is very high which (a) bauxite is heated with lime
will not be compensated by lattice energies or (b) bauxite is heated with NaOH solution at 150oC
hydration energies of such ion
(c) bauxite is heated with Na2CO3 solution
(b) boron is non-metal
(d) bauxite is heated at 1800oC in an atmosphere of N2
(c) boron is semi-metal
19. (S) Which of the following minerals does not contain alu-
(d) none of the above minium?
11. (S) Na2B4O7.10H2O is correctly represented as: (a) Cryolite (b) Mica
(a) 2NaBO2.Na2B2O3.10H2O (c) Feldspar (d) Fluorspar
(b) Na2[B4O5(OH)4].8H2O 20. (S) The function of fluorspar in the electrolytic reduction of
(c) Na2[B4(H2O)4O7].6H2O alumina dissolved in fused cryolite (Na3AlF6) is:
(d) all of these (a) as a catalyst
12. (S) Orthoboric acid behaves as a weak monobasic acid (b) to lower the temperature of melt and to make the
giving H3O+ and fused mixture very conducting
(a) [B(OH)4]+ (b) H2BO2+ (c) to decrease the rate of oxidation of carbon anode
– –
(c) [B(OH)4] (d) H2BO 2
(d) none of the above
13. (S) Amorphous boron is extracted from borax by following 21. (S) Al2O3 can be converted to anhydrous AlCl3 by heating:
steps: (a) a mixture of Al2O3 and carbon in dry Cl2 gas
(A ) Heat (B)
Borax  H 3 BO 3 
 B2 O 3  Boron. (b) Al2O3 with Cl2 gas
(A) and (B) are: (c) Al2O3 with HCl gas
(a) H2SO4, Al (b) HCl, carbon (d) Al2O3 with NaCl in solid state
(c) H2SO4, Mg (d) HCl, Fe Passage (Ques 22 to 27)
14. (S) Aqueous solution of borax acts as a buffer because: Aluminium is stable in air and water in sipite of the fact
(a) it contains weak acid and its salt with strong base that it is reactive metal. The reason is that a thin film of
its oxide is formed on its surface which it passive for
(b) it contains tribasic acid and strong base further attack. The layer is so useful that in industry, it is
(c) it contains number of neutral water molecules purposely deposited by an electrolytic process called
(d) none of the above anodising
Reaction of aluminium with oxygen is highly exothermic
15. (S) and is called thermite reaction.
3
2Al(s) + O 2 (g)  Al 2 O3 (s); ΔH = –1670 kJ
Which of the statement is true for the above sequence 2
of reactions?
Thermite reaction finds applications in the metallurgical
(a) Z is hydrogen (b) X is B2H6 extraction of many metals from their oxides and for weld-
(c) Z and Y are F2 and B2H6 respectively ing of metals. The drawback is that to start the reaction,
high temperature is required for which an ignition mix-
(d) Z is potassium hydroxide ture is used.
16. (S) The structure of diborane (B2H6) contains : 22. (C) Anodising can be done by electrolysing dilute H2SO4
(a) four 2C-2e bonds and two 3C-2e bonds with aluminium as anode. This results in :
(b) two 2C-2e bonds and four 3C-2e bonds (a) the formation of Al 2(SO 4 ) 3 on the surface of
aluminium anode
(c) two 2C-2e bonds and two 3C-3e bonds
(b) the formation of oxide film (Al2O3) on the surface of
(d) four 2C-2e bonds and four 3C-2e bonds aluminium anode
17. (S) Alumina is: (c) the formation of polymeric aluminium hydride film
(a) acidic (b) basic on the surface of aluminium anode
(c) neutral (d) amphoteric (d) none of the above
 44 p-block (GROUP 13 & 14) MEDICAL

23. (C) The reaction which is not involved in thermite process: 28. (C) Compound (A) is:

(a) 3Mn 3 O 4  8Al 

 9Mn  4Al2 O3 (a) NaBO2 (b) Na2B4O7

(b) Cr2 O3  2Al 

 2Cr  Al2 O3 (c) Na3BO3 (d) NaOH
29. (C) Compound (B) is:
(c) 2Fe  Al2 O3 
 Fe 2 O3  2Al
(a) NaBO2 (b) Na2B4O7
(d) B2 O3  2Al 
 2B  Al2 O3
(c) Na3BO3 (d) NaOH
24. (C) Thermite a mixture used for welding is:
30. (C) Compound (C) is:
(a) Fe and Al (a) H2B4O7 (b) HBO2
(b) BaO and Mg powder (c) H3BO3 (d) HB3O5

(c) Cu and Al 31. (C) Compound (D) is:

(d) Fe2O3 and Al powder (a) H3BO3 (b) B2O3

(c) B (d) none of these
25. (C) Anodised aluminium is :
32.(C) Compound (E) is:
(a) Al obtained at anode
(a) Cu2O (b) CuS
(b) Al prepared electrolytically
(c) CuSO3 (d) Cu(BO2)2
(c) alloy of Al containing 95% Al Passage (Ques. 33 to 37)
(d) Al electrolytically coated with aluminium oxide o
700 C
(i) Boron  O 2  (X)
26. (C) Which one of the following metals cannot be extracted
by using Al as a reducing agent? (ii) (X)  C(carbon)  Cl2  (Y)  CO

(a) Na from Na2O (b) Cr from Cr2O3 (iii) (Y)  LiAlH 4  (Z)  LiCl  AlCl3

(c) W from WO3 (d) Mn from Mn3O4 Heat

(iv) (Z)  NH 3  (A)  (B)
27. (C) Aluminium becomes passive in:
(v) (Z)  NaH  (D)
(a) conc. HNO3 (b) H2CrO4 33. (C) Compound (Z) is:
(c) HClO4 (d) all of these (a) an ionic compound

Passage (Ques. 28 to 32) (b) an electron deficient compound

(c) 3C—2e compound
Fused
Ca 2 B6 O11  Na 2 CO3   (A)  (B)  CaCO 3
(d) having ethane like structure

(A)  CO 2 
(B)  Na 2 CO 3 34. (C) Compounds (X) and (Y) are :
Solution
(a) (X) = BO2, (Y) = BCl2

(B)  Conc.HCl  H2 O
 NaCl  Acid   Acid (C) (b) (X) = BO3, (Y) = BCl4
(c) (X) = B2O3, (Y) = BCl3
Strongly
(C) (D) heated (d) (X) = BO3, (Y) = B4C

Heated
(D)  CuSO 4 
inflame
 Blue coloured (E) compound
 45 p-block (GROUP 13 & 14) MEDICAL

35. (C) Compound (B) is : 40. (C) Compound S is:

(a) borazole (b) inorganic benzene (I) an odd-e– compound
(c) borazon (d) boron nitride (II) (2c-3e–) compound
36. (C) Compound (D) is used as a/an : (III) a electron deficient compound
(a) oxidising agent (b) complexing agent (IV) a sp2 hybridized compound
(c) buffer agent (d) reducing agent
Choose the correct code:
37. (C) Compound Y:
(a) III (b) I, III
(a) has boron in sp2 hybridized state
(c) II, III, IV (d) I, II, IV
(b) is a planar molecule
41. (A) Assertion : Borax bead test is not suitable for Al (III)
(c) has zero dipole moment
Reason : Al2O3 is insoluble in water.
(d) is a Lewis base
(a) A (b) B
Passage (Ques. 38 to 40)
(c) C (d) D
(i) P  C(carbon)  Cl 2  Q  CO 

(ii) Q  H 2 O  R  HCl
42. (A) Assertion : BF3 is a weaker Lewis acid than BCl3.
(iii) BN  H 2 O  R  NH 3 
Reason : BF3 molecule is stabilized to a greater extent
(iv) Q  LiAlH 4  S  LiCl  AlCl3 than BCl3 by B—F -bonding.
(a) A (b) B
(v) S  H 2  R  H 2 
(c) C (d) D
(vi) S  NaH  T
43. (A) Assertion : In water, orthoboric acid behaves as a weak
(P, Q, R, S and T do not represent their chemical monobasic acid.
symbols)
Reason : In water, othoboric acid acts as a proton donor.
38. (C) Compound Q has:
(a) A (b) B
(I) Zero dipole moment
(c) C (d) D
(II) a planar trigonal structure
44. (A) Assertion : Boron always forms covalent bond.
(III) an electron deficient compound
Reason : The small size of B3+ favours formation of co-
(IV) a Lewis base valent bond.

Choose the correct code: (a) A (b) B

(a) I, IV (b) I, II, IV (c) C (d) D

(c) I, II, III (d) I, II, III, IV 45. (A) Assertion : If aluminium atoms replace a few silicon at-
oms in three dimensional network of silicon dioxide, the
39. (C) Compound T is used as a/an : overall structure acquires a negative charge.
(a) oxidising agent (b) complexing agent Reason : Aluminium is trivalent while silicon is tetrava-
(c) bleaching agent (d) reducing agent lent.
(a) A (b) B
(c) C (d) D
 46 p-block (GROUP 13 & 14) MEDICAL

46. (A) Assertion : AlF3 is soluble in KF but addition of BF3 51. (S) The blue coloured mineral ‘Lapis Lazuli’ which is used
brings its precipitation. as a semi-precious stone is a mineral of the following
Reason BF3 is more acidic than AlF3. class:

(a) A (b) B (a) sodium alumino silicate

(c) C (d) D (b) zinc cobaltate

(c) basic copper carbonate

47. (S)  Na  B(OH) 4 
B(OH)3  NaOH 
(d) prussian blue
How can this reaction be made to proceed in forward
direction? 52. (S) The chemical formula of feldspar is:

(a) Addition of cis-1,2-diol (a) KAlSi3O8 (b) Na3AlF6

(b) Addition of borax (c) NaAlO2

(c) Addition of trans-1, 2-diol (d) K2SO4.Al2(SO4)3.4Al(OH)3

(d) Addition of Na2HPO4 53. (S) A metal X reacts with aqueous NaOH solution to form Y
and a highly inflammable gas. Solution Y is heated and
48. (S) Select the true statements: CO2 is poured through it. Z precipitates out and Na2CO3
is formed. Z on heating gives Al2O3, Identify X, Y and Z.
(i) The formula of the corundum is Al2O3.
X Y Z
(ii) The formula of cryolite is Na3AlF6.
(a) Al NaAlO2 Al(OH)3
(iii) The formula of borax is Na2B4O7.10H2O
(b) Al2O3 NaAlO2 Al2CO3
(iv) The formula of bauxite is Al2O3.H2O
(c) Al2O3 [Na2AlO2]+OH– Al(OH)3
(a) (i), (ii) and (iii) (b) (ii), (iii) and (iv)
(d) Al Al(OH)3 Al2O3
(c) (i), (ii) and (iv) (d) (iii) and (iv)
54. (S) Which of the following will be formed, if we heat an
49. (S) Alumina is: aqueous solution of AlCl3 to dryness ?
(a) a bad conductor of electricity
(a) Solid AlCl3 (b) Dimer Al2Cl6
(b) a good conductor of electricity
(c) Al(OH)3 (d) Al2O3
(c) a dehydrating agent
55. (S) Identify X and Y in the following reaction.
(d) soluble in water
140ºC 4 NaBH
BCl3  NH4 Cl 
C H Cl  X  Y
50. (S) Alumina is not used as: 6 5

(a) refractory material (a) X = NaBO2, Y = B2O3

(b) a medium in chromatography (b) X = Na2B4O7, Y = H3BO3
(c) abrasive
(c) X = BN, Y = [NH 4 ] [BCl4 ]
(d) a white pigment
(d) X = B3N3H3Cl3, Y = B3N3H6
 47 p-block (GROUP 13 & 14) MEDICAL

56. (S) Match the column I with column II and mark the H O Heat NaOH
58. (S) 2
SiCl4  X   Y  Z
appropriate choice.
Column I Column II X, Y and Z in the above reaction are

(A) Galena (i) Abrasive X Y Z

(B) Diamond (ii) Metal carbonyls (a) SiO2 Si NaSi

(C) Carbon monoxide (iii) Hydrides of Si (b) Si(OH)4 SiO2 Na2SiO3

(D) Silanes (iv) An ore of lead (c) Si(OH)4 Si SiO2
(a) (A)  (iv), (B)  (ii), (C)  (i), (D)  (iii) (d) SiO2 SiCl4 Na2SiO3
(b) (A) (iv), (B)  (i), (C)  (ii), (D)  (iii)
(c) (A)  (ii), (B)  (i), (C)  (iii), (D)  (iv)
(d) (A)  (i), (B)  (ii), (C)  (iii), (D)  (iv)
57. (S) Which of the following is not matched correctly with its
use ?
(a) Piezoelectric material - Quartz
(b) Ion-exchangers - Graphite
(c) Filtration platns - Silica
(d) Electrical insulators - Silicones
 48 p-block (GROUP 13 & 14) MEDICAL

EXERCISE - 4 : PREVIOUS YEAR OTHER COMPETITION QUESTIONS

AIEEE Questions 6. Among the following substituted silanes the one

which will give rise to cross linked silicone polymer
1. When metal ‘M’ is treated with NaOH, a white gelatinous on hydrolysis is (2008)
precipitate ‘X’ is obtained, which is soluble in excess of (a) R3SiCl (b) R4Si
NaOH. Compound ‘X’ when heated strongly gives an (c) RSiCl3 (d) R2SiCl2
oxide which is used in chromatography as an adsorbent. 7. Which one of the following is the correct statement ?
The metal ‘M’ is: (2018) (2008)
(a) Fe (b) Zn (a) B 2H6  2NH3 is known as ‘inorganic benzene’.
(c) Ca (d) Al (b) Boric acid is a protonic acid.

2. Which one of the following alkaline earth metal (c) Beryllium exhibits coordination number of six.
(d) Chlorides of both beryllium and aluminium have
sulphates has its hydration enthalpy greater than its
bridged chloride structures in solid phase.
lattice enthalpy? (2015) 8. The stability of dihalides of Si, Ge, Sn and Pb
(a) BaSO4 (b) SrSO4 increases steadily in the sequence
(c) CaSO4 (d) BeSO4 (a) PbX2  SNX2  GeX2  SiX2
3. Which of the following exists as covalent crystals in (b) GeX2  SiX2  SnX2  PbX2
the solid state ? (2013) (c) SiX2  GeX2  PbX2  SnX2
(a) Phosphorus (b) Iodine
(d) SiX2  GeX2  SnX2  PbX2.
(c) Silicon (d) Sulphur
9. In silicon dioxide (2005)
4. Boron cannot form which one of the following anions ?
(a) Each silicon atom is surrounded by four oxygen
(2011) atoms and each oxygen atom is bonded to two
silicon atoms
(a) BF63 (b) BH4
(b) Each silicon atom is surrounded by two oxygen
(c) B(OH)4 (d) BO2 atoms and each oxygen atom is bonded to two
silicon atoms
5. The bond dissociation energy of B – F in BF 3 is
(c) Silicon atom is bonded to two oxygen atoms
646 kJ mol –1 whereas that of C – F in CF 4 is
(d) There are double bonds between silicon and
515 kJ mol–1. The correct reason for higher B – F
oxygen atoms
bond dissociation energy as compared to that of
C – F is (2009) 10. Heating an aqueous solution of aluminium chloride to
(a) smaller size of B-atom as compared to that of dryness will give (2005)
C-atom (a) AlCl3 (b) Al2Cl6
(b) Stronger  bond between B and F in BF 3 as (c) Al2O3 (d) Al(OH)Cl2
compared to that between C and F in CF4 11. The structure of diborane (B2H6) contains (2005)
(c) significant p-p interaction between B and F in (a) Four 2c-2e bonds and two 3c-2e bonds
BF 3 whereas there is no possibility of such
(b) Two 2c-2e bonds and four 3c-2e bonds
interaction between C and F in CF 4.
(c) Two 2c-2e bonds and two 3c-3e bonds
(d) lower degree of p-p interaction between B and
(d) Four 2c-2e bonds and four 3c-2e bonds
F in BF3 than that between C and F in CF4.
 49 p-block (GROUP 13 & 14) MEDICAL

12. Aluminium chloride exists as dimer, Al2Cl6 in solid state 19. H3BO3 is (2003)
as well as in solution of non-polar solvents such as
benzene. When dissolved in water, it gives (2004) (a) monobasic acid and weak Lewis acid
(a) [Al(OH)6]3– + 3HCl (b) [Al(H2O)6]3+ + 3Cl– (b) monobasic and weak Bronsted acid
(c) Al3+ + 3Cl– (d) Al2O3 + 6HCl (c) monobasic and strong Lewis acid
13. The soldiers of Napolean army while at Alps during (d) tribasic and weak Bronsted acid
freezing winter suffered a serious problem as regards
to the tin buttons of their uniforms. White metallic 20. (Me)2 SiCl2 on hydrolysis will produce (2003)
tin buttons got converted to grey powder. This
(a) (Me)2 Si(OH)2 (b) (Me)2 Si = O
transformation is related to (2003)
(a) A change in the partial pressure of oxygen in the (c) [—O—(Me)2 Si—O—]n
air
(d) Me2SiCl(OH)
(b) A change in the crystalline structure of tin
21. Name the structure of silicates in which three oxygen
(c) An interaction with nitrogen of the air at very low
to temperatrure atoms of [SiO4]4– are shared is (2005)

(d) An interaction with water vapour contained in the (a) pyrosilicate (b) sheet silicate
humid air. (c) linear chain silicate (d) three dimensional silicate
IIT Questions 22. B (OH)3 + NaOH  NaBO2 + Na [B (OH)4] + H2O
Only One Correct Option How can this reaction is made to proceed in forward
14. Moderate electrical conductivity is shown by (1982) direction ? (2006)

(a) silica (b) graphite (a) Addition of cis 1, 2 diol

(c) diamond (d) None of these (b) Addition of borax

15. Which of the following halides is least stable and has (c) Addition of trans 1, 2 diol
doubtful existence ? (1996) (d) Addition of Na2HPO4
(a) CCl4 (b) GeI4 Assertion and Reason
(c) SnI4 (d) PbI4 Read the following question and answer as per the
16. Which one of the following oxides is neutral ? (1996) direction given below :

(a) CO (b) SnO2 (a) Statement I is true; Statement II is true; Statement II

is the correct explanation of Statement I.
(c) ZnO (d) SiO2
(b) Statement I is true; Statement II is true; Statement II
17. In compounds of type ECl3, where E = B, P, As or Bi, the is not the correct explanation of Statement I.
angles Cl—E—Cl for different E are in the order (1999)
(c) Statement I is true; Statement II is false.
(a) B > P = As = Bi (b) B > P > As > Bi
(d) Statement I is false; Statement II is true.
(c) B < P = As = Bi (d) B < P < As < Bi
23. Statement I : Al(OH)3 is amphoteric in nature.
18. Identify the correct order of acidic strength of CO2, CuO,
Statement II : Al—O and O—H bonds can be broken
CaO, H2O. (2002)
with equal ease in Al (OH)3. (1998)
(a) CaO < CuO < H2O < CO2
24. Statement I : Between SiCl4 and CCl4, only SiCl4 reacts
(b) H2O < CuO < CaO < CO2 with water.
(c) CaO < H2O < CuO < CO2 Statement II : SiCl4 is ionic and CCl4 is covalent.
(d) H2O < CO2 < CaO < CuO (1997)
 50 p-block (GROUP 13 & 14) MEDICAL

25. Statement I : Boron always forms covalent bond. 27. Statement I : Pb4+ compounds are stronger oxidising
Statement II : The small size of B3+ favours formation of agents than Sn2+ compounds.
covalent bond. Statement II : The higher oxidation states for the group
(1996) 14 elements are more stable for the heavier members of
the group due to ‘inert pair effect’. (2008)
26. Statement I : In water, orthoboric acid behaves as a weak
monobasic acid.
Statement II : In water, orthoboric acid acts as a proton
donor.
(1995)
 51 p-block (GROUP 13 & 14) MEDICAL

ANSWER KEY
Exercise-1 (Basic Objective Questions)

1. (d) 2. (d) 3. (b) 4. (a) 5. (b) 6 (c) 7. (a) 8. (d) 9. (b) 10. (d)

11. (b) 12. (d) 13. (b) 14. (d) 15. (b) 16. (c) 17. (c) 18. (b) 19. (c) 20. (b)

21. (c) 22. (a) 23. (a) 24. (d) 25. (d) 26. (b) 27. (a) 28. (b) 29. (d) 30. (d)

31. (d) 32. (d) 33. (d) 34. (c) 35. (c) 36. (d) 37. (d) 38. (a) 39. (a) 45. (d)

41. (d) 42. (d) 43. (c) 44. (a) 45. (d) 46. (b) 47. (d) 48. (c) 49. (b) 56. (b)

51. (b) 52. (c) 53. (d) 54. (a) 55. (d) 56. (b) 57. (b) 58. (a) 59. (a) 68. (c)

61. (a) 62. (d) 63. (c) 64. (b) 65. (a) 66. (b) 67. (b) 68. (a) 69. (b) 79. (a)

71. (c) 72. (a) 73. (b) 74. (b) 75. (d) 76. (a) 77. (c) 78. (a) 79. (a) 89. (a)

81. (a) 82. (d) 83. (b) 84. (a) 85. (d) 86. (b) 87. (c) 88. (a) 89. (b) 90. (c)

91. (d) 92. (a) 93. (d) 94. (b) 95. (d) 96. (d) 97. (b) 98. (d) 99. (d) 100. (c)

101. (c) 102. (c) 103. (d) 104. (c) 104. (c) 106. (c) 107. (a) 108. (b) 109. (c) 110. (a)

111. (b) 112. (b) 113. (d) 114. (b) 115. (a) 116. (d) 117. (b) 118. (d) 119. (a) 120. (a)1

121. (d) 122. (a) 123. (c) 124. (c) 125. (a) 126. (d) 127. (a) 128. (b) 129. (c) 130. (d)

131. (c) 132. (b)

Exercise-2 (Previous Year Competition Questions)

1. (c) 2. (c) 3. (a) 4. (a) 5. (c) 6. (a) 7. (c) 8. (a) 9. (b) 10. (a)

11. (a) 12. (c) 13. (b) 14. (b) 15. (b) 16. (c) 17. (d) 18. (a) 19. (d) 20. (b)

21. (a) 22. (b) 23. (d) 24. (a) 25. (c) 26. (b) 27. (d) 28. (a) 29. (d) 30. (d)

31. (a) 32. (c) 33. (a) 34. (c) 35. (c) 36. (a) 37. (a) 38. (d) 39. (d) 40. (b)

41. (c) 42. (c) 43. (d) 44. (b) 45. (b) 46. (a) 47. (b) 48. (b) 49. (a) 50. (a)

51. (d) 52. (b) 53. (d)

 52 p-block (GROUP 13 & 14) MEDICAL

Exercise-3 (Advanced Objective Questions)

1. (a) 2. (c) 3. (a) 4. (a) 5. (d) 6. (a) 7. (c) 8. (d) 9. (d) 10. (a)

11. (b) 12. (c) 13. (c) 14. (a) 15. (c) 16. (a) 17. (d) 18. (c) 19. (d) 20. (b)

21. (a) 22. (b) 23. (c) 24. (d) 25. (d) 26. (a) 27. (d) 28. (a) 29. (b) 30. (c)

31. (b) 32. (d) 33. (bc) 34. (c) 35. (ab) 36. (d) 37. (abc) 38. (c) 39. (d) 40. (a)

41. (b) 42. (a) 43. (c) 44. (a) 45. (a) 46. (a) 47. (a) 48. (a) 49. (a) 50. (d)

51. (a) 52. (b) 53. (a) 54. (a) 55. (d) 56. (b) 57. (b) 58. (b)

Exercise-4 (Previous Year Other Competition Questions)

1. (d) 2. (b) 3. (c) 4. (a) 5. (c) 6. (c) 7. (d) 8. (c) 9. (a) 10. (c)

11. (a) 12. (a) 13. (b) 14. (b) 15. (d) 16. (a) 17. (b) 18. (a) 19. (a) 20. (c)

21. (b) 22. (a) 23. (a) 24. (c) 25. (a) 26. (c) 27. (a)