HINTS & SOLUTIONS TO WORKBOOK-4

p-Block Elements
Daily Tutorial Sheet Level-0

Very Short Answer Type (1 Mark)

1. Anhydrous aluminium chloride is partially hydrolysed by atmospheric moisture to liberate HCl gas. Moist
HCl appears white in colour. AlCl 3 (s)  3H 2O( )  Al(OH)3 (s)  3HCl(g)

2. Due to absence of d-orbitals, B at the maximum can accommodate 8 electrons or 4 pairs of electrons in
its valence shell. In other words, B can have a maximum covalency of 4. Therefore, BF3 can form BF4 but

never BF63  in which the covalency of B is 6.

3. The main reasons are :

(i) Due to smaller size of F as compared to Cl, six small F atoms can be easily accommodated
around Si atom but six large Cl atoms cannot.
(ii) The lone pair of electrons on F are present in a smaller 2p-orbital but in Cl they are present on a
larger 3p-orbital. Therefore, interaction of F lone pair electrons with d-orbitals of silicon are
stronger than that of Cl lone pairs.

4. Diamond has three dimensional network structure involving strong C—C bonds. These bonds are difficult
to break and hence the melting point of diamond is very high.
 R 
 | 
5. Silicones are synthetic organosilicon compounds containing repeated  O  S i   units held by
 | 
 R  n
Si  O  Si linkages. They are hydrophobic (water-repellent) in nature.

6. The B atom in BF3 has only 6 electrons in the valence shell and thus need two more electrons to complete
its octet. Therefore, it easily accepts a pair of electrons from nucleophiles such as F  , (C2H5 )2 O, RCH2OH
etc. and thus behaves as a Lewis acid.

Short Answer Type-I (2 Marks)

7. It is a not a protic acid since it does not ionize in H2O to give a proton :
H2 BO3  H2O  H2BO3  H3O 
Instead because of the small size of boron atom and presence of only six electrons in its valence shell,
B(OH)3 accepts a lone pair of electrons from the oxygen atom of the H2O molecule to form a hydrate
The +ve charge on the O-atom, in turn, pulls the electrons of the O—H bond towards itself thereby
facilitating the release of a proton. As a result, B(OH)3 acts as a weak monobasic Lewis acid and thus
reacts with NaOH solution to form sodium metaborate.
B(OH)3  NaOH  Na  [B(OH)4 ] or Na BO2  2H2O
Sod. metaborate

VMC | Chemistry 24 Solutions | p-Block Elements-I

8. Boric acid, on heating, loses water in three different stages at different temperatures ultimately giving
boron trioxide or boric anhydride.
370 K
H3BO3  HBO2  H2O
Boric acid Metaboric acid

410K Re d heat
4HBO2  H2B4O7 
 2B2O3  H2O
 H2 O
Metaboric acid Tetraboric acid Boron trioxide

9. In BF3, boron is sp2-hybridized and, therefore, BF3 is

a planer molecule. On the other hand, in [BH4]–
3 –
boron is sp -hybridized and hence [BH4] is a
tetrahedral species.

10. It dissolves both in acids and alkalies evolving dihydrogen.

2Al(s)  3H2SO4 (aq)  Al 2 (SO4 )3 (aq)  3 H2 (g)

2Al(s)  2 NaOH(aq)  6H2O( )  2Na  [Al(OH)4 ] (aq)  H2 (g)

Sod. tetrahydroxoaluminate (III)

11. Resonance structures of ion :

Resonance structures of ion :

12. (a) sp2 (b) sp3 (c) sp2

13. BF3 is a planer molecule in which B is sp2 -hybridized. It has an empty 2p-orbital.F-atom has three lone
pairs of electrons in the 2p-orbitals. Because of similar sizes, p - p back bonding occurs in which a lone
pair of electrons is transferred from F to B:
As a result of this back bonding, B–F bond acquires some double bond character. In contrast, in [BF4]–
ion, B is sp3-hybridized and hence does not have an empty p-orbital available to accept the electrons
donated by the F atom. Consequently, [BF4]–, B–F is a purely single bond. Since double bonds are shorter
than single bonds, therefore, the B–F bond length in BF3 is shorter (130 pm) than B–F bond length (143
pm) in [BF4]–.

Short Answer Type-II (3 Marks)

14. Due to electronegativity difference between B (E. N. = 2.0) and Cl (E. N. = 3.0), the B—Cl bond is polar
and hence has a finite dipole moment. The overall dipple moment of a molecule, however, depends upon
its geometry. Now BCl3 is a planer molecule in which the three B—Cl bonds are inclined at an angle of
120°. Therefore, the resultant of two B–Cl bonds is cancelled by equal and opposite dipole moment of the
third B–Cl bond as shown below : As a result, overall dipole moments of BCl3 is zero.

15. (i) Borax : Na 2 B4 O7  10H 2O (ii) Metaboric acid : HBO 2

(iii) Boric acid : H3BO3 (iv) Sodium metaboric : NaBO2

(v) Inorganic Benzene : B3 N 3 H6 (vi) Inorganic graphite : BN

VMC | Chemistry 25 Solutions | p-Block Elements-I

16. Due to poor shielding of the valence electrons of Ga by the inner 3d-electrons, the effective nuclear
charge of Ga is greater in magnitude than that of Al. As a result. the electrons in gallium experience
greater force of attaraction by the nucleus than in Al and hence atomic size of Ga (135 pm) is slightly less
than that of Al (143 pm).

17. Boron does not form B3 ion due to very high IE.

18. Absence of vacant d orbital and small size.

19. Catenation means self linking tendency of atoms to form chain. Carbon has maximum catenation power.

Long Answer Type (5 Marks)

20. In the lungs, haemoglobin present in red blood cells combines with molecular oxygen loosely and
reversibly to form oxyhaemoglobin.
Haemoglobin  O2  Oxyhaemoglobin
Oxyhaemoglobin thus formed in the lungs then travels to different parts of the body through blood
stream and delivers O2 to the various tissues of the body. However, CO combines with haemoglobin
irreversibly to form carboxyhaemoglobin which is about 300 times more stable than the oxyhaemoglobin.
Haemoglobin  CO  Carboxyhaemoglobin
As a result, the oxygen carrying capacity of haemoglobin is destroyed and the man dies of suffocation.
Thus, the highly poisonous nature of CO arise due to its ability to form a complex with haemoglobin
which is about 300 times more stable than the oxygen-haemoglobin complex.

21. (i) Due to inert pair effect, Pb is more stable in +2 than in +4 oxidation state. Therefore, lead (II)
chloride is more stable than lead (IV) chloride and hence lead (II) chloride does not react with Cl2
to form lead (IV) chloride
(ii) Due to greater stability of +2 over +4 oxidation state because of inert pair effect, lead (IV) chloride
on heating decomposes to give lead (II) chloride and Cl2
(iii) Due to oxidising power of Pb4+ ion and reducing power of I– ion, Pbl4 does not exist.
Alternatively, Pb-I bond initially formed during the reaction does not release enough energy to
unpair 6s2 electrons and excite one of them to the higher 6p-orbital to have four unpaired
electrons around lead atom needed for formation of PbI4.

22. (a) When borax is heated strongly, a transparent glassy bead which consists of sodium metaborate
(NaBO2) and boric anhydride is formed.
(b) Boric act acts as a weak Lewis acid by accepting a hydroxide ion of water and releasing a proton
into the solution.
(c) Dihydrogen is evolved.
(d) BF3 being a Lewis acid accepts a pair of electrons from NH3 to form the corresponding
complex.
23. (i) 2 BF3  6LiH  B2H6  6LiF
Diborane

(ii) B2H6  6H2O  2H3BO3  6H2

Diborane Orthoboric acid

(iii) 2NaH  B2H6  2Na  [BH4 ]

Sod. boronhydride

VMC | Chemistry 26 Solutions | p-Block Elements-I

 
(iv) H3BO3 
 HBO2  H2O
Orthoboric acid Metaboric acid

 
4HBO2  H2B4O7 
 2B2O3  H2O
 H2 O
Metaboric acid Tetraboric acid Boron trioxide

(v) 2Al  2NaOH  6H2O  2Na  [Al(OH)4 ]  3H2

Sod. tetrahydroxoalu min ate (III)

(vi) B2 H6  2NH 3  [BH2 (NH 3 )2 ] [BH 4 ]

24. CO2 is produced during combustion. It is utilized by plants during photosynthesis and O2 is released into
the atmosphere. As a result of this CO2 cycle, a constant percentage of 21% O2 is maintained in the
atmosphere. However, if the concentration of CO2 increases beyond 0.03% by volume in the atmosphere
due to excessive combustion, some of the CO2 will always remain unutilized. This excess CO2 absorbs
heat radiated by the earth. Some of it is dissipated into the atmosphere while the remaining part is
radiated back to the earth and other bodies present on the earth. As a result, temperature of the bodies
on the earth increases. This is called green house effect and CO2 is called a greenhouse gas. As a result of
greenhouse effect, global warming occurs which has serious consequences.

25. (i) Al reacts with conc. HNO3 to form a very thin film of aluminium oxide on its surface which
protects it from further action.
Thus, Al becomes passive and hence aluminium containers can be used to transport conc. HNO3
(ii) NaOH reacts with Al to evolve dihydrogen gas. The pressure of the gas thus produced can be
used to open clogged drains.
(iii) Graphite has layered structure in which the different layers are held together by weak vander
Waals forces and hence can be made to slip over one another. Therefore, graphite acts as a
lubricant.
(iv) Diamond is very hard and hence can be used as an abrasive.
(v) Aluminium alloys such as duralumin is light, tough and resistant to corrosion and hence is used
to make aircraft body.
(vi) Al reacts with H2O and dissolved O2 to form a thin film of aluminium oxide.
A very very small amount of Al2O 3 may dissolve to give a few ppm of Al3+ ions in the solution.
Since Al3+ ions are injurious to health, therefore, drinking water should not be kept in alumium
utensils overnight.
(vii) On weight to weight basis, aluminium conducts twice as Cu. Therefore, it is used in transmission
cables

VMC | Chemistry 27 Solutions | p-Block Elements-I

 Daily Tutorial Sheet-1 Level-1

1.(A)

Due to absence of 2d-orbital, maximum covalency is four. Thus BF63  is not formed.
Thus (A) is not formed.
BH4 (BH3  H  )

B(OH)4 and BO2 are formed.

2.(D) The enthalpy of formation of Al2O3 is very high and hence, it is not possible to reduce it by carbon.

3.(C) Borazine, B3N3H6 is also known as inorganic benzene due to its resemblance in structure and properties
with benzene.

4.(C)

5.(C) Chlorides of both beryllium and aluminium have bridged structures in solid phase.

Boric acid is not a protonic acid

6.(C) According to Lewis, the compound which can accept a lone pair of electron, are called acids. Boron
halides, being electron deficient compounds, can accept a lone pair of electrons, so termed as Lewis acid.

7.(C) The outer electronic structure of X is s2p1, hence, element X belongs to third group. It will be non-metal
because it is present in the first short period of third group. Its valency is +3 because it belongs to third
group. Hence, formula of its oxide will be X2O3. The oxide will be acidic in nature because it is oxide of
non-metal.

VMC | Chemistry 28 Solutions | p-Block Elements-I

8.(D) B2H6 has structure

9.(B) BCl 3  3H2O  B(OH)3  3HCl

Thus, the products are B(OH)3 or H3 BO3 and HCl.

10.(B) AlCl3 is covalent but in water, it becomes ionic due to large hydration energy of Al3+.
 3 
AlCl 3  6H2O  [Al(H 2O)6 ]  3Cl

11.(B) Al 4C3 is Methanide as on hydrolysis, it gives CH4 . Al 4Cl3  H2O  Al(OH)3  CH4

12.(A) Boric acid is used in carom boards for smooth gliding of pawns because H-bonding in H3BO3 gives it a
layered structure.


13.(B) Aqueous solution of AlCl 3 is acidic due to hydrolysis. AlCl 3  3H2O  Al(OH)3  3HCl ;

On strongly heating Al(OH)3 is converted into Al2O3. 2Al(OH)3 
 Al2O3  3H2O

14.(A) Boron trihalides are Lewis acid. The order of their acidic strength is as
BF3  BCl3  BBr3  BI3
In the boron halides, a p   p back bonding arises due to empty orbital
of boron and filled orbitals of halogens.
This p   p back bonding has maximum effect in BF3 as the size of B
and F-atoms are comparative and this effect decreases as the size of
halogen increases.
Due to this effect, tendency of accepting lone pairs of electron of boron decreases i.e., Lewis acid
character decreases.

15.(B) In diborane, H  B  H (H-terminal) and H  B  H (H-bridged) bond angles are 120 and 97 respectively.

Daily Tutorial Sheet-2 Level-1

16.(B) In III group, Tl (thalium) show +1 oxidation state due to inert pair effect. The outer shell ‘s’ electrons (ns2)
penetrate to (n – 1) d electrons and thus, becomes closer to nucleus and are more effectively pulled
towards the nucleus. This results in less availability of ns2 electron pair for bonding or ns2 electron pair
becomes inert.

17.(B)

Bond angle :   
Due to stearic repulsion by two chlorine atoms.

18.(D) Atomic size increases in a group from top to bottom. But in IIIA group, gallium (Ga, 1.35 Å ) has size
smaller than aluminium (1.43 Å ) . Because of poor shielding effect of d electrons in Ga.

19.(D) When two oxygen of each SiO44  tetrahedron are shared with others, cyclic or ring structures are
obtained. These silicates are known as cyclosilicates or cyclic silicates. [Si 6O18 ]12 is an example of
cyclosilicate. In this silicate six SiO4 tetrahedra linked together.

VMC | Chemistry 29 Solutions | p-Block Elements-I

20.(B) A recently discovered family of carbon allotropes is Buckminster fullerene. The most common fullerene
has the formula C60 and contains hexagonal and pentagonal rings of carbon atoms.

21.(C) The basic structural unit in silicates is SiO 44  tetrahedron. In SiO 44  unit, silicon atom is bonded to four
oxide ions tetrahedrally.

22.(B) 2Pb(NO3 )2 
 2PbO  4NO 2  O2

23.(B) RSiCl3 gives cross linked silicon polymer on hydrolysis.

24.(D) Orthosilicic acid (H4SiO4), on heating at high temperature, loses two water molecules and gives silica
(SiO2) which on reduction with carbon gives carborundum (SiC) and CO.
1000C C
H4SiO4 
 SiO2 
 SiC  CO
2H2O  Carborundum

25.(B) In silica, silicon has large size, so the 3p-orbitals of Si does not
overlap effectively with 2p-orbitals of oxygen. Therefore,
Si  O  are not formed. The tetravalency of Si is satisfied by
the formation of Si – O bonds, thus it is surrounded by the
four oxygen atoms.

26.(C) II. Pb2+ is more stable than Pb4+, due to inert pair effect.
III. In carbonate ion (CO23  ) all the three C – O bonds are identical due to resonance.

IV. B2 is paramagnetic due to presence of 2 unpaired electrons while C2 is diamagnetic.

(Refer MOT)
27.(C) Due to inert pair effect, the stability of +2 oxidation state increases as we move down this group.
 SiX 2  GeX 2  SnX 2  PbX 2

28.(A) In SiF62  and SiCl 26 , former is known due to the small size of F atoms. The small six F atoms can be
easily accommodated around Si atom to form SiF62  while SiCl26  , six large Cl atoms cannot be
accommodated around Si atom.
29.(D) C-atoms form covalently bonded plates (layers). Layers are bonded weakly together, that’s why one layer
can slide over other cause lubricacy. Cannot be melted easily as large number of atoms being bonded
strongly in the layer form big entity.

30.(B) Water glass is sodium silicate Na 2SiO3

VMC | Chemistry 30 Solutions | p-Block Elements-I

 Daily Tutorial Sheet-3 Level-1

31.(A) Monosilane  e.g., SiH4  on coming in contact with air burns with a luminous flame producing vortex

ring. These rings are of silica. SiH4  2O2  SiO2  2H2O

32.(D) Zeolites are alumino silicates having three dimensional open structure with four or six membered ring,
they have cavities and can take up water and other small molecules

33.(C) MCl 2 oxidation state of M   2 ; MCl4 oxidation state of M   4

Higher the oxidation state, smaller the size, greater the polarizing power and greater the covalent
character. Hence, MCl4 is more covalent and MCl 2 is more ionic.
34.(AB) C-60 is called Buckminster fullerene. It is discovered in 1990 as a constituent of soot. Its shape is like a
soccer ball. It is a covalent network solid.
35.(B) Stannous chloride  SnCl 2  is a good reducing agent. It reduces HgCl 2 into Hg (grey precipitate), in two

steps. SnCl2  2HgCl2  SnCl 4  Hg 2Cl2  ; SnCl2  Hg 2Cl 2  SnCl 4  2Hg 
white grey

36.(B) The structure of silicates has been found with the help
of X-ray diffraction technique. All silicates have
tetrahedral SiO44  ion as a basic building unit i.e., all
silicates are composed of many units. Tetrahedral
shape of SiO4 4  ion is due to sp3 - hybridisation of
Si-atom. Sheet silicates are formed when three oxygen
(bridging O-atoms) of each  SiO4 4  unit are shared.
2n 
Hence general formula of sheet silicates is  Si 2O5 

37.(D)

38.(A) The correct decreasing order of catenation property of group 14 elements is as C  Si  Ge  Sn  Pb

Catenation property is directly proportional to the bond energy.

39.(A) R 3SiCl on hydrolysis can only form a dimmer.

40.(C) Tin is oxidised to meta stannic acid when it is treated with nitric acid
Sn  4HNO3  H2SnO3  4NO2  H2O

41.(A)  CH3 2 SiCl2 undergoes hydrolysis but  CH3 2 CCl2 does not because in Si, low lying d-orbital is

present.

42.(C) The reluctance of the s-electrons of the valence shell to take part in bonding is called inert pair.
It increases on moving down in a group. Hence, Pb shows most pronounced inert pair effect.

43.(C) B2H6 is electron deficient due to presence of 3c-2e bond.

VMC | Chemistry 31 Solutions | p-Block Elements-I

44.(D) Cyclic silicone is obtained by hydrolysis od R 2SiCl2.
hydrolysis
R 2 SiCl 2  H2O 
 R 2 Si(OH)2  2HCl
Polymerization
R 2 Si(OH)2 
 Cyclic silicon

45.(B) Due to smaller size and better overlapping.

Daily Tutorial Sheet-4 Level-1

46.(D) Oxides of Trivalent metal are also called sesqui oxide

47.(A) BF63  does not exist due to unavailability of vacant orbitals.

48.(D) Co-ordinate bond is called Dative Bond

49.(B) NaBH 4  I2  B2 H6  NaI  H 2

50.(C) Refer NCERT

51.(C) 
H3BO3  H2O  
 B(OH)4  H


52.(B) BH3 exist as dimer due to absence of Back Bonding

53.(A) BI4 does not exist due to stearic hinderance

54.(B) BCl 3  LiAlH 4  B2 H6  LiCl  AlCl3

B2 H6  H 2O  H 3BO 3  H 2
Red
H3 BO3 
 B 2O 3
heat

55.(C) Oxides of Be and Al are amphoteric. 56.(A) Refer NCERT

2 3 Na CO
57.(D) SiF4  H2O  Si(OH)4 
 SiO2   Na 2SiO3
(A) (B) (C)

3
58.(D) SiO2 : sp hybridised
CO2 : sp hybridised

Graphite : sp2 hybridised

% P character SiO 2  Graphite  CO2

59.(C) R 3SiCl act as chain stopping unit

60.(D) BCl 3  H 2O  B(OH)3  3HCl

COCl 2  H 2O  CO2  HCl

SiCl 4  H2O  Si(OH)4  HCl

VMC | Chemistry 32 Solutions | p-Block Elements-I

 Daily Tutorial Sheet-5 Level-1


61.(D) Refer NCERT 62.(B) SiCl 4  H2O  Si(OH)4 
 SiO2

63.(B) Co reacts with haemoglobin in blood to form carbxyhaemoglobin.

64.(A) Boiling point of hydrides increases down the group due to increase in Vander Waal radius.

65.(D) C H3 is sp2 hybridised. Here unpaired electrons is present in pure p-orbital.

66.(C) CO 2 : Bond angle is 180°

SiO 2 exist in form of 3D silicate. Bond angle = 109°28°

67.(B) Na 2B4O7  10H2O  HCl  2NaCl  H3BO3

68.(B) BF3 is covalent

69.(B) Al 2O3 is insoluble in water due to high Lattice energy and low hydration energy.

70.(A) H2C2O4 
 H2O  CO  CO2
Conc. H2SO4

71.(D) (AlCl3 )n sublime on heating

72.(B) Number of shared oxygen in sheet silicate is 3.

73.(A) Silicones repel water due to presence of alkyl group

74.(D) H2C2O4  CO CO2  H2O

( y) ( x) ( z)
3
x : sp , y : sp , z : sp

75.(D) Bond order

CO : 3
CO2 : 2
CO 23  : 1.33
1
Bond length 
Bond order

Daily Tutorial Sheet-6 Level-2

76.(A) B(OH)3 is accepting electron pair from (OH ). So it is acting as Lewis acid.

77.(C) In case of N(CH3)3, geometry is pyramidal, but in case of N(SiH3)3 it is planer. It is due to the fact that in
the latter, the lone pair of N-atom is transferred to empty d-orbital of silicon (p -d overlapping)

78.(C) NH3 > NF3 > BF3

This is due to different directions of the bond moments of N – H and N – F bonds. In NH3, N is more
electronegative but in NF3, F is more electronegative. Thus in NH3, the dipole-moment of N – H bond are
in the same direction as that of the lone pair. BF3 being trigonal planer has zero dipole moment.

79.(C) The ability of boron carbide to absorb neutrons without forming long lived nuclides makes it attractive as
an absorbent for neutron radiation arising in nuclear power plants. It is one of the hardest materials
known, behind boron nitride and diamond.

VMC | Chemistry 33 Solutions | p-Block Elements-I

80.(D) When borax is dissolved in water, both B(OH)3 and [B (OH4 )] are formed, but only [B(OH4 )] reacts with
HCl.

Na 2B 4 O7  7H2O  2Na[B  OH 4 ]  2H 3BO 3

2Na B  OH 4   2HCl  2NaCl  2H3BO3  2H2O

2 4 H SO
81.(B) H3BO3  3MeOH   B  OMe 3  3H2O
Green Colour

82.(C) 3NaBH4  BF3  2B 2H 6  g   3NaF 

83.(A) Boron nitride (inorganic graphite) had diamond like structure. (Non planer)
Borazole : Inorganic Benzene (planer)

B3O63 :

84.(A) Borax dissolves in water and gives an alkaline solution.


Na 2B4O7  7H2O  2NaOH  4H3BO3
strong base weak acid

85.(C) B2 H6  3O2  B2O 3  3H2O

B2H6  6H2O  2H3BO3  6H2


B2 H6  2NH 3  B 2H 6  2NH 3 
 B 3 N 3 H6

Daily Tutorial Sheet-7 Level-2

86.(D) To obtain crystalline boron in small amounts, B2O3 is reduced with aluminium powder.

87.(ABD) Refer theory

88.(ABC) Boranes are easily hydrolysed due to presence of vacant p-orbital. Because of incomplete octet BH3 acts
as Lewis acid. All B–H distances in diboranes are not equal [Refer NCERT class XI part II Page No.313]

89.(B) PbCl 2 is insoluble in cold water and soluble in hot water.

90. [A-s] [B-r] [C-p] [D-q]


Na 2B 4 O7 
 2NaBO2  B2O 3

B2H6  6H2O  2H3BO3  6H2


3B 2H 6  6NH3 
 2B3 N 3H6  12H2

4BCl3  3LiAlH 4  2B2 H6  3LiCl  3AlCl3

VMC | Chemistry 34 Solutions | p-Block Elements-I

91. [A-t] [B-r] [C-p] [D-q]
Inorganic benzene - B3 N 3 H6
Jeweller’s borax - Na 2B 4 O7  5H2O
Borax - Na 2B4O7  10H2O
Diborane - B2H6

92.(C) In Borax, 2 boron atoms are sp3 and 2 boron atoms are sp2 hybridised

93.(ABCD)Borax when dissolved in water because of partial hydrolysis form solution having weak acid and its
salt with strong base Na 2B4O7  7H2O  2NaOH  4H3BO3
(Salt) ( Weak acid)

Borax and HCl reacts in 1 : 2 molar ratio as per given reaction

Na 2B4O7  2HCl  5H2O  2NaCl  4H3BO3
In borax bead test coloured bead is due to formation of metaborate.
CuSO4  Na 2B4O7  Cu(BO2 )2  2NaBO2
SO3
(Coloured bead)

94.(AB) d and f orbital electrons are more diffused and so less repulsive. This causes increase in effective
nuclear charge. Therefore, atomic size decreases.

95.(ABD) 2 bridged hydrogen and 2 boron lie in perpendicular plane. Bridged hydrogen are described as 3
centre-2 electron bond

Daily Tutorial Sheet-8 Level-2

96.(A) A silica garden or chemical garden is an experiment normally performed by adding coloured metals
salt such as CuSO4 or CoCl2 to an aqueous solution of sodium silicate. This results in growth of plant
like forms.

97.(A) Silicon carbide is a hard substance hence used as abrasive.

98.(D) Because some of the silicon chloride get hydrolyzed.

99.(A) Silicone have large number of alkyl groups which have appreciable hydrophobic properties.
As general representation of linear silicon is
R R  R
   
H – O  Si  O  Si  O Si  O  H n  very large
 
   
R  R n R
Or

VMC | Chemistry 35 Solutions | p-Block Elements-I

 R R  R R
    
R  Si  O  Si  O Si  O  Si  R
 
    
R R n R R

100.(BC) O–C–O bond angle is 180° due to sp hybridization of carbon and also due to p –p bonding between
carbon and oxygen.

101.(AB) Organic silicone chain shows condensation reaction with R3SiCl as shown
R R R R
   
    Si  O  H  Cl  Si  R      Si  O  Si  R  HCl
   
R R R R

102.(BC) Graphite is more stable than Diamond because of

(1) Double bond character in C – C bond
(2) e– are delocalized in a manner that aromaticity get introduced in graphite sheets.
(3) Graphite has greater entropy which is because of
(a) Layers are much distant
(b) Layers can get align randomly is AB AB packing and ABC ABC packing [XIIth Class Solid
State]

 
103.(BD) O  C  O 
O  C – O 104.(D) Si  HF  H2SiF6
excess

105.(B) Gas P and Q are same i.e. H2

Daily Tutorial Sheet-9 Level-2

106.(B) Conversion of white tin to grey tin is called plague of tin.


107.(C) H2C 2O 4 
 CO  CO2  H 2O
(A ) (B)

CO burns with a blue flame

3 NH
CO  Cl 2  COCl 2  NH2CONH2
(D)

108.(B) Number of shared oxygen per tetrahedron in Amphibole Silicate is 2.5.

109.(C) Kinoite mineral contain silicate anion. Silicate is a chain of three SiO4 tetrahedral shares corners with
4
adjacent anion. Silicate share corners with adjacent tetrahedral. Therefore formula is Si3O10 .

110.(D) H will be maximum for BI3 because of its maximum Lewis acidic character.

111.(A) As Al (OH)3 is amphoteric in nature hence, it can act as an antacid.

Al(OH)3  3H  Al3  3H2O

112.(C) SiCl4 reacts with water due to vacant d-orbitals available with Si and no such vacant d-orbitals are
available with carbon, hence CCl4 does not react with water. Although, both SiCl4 and CCl4 are
covalent.

VMC | Chemistry 36 Solutions | p-Block Elements-I

113.(C) In group 13, 14, 15 as we descend down in group, the higher oxidation state becomes less stable due to
inert pair effect. Therefore, lead show +2 as stable oxidation state. Hence, Pb4+ act as a strong oxidizing
agent, itself reduced to Pb2+ very easily. Only statement I is correct and statement II is a incorrect.

114.(CD) K a  K a
a K

CHF3   
 CF3  H


aK

CHCl3   
 CCl3  H


115.(B) Bond length is inversely proportional to bond order.

Daily Tutorial Sheet-10 Level-2

116.(B) Thermal stability  bond strength.

117.(C) SiH3  O  SiH3 is weaker Lewis base due to presence of back bonding.

OH  HF 2H O
118.(AD)  Al(OH)4 ; SiF4  H2SiF6 ;
Al(OH)3  SiCl 4  H4SiO4
sp2 sp3 sp3 sp3d 2 sp3 sp3

119.(A) Zeolites are used as ion exchange material and remove hardness of water.

120.(ABC) Carbogen is a mixture of 90% CO2 and 10% O 2

121.(BD) Anhydride of malonic acid is C3O2.

CaC2 exist in form of Ca 2 and C22 

C22  has 2 bond and 1 bond

Trisilylamine is planer due to back Bonding.

122.(ABC) SiO2  Na 2CO3  Na 2 SiO3

SiO2  C  Si  CO

SiO2  HF  H2SiF6

123-125. 123.(A), 124.(A), 125.(A)

VMC | Chemistry 37 Solutions | p-Block Elements-I

 Daily Tutorial Sheet-11 Level-3

126.(C)

127.(C) A is B2 H6

B2 H6  6NaOH  2Na 3 BO3  6H2

B is Na 3 BO3 a salt of SBWA

128.(C) Na 2B 4 O7  H 2SO 4  5H 2O  Na 2 SO 4  4H3 BO3


2H3BO3 
 B2O3  3H2O

B2O 3  3Mg  2B  3MgO

129.(B) Being containing one halogens atom it is utilized for terminating the chain.

130.(C) 4(CH 3 )2 SiCl2  8H 2O  [P]  8HCl  4H 2O

8
131.(B) Chain of three SiO 4 tetrahedral exist as Si 3O10

Charge of silicate anion is 8.

Daily Tutorial Sheet-12 Level-3

132.(A) CoO  B 2O3  Co(BO2 )2 (blue bead)

133.(B) As Al (OH)3 is amphoteric in nature and thus form [Al (H2O)2 (OH)4  ].

134.(D) Due to small size of boron, the sum of its first three ionization enthalpies is very high. This prevents it
to form  3 ions.

135.(C) The reactions involved are

3B2H6  6NH3  2B3 N3H6  12H2 ;

(X) ( Y)

at higher temperature
B2H6  NH3 (excess)  (BN)n  H2

136.(B) Very pure silicon is obtained by zone refining.

137.(D) In TIX 3 , TI exist as TI due to inert pair effect

VMC | Chemistry 38 Solutions | p-Block Elements-I

 Daily Tutorial Sheet-13 Level-3

138.(D) In group 14, stability of +4 oxidation state decreases down the group and +2 oxidation state increases
down the group due to inert pair effect.

139.(C)

140.(B) E is ' C'

C  O 2  CO2

CO2  2 NaOH  Na 2CO3  H2O

C  H2O (g) CO  H2

 
can be used as fuel

141.(B) E is 'B'

2 B  N2  2 BN

BN  H2O  H3BO3  NH3

( Acid) (Pungent gas)

142.(BC) Al 2O3  3 C  3 Cl2  3 CO  2 AlCl 3

(X) (Y )

AlCl 3  6 H2O 
 Al 2O3  HCl  H2O
(Z )
X Z

AlCl 3  LiH  LiAlH4

( used as reducing agent)
AlCl3 is used as catalyst in Friedal craft reaction.
C.N of anhyd. AlCl3 is 4

143.(C) NaOH  Potash alum 

(excess)

OH 
NaOH  K 2SO4  Al 2  SO4 3  24 H2O  Al  OH 3 
  Al  OH 4 
white ppt clear solution

Daily Tutorial Sheet-14 Level-3

144.(9) Diamond, H3BO3 (aq), Zeolites, Silicones, Chlorosilanes, Borax, Al 2Cl6 , B2H6 , SiO2

145.(5) Mg  B  Mg 3 B2  xy 5

146.(5) H3 PO 4 , H3 BO3 , H3 P3 O9 , BF3 , SiCl 4 147.(4) H3 BO3 , AlCl 3 , BF3 , B2 H6

148.(C) Al 2Cl 6 : complete octet

AlF3 is ionic  complete octet
149.(A) BF3
Bond order is maximum in BF3 due to Back Bond.

VMC | Chemistry 39 Solutions | p-Block Elements-I

 Miscellaneous Exercise Question Bank
Single correct option Type
150.(D) Catenation means chain linking tendency of atoms of the same element.
Catenation tendency is maximum in C due to strong C  C bond strength.

151.(D)

152.(C) Aluminium vessel should not be washed with materials containing washing soda because it reacts with
aluminium to form soluble aluminate.
Al  Na 2CO3  H2 O(aq)  NaAlO2  NaHCO 3  H 2


153.(D) B2H6  NH3  2 BH3  NH3   B3N3 H6  H2
450 K

154.(C) NH4OH is used as a precipitating reagent for Al 3 ion as Al  OH 3 rather than aqueous NaOH

because Al  OH 3 being amphoteric dissolve in NaOH to form soluble aluminate

Al  OH 3  NaOH  NaAlO2


155.(C)  Na  B  OH 4 
H3 BO3  NaOH (aq ) 

H3BO3  molten NaOH 
 Na 3BO3  H2O

156.(C) BF3  H2O  H3BO3  HBF4

157.(D) BCl 3  C5H5 N  BCl 3  C5H5 N 

158.(D) These reactions lead to chemical inertness due to formation of inert oxide layer on its surface.

159.(D) H3 BO3 is monobasic acid.


H3 BO3  H2O  B  OH 4  H 

160.(C) III is incorrect. Potash alum is K 2SO4  Al2 (SO4 )3  24H2O

161.(A) TI3 exist as T  and I3

Red hot
162.(D) Na 2B 4 O7  2NH 4 Cl 
 BN  NaCl  H 2O  HCl

163.(A) H3BO3  C2H5OH  B  OEt 3

green flame

164.(D) Li does not form alums.

165.(D) B  A  Ga  In  T
Here IE of Ga is more than that of A due to high zeff and poor shielding effect of d-electrons. It of
T  In due to poor shielding effect of f electrons and high zeff.

VMC | Chemistry 40 Solutions | p-Block Elements-I

166.(D) Tl3  is good oxidising agent because its stable oxidation state is + 1 due to inert pair effect.

2 H O
167.(B) Al  N 2  AlN  Al  OH 3  NH3 
Y Z

Y is insoluble in water.

168.(D) All are correct

169.(A) Si forms covalent compounds due to its high ionization energy.

170.(A) Silicons are not combustible

171.(B) CH 2 (COOH)2  P4 O10  C3O 2  H 2O

172.(B) Silyl Isocyanate is linear due to presence of back bonding.

173.(C) Water in presence of oxygen reacts with Pb pipes to from soluble Pb (OH)2 which gives poisnous Pb2
ions in solution.

174.(B) C  H2O  CO  H2

water gas

175.(A) Bond length of B  F in Me3N  BF3 is greater than that in BF3 due to back bonding in BF3.

176.(D) Al  KOH  K 2 AlO2  H2

(soluble)

More than One Correct Type

177.(BCD) Graphite is paramagnetic due to presence of unpaired electron.

178.(BCD) B2O 3  C  CO  B4C

179.(ABC) SiO2  C  Si  CO

180.(ABC) Al can form covalent compound also like AlCl 3, Al2O3

181.(AB) PbO and PbO 2 are amphoteric as they reacts with acid as well as base.

182.(ABC) Diborane undergoes unsymmetrical cleavage with lewis base like NH 3 , RNH2 and Me2 NH due to
small size of Base

183.(BD) B2 H6 is also lewis acid due to 3c-2e bond.

It is used as reducing agent for reduction of organic compounds.

184.(BCD) Borax bead test is given by transition metal salts.

conc.
185.(AB) HCOOH 
 H2O  CO
H2 SO4
4 10 P O
Malonic acid   C3O2
2H O
Mg 2C3  Mg(OH)2  C3H4
K 4 [Fe(CN)6 ]  H2 SO 4  K 2 SO 4  FeSO 4  CO  (NH 4 )2 SO 4

VMC | Chemistry 41 Solutions | p-Block Elements-I

186.(B) Bond angle in graphite is 120°.

187.(ABCD) B2 H6 is not polar. It undergoes symmetrical cleavage with large Nucleophiles due to stearic strain.

188.(C) Hydrolysis of R 3 SiCl will form dimer hydrolysis of SiCl 4 will for Si(OH)4

189.(ABCD) Refer module

190(A) C 4 H 4 cannot be prepared by hydrolysis. CH 4 , C2 H 2 and C 3H 4 are prepared by hydrolysis of

Al 4 C3 , CaC2 and Mg 2C 3 respectively.

191.(ABD) NO BF4
It has 5  and 2  bond
Nitrogen-oxygen bond length in NO is higher than that in Nitric oxide.

Comprehension Type
192 to 197
NaBH4  I2  B2H6  H2  NaI
(A ) (B)

B2 H6 
 H 3BO 3  H2
H2 O
(B) (C)
H2 O
 B(OH)4  H
H3BO3 
(C)
2 O
B2H6   B2O3


192.(C)

193.(ACD) Cis-1, 2-diol is used in titration of NaOH with H3BO3 . Cis-1, 2-diol forms stable complex with
H3BO3 .

NH
194.(B) 3
B2 H6  [B(NH 3 )2 H2 ] [BH 4 ]

H O
195.(A) 2
Al 2Cl6  [Al(H2O)6 ]3 3Cl 

196.(C) AlX 3 forms dimer to overcome electron deficiency.

197.(C) AlCl3 can be sublimed easily. It is lewis acid and covalent compound.

198 to 202
A is Na 2B4O7  10H2O
aq. Solution of A is alkaline due to formation of NaOH
 NiO
Na 2B4O7  10H2O 
 Na 2B4O7  B2O3  NaBO2  Ni(BO2 )2
(B) (C)

198.(A) Water of crystallisation present per mole Borax is 10.

199.(C) Aqueous solution of Borax is alkaline due to hydrolysis of anion i.e. B4O72

Ca 2 
200.(A) Na 2B4O7 
 CaB4O7

201.(A) Na 2B 4 O7 202.(B) C is B2O 3 and D is Ni (BO2 )2

VMC | Chemistry 42 Solutions | p-Block Elements-I

203-204. 203.(C) 204.(C)

205-207. 205.(D) 206.(D) 207.(B)

205.(D) R 2SiCl2 on hydrolysis gives linear polymer and R 3SiCl act as chain terminating unit.

207.(B)

208-210. 208.(A) 209.(B) 210.(D)

3 NH
211.(C) B2H6   (B3N 3H6 )
450K
Borazine

212.(A) CO undergoes symmetrical cleavage with B2 H6

2CO
B2 H6 
 2[BH3  CO]

213.(C) B2H6  HCl  B2H5Cl  H2

Assertion & Reason Type

214.(A) Al shows passivity with conc. HNO 3 due to formation of inert layer of Al 2O3.

215.(A) Al + HCl  AlCl3  H2

Al  NaOH  NaAlO2  H2

Amphoteric substance react with acid as well as base.

216.(A) Alums like K 2SO4  Al2 (SO4 )3  24H2O are acidic due to cationic hydrolysis.

217.(A) Boron has highest melting point due to Icosahedral structure.

218.(A) Stable O.S of Tl is +1 due to inert pair effect.

219.(C) BF3 undergoes partial hydrolysis due to formation of HBF4 .

BF3  H 2O  H3 BO3  HBF4

220.(A) Diborane has two types of hydrogen i.e. Bridging hydrogen and terminal hydrogen.
Bridge bond is stronger than terminal bond.

VMC | Chemistry 43 Solutions | p-Block Elements-I

221.(A) Borazole i.e B3N 3H6 is isostructural to Benzene

222.(B) Borazole is known as inorganic benzene. When diborane is heated with NH3 at 200°C, Borazole is
formed. Here NH 3 act as Lewis base and diborane as Lewis acid.

223.(A) Refer 219

224.(A) Lewis acid character in boron trihalide is BF3  BCl3  BBr3  BI3 because electron deficiency in
BF3 is overcome by back bonding. As we move down the group, extent of back bonding decreases
due to poor overlap.

225.(A) BCl 3  H2O  H3BO3  HCl

226.(D) Boric acid is Monobasic acid

227.(A) OCCCO

228.(C) CCl 4 does not react with water due to absence of Vacant d orbital.
Both SiCl 4 and CCl 4 are covalent.

229.(B) SiO2 is solid (3D silicate)

230.(C)

231.(D) Carbonates are sp2 and silicates are sp3 hybridised.

232.(B) Bond dissociation energy of B  F bond in BF3 molecule is more than C  F bond in CF4 due to
back bonding in BF3 .
233.(A) Diamond is extremely hard to strong C  C bond in the structure.
x
234.(A) In Si 4O12 , x is 8. Here number of shared oxygen is 2.

Matrix Matching Type

235.(D) Al 2O3 : Amphoteric i.e. can react with acid as well as base
AlCl3 : exist as dimer
B : Non metal
B2O 3 : Non-metallic oxide so acidic

236.(D) BN (Boron Nitride) : Inorganic graphite

B3 N3 H6 (Borazole) : Inorganic Benzene
Ruby : Contain Al 2O3
Black lead : Graphite

237.(A) Water gas : CO  H2

Producer gas : CO  N 2
Coal gas : Mixture of CO, H2 , CH4 and CO2
Natural gas : CH4

VMC | Chemistry 44 Solutions | p-Block Elements-I

238. [A-p, r] [B-p, r] [C-q, r] [D-q, s]
B2H6 : Dimer of BH3 and each B is sp3 hybridised

Al 2Cl 6 : Dimer of AlCl3 and each Al is sp3 hybridised

BeCl 2 (Solid) : sp2 hybridised

(SiH 3 )3 N : Planer (sp2 ) due to Back bonding

239. [A-p, q] [B-p, r] [C-q, r, s] [D-q]

Al 2 (C2 )3  H2O  Al(OH)3  C2H2

CH2(COOH)2  P4O10  C3O2  H3PO4

CH3SiCl 3  H2O  CH3Si(OH)3  Cross linked polymer

on
SnCl2  2H2O 
 SnO2
Standary

Integer Answer Type

240.(2) In Borax, 2 Boron are sp2 hybridised and 2 Boron are sp3 hybridised.

241.(5) B2 H6 , BCl3 , CO 2 , Al2Cl6 , SiCl 4

3
242.(4) Al 2 (Me)6 , all atoms are sp hybridized
No. of 3c-2e bonds = 2

x
x = 8, y = 2, 4
y
243.(3) (CH3 )3 N, CO, (CH3 )2 O 244.(3) i, iv, vi

245.(7) PbO, PbO2, SnO, SnO2, Al2O3, BeO, Ga 2O3

246.(4) Refer structure of borax in solution 252. 247.(2) SiO 2 , B2O 3

248.(4) Al 4 C3 , Mg 2C 3 ,B2 H6 , BaC 2

2 H O
Al 4 C3  Al(OH)3  CH 4(g )
2 H O
Mg 2C3  Mg(OH)2  C3H2 (g)
2 H O
B2H6  H3BO3  H2 (g)
2 H O
BaC2  Ba(OH)2  C2H2 (g)

2 H O
249.(7) SiF4  Si(OH)4  H2SiF6
x y

H2 O
SiCl 4 
 Si(OH)4  HCl
x z

Covalency of Si in y is 6 = C1
Covalency of Cl in z is 1 = C 2

VMC | Chemistry 45 Solutions | p-Block Elements-I

 Daily Tutorial Sheet-1 Past year Exams

250.(ABC) A, B, D are correct

D is incorrect because AlCl3 has three centre four electron bonds in its dimeric structure.

251.(B) B < Ga < Al < ln < Tl (size)

252.(ACD) Borax :

(a) Borax can be represented as [B4O5 (OH)4 ]2

(b) Incorrect as two Boron are sp2 hybridised and two Boron are sp3 hybridised.
(c) Correct (d) Correct

253.(A) Orthoboric acid is a very weak acid, direct neutralization does not complete. However, addition of cis-
diol allow the reaction to go completion by forming a stable complex with [B(OH)4 ] as :

254.(A) Orthoboric acid is a weak, monobasic, Lewis acid.

p  p back bonding between ‘B’ and ‘O’ decreases acid strength greatly :

It does not undergo self ionization in water but accepts an electron pair from water, so it behaves as
weak monobasic acid.

H3 BO  H2O  
 B(OH)4  H


255.(B) In BCl3, bond angle = 120°.

In PCl3, AsCl3 and BiCl3, central atom is sp3 hybridised. Since P, As and Bi are from the same group,
bond angle decreases down the group. Hence, overall order of bond angle is :
B > P > As > Bi

256.(B) Graphite has layered structure and conducted electricity moderately. Silica and diamond have 3-
dimensional network structure and non-conducting.

VMC | Chemistry 46 Solutions | p-Block Elements-I

257.(BD) (a) It does not undergo self ionization in water but accepts an electron pair from water, so it behaves
as weak monobasic acid.

H3BO  H2O  
 B(OH)4  H


Hence, (a) is incorrect.

(b) When treated with 1, 2-dihydroxy or polyhydroxy compounds, they form chelate (ring
complex) which effectively remove [B(OH)4 ] species from solution and thereby produce

maximum number of H3O or H ions. i.e. results in increased acidity.

(c) Boric acid crystallizes in a layer structure in which planar triangular BO33 ions are bonded
together through hydrogen bonds.

(d) In water the pK a value of H3BO3 is 9.25.


H3 BO3  H2O   
 B(OH)4  H ; pK a  9.25
So, it is a weak electroyte in water.

258.(ABC) Diborane (B2H6) undergoes unsymmetric cleavage with NH3, primary and secondary amine while
tertiary amine brings about symmetrical cleavage of B2H6 as:

259.(A) Small size and high charge on B3  makes it highly polarizing. Therefore, in most of its compounds,
boron forms covalent bonds.
Hence, both statement I and statement II are correct and statement II is a correct explanation of
statement I

260.(A) Orthoboric acid is a weak, monobasic, Lewis acid and the poor acidic character is due to p  p
backbonding as :

VMC | Chemistry 47 Solutions | p-Block Elements-I

261.(A) Due to small size and high charge on Al in Al(OH)3 the fission ability of Al – O and O – H bonds become
comparable and compound can give both H+ and HO– under appropriate reaction conditions as :
Al(OH)3  3HCl  AlCl 3  3H2O ; Al(OH)2  NaOH  Na[Al(OH)4
Base Acid

Therefore, both statements are correct and statement-II is a correct explanation of statement I.

262. [A-q] [B-r] [C-p] [D-q, r]

(A) Bi 3 hydrolysis to (BiO)  q

(B) [AlO2 ] exist in basic medium, on acidification gives Al(OH)3  r

(C) Orthosilicate (SiO44  ) on heating changes into pyrosilicate Si2O76   p

(D) Tetraborate ion [B 4 O72  ] on treatment with dil. Acid hydrolysis gradually to orthoboric acid
 q, r

263. Three centred two electron bonds.

264.(F) The basic nature of hydroxide of group –13 increases form top to bottom due to increase in
electropositive character.

Daily Tutorial Sheet-2 Past year Exams

265.(F) In Al2Cl6, Al – Cl bonds are not equivalent :

266.(3) BenAl2Si6O18. 2n + 6 + 24 – 36 = 0  n=3

267. 3KF + AlF3 

 K3AlF6
K 3 AlF6  3BF3  AlF3   3KBF4
ppt

268. (i) Na 2B 4 O7  HCl  NaCl  H 3 BO 3


H3 BO3  HCl 
 BCl 3  H 2O

BCl 3  Al 
 B  AlCl3

(ii)

It has 4 terminal B – H bonds. There are two B – H – B, three centred two electron bridge bonds.

B2 H6  HCl  B2 H5Cl  H 2

VMC | Chemistry 48 Solutions | p-Block Elements-I

 4 LiAiH
269. Compound X Y a hydride + other compound. Hydride Y contains 21.72% hydrogen.

Y  O2 
 B2O3  H2O
Therefore, Y is a hydride of boron and it is obtained by reduction of X with LiAlH4. So, X is either BCl3
or BF3.
4 BCl 3  LiAlH4  B2H6  3AlCl 3  3LiCl
X Y
  
Other products
6
Molar mass of B2 H6  2  11  6  28 ; % of H in B2H6   100  21.5  28
28
B2 H6  3O 2  B2O 3  3H2O  Heat
Structure of Y (B2H6)

(a) There are 4 terminal B – H bonds

(b) There we two 3-center-2-electron B – H – B bridged bonds.
(c) Terminal H – B – H planes are perpendicular to bridged B – H – B bonds :

270. Al 2S3  6H2O  2Al(OH)3   3H2S(g) 

Foul odour

Foul odour on damping of Al2S3 is due to the formation of H2S gas as shown above.

271. The total hydration energy of AlCl3

= Hydration energy of Al 3  3  Hydration energy of Cl 
  4665  3 ( 381)kJ/ mol   5808 kJ / mol
The above hydration energy is more than the energy required for ionization of AlCl3 into Al3+ and 3Cl  .
Due to this reason, AlCl3 becomes ionic in aqueous solution. In aqueous solution, it undergoes
ionization completely as
AlCl 3  6H2O  [Al(H2O)6 ]3   3Cl 

272.(A) In sheet silicates, three out of four oxygen of SiO44  unit are shared as shown below

In pyrosilicates, there is only one shared oxygen, in linear chain silicates, two oxygen per tetrahedral
are shared while in three-dimensional silicates, all four oxygen’s are shared.

VMC | Chemistry 49 Solutions | p-Block Elements-I

273.(C) Me2SiCl2 on hydrolysis yields a linear chain silicone as :
CH3 CH3
| |
Cl — S i — Cl  2H2O  HO — S i — OH  2HCl
| |
CH3 CH3
CH3 CH3
| Polymerisation |
nHO  S i  OH   [ O S i O ]
| |
CH3 CH3

274.(A) CO2 is acidic oxide, H2O is neutral, CaO is strongly basic and CuO is weakly basic. Therefore, order of
acid strength is : CaO  CuO  H2O  CO 2

275.(A) Carbon monoxide is a neutral oxide, all other are amphoteric :

SnO2  4HCl  SnCl 4  2H2O

ZnO  2HCl  ZnCl 2  H2O

SiO2  2NaOH  Na 2SiO3  H2O

SnO2 and ZnO also react with NaOH. SiO2 is also attached by H3PO4.

276.(D) PbI4 is least stable, has doubtful existence. It is due to inert pair effect, the stable oxidation state of
lead is +2.

277.(BD) Diamond has a three-dimensional network structure, a hard substance where graphite is soft due to
layered structure.
In graphite, only three valence electrons are involved in bonding and one electron remain free giving
electrical conductivity. In diamond, all the four valence electrons are covalently bonded hence,
insulator.
Diamond is better thermal conductor than graphite. Electrical conductivity is due to availability of free
electrons, thermal conduction is due to transfer of thermal vibration energy from one atom to another
atom. A compact and precisely aligned crystals like diamond thus facilitate better movement of heat.
In graphite C – C bond acquire some double bond character, hence, higher bond order than in
diamond.

278.(C) In group 13, 14, 15 as we descend down in group, the higher oxidation state become less stable due to
inert pair effect. Therefore, lead show +2 as stable oxidation state. Hence, Pb4  act as a strong
oxidizing agent, itself reduced to Pb2 very easily. Both statement I and statement II are correct and
statement II is a correct explanation of statement I.

Daily Tutorial Sheet-3 Past year Exams

279.(C) SiCl4 reacts with water due to vacant d-oribitals available with Si as :

No such vacant d-orbitals are available with carbon, hence CCI4 does not react with water. Otherwise,
both SiCl4 and CCl4 are covalent.
Statement I is correct but statement II is incorrect.

VMC | Chemistry 50 Solutions | p-Block Elements-I

280. (Glass)
Glass is commonly known as supercooled liquid.

281. (Buckminster)
Buckminster fullerene is the name of recently discovered allotrope of carbon.

282. (R2SiO)2
After dimersation, no reactive function group remains.
R R R
| | |
R 3SiCl  H2O   R  S i  OH  R  S i O  S i R
 HCl
| | |
R R R
Dimeric silicone

283. (Silicones)
Silicones are organosilicon polymers, obtained by hydrolysis of alkyl substituted chlorosilanes.

284.(T) Due of smaller size of carbon than silicon, C –C bond is stronger than Si–Si bond, hence former is more
likely to extend than later.

285.(T) Graphite has a layer structure of hexagonal carbon rings stacked one over other which makes it
slippery.
On the other hand, in diamond, each carbon is tetrahedrally bond to other four carbons extended in
three dimensional space, giving a giant, network structure. Due to this reason, diamond is harder than
graphite.

286.(T) Graphite is better lubricant on moon than on earth because of absence of gravitational pull on the
moon.

287.(F) Phosgene gas is obtained by treatment of CCl4 with superheated steam :

CCl 4  H2O(vapour)  COCl 2  2HCl


288. (i)  3Si  4 AlCl 3 .
2SiCl 4  4 Al 
Mg or Zn can also be used.
(ii) SiCl 4  2CH3 MgCl  (CH 3 )2 SiCl2  2MgCl2
OH CH3 CH3 CH3
|
 HCl | | |
(CH3 )2 SiCl2  H2O 
 CH3  S i  CH3  — O — S i — O — S i — O — S i  O
| | | |
OH CH3 CH3 CH3

(iii) SiCl 4  4H2O  Si(OH)4  4HCl

Unstable
heat
Si(OH)4 
 SiO2  2H2O

SiO2  Na 2CO3 
 Na 2SiO3  CO2

289.

VMC | Chemistry 51 Solutions | p-Block Elements-I

 
290. 3SiCl 4  4 Al  4AlCl 3 
 3Si
Vapour Molten Voltilizes Crystalline

291.

X Y Z

Yeast Fermentation Ethanol

Mica Layered structure Insulator

Superphosphate Bone ash Fertiliser

Carbon fibers Graphite Reinforced plastics

Rock salt Crystalline cubic Preservative

Carborundum Diamond structure Abrasive

292. Graphite has layered structure and the adjacent layers are weakly associated giving slippery nature,
used as solid lubricant.

293. Carbon dioxide solidifies at very low temperature, hence solid CO2 is very cold, commonly known as dry
ice. Also solid carbon dioxide sublimes without passing through liquid state.

294. The two common allotropes of carbon are diamond and graphite. Diamond is the hardest, natural,
substance, used as an abrasive while graphite is soft, used as a lubricant.

295.(A) Silica gel - Drying agent

Silicon - Transistor
Silicone - Sealant
Silicate - Ion-exchanger

296.(B) In graphite, only three valence electrons are involved in bonding and one electron remain free giving
electrical conductivity. In diamond, all the four valence electrons are covalently bonded hence,
insulator.
Diamond is better thermal conductor than graphite. Electrical conductivity is due to availability of free
electrons, thermal conduction is due to transfer of thermal vibration energy from one atom to another
atom. A compact and precisely aligned crystals like diamond thus facilitate better movement of heat.

297.(6) 1B 2H 6  6 CH3OH  2B(OCH3 )3  6H 2

1 mole 2 moles
3 moles 6 moles

298.(B)  CH3 2 SiCl2 is used for preparation of linear polymer and  CH3 3 SiCl is used for chain termination.

VMC | Chemistry 52 Solutions | p-Block Elements-I