NCERT Solutions for Class 11 Chemistry The p – Block Elements [Old Book]

NCERT
SOLUTIONS
CLASS - 11th

aglase .co

Class : 11th
Subject : Chemistry
Chapter : 11
Chapter Name : The p-Block element

Q11.1 Discuss the pattern of variation in the oxidation states of (i) B to Tl and (ii) C to Pb.

Answer. (i) B to Tl
The electric con guration of group 13 elements is ns np . Therefore, the most common
2 1

oxidation state exhibited by them should be +3. However, it is only boron and aluminium
which practically show the +3 oxidation state. The remaining elements, i.e., Ga, In, Tl, show
both the +1 and +3 oxidation states. On moving down the group, the +1 state becomes more
stable. For example, Tl (+1) is more stable than Tl (+3). This is because of the inert pair effect.
The two electrons present in the s-shell are strongly attracted by the nucleus and do not
participate in bonding. This inert pair effect becomes more and more prominent on moving
down the group. Hence, Ga (+1) is unstable, In (+1) is fairly stable, and Tl (+1) is very stable.

The stability of the +3 oxidation state decreases on moving down the group.

(ii) C to Pb
The electronic con guration of group 14 elements is ns np . Therefore, the most common
2 2

oxidation state exhibited by them should be +4. However, the +2 oxidation state becomes more
and more common on moving down the group. C and Si mostly show the +4 state. On moving
down the group, the higher oxidation state becomes less stable. This is because of the inert
pair effect. Thus, although Ge, Sn, and Pb show both the +2 and + 4 states, the stability of the
lower oxidation state increases and that of the higher oxidation state decreases on moving
down the group.

Page : 331 , Block Name : Exercise

Q11.2 How can you explain higher stability of BCl as compared to TICl ?
3 3

Answer. Boron and thallium belong to group 13 of the periodic table. In this group, the +1
oxidation state becomes more stable on moving down the group.
BCl is more stable than TICl because the +3 oxidation state of B is more stable than the +3
3 3

oxidation state of Tl. In Tl, the +3 state is highly oxidising and it reverts back to the more
stable +1 state.

Page : 331 , Block Name : Exercise

Q11.3 Why does boron tri uoride behave as a Lewis acid ?

Answer. The electric con guration of boron is ns np . It has three electrons in its valence
2 1

shell. Thus, it can form only three covalent bonds. This means that there are only six electrons
around boron and its octet remains incomplete. When one atom of boron combines with three
uorine atoms, its octet remains incomplete. Hence, boron tri uoride remains electron-
de cient and acts as a Lewis acid.

Page : 331 , Block Name : Exercise

Q11.4 Consider the compounds,BCl and CCl . How will they behave with water ? Justify.
3 4

Answer. Being a Lewis acid, BCl3 readily undergoes hydrolysis. Boric acid is formed as a result
BCl3 + 3H2 O ⟶ 3HCl + B(OH)3

CCl4 completely resists hydrolysis. Carbon does not have any vacant orbital. Hence, it cannot
accept electrons from water to form an intermediate. When
CCl and water are mixed, they form separate layers.
4

CCl + H O -> NO reaction
4 2

Page : 331 , Block Name : Exercise

Q11.5 Is boric acid a protic acid ? Explain.

Answer. Boric acid is not a protic acid. It is a weak monobasic acid, behaving as a Lewis acid.

− +
B(OH)3 + 2HOH ⟶ [B(OH)4 ] + H3 O

It behaves as an acid by accepting a pair of electrons from –OH ion.

Page : 331 , Block Name : Exercise

Q11.6 Explain what happens when boric acid is heated .

Answer. On heating orthoboric acid H BO at 370 K or above, it changes to metaboric acid
3 3

HBO . On further heating, this yields boric oxide B O .
2 2 3

Δ Δ
H3 BO3 HBO2 B2 O3
370K redhot

Page : 331 , Block Name : Exercise

Q11.7 Describe the shapes of BF3 and BH4 – . Assign the hybridisation of boron in these

species.

Answer. (i) BF 3

As a result of its small size and high electronegativity, boron tends to form monomeric
covalent halides. These halides have a planar triangular geometry. This triangular shape is
formed by the overlap of three sp hybridised orbitals of boron with the sp orbitals of three
2

halogen atoms. Boron is sp hybridised in
2

BF .
3

(ii) BH −

4

Boron-hydride ion BH is formed by the sp3 hybridisation of boron orbitals. Therefore, it is
−

4

tetrahedral in structure.

Page : 331 , Block Name : Exercise

Q11.8 Write reactions to justify amphoteric nature of aluminium.

Answer. A substance is called amphoteric if it displays characteristics of both acids and bases.
Aluminium dissolves in both acids and bases, showing amphoteric behaviour.

(i) 2Al -> 2Al
3+ −
(s) + 6HCl(aq) + 6Cl + 3H2(g)
(aq) (αq)

(ii)2Al -> 2Na [Al(OH) ]
+ −

(s)
+ 2NaOH(aq) + 6H2 O(l) 4 (aq) + 3H2(g)

Page : 331 , Block Name : Exercise

Q11.9 What are electron de cient compounds ? Are BCl and 3

SiCl electron de cient species ? Explain.
4

Answer. In an electron-de cient compound, the octet of electrons is not complete, i.e., the
central metal atom has an incomplete octet. Therefore, it needs electrons to complete its
octet.

(i) BCl 3

BCl is an appropriate example of an electron-de cient compound. B has 3 valence electrons.
3

After forming three covalent bonds with chlorine, the number of electrons around it increases
to 6. However, it is still short of two electrons to complete its octet.

(ii) SiCl 4

The electronic con guration of silicon is ns np . This indicates that it has four valence
2 2

electrons. After it forms four covalent bonds with four chlorine atoms, its electron count
increases to eight. Thus, SiCl is not an electron-de cient compound.
4

Page : 332 , Block Name : Exercise

Q11.10 Write the resonance structures of CO and HCO .
2− −

3 3

Answer.

There are only two resonating structures for the bicarbonate ion.

Page : 332 , Block Name : Exercise

Q11.11 What is the state of hybridisation of carbon in
(a) CO
2−

3

(b) diamond
(c) graphite?

Answer. (a) CO
2−

3

C in CO is sp hybridised and is bonded to three oxygen atoms.
2− 2
3

(b) Diamond
Each carbon in diamond is sp hybridised and is bound to four other carbon atoms.
3

(c) Graphite
Each carbon atom in graphite is sp hybridised and is bound to three other carbon atoms.
2

Page : 332 , Block Name : Exercise

Q11.12 Explain the difference in properties of diamond and graphite on the basis of their
structures.

Answer.

Page : 332 , Block Name : Exercise

Q11.13 Rationalise the given statements and give chemical reactions :

Lead(II) chloride reacts with Cl to givePbCl .
2 4

Lead(IV) chloride is highly unstable towards heat.

Lead is known not to form an iodide,PbCl . 4

Answer. (a) Lead belongs to group 14 of the periodic table. The two oxidation states
displayed by this group is +2 and +4. On moving down the group, the +2 oxidation state
becomes more stable and the +4 oxidation state becomes less stable. This is because of
the inert pair effect. Hence, PbCl is much less stable than PbCl . However, the
4 2

formation of PbCl takes place when chlorine gas is bubbled through a saturated
4

solution of PbCl .
2

PbCl2(s) + Cl2(g) ⟶ PbCl4(l)

(b) On moving down group IV, the higher oxidation state becomes unstable because of
the inert pair effect. Pb(IV) is highly unstable and when heated, it reduces to Pb(II).
Δ
PbCl4(1) ⟶ PbCl2(s) + Cl2(g)

(c) Lead is known not to form PbCl . Pb (+4) is oxidising in nature and I– is reducing in
4

nature. A combination of Pb(IV) and iodide ion is not stable. Iodide ion is strongly
reducing in nature. Pb(IV) oxidises I– to I2 and itself gets reduced to Pb(II).
PbI4 ⟶ PbI2 + I2

Page : 332 , Block Name : Exercise

Q11.14 Suggest reasons why the B–F bond lengths in BF (130 pm) and BF (143 pm)
3
−

4

differ.

Answer. The B–F bond length in BF3 is shorter than the B–F bond length in BF , BF is
−

4 3

an electron-de cient species. With a vacant p-orbital on boron, the uorine and boron
atoms undergo pπ − pπ back-bonding to remove this de ciency. This imparts a double-
bond character to the B–F bond.

This double-bond character causes the bond length to shorten in BF3(130 pm). However,
when BF3 coordinates with the uoride ion, a change in hybridisation from
sp (in BF ) to sp (in BF ) occurs. Boron now forms 4σ bonds and the double-bond
2 3 −
3 4

character is lost. This accounts for a B–F bond length of 143 pm in BF ion.
−

4

Page : 332 , Block Name : Exercise

Q11.15 If B–Cl bond has a dipole moment, explain why BCl molecule has zero dipole
3

moment.

Answer. As a result of the difference in the electronegativities of B and Cl, the B–Cl bond
is polar in nature. However, the BCl molecule is non-polar. This is because BCl is
3 3

trigonal planar in shape. It is a symmetrical molecule. Hence, the respective dipole-
moments of the B–Cl bond cancel each other, thereby causing a zero-dipole moment.

Page : 332 , Block Name : Exercise

Q11.16 Aluminium tri uoride is insoluble in anhydrous HF but dissolves on addition of
NaF. Aluminium tri uoride precipitates out of the resulting solution when gaseous BF 3

is bubbled through. Give reasons.

Answer. Hydrogen uoride (HF) is a covalent compound and has a very strong
intermolecular hydrogen-bonding. Thus, it does not provide ions and aluminium

uoride (AlF) does not dissolve in it. Sodium uoride (NaF) is an ionic compound and
when it is added to the mixture, AlF dissolves. This is because of the availability of free
F–. The reaction involved in the process is:
AlF3 + 3NaF ⟶ Na3 [AlF6 ]

When boron tri uoride ( BF ) is added to the solution, aluminium uoride precipitates
3

out of the solution. This happens because the tendency of boron to form complexes is
much more than that of aluminium. Therefore, when
BF is added to the solution, B replaces Al from the complexes according to the
3

following reaction:
Na3 [AlF6 ] + 3BF3 ⟶ 3Na [BF4 ] + AlF3

Page : 332 , Block Name : Exercise

Q11.17 Suggest a reason as to why CO is poisonous.

Answer. Carbon monoxide is highly-poisonous because of its ability to form a complex
with haemoglobin. The CO–Hb complex is more stable than the O –Hb complex. The
2

former prevents Hb from binding with oxygen. Thus, a person dies because of
suffocation on not receiving oxygen. It is found that the CO–Hb complex is about 300
times more stable than the O –Hb complex.
2

Page : 332 , Block Name : Exercise

Q11.18 How is excessive content of co responsible for global warming ?
2

Answer. Carbon dioxide is a very essential gas for our survival. However, an increased
content of co in the atmosphere poses a serious threat. An increment in the combustion
2

of fossil fuels, decomposition of limestone, and a decrease in the number of trees has led
to greater levels of carbon dioxide. Carbon dioxide has the property of trapping the heat
provided by sun rays. Higher the level of carbon dioxide, higher is the amount of heat
trapped. This results in an increase in the atmospheric temperature, thereby causing
global warming.

Page : 332 , Block Name : Exercise

Q11.19 Explain structures of diborane and boric acid.

Answer. (a) Diborane
B H is an electron-de cient compound. B H has only 12 electrons – 6 e– from 6 H
2 6 2 6

atoms and 3 e– each from 2 B atoms. Thus, after combining with 3 H atoms, none of the
boron atoms has any electrons left. X-ray diffraction studies have shown the structure of
diborane as:

2 boron and 4 terminal hydrogen atoms (Ht) lie in one plane, while the other two
bridging hydrogen atoms (Hb) lie in a plane perpendicular to the plane of boron atoms.
Again, of the two bridging hydrogen atoms, one H atom lies above the plane and the
other lies below the plane. The terminal bonds are regular two-centre two-electron (2c –
2e–) bonds, while the two bridging (B–H–B) bonds are three-centre two-electron (3c –
2e–) bonds.

(b) Boric acid
Boric acid has a layered structure. Each planar BO unit is linked to one another
3

through H atoms. The H atoms form a covalent bond with a BO unit, while a hydrogen
3

bond is formed with another BO unit. In the given gure, the dotted lines represent
3

hydrogen bonds.

Page : 332 , Block Name : Exercise

Q11.20 What happens when
(a) Borax is heated strongly,
(b) Boric acid is added to water,
(c) Aluminium is treated with dilute NaOH,
(d) BF3 is reacted with ammonia ?

Answer. (a) When heated, borax undergoes various transitions. It rst loses water
molecules and swells. Then, it turns into a transparent liquid, solidifying to form a glass-
like material called borax bead.
Δ Δ

Na2 B4 O7 ⋅ 10H2 O ⟶ Na2 B4 O7 ⟶ 2NaBO2 + B2 O3

(b) When boric acid is added to water, it accepts electrons from –OH ion.
− +
B(OH)3 + 2HOH ⟶ [B(OH)4 ] + H3 O

(c) Al reacts with dilute NaOH to form sodium tetrahydroxoaluminate(III). Hydrogen gas
is liberated in the process.
-> 2Na [Al(OH) ]
+ −
2Al (s) + 2NaOH (oq) + 6H O2 (l) + 3H 4 (aq) 2(g)

(d) BF (a Lewis acid) reacts with NH (a Lewis base) to form an adduct. This results in a
3 3

complete octet around B in BF . 3

F3 B+ : NH3 ⟶ F3 B ←: NH3

Page : 332 , Block Name : Exercise

Q11.21 Explain the following reactions
(a) Silicon is heated with methyl chloride at high temperature in the presence of copper;
(b) Silicon dioxide is treated with hydrogen uoride;
(c) CO is heated with ZnO;
(d) Hydrated alumina is treated with aqueous NaOH solution.

Answer. (a) When silicon reacts with methyl chloride in the presence of copper (catalyst)
and at a temperature of about 537 K, a class of organosilicon polymers called methyl-
substituted chlorosilanes (MeSiCl , Me SiCl , Me SiCl, and Me Si) are formed.
3 2 2 3 4

(b) When silicon dioxide (SiO ) is heated with hydrogen uoride (HF), it forms silicon
2

tetra uoride (\mathrm{SiF}_{2}\). Usually, the Si–O bond is a strong bond and it resists
any attack by halogens and most acids, even at a high temperature. However, it is
attacked by HF.
SiO2 + 4HF ⟶ SiF4 + 2H2 O

The SiF formed in this reaction can further react with HF to form hydro uorosilicic
4

acid.
SiF4 + 2HF ⟶ H2 SiF6

(c) When CO reacts with ZnO, it reduces ZnO to Zn. CO acts as a reducing agent.
Δ

ZnO(s) + CO(g) ⟶ Zns + CO2(g)

(d) When hydrated alumina is added to sodium hydroxide, the former dissolves in the
latter because of the formation of sodium meta-aluminate.
Al2 O3 ⋅ 2H2 O + 2NaOH ⟶ 2NaAlO2 + 3H2 O

Page : 332 , Block Name : Exercise

Q11.22 Give reasons :
(i) Conc.HNO can be transported in aluminium container.
3

(ii) A mixture of dilute NaOH and aluminium pieces is used to open drain.
(iii) Graphite is used as lubricant.
(iv) Diamond is used as an abrasive.
(v) Aluminium alloys are used to make aircraft body.
(vi) Aluminium utensils should not be kept in water overnight.
(vii) Aluminium wire is used to make transmission cables.

Answer. (i) Concentrated .HNO can be stored and transported in aluminium containers
3

as it reacts with aluminium to form a thin protective oxide layer on the aluminium
surface. This oxide layer renders aluminium passive.
(ii) Sodium hydroxide and aluminium react to form sodium tetrahydroxoaluminate(III)
and hydrogen gas. The pressure of the produced hydrogen gas is used to open blocked
drains.
+
2Al + 2NaOH + 6H2 O ⟶ 2Na [Al(OH)4 ] + 3H2

(iii) Graphite has a layered structure and different layers of graphite are bonded to each
other by weak van der Waals’ forces. These layers can slide over each other. Graphite is
soft and slippery. Therefore, graphite can be used as a lubricant.
(iv) In diamond, carbon is sp hybridised. Each carbon atom is bonded to four other
3

carbon atoms with the help of strong covalent bonds. These covalent bonds are present
throughout the surface, giving it a very rigid 3-D structure. It is very dif cult to break
this extended covalent bonding and for this reason, diamond is the hardest substance
known. Thus, it is used as an abrasive and for cutting tools.
(v) Aluminium has a high tensile strength and is very light in weight. It can also be
alloyed with various metals such as Cu, Mn, Mg, Si, and Zn. It is very malleable and
ductile. Therefore, it is used in making aircraft bodies.

(vi) The oxygen present in water reacts with aluminium to form a thin layer of
aluminium oxide. This layer prevents aluminium from further reaction. However, when
water is kept in an aluminium vessel for long periods of time, some amount of
aluminium oxide may dissolve in water. As aluminium ions are harmful, water should
not be stored in aluminium vessels overnight.
(vii) Silver, copper, and aluminium are among the best conductors of electricity. Silver is
an expensive metal and silver wires are very expensive. Copper is quite expensive and is
also very heavy. Aluminium is a very ductile metal. Thus, aluminium is used in making
wires for electrical conduction.

Page : 332 , Block Name : Exercise

Q11.23 Explain why is there a phenomenal decrease in ionization enthalpy from carbon
to silicon ?

Answer. Ionisation enthalpy of carbon (the rst element of group 14) is very high (1086
kJ/mol). This is expected owing to its small size. However, on moving down the group to
silicon, there is a sharp decrease in the enthalpy (786 kJ). This is because of an
appreciable increase in the atomic sizes of elements on moving down the group.

Page : 332 , Block Name : Exercise

Q11.24 How would you explain the lower atomic radius of Ga as compared to Al ?

Answer.

Although Ga has one shell more than Al, its size is lesser than Al. This is because of the
poor shielding effect of the 3d-electrons. The shielding effect of d-electrons is very poor
and the effective nuclear charge experienced by the valence electrons in gallium is much
more than it is in the case of Al.

Page : 332 , Block Name : Exercise

Q11.25 What are allotropes? Sketch the structure of two allotropes of carbon namely
diamond and graphite. What is the impact of structure on physical properties of two
allotropes?

Answer. Allotropy is the existence of an element in more than one form, having the same

chemical properties but different physical properties. The various forms of an element
are called allotropes.

Diamond:

The rigid 3-D structure of diamond makes it a very hard substance. In fact, diamond is
one of the hardest naturally-occurring substances. It is used as an abrasive and for
cutting tools.
Graphite:

It has sp hybridised carbon, arranged in the form of layers. These layers are held
2

together by weak van der Walls’ forces. These layers can slide over each other, making
graphite soft and slippery. Therefore, it is used as a lubricant.

Page : 332 , Block Name : Exercise

Q11.26 (a) Classify following oxides as neutral, acidic, basic or amphoteric:
cO, B2 O3 , SiO2 , CO2 , Al2 O3 , PbO2 , TI2 O3

(b) Write suitable chemical equations to show their nature.

Answer. (1) CO = Neural

(2) B O = Acidic
2 3

Being acidic, it reacts with bases to form salts. It reacts with NaOH to form sodium
metaborate.
B2 O3 + 2NaOH ⟶ 2NaBO2 + H2 O

(3) SiO = Acidic
2

Being acidic, it reacts with bases to form salts. It reacts with NaOH to form sodium
silicate.
SiO2 + 2NaOH ⟶ 2Na2 SiO3 + H2 O

(4) CO = Acidic
2

Being acidic, it reacts with bases to form salts. It reacts with NaOH to form sodium
carbonate.
CO2 + 2NaOH ⟶ Na2 CO3 + H2 O

(5) Al O = Amphoteric
2 3

Amphoteric substances react with both acids and bases. Al O reacts with both NaOH
2 3

and H SO .
2 4

Al2 O3 + 2NaOH ⟶ NaAlO2

Al2 O3 + 3H2 SO4 ⟶ Al2 (SO4 ) + 3H2 O
3

(6) PbO = Amphoteric
2

Amphoteric substances react with both acids and bases. PbO reacts with both NaOH
2

and H SO .
2 4

PbO2 + 2NaOH ⟶ Na2 PbO3 + H2 O

2PbO2 + 2H2 SO4 ⟶ 2PbSO4 + 2H2 O + O2

(7) TI O = Basic
2 3

Being basic, it reacts with acids to form salts. It reacts with HCl to form thallium
chloride.
TI2 O3 + 6HCl ⟶ 2TICl3 + 3H2 O

Page : 333 , Block Name : Exercise

Q11.27 In some of the reactions thallium resembles aluminium, whereas in others it
resembles with group I metals. Support this statement by giving some evidences.

Answer. Thallium belongs to group 13 of the periodic table. The most common oxidation
state for this group is +3. However, heavier members of this group also display the +1
oxidation state. This happens because of the inert pair effect. Aluminium displays the +3
oxidation state and alkali metals display the +1 oxidation state. Thallium displays both
the oxidation states. Therefore, it resembles both aluminium and alkali metals.

Thallium, like aluminium, forms compounds such as TICl and Tl O . It resembles
3 2 3

alkali metals in compounds Tl O and TlCl.
2

Page : 333 , Block Name : Exercise

Q11.28 When metal X is treated with sodium hydroxide, a white precipitate
(A) is obtained, which is soluble in excess of NaOH to give soluble complex
(B) Compound (A) is soluble in dilute HCl to form compound
(C) The compound (A) when heated strongly gives
(D) which is used to extract metal. Identify (X), (A), (B), (C) and (D). Write suitable
equations to support their identities.

Answer. The given metal X gives a white precipitate with sodium hydroxide and the
precipitate dissolves in excess of sodium hydroxide. Hence, X must be aluminium.
The white precipitate (compound A) obtained is aluminium hydroxide. The compound B
formed when an excess of the base is added is sodium tetrahydroxoaluminate(III).
+
2Al + 3NaOH ⟶ Al(OH)3 ↓ +3Na
+ −
Al(OH)3 + NaOH ⟶ Na [Al(OH)4 ]

Now, when dilute hydrochloric acid is added to aluminium hydroxide, aluminium
chloride (compound C) is obtained.
Al(OH)3 + 3HCl ⟶ AlCl3 + 3H2 O

Also, when compound A is heated strongly, it gives compound D. This compound is used
to extract metal X. Aluminium metal is extracted from alumina. Hence, compound D
must be alumina.
Δ

2Al(OH)3 ⟶ Al2 O3 + 3H2 O

Page : 333 , Block Name : Exercise

Q11.29 What do you understand by
(a) inert pair effect
(b) allotropy and
(c) catenation?

Answer. (a) Inert pair effect
As one moves down the group, the tendency of s-block electrons to participate in
chemical bonding decreases. This effect is known as inert pair effect. In case of group 13
elements, the electronic con guration is ns np and their group valency is +3. However,
2 1

on moving down the group, the +1 oxidation state becomes more stable. This happens
because of the poor shielding of the ns electrons by the d– and f– electrons. As a result
2

of the poor shielding, the ns electrons are held tightly by the nucleus and so, they
2

cannot participate in chemical bonding.
(b) Allotropy
Allotropy is the existence of an element in more than one form, having the same
chemical properties but different physical properties. The various forms of an element

are called allotropes. For example, carbon exists in three allotropic forms: diamond,
graphite, and fullerenes.
(c) Catenation
The atoms of some elements (such as carbon) can link with one another through strong
covalent bonds to form long chains or branches. This property is known as catenation. It
is most common in carbon and quite signi cant in Si and S.

Page : 333 , Block Name : Exercise

Q11.30 A certain salt X, gives the following results.
(i) Its aqueous solution is alkaline to litmus.
(ii) It swells up to a glassy material Y on strong heating.
(iii) When conc. H2SO4 is added to a hot solution of X,white crystal of an acid Z
separates out. Write equations for all the above reactions and identify X, Y and Z.

Answer. The given salt is alkaline to litmus. Therefore, X is a salt of a strong base and a
weak acid. Also, when X is strongly heated, it swells to form substance Y. Therefore, X
must be borax.
When borax is heated, it loses water and swells to form sodium metaborate. When
heating is continued, it solidi es to form a glassy material Y. Hence, Y must be a mixture
of sodium metaborate and boric anhydride.

When concentrated acid is added to borax, white crystals of orthoboric acid (Z) are
formed.

Δ

Na2 B4 O7 ⋅ 10H2 O + H2 SO4 ⟶ Na2 SO4 + 4H3 BO3 + 5H2 O

Page : 333 , Block Name : Exercise

Q11.31 Write balanced equations for:
(i) BF3 + LiH →

(ii) B2 H6 + H2 O →

(iii) NaH + B2 H6 →

(iv) H3 BO3

Δ

⟶

(v) Al + NaOH →

(vi) B2 H6 + NH3 →

2BF3 + 6LiH ⟶ B 2 H6 + 6liF
Answer. (i)
Boron trifluoride Lithium hydride Diborane Lithium fluoride

B2 H6 + 6H2 O ⟶ 2H3 BO3 + 6H2
(ii)
Diborane Water Orthoboric acid Hydrogen

B 2 H6 + 2NaH ⟶ NaBH4
(iii)
Diborane Sodium hydride Sodium borohydride

(iv)

(v) 2Al + 2NaOH + 6H O ⟶ + −
2 2Na [Al(OH)4 ] + 3H2
(ea)

′
(vi) 3B H
−
2 6 + 6NH3 ⟶ 3[BH2 (NH3 ) ] [BH4 ] ⟶ 2B3 N3 H6 + 12H2
2

Page : 333 , Block Name : Exercise

Q11.32. Give one method for industrial preparation and one for laboratory preparation
of CO and CO2 each.

Answer. Carbon dioxide
In the laboratory, co can be prepared by the action of dilute hydrochloric acid on
2

calcium carbonate. The reaction involved is as follows:
CaCO3 + 2HCl(aq) ⟶ CaCl2(aq) + CO2(g) + H2 O(i)

co2 is commercially prepared by heating limestone. The reaction involved is as follows:
Δ

CaCO3 ⟶ CaO + CO2 ↑

Carbon monoxide
In the laboratory, CO is prepared by the dehydration of formic acid with conc. H SO , at
2 4

373 K. The reaction involved is as follows:
373k
HCOOH H2 O + CO ↑
conc⋅H3 SO4

CO is commercially prepared by passing steam over hot coke. The reaction involved is as
follows:
C(s) + H O
2 -> CO + H
(g) (g) water gas
2(g)


Page : 333 , Block Name : Exercise

Q11.33 An aqueous solution of borax is
(a) neutral
(b) amphoteric
(c) basic
(d) acidic

Answer. (c) Borax is a salt of a strong base (NaOH) and a weak acid (H BO ). It is,
3 3

therefore, basic in nature.

Page : 333 , Block Name : Exercise

Q11.34 Boric acid is polymeric due to
(a) its acidic nature
(b) the presence of hydrogen bonds
(c) its monobasic nature
(d) its geometry

Answer. (b) Boric acid is polymeric because of the presence of hydrogen bonds. In the
given gure, the dotted lines represent hydrogen bonds.

Page : 333 , Block Name : Exercise

Q11.35 The type of hybridisation of boron in diborane is

(a) sp
(b) sp2
(c) sp3
(d) dsp2

Answer. Boron in diborane is sp hybridised.
3

Page : 333 , Block Name : Exercise

Q11.36 Thermodynamically the most stable form of carbon is
(a) diamond
(b) graphite
(c) fullerenes
(d) coal

Answer. (b) Graphite is thermodynamically the most stable form of carbon.

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Q11.37 Elements of group 14
(a) exhibit oxidation state of +4 only
(b) exhibit oxidation state of +2 and +4
(c) form M and M ions
2− 4+

(d) form M and M ions.
2+ 4+

Answer. (b)The elements of group 14 have 4 valence electrons. Therefore, the oxidation
state of the group is +4. However, as a result of the inert pair effect, the lower oxidation
state becomes more and more stable and the higher oxidation state becomes less stable.
Therefore, this group exhibits +4 and +2 oxidation states.

Page : 333 , Block Name : Exercise

Q11.38 If the starting material for the manufacture of silicones is RSiCl3 , write the
structure of the product formed.

Answer. RSiCl 3 + 3H2 O ⟶ RSi(OH)3 + 3HCl

Page : 333 , Block Name : Exercise