7.1 Why are
pentahalides of P, As, Sb and Bi more covalent than their trihalides?
Ans: In pentahalides, the oxidation
state is +5 which is more than the oxidation state of trihalides (+3). According
to Fazan’s rule, the cations with higher positive charge have larger polarizing
power and can polarise the halide ion to a greater extent. Since larger
polarization, larger is the covalent character, therefore, pentahalides are
more covalent than trihalides.
7.2 Why is BiH3 the strongest reducing agent
amongst all the hydrides of Group 15 elements ?
Ans: Among the hydrides of group
15, BiH3 is the least stable molecule because Bi has largest size in
the group and has least tendency to form covalent bond with small hydrogen
atom. Therefore, it can readily lost H atom has strongest tendency to act as
reducing agent.
7.3 Why is N2 less reactive at room
temperature?
Ans: N2 has triple bond
(N≡N) between two nitrogen atoms
and it is non polar in character. Due to the presence of triple bond, it has
very high bond dissociation energy (941.4 kJ/mol) and therefore, it does not
reacts with other elements under normal conditions and is very unreactive.
7.4 Mention the
conditions required to maximise the yield of ammonia.
Ans: The conditions to maximise the
yield of ammonia are:
i.
Low temperature of about 700
K
ii.
High pressure of 200 x 105
Pa (about 200 atm)
iii.
Presence of catalyst such as
iron oxide with small amount of K2O and Al2O3.
7.5 How does ammonia
react with a solution of Cu2+
?
Ans: Ammonia reacts with a solution
of Cu2+ to form deep blue coloured complex, tetraamminecopper (ii)
ion:
Cu2+ (aq) + 4 NH3 (aq) ↔ [Cu(NH3)4]2+ (aq)
7.6 What is the
covalence of nitrogen in N2O5 ?
Ans: N2O5 has
the structure
The covalence of N in N2O5
is four because it has four shared pair of electrons.
7.7 (a) Bond angle in
PH4+ is higher than that in PH3. Why?
Ans: Both PH4+
and PH3 involve sp3 hybridisation of P atom. In PH4+
all the four orbitals are bonded, whereas in PH3 there is a lone
pair of electrons on P. In PH4+, the HPH bond angle is
tetrahedral angle of 109.5O. But in PH3, lone pair-bond
pair repulsion is more than bond pair-bond-pair repulsion so that bond angles
become less than normal tetrahedral angle of 109.5O. The bond angle
in PH3 has been found to be about 93.6O.
(b)
What is formed when PH3 reacts
with an acid?
Ans:
Phosphine is weakly basic and like ammonia, gives phosphonium compounds with
acids e.g., PH3 + HBr ® PH4Br
7.8 What happens when
white phosphorus is heated with concentrated NaOH solution in an inert
atmosphere of CO2 ?
Ans: When white phosphorus is
heated with concentrated NaOH solution in an inert atmosphere of CO2 phosphine is formed.
7.9 What happens when
PCl5 is heated?
Ans: On heating, PCl5
decomposes into PCl3 and Cl2
7.10 Write a balanced
equation for the reaction of PCl5 with water.
Ans:
7.11 What is the
basicity of H3PO4?
Ans: H3PO4
contains three P-OH bonds and therefore, its basicity is three.
7.12 What happens when
H3PO3 is heated?
Ans: On heating, H3PO3
disproportionate to give orthophosphoric acid and phosphine.
7.13 List the important sources of sulphur.
Ans: Sulphur occurs in the combined
form as sulphide ores and sulphate ores. The common sulphide ores in which
sulphur occurs are galena (PbS), zinc blend (ZnS), copper pyrites (CuFeS2)
and sulphate ores are gypsum (CaSO4.2H2O), Epsom salt
(MgSO4.7H2O) and barite (BaSO4). Traces of
sulphur occurs as hydrogen sulphide in volcano. Organic material such as eggs,
proteins, onion, garlic, mustard, hair and wool also contain sulphur.
7.14 Write the order
of thermal stability of the hydrides of Group 16 elements.
Ans: The thermal stability of the
hydrides of group 16 elements decreases from H2O to H2Te
as:
H2O > H2S > H2Se > H2Te.
7.15 Why is H2O a liquid and H2S a gas ?
Ans: Due to high electronegativity
and small size of oxygen, there are strong intermolecular hydrogen bonding
between water molecules. As a result the water molecules are associated and in
liquid at room temperature. Whereas in H2S, molecules do not form
hydrogen bonding therefore, these molecules are gas at room temperature.
7.16 Which of the
following does not react with oxygen directly?
Zn, Ti, Pt, Fe
Ans: Pt, because Pt is a noble
metal.
7.17 Complete the
following reactions:
(i) C2H4 + O2 ®
(ii) 4Al + 3 O2 ®
Ans: (i)
C2H4 + 3 O2 ® 2 CO2
+ 2 H2O
(ii) 4Al + 3 O2 ® 2 Al2O3.
7.18 Why does O3 act as a powerful oxidising
agent?
Ans: Ozone acts as a powerful
oxidising agent because it decomposes readily to give atomic oxygen as:
O3 ® O2 + O
Therefore, ozone can oxidise a number of non-metals and other
compounds. For example:
PbS (s) + 4 O3 (g) ® PbSO4 (S) + 4 O2 (g)
2 I- (aq) + H2O (l) + O3 (g) ® 2 OH- (aq) + I2
(s) + O2 (g)
7.19 How is O3 estimated quantitatively?
Ans: When ozone reacts with an excess of
potassium iodide solution buffered with a borate buffer (pH 9.2), iodine is
liberated which can be titrated against a standard solution of sodium
thiosulphate.
PbS(s)
+ 4O3(g) ® PbSO4(s)
+ 4O2(g)
2I–(aq)
+ H2O(l) + O3(g)
® 2OH–(aq)
+ I2(s) + O2(g)
7.20 What happens when
sulphur dioxide is passed through an aqueous solution of Fe(III) salt?
Ans: Sulphur dioxide
converts iron(III) ions to iron(II) ions.
2
Fe3+ + SO2 + 2 H2O ®
2 Fe2+ + SO42- + 4 H+.
7.21 Comment on the nature
of two S–O bonds formed in SO2 molecule. Are the two S–O bonds in this molecule equal ?
Ans:
Sulphur in SO2 involves sp2 hybridisation. Two of these
sp2 hybrid orbitals form σ bonds with oxygen atoms. The remaining two
unhybridised orbitals of S form pπ-pπ and pπ-dπ double bonds with oxygen atoms.
Though
the two S-O bonds appear different because of pπ-pπ and pπ-dπ double bonds. However, both the S-O bonds
are equal (bond length 143 pm) because of resonance between two structures.
7.22 How is the
presence of SO2 detected ?
Ans: SO2 can be detected
by the following tests:
a) It
decolourises acidified potassium permanganate solution.
b) It
turns acidified potassium dichromate solution green.
c) It
turns blue litmus red.
7.23 Mention three
areas in which H2SO4 plays an important role.
Ans:
i.
Manufacture of fertilisers (e.g., ammonium
sulphate, superphosphate).
ii.
Metallurgical applications (e.g. cleansing
metals before enameling, electroplating and galvanising).
iii.
Petroleum refining
7.24 Write the
conditions to maximise the yield of H2SO4 by Contact process.
Ans:
i.
Low temperature (optimum
temperature 720 K)
ii.
High pressure (optimum pressure
2 bar)
iii.
Presence of catalyst (V2O5)
7.25 Why is Ka2
<< Ka1 for H2SO4 in water ?
Ans: H2SO4 is a strong acid in water because of its first
ionisation to H3O+ and HSO4-.
H2SO4 + H2O ↔ H3O+ + HSO4-.
It
is almost completely ionised so that its Ka1 is very large (Ka1
> 10). However, dissociation of HSO4- in water to H3O+ and SO42-
is very small (Ka2 = 1.2 x 10-2). That is why its Ka2 << Ka1.
HSO4-+ H2O ↔ H3O+ + SO42-.
7.26 Considering the
parameters such as bond dissociation enthalpy, electron gain enthalpy and
hydration enthalpy, compare the oxidising power of F2 and Cl2.
Ans: The oxidising powers of both the members of halogen family
are expressed in terms of their electron accepting tendency and can be compared
as their standard reduction potential values.
F2 + 2e– → 2F–;
E° = 2.87 V,
Cl2 + 2e– → 2Cl– ;
E° = 1.36 V
Since the E° of fluorine is more than that of
chlorine, it is a stronger oxidising agent.
Explanation : Three factors contribute towards
the oxidation potentials of both the halogens. These are :
(i) Bond dissociation enthalpy: Bond dissociation enthalpy of F2 (158
kJ mol-1) is less compared to that of Cl2 (242·6 kJ
mol-1).
(ii) Electron gain enthalpy: The negative electron gain enthalpy of F (-
332·6 kJ mol-1) is slightly less than of Cl (-348·5 kJ mol-1).
(iii) Hydration enthalpy: The hydration enthalpy of F- ion (515
kJ mol-1) is much higher than that of Cl- ion (381 kJ mol-1)
due to its smaller size.
From the available data, we may conclude that
lesser bond dissociation enthalpy and higher hydration enthalpy compensate
lower negative electron gain enthalpy of fluorine as compared to chlorine.
Consequently, F2 is a more powerful oxidising agent than Cl2.
7.27 Give two examples
to show the anomalous behaviour of fluorine.
Ans:
1. Fluorine is the most electronegative element and it does not show
any positive oxidation state except in HOF.
2.
Ionisation enthalpy,
electro-negativity and electrode potential are higher for fluorine than the
expected trends of other halogen.
7.28 Sea is the greatest source of some halogens. Comment.
Ans: Sea water contains
chlorides, bromides and iodides of sodium, potassium, magnesium and calcium but
sodium chloride being the maximum makes sea water saline. Various sea weeds
contain upto 0.5% iodine. The deposits of dried up sea contains sodium
chloride, carnallite (KCl, MgCl2.6H2O).
7.29 Give the reason for bleaching action of Cl2.
Ans: Bleaching action of chlorine is due to its oxidation. In the
presence of moisture, chlorine gives nascent oxygen. This nascent oxygen
bleaches colouring substances to colourless substances.
Cl2
+ H2O → 2 HCl + O
Colouring
substance + O → Colourless substance.
7.30 Name two poisonous gases which can be prepared from chlorine gas.
Ans:
1. Phosgene (COCl2)
2. Tear gas (CCl3.NO2)
7.31 Why is ICl more reactive than I2?
Ans: Interhalogen
compounds are more reactive than halogens (except fluorine). This is because
X–X¢ bond in interhalogens is weaker than X–X
bond in halogens except F–F bond. I-Cl is weaker than Cl-Cl and I-I and
therefore, ICl is more reactive than I2.
7.32 Why is helium
used in diving apparatus?
Ans: Helium along with oxygen is used in the diving apparatus by
the sea divers. Since it is very less soluble in blood, it reduces
decompression and causes less discomfort to the diver in breathing. A mixture
of helium and oxygen does not cause pain due to very low solubility of helium
in blood as compared to nitrogen.
7.33 Balance the
following equation: XeF6 + H2O ® XeO2F2 + HF
Ans: XeF6 + 2 H2O ® XeO2F2
+ 4 HF
7.34 Why has it been
difficult to study the chemistry of radon?
Ans: Radon is a radioactive element
and has very short half-life period. Therefore, it is difficult to study the
chemistry of radon.
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