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1. Discuss the general characteristics of Group 15 elements with
reference to their electronic configuration, oxidation state, atomic size,
ionisation enthalpy and electronegativity.
•
Solution:
(i) Electronic configuration
The valence shell electronic configuration of these elements is ns2 np3.The s orbital in these elements
is completely filled and p orbitals are half filled, making their electronic configuration extra stable.
(ii) Oxidation state
The common oxidation state of the elements are -3, +3 and +5. The tendencies to exhibit -3
oxidation state decreases down the group due to increase in size and metallic group. In the last
member of the group, bismuth hardly forms any compound in -3 oxidation state. The stability of +5
oxidation state decreases down the group. The stability of +5 oxidation state decreases and that of
+3 state increases (due to inert pair effect) down the group. Nitrogen exhibits +1, +2, +4 oxidation
states also when it reacts with oxygen. Phosphorus also shows +1 and +4 oxidation states in some
oxo acids.
(iii) Atomic size:
Covalent and ionic (in a particular state) radii increase in size down the group. There is a
considerable increase in size down the group. There is a considerable increase in covalent radius
from N to P. However, from As to Bi only a small increase in covalent radius is observed. This is due
to the presence of completely filled d and f orbitals in heavier members.
(iv) Ionization enthalpy:
Ionisation enthalpy decreases down the group due to gradual increase in atomic size. Because of the
extra stable half filled p orbitals electronic configuration and smaller size, the ionization enthalpy of
the group 15 elements is much greater than that of group 14 elements in the corresponding periods.
The order of successive ionization ehthalpies are expected as
H1 <
H2 <
H 3.
(v) Electronegativity:
in
The electronegativity value, in general, decreases down the group with increasing atomic size.
However, amongst the heavier elements, the different is not that much pronounced.
2. Why does the reactivity of nitrogen differ from phosphorus?
•
Solution:
The reactivity of nitrogen is different from phosphorus because of the following reasons
(i) Nitrogen has a small size, high electronegativity, and high ionization enthalpy as compared to
phosphorus.
1
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(ii) Nitrogen does not contain vacant d – orbitals in its valence shell whereas phosphorus contains
vacant d – orbitals in its valence shell.
(iii) Nitrogen has ability of form (N≡N) triple bond as a result of which its bond enthalpy (941.4 kJ
mol-1) is very high making it less reactive.
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3. Discuss the trends in chemical reactivity of group 15 elements.
•
Solution:
The elements of Group 15 differ from one another appreciably in their chemical reactivity. Nitrogen
has a very high dissociation energy (941.4 kJ mol-1) and is practically inert, which is why it has
accumulated in large amounts in the atmosphere. Phosphorus, in one of its allotrophic forms i.e.
white phosphorus is extremely reactive. The strained structure of P4 is responsible for high chemical
activity. It catches fire when exposed to air forming P4O10. The other allotrophe, red phosphorus is
stable in air at room temperature, though it reacts on heating. The heavier elements, As, Sb and Bi
are less reactive. Arsenic is stable in dry air. When heated in air it sublimes at 615˚C forming
As4O6.Antimony is less reactive and stable towards water and air. On heating in air it forms Sb4O6,
Sb4O8 or Sb4O10. Bismuth is also known to form Bi2O3 on heating.
4. Why does NH3 form hydrogen bond but PH3 does not?
•
Solution:
In ammonia, the nitrogen atom forms hydrogen bond because of the following reasons:
(i) small size of nitrogen
(ii) High electronegativity of nitrogen (3.0)
Due to more difference of electronegativity between N and H atom then N-H bond is polar forming H
bond. On the contrary, Phosphoru due to its larger size and lesser electronegativity (2.1) does not
form hydrogen bond because P-H bond is almost purely covalent and no hydrogen bond is formed.
5. How is nitrogen prepared in the laboratory? Write the chemical
equationsof the reactions involved.
•
Solution:
Dinitrogen is prepared in the laboratory by treating an aqueous solution of ammonium chloride with
sodium nitrite.
NH4Cl (aq) + NaNO2 (aq)
N2 (g) + 2H2O (l) + NaCl (aq).
The impurities of NO and HNO3 obtained in small amounts is removed by passing the gas through
aqueous sulphuric acid containing potassium dichromate.
•
in
6. How is ammonia manufactured industrially
Solution:
Haber Process for the Production of Ammonia In 1909 Fritz Haber established the conditions under
which nitrogen, N2(g), and hydrogen, H2(g), would combine using
• medium temperature (~500oC)
• very high pressure (~250 atmospheres, ~351kPa)
• a catalyst (a porous iron catalyst prepared by reducing magnetite, Fe3O4).
Osmium is a much better catalyst for the reaction but is very expensive. This process produces an
ammonia, NH3(g), yield of approximately 10-20%. The Haber synthesis was developed into an
industrial process by Carl Bosch. The reaction between nitrogen gas and hydrogen gas to produce
ammonia gas is exothermic, releasing 92.4kJ/mol of energy at 298K (25oC).
2
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N2(g)
nitrogen
•
•
+
3H2(g)
hydrogen
heat, pressure, catalyst
2NH3(g)
ammonia
∆H = -92.4 kJ mol-1
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By Le Chetalier's Principle:
•€increasing the pressure causes the equilibrium position to move to the right resulting in a higher
yeild of ammonia since there are more gas molecules on the left hand side of the equation (4 in
total) than there are on the right hand side of the equation (2). Increasing the pressure means the
system adjusts to reduce the effect of the change, that is, to reduce the pressure by having fewer
gas molecules.
• decreasing the temperature causes the equilibrium position to move to the right resulting in a
higher yield of ammonia since the reaction is exothermic (releases heat). Reducing the temperature
means the system will adjust to minimise the effect of the change, that is, it will produce more heat
since energy is a product of the reaction, and will therefore produce more ammonia gas as well
However, the rate of the reaction at lower temperatures is extremely slow, so a higher temperature
must be used to speed up the reaction which results in a lower yield of ammonia.
•
7. Illustrate how copper metal can give different products on
reaction with HNO3.
•
Solution:
The reaction of copper with nitric acid produces different t products depending upon the
concentration of nitric acid.
(i) Reaction with cold dilute HNO3 forms nitric oxide
2HNO3
[Cu + [O]
H2O + 2NO + 3(O)
CuO] x 3
[CuO + 2HNO3
Cu(NO3)2 + H2O] x 3
________________________________________
3Cu +8HNO3
3Cu(NO3)2 + 2NO + 4H2O
(ii) Reaction with hot conc. HNO3 forms nitrogen dioxide
2HNO3
Cu + [O]
H2O + 2NO2 + (O)
CuO
CuO + 2HNO3
Cu(NO3)2 + H2O
________________________________________
Cu(NO3)2 + 2NO2 + 2H2O
in
Cu +4HNO3
8. Give the resonating structures of NO2 and N2O5.
•
Solution:
The resonating structures of NO2 and N2O5 are as follows
NO2
3
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N2O5:
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9. The HNH angle value is higher than HPH, HAsH and HSbH angles.
Why?
[Hint: Can be explained on the basis of sp3 hybridisation in NH3 and only
s–p bonding between hydrogen and other elements of the group].
•
Solution:
The bond angle of the hydrides of group 15 are given below:
Hydride
Bond angle
•
NH3
106.5˚
PH3
93.5˚
AsH3
91.5˚
SbH3
91.3˚
The HNH angle in ammonia is higher as compared to other hydrides of the group. It is because
nitrogen in NH3 is sp3 hybridised. However due to lone pair of electrons the bond angle contracts
from 109˚28’ to 106.5˚. The decreased bond angle in other hydrides is because of the fact that the
sp3 hybridisation becomes less and le ss distinct with increasing size of the central atom i.e. pure porbitals are utilized in M-H bonding or in simple words the s-orbital of H atom overlaps with orbital
having almost pure p-character.
10. Why does R3P = O exist but R3N = O does not (R = alkyl group)?
•
Solution:
Compounds like R3N = O does not exist because of the absence of d-orbitals in the valence shell of
nitrogen atom. As a result of this nitrogen cannot expand its covalency beyond 4 (lack of formation
of d
-p
). On the contrary R3P = O .exists because phosphorus has vacant d-orbitals in its
valence shell and can expand it covalency beyond 4 by forming d
-p
.
•
in
11. Explain why NH3 is basic while BiH3 is only feebly basic.
Solution:
The hydrides of group 15 are basic in character because of the presence of lone pairs of electrons on
the nitrogen atom (Lewis bases). Ammonia is strongly basic in character because it can easily donate
its electron pair due to small size of nitrogen atom. As result of which electron density of lone pair is
concentrated over a small region. BiH3 on the contrary is feebly basic because of its larger size. Due
to increase in size the electron density gets diffused over a larger region and hence the ability to
donate the electron pair decreases.
4
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12. Nitrogen exists as diatomic molecule and phosphorus as P4. Why?
•
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Solution:
Nitrogen exist6s as diatomic molecule (N") because due to small size of nitrogen atom and absence
of vacant d-orbitals in its valence shell it has strong ability to form multiple bonds (N≡N). The other
members of group 15 i.e. phosphorus, arsenic and antimony all exists as discrete tetra atomic
tetrahedral molecules viz, P4, As4 and Sb4 because these are not capable of forming multiple bonds
to their bigger size and repulsions between non bonded electrons of the inner core. Now, since the
angles P-P-P is 60˚, the p
-p
bonding is therefore not possible.
13. Write main differences between the properties of white phosphorus
and red phosphorus.
•
Solution:
Properties
1.Colour
2.State and melting point
3. Density
4. Ignition temperature
5. Stability
6. Chemical reactivity
7. Toxic nature
8. Structure
Structure of white and red phosphorus are shown below:
Red Phosphorus
Dark red and no effect of
exposure to air.
Brittle solids, high melting point
2.16 g/cm3
543 K
More stable at ordinary
temperature
Less reactive
Non toxic
P4 molecules are linked by
covalent bonds forming giant
(P4)n molecule.
in
•
White Phosphorus
White but turns yellow on
exposure to air.
Waxy solids, low melting point
1.84 g/cm3
307 K
less stable at ordinary
temperature
very reactive
highly toxic
The four P- atoms lie at the
corners of a regular
tetrahedron. Each is
phosphorus is bonded to other
three P-atoms, by covalent
bonds, so that each P-atom
completes its valence shell.
5
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•
•
14. Why does nitrogen show catenation properties less than
phosphorus?
•
Solution:
The tendency of catenation of nitrogen is less than phosphorus. It is because the single N-N bond is
weaker than the single P-P bond due to high inter electronic repulsion of the non bonding electrons in
the case of nitrogen atom (smaller size).
15. Give the disproportionation reaction of H3PO3.
•
Solution:
The phosphorous acid (H3PO3) when heated at 473 K disproportionate to form Ortho phosporic acid
and Phosphine
4H3PO3
3H3PO4 + PH3.
in
16. Can PCl3 act as an oxidising as well as a reducing agent? Justify.
•
Solution:
PCl3 can act both as an oxidizing agent as well as reducing agent.
(a) As a reducing agent:
The following reactions support the reducing character
PCl3 + SO2Cl2
PCl5 + SO2
PCl3 + SO3
POCl3 + SO2
(b) As an oxidizing agent:
It oxidizes metals to their respective chlorides.
6
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12Ag + 4PCl3
6Na + PCl3
12AgCl + P
3NaCl + Na3P
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17. Justify the placement of O, S, Se, Te and Po in the same group of the
periodic table in terms of electronic configuration, oxidation state and
hydride formation.
•
Solution:
Electronic configuration:-All these elements have the same valence shell electronic configuration of
ns2 np4.and hence must be placed in the same group.
Oxidation states:-Oxygen predominantly shows -2 oxidation state and sulphur exhibits -2 to some
extent. Except oxygen all the elements exhibit maximum oxidation state of +6. They also exhibit +2
and+4 states also.
Hydrides:- All elements of the group form oxides of formula E2O
On the basis of above facts it is justified to place them in the same group.
18. Why is dioxygen a gas but sulphur a solid?
•
Solution:
Dioxygen is a gas because due to small size of oxygen atom and absence of d-orbitals in its valence
shell. It has strong tendency to form p
-p
bonds multiple bonds (O=O) so as to complete its
octet and hence can exists as discrete O2 molecule. On the contrary sulphur is a solid and exists as
staggered 8 atom ring. It is because the tendency of S=S bond formation is missing in sulphur atom
due to its larger size and low S-S double energy. As a result sulphur atoms complete their octet
through formation of S-S bonds in S8 molecule. This leads to increase the forces of attraction and
hence its physical state is solid.
19. Knowing the electron gain enthalpy values for O
O– and O
–1
O as –141 and 702 kJ mol respectively, how can you account for the
formation of a large number of oxides having O2– species and not O–?(Hint:
Consider lattice energy factor in the formation of compounds).
2–
Solution:
in
•
The second electron gain enthalpy (O- + eO2-) is positive (702 kJ mol-1) at a large number
of oxides have O2- species. This is attributable to high lattice energy released during formation of
such oxides. As a result off this the lattice energy realeased by the process compensates for the
second electron gain enthalpy.
2M+ + O2M + O2-
20. Which aerosols deplete ozone?
7
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•
Solution:
The aerosols responsible largely for the depletion of ozone layer are
(i) nitric oxide emitted from the exhaust systems of supersonic jet aeroplanes.
NO (g) + O3
NO2 (g) + O2 (g)
(ii) Freons (Chlorofluoro hydrocarbons) which are used in aerosol sprays and refrigerants.
CCl2F2
CF2Cl + Cl
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Cl + O3
ClO + O2
21. Describe the manufacture of H2SO4 by contact process?
•
Solution:
The Contact Process
• makes sulphur dioxide;
• convers the sulphur dioxide into sulphur trioxide (the reversible reaction at the heart of the
process);
• converts the sulphur trioxide into concentrated sulphuric acid.
Making the sulphur dioxide
This can either be made by burning sulphur in an excess of air:
. . . or by heating sulphide ores like pyrite in an excess of air:
In either case, an excess of air is used so that the sulphur dioxide produced is already mixed with
oxygen for the next stage.Converting the sulphur dioxide into sulphur trioxideThis is a reversible
reaction, and the formation of the sulphur trioxide is exothermic.
in
8
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The reasons for all these conditions will be explored in detail further down the page.Converting the
sulphur trioxide into sulphuric acid This can't be done by simply adding water to the sulphur trioxide
- the reaction is so uncontrollable that it creates a fog of sulphuric acid. Instead, the sulphur trioxide
is first dissolved in concentrated sulphuric acid:
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The product is known as fuming sulphuric acid or oleum.This can then be reacted safely with water to
produce concentrated sulphuric acid - twice as much as you originally used to make the fuming
sulphuric acid.
22. How is SO2 an air pollutant?
•
Solution:
Sulphur dioxide and sulphur trioxide are harmful gaseous pollutants.SO2 is released into atmosphere
through volcanic eruptions and accounts for 67% of total SO2 present in atmosphere. The remaining
33% comes from human activity.The SO2 present in atmosphere undergoes photolytic and catalytic
oxidation to SO3 which gets converted to H2SO4 in the presence of moisture. This comes down in the
form of acid rain.SO2 is strongly irritating to respiratory of humans and animals. It causesacute
irritation to eyes. It also produces leaf injuries (called necrotic blotching).
23. Why are halogens strong oxidising agents?
•
Solution:
Halogens act as strong oxidizing agents because of their high electronegativity values and high
electron affinity values. The oxidizing power of the halogens is comparable in terms of their reduction
potential values given below:
Element: F Cl Br I
Red. Potential (E0 Volt) +2.87 +1.36 +1.06 +0.54
Oxidizing nature decreases As the reduction potential values decrease from fluorine to iodine, the
oxidizing power also decreases.
24. Explain why fluorine forms only one oxoacid, HOF.
•
in
Solution:
Flourine is known to form only one oxoacid, HOF which is highly unstable. Other halogens form
oxoacids of the type HOX, HXO2, HXO3, and HXO4(X=Cl,Br,I). Flourine due to its small size and high
electro negativity cannot act as central atom in higher oxoacids and hence do not form higher
oxoacids.
25. Explain why inspite of nearly the same electronegativity, oxygen
forms hydrogen bonding while chlorine does not.
•
Solution:
Oxygen atom can form H bonds whereas chlorine does not. The tendency for H bonding depends
upon small size and high electronegativity value. Although the electreonegativity of O and Cl is
nearly same yet chlorine does not form H bond due to its larger size (99pm) as compared to oxygen
(66pm).
9
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26. Write two uses of ClO2.
•
Solution:
Chlorine dioxide is used as
(i) oxidizing agent
(ii) bleaching agent.
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27. Why are halogens coloured?
•
Solution:
All the halogens are coloured. The colour of halogens deepens with raise in atomic number from
fluorine to iodine.
Elements
Colour
•
F
Light yellow
Cl
Yellow green
Br
Reddish
I
Deep violet
The colour is due to absorption of energy from visible light by their molecules for excitation of outer
electrons of higher energy levels (the gap of energy between valence shell of halogens and higher
energy shells is less). Flourine absorbs violet portion of the light and appears yellow while iodine
absorbs yellow and green portions of the light and thus appears violet.Change in colour on moving
from F to I is calkled blue shift or bathochromic shift.
28. Write the reactions of F2 and Cl2 with water.
•
Solution:
(i) Flourine is highly reactive and decomposes water very readily even at low temperature and in
dark forming a mixture of O2 And O3.
3F2 + 3H2O
6HF + O3
2F2 + 2H2O
4HF + O2
___________________________
5F2 + 5H2O
10HF + O3 + O2
(ii) Chlorinbe decomposes water in the presence of sunlight forming halogen acid and oxoacid.
Cl2 + H2O
HCl + HClO
in
29. How can you prepare Cl2 from HCl and HCl from Cl2? Write reactions
only.
•
Solution:
(i) Chlorine can be obtained from HCl by heating with oxidizing agents like MnO2, KMnO4,
K2Cr2O7 etc.
MnO2 + 4HCl
MnCl2 + 2H2O + Cl2
2KMnO4 + 16 HCl
2KCl + 2MnCl2 + 8H2O + 5 Cl2
(ii) HCl can be obtained from chlorine by burning it in excess of hydrogen
H2 + Cl2
2HCl
10
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30. What inspired N. Bartlett for carrying out reaction between Xe and
PtF6?
•
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Solution:
N. Bartlett in 1962, prepared a compound for reacting oxygen with PtF6, a powerful oxidizing agent.
The X-Ray examination of solid compound, O2PtF6, showed that it consisted of O2+ and PtF6- ions.
Bartlett thought that a similar could be prepared with xenon because the ionization enthalpy of
xenon is quite close to the ionization enthalpy of oxygen (1314 kJ mol-1). Accord8ingly he reacted
xenon with PtF6 and got a red solid compound
Xe + PtF6
Red solid
XePtF6
31. What are the oxidation states of phosphorus in the following:
(i) H3PO3 (ii) PCl3 (iii) Ca3P2 (iv) Na3PO4 (v) POF3?
•
Solution:
Compound
H3PO3
PCl3
Ca3P2
Na3PO4
POF3
Oxidation state of phosphorus
+3
+3
-3
+5
+5
32. Write balanced equations for the following:
(i) NaCl is heated with sulphuric acid in the presence of MnO2.
(ii) Chlorine gas is passed into a solution of NaI in water.
•
Solution:
(i) Chlorine gas is produced.
2NaCl + MnO2 + 2H2SO4
Na2SO4 + MnSO4 + 2H2O + Cl2↑
(ii) Iodine is liberated
2NaI + Cl2
2NaCl + I2.
33. How are xenon fluorides XeF2, XeF4 and XeF6 obtained?
in
•
Solution:
All three binary fluorides of Xenon are formed by direct union of elements under approximate
experimental conditions. XeF2 can also be prepared by irradiating a mixture of Xenon and fluorine
with sunlight or light from a pressure-mercury arc lamp.
Xe (g) + F2 (g)
XeF2 (g)
1 vol 2 vol sealed Ni vessel Xenon diflouride
Xe (g) + F2 (g)
1 vol 2 vol sealed Ni vessel
XeF4 (g)
Xe (g) + F2 (g)
1 vol 20 vol sealed Ni vessel
XeF6 (g)
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34. With what neutral molecule is ClO– isoelectronic? Is that molecule a
Lewis base?
•
Solution:
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ClO- species has 26 electrons. The neutral molecule which is isoelectronic with it is ClF
Since it can form ClF3 and ClF5, it is a Lewis base.
35. How are XeO3 and XeOF4 prepared?
•
Solution:
6XeF4 + 12H2O
XeF6 + 3H2O
XeF6 + H2O
4Xe + 2XeO3 + 24HF + 3O2.
XeO3 + 6HF
XeOF4 + 2HF
36. Arrange the following in the order of property indicated for each set:
(i) F2, Cl2, Br2, I2 - increasing bond dissociation enthalpy.
(ii) HF, HCl, HBr, HI - increasing acid strength.
(iii) NH3, PH3, AsH3, SbH3, BiH3 – increasing base strength.
•
Solution:
(i) Cl-Cl > Br-Br > F-F > I-I - Bond energy is highest for chlorine molecule.
(ii) HI > HBr > HCl > HF - HI is the strongest acid among the halogen acids.
(iii) NH3 > PH3 > AsH3 > SbH3 > BiH3 – Ammonia is the most basic hydride.
37. Which one of the following does not exist?
(i) XeOF4 (ii) NeF2 (iii) XeF2 (iv) XeF6
•
Solution:
The sum of the first two ionization energies of Ne is too high to be compensated.
in
38. Give the formula and describe the structure of a noble gas species
which is isostructural with:
(i) ICl4(ii) IBr2(iii) BrO3-
•
Solution:
(i) XeF4 (ii) XeF2 (iii) XeO3.
39. Why do noble gases have comparatively large atomic sizes?
12
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•
Solution:
The atomic size in the case of noble gases is expressed in terms Vander waal’s radii whereas the
atomic size of other members of the period is either metallic radii or covalent radii. As the Vander
waal’s radii is larger than both metallic as well as covalent radii, ther3efore the atomic size of the
noble gas is quite large. Among the noble gases, the atomic size increases down the group due to
addition of new electronic shells.
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40. List the uses of neon and argon gases.
•
Solution:
Uses of neon:
It is used in neon discharge lamps and signs which are used for advertising purposes.It is used in
safety devices for protecting electrical instruments because it has a property of carrying exceedingly
high currents under high voltage.
Uses of Argon:
a. It is widely used for filling in candescent metal filament electric bulbs.
b. It is used for filling radio-valves, rectifiers and fluorescent tubes.
c. It is used in producing inert atmosphere during welding and extraction of various metals.
41. Why are pentahalides more covalent than trihalides?
•
Solution:
Higher the positive oxidation state of central atom, more will be its polarising power, which, in turn,
increases the covalent character of bond, formed between the central atom and the other atom.
42. Why is BiH3 the strongest reducing agent amongst all the hydrides of
Group 15 elements?
•
Solution:
Because BiH3 is the least stable among the hydrides of Group 15.
43. Why is N2 less reactive at room temperature?
•
Solution:
Because of strong pπ–pπ overlap resulting into the triple bond, N≡N.
44. Mention the conditions required to maximise the yield of ammonia.
•
in
Solution:
The conditions required to maximize the yield of ammonia:
1. An optimum temperature of about 750 K
2. A pressure of about 200 atmospheres
3. Using finely divided iron as catalyst and molybdenum as promoter.
45. How does ammonia react with a solution of Cu2+?
•
Solution:
It gives a deep blue solution with a solution of copper (II) sulphate due to the formation of
[Cu(NH3)4]SO4.
13
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46. What is the covalence of nitrogen in N2O5?
•
Solution:
From the structure of N2O5 it is evident that covalence of nitrogen is four
47. Bond angle in PH4+ is higher than that in PH3. Why?
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•
Solution:
Both are sp3 hybridised. In PH4+ all the four orbitals are bonded whereas in PH3 there is a lone pair of
electrons on P, which is responsible for lone pair-bond pair repulsion in PH3 reducing the bond angle
to less than109° 28′.
48. What happens when white phosphorus is heated with concentrated
NaOH solution in an inert atmosphere of CO2 ?
•
Solution:
When white phosphorus is heated with concentrated NaOH solution in an inert atmosphere of CO2,
Phosphine gas is produced.
P4 + 3NaOH + 3H2O → 3NaH2PO2 + PH3.
49. What happens when PCl5 is heated?
•
Solution:
On heating phosphorus pentachloride sublimes at 160˚C. Whenheated it dissociates into PCl3 and Cl2.
PCl5 ↔ PCl3 + Cl2
50. Write a balanced equation for the hydrolytic reaction of PCl5 in
heavy water.
•
Solution:
PCl5 + D2O → POCl3 + 2DCl
51. What is the basicity of H3PO4?
•
Solution:
Three P–OH groups are present in the molecule of H3PO4. Therefore, itsbasicity is three.
•
in
52. What happens when H3PO3 is heated?
Solution:
When H3PO3 is heated, it undergoes auto oxidation and reduction to form phosphoric acid and
phosphine.
4H3PO3 → H3PO4 + PH3.
53. List the important sources of sulphur.
•
Solution:
Sulphur occurs mainly as the sulphides and sulphates in the combined state.
For example: lead sulphide – galena, copper pyrites, copper sulphide – copper glance, calcium
14
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sulphate – gypsum etc.
Traces of sulphur also occur as hydrogen sulphide gas coming out of the volcanoes. Organic
substances such as eggs, proteins, garlic, onion, hair and wool, etc., also contain a small percentage
of sulphur..
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54. Write the order of thermal stability of the hydrides of Group 16
elements.
•
Solution:
The members of Group 16 family forms volatile hydrides such as H2O,H2S,H2Se,H2Te and H2Po.The
energy needed to dissociate these molecules decreases down the group. Hence the thermal stability
of these hydrides follows the order:
H2O > H2S > H2Se > H2Te > H2Po.
55. Why is H2O a liquid and H2S a gas?
•
Solution:
Because of small size and high electronegativity of oxygen, molecules of water are highly associated
through hydrogen bonding resulting in its liquid state.
56. Which of the following does not react with oxygen directly?Zn, Ti, Pt,
Fe
•
Solution:
Platinum does not react with oxygen directly.
57. Complete the following reactions:
(i) C2H4 + O2 →
(ii) 4Al + 3 O2 →
•
Solution:
(i) C2H4 + O2
(ii) 4Al + 3 O2
2CO2 + 2H2O.
2Al2O3.
58. Why does O3 act as a powerful oxidising agent?
•
O3 → O2 + O.
in
Solution:
Ozone is a powerful oxidizing agent because on decomposition it forms atomic oxygen.
59. How is O3 estimated quantitatively?
•
Solution:
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. This
is a quantitative method for estimating O3 gas.
15
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60. What happens when sulphur dioxide is passed through an aqueous
solution of Fe(III) salt?
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Solution:
Sulphur dioxide acts as a fairly strong reducing agent in the presence of moisture. Hence it reduces
ferric sulphate to ferrous sulphate.
Fe2(SO4)3 + SO2 + 2H2O
2FeSO4 + 2H2SO4.
61. Comment on the nature of two S–O bonds formed in SO2 molecule.
Are the two S–O bonds in this molecule equal ?
•
Solution:
Both the S–O bonds are covalent and have equal strength due to resonatingstructures.
62. How is the presence of SO2 detected?
•
Solution:
SO2 gas when passed through acidified potassium permanganate decolourises it due to its reducing
property.
2MnO4- +5SO2 +2H2O
2Mn2+ +5SO42- +4H+
63. Mention three areas in which H2SO4 plays an important role.
•
Solution:
1. It is used in the manufacture of fertilizers.
2. It is used in storage batteries
3. IT is used in petroleum refining and making of paints, detergents etc.
•
in
64. Write the conditions to maximise the yield of H2SO4 by Contact
process.
Solution:
The conditions to maximise the yield of H2SO4 by Contact process are:
1. an optimum temperature of 673 – 723 K.
2. an optimum pressure of 1.5 – 2 atm.
3. using platinised asbestos or divanadium pentoxide as catalyst.
4. using oxygen is excess.
5. preventing poisoning of catalyst, by removing impurities like arsenic oxide, dust particles and
moisture.
65. Why is Ka2 << Ka1 for H2SO4 in water?
16
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•
Solution:
H2SO4 is a very strong acid in water largely because of its first ionisation to H3O+ and HSO4–. The
ionisation of HSO4–to H3O+ and SO42– is very very small. That is why Ka2 << Ka1.
HSO4- does not have the tendency to lose the proton readily.
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66. Considering the parameters such as bond dissociation enthalpy,
electron gain enthalpy and hydration enthalpy, compare the oxidising
power of F2 and Cl2.
•
Solution:
∆dissH of fluorine (79.1 kJ) is smaller than that of chlorine (122kJ). Electron gain enthalpy of chlorine
(-384 kJ) is higher than that of fluorine (-333kJ). The value of ∆hydH of fluoride ion (-460 kJ) is much
higher than that of chloride ion (-383 kJ). It is due to small size and high charge density of fluoride
ion. Thus net energy released for
½ F2 (g)
for ½ Cl2 (g)
F- (aq) is -752 kJ whereas that
Cl- (aq) is - 708 kJ, which accounts for higher oxidizing power of F2.
67. Give two examples to show the anomalous behaviour of fluorine.
•
Solution:
Ionisation enthalpy, electro negativity, enthalpy of bond dissociation and electrode potentials are all
higher for fluorine than expected from the real trends set by other halogens. Also, ionic and covalent
radii, melting point and boiling point and electron gain enthalpy are quite lower than expected.
Fluorine can exhibit only -1 oxidation state and cannot show positive oxidation states like other
halogens.
Fluorine can form hydrogen bonding and so HF is much less acidic than other hydrogen halides.
68. Sea is the greatest source of some halogens. Comment.
•
Solution:
in
Sea water contains some of the halogens like chlorine, bromine and traces of fluorine. It is evident
from the electrolysis of sea water which yields chlorine and bromine. Sea water contains chlorides,
bromides and iodides of sodium, magnesium,calcium. Sea weeds contain iodide.
69. Give the reason for bleaching action of Cl2.
•
Solution:
In the presence of moisture, chlorine acts as an oxidizing and a bleaching agent. Chlorine reacts with
water forming HCl and HClO (hypochlorous acid).
HClO is not so stable and decomposes giving nascent oxygen, which is responsible for oxidizing and
bleaching properties of chlorine.
Cl2 + H2O
HCl + HClO
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HClO
HCl + O
_________________________
Cl2 + H2O
2HCl + O
Coloured matter + Nascent oxygen
colourless matter
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70. Name two poisonous gases which can be prepared from chlorine
gas..
•
Solution:
Phosgene (COCl2) and tear gas (CCl3.NO2) are the poisonous gases that can be prepared from
chlorine gas.
71. Why is ICl more reactive than I2?
•
Solution:
In general, interhalogen compounds are more reactive than halogens due to weaker X–X′ bonding
than X–X bond. Thus, ICl is more reactive than I2.
72. Why is helium used in diving apparatus?
•
Solution:
Helium – oxygen mixture is used by deep-sea divers in preference to nitrogen – oxygen mixtures. It
is much less soluble in blood than nitrogen. This prevents ‘bends’ which is the pain caused by
formation of nitrogen bubbles in blood veins when a diver comes to the surface.
73. Balance the following equation: XeF6 + H2O → XeO2F2 + HF
•
Solution:
XeF6 + 2H2O → XeO2F2 + 4HF
74. Why has it been difficult to study the chemistry of radon?
•
Solution:
Radon is radioactive with very short half-life which makes the study of chemistry of radon difficult.
in
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