The p-Block Elements
 Give Reason:
 Group 15 (N, P, As, Sb, Bi): ns2 np3
1. Nitrogen shows anomalous properties.
A. Due to its small size, high electronegativity, high I.E., non-availability of d-orbitals in its
valence shell and its ability to form pπ-pπ multiple bonds with itself, and with other
elements like C and O.
2. Nitrogen exists as diatomic molecule N2 gas at room temepertaure.
A. Due to the ability of the N atom to form pπ-pπ triple bond with another N atom.
3. N2 gas is chemically inert at room temperature.
A. Due to very high bond enthalpy of N≡N.
4. I.E. of Group 15 elements is much greater than that of the Group 14 and Group 16
elements in the corresponding periods.
A. Because of the extra stability of a half-filled p-orbital configuration.
5. Atomic radius of elements increases in a group, but there is only a small increase from As
to Bi.
A. This is due to the presence of completely filled d-orbitals and/or f-orbitals in them, which
have weak screening effect.
6. Metallic character increases down the group.
A. Due to decrease in I.E. and increase in atomic size.
7. The stability of +5 state decreases and that of +3 state increases down the group.
A. Due to inert pair effect. (explain inert pair effect)
8. Nitrogen does not exhibit covalency greater than 4.
A. Due to the absence of d-orbitals in the valence shell of N.
9. PCl5 exists, but NCl5 does not.
A. Due to the absence of d-orbitals in the valence shell of N, N cannot expand its valence
shell.
10. Tendency for catenation is less in N as compared to the other members of the group.
A. N-N single bond is weaker than those of the other elements. This is due to the high interelectronic repulsion of the non-bonding electrons, owing to the small N-N bond length.
11. Thermal stability of hydrides decreases from NH3 to BiH3.
A. E-H bond enthalpy (bond strength) decreases down the group. (E  Group 15 element)
12. Reducing character of EH3 increases from NH3 to BiH3.
A. E-H bond strength decreases, so H is easily supplied for reduction, on moving down the
group.
13. Basic character decreases down the group for EH3.
A. Due to the presence of a lone pair of electrons on the central atom, they behave as Lewis
bases. N being the smallest in size, electron-density on N is high. Thus, it is the strongest
base. Base strength decreases with increase in the size of the central atom.
14. NH3 is a stronger base than PH3. (phosphine).
A. Due to smaller size of N, electron-density on N is higher. Thus, it is a strong Lewis base.
15. NH3 has exceptionally high b.p.
A. Due to the high electronegativity of N, intermolecular H bonds exist between NH3
molecules.
16. Pentahalides of Group 15 elements are more covalent than trihalides.
A. Higher the positive oxidation state of the central atom, the more is its polarizing power
and, therefore, more the covalent nature of the bond.
17. Among halides of Group 15 elements, BiH3 is the strongest reducing agent.
A. BiH3 is the least stable, because Bi-H bond is the weakest (least bond enthalpy) compared
to the E-H bond of the other elements of the group.
18. NH3 molecule is triagonal pyramidal in shape.
A. Due to the presence of 3 bp and 1 lp of electrons on the central N atom in the molecule.
19. NH3 is used in the detection of metal ions such as Cu2+, Ag+ etc.
A. Due to the presence of a lone pair on the central N atom, NH3 is a Lewis base. Thus, it
donates electron pairs to metal ions and hence forms coloured complexes which help in
the detection of metal ions.
e.g Cu2+(aq) + 4 NH3 → [Cu(NH3)4]2+(aq)
deep blue solution
20. NO2 dimerises.
A. NO2 contains an odd no. of valence electrons. Thus, it behaves as a typical odd molecule.
On dimerisation, it is converted into a more stable N2O4 molecule with an even no. of
electrons.
21. Conc. nitric acid is a strong oxidizing agent and oxidises metals like Cu and Zn.
However, metals like Cr and Al do not dissolve in conc. HNO3.
A. Due to the formation of passive oxide films on the surface of Cr and Al.
22. PH3 is a Lewis base.
A. Due to the presence of a lone pair of electrons on the P atom.
23. Bond angle in PH4+ is higher than that of PH3.
A. In PH4+, there are 4 bond pairs on the central P atom. Thus, it has a tetrahedral shape with
a bond angle of 109.5. However, in PH3, there is a lone pair on the central P atom. Due
to lp-bp repulsion, bond angle in PH3 is slightly less than 109.5.
24. PCl3 fumes in the presence of moisture.
A. This is due to the formation of HCl.
PCl3 + 3H2O  H3PO4 + 3HCl
25. All the 5 P-Cl bonds in PCl5 are not equivalent.
A. PCl5 molecule is trigonal bipyramidal in shape. The 3 equatorial bonds are of the same
length but the 2 axial bonds are longer due to repulsion between bond pairs of the axial
and equatorial bonds.
26. PCl5 in the gaseous and solid states, respectively, do not have the same geometry.
A. PCl5 (g) is trigonal bipyramidal in shape. In the solid state, PCl5 is ionic with tetrahedral
[PCl4]+ cation and octahedral [PCl6]- anion.
27. Orthophosphorous acid H3PO3 undergoes disproportionation.
A. In H3PO3, P is in the +3 state. ─3 and +5 states of P are more stable. So, it undergoes
disproportionation.
4H3PO3  3H3PO4 + PH3
28. Ortho phosphorous acid H3PO3 (phosphonic acid) is diprotic (or dibasic) while ortho
phosphoric acid H3PO4 is triprotic (or tribasic).
A. Only those H atoms which are attached to oxygen in P-OH form are ionisable and cause
basicity. Thus, H3PO3 with only 2 P-OH bonds is dibasic and H3PO4 with 3 P-OH bonds
is tribasic.
29. Hypophosphoric acid H3PO2 (phosphinic acid) is monobasic or monoprotic.
A. There is only one P-OH bond. Thus, there is only 1 ionisable H atom and, hence only one
H+ ion can be liberated.
30. H3PO2 is a strong reducing agent.
A. Oxoacids with P-H bonds have reducing properties and H3PO2 (above fig.) has 2 P-H
bonds in it.
31. NH3 forms hydrogen bonds but PH3 does not.
A. Due to the small size and high electronegativity of the central N atom, NH3 is able to
form intermolecular H-bonds.
32. R3P=O exists, but R3N=O does not. (R  alkyl group)
A. P can form dπ-pπ bond with O. However, N, due to the absence of d-orbitals, cannot
form dπ-pπ bond with O.
33. P(C2H5)3 and As(C6H5)3 act as ligands and bond with transition metals.
A. Due to the ability of P and As to form dπ-dπ bond with transition metals.
34. Nitrogen exists as N2 gas while phosphorous exists as P4 solid.
A. N can form pπ-pπ triple bond with another N atom while P, due to its larger size, cannot
form pπ-pπ bond with another P atom.
35. The HNH bond angle in NH3 is higher than HPH, HAsH and HSbH bond angles in their
respective hydrides.
A. Due to the small size of the central N atom and the small length of the N-H bond, bp-bp
repulsion in NH3 is greater than that of the other halides of the group. Moreover, in NH3,
hybridisation is sp3 while in heavier elements the bond is between s and p orbitals of
hydrogen and the element respectively.

Group 16 (O, S, Se, Te, Po): ns2 np4
36. Oxygen shows anomalous behavior.
A. Due to its small size, high electronegativity, high I.E., non-availability of d-orbitals in its
valence shell and its ability to form pπ-pπ multiple bonds with itself, and with other
elements of small size & high electronegativity such as C and N.
37. O2 exists as a gas while other members are solids.
A. Due to the ability of O to form pπ-pπ double bond with another O atom.
38. I.E. decreases down the group.
A. Due to increase in size.
39. Atomic and ionic radii increase from top to bottom.
A. Due to increase in the no. of shells.
40. Electron gain enthalpy of O is less negative than that of S.
A. Due to the very small size of the O atom there is greater inter-electronic repulsion in the
relatively smaller 2p sub-shell hence the incoming electron experiences less attraction in
oxygen atom.
41. Metallic character increases from O to Po.
A. Electronegativity decreases from O to Po, as does I.E.
42. Elements of Group 16 generally show lower value of first ionization enthalpy (I.E.)
compared to the elements of Group 15 belonging to the corresponding periods.
A. Due to the extra stable half-filled p-orbital configuration of Group 15 elements, larger
amount of energy is required to remove their electron.
43. O shows only negative oxidation state, except in OF2.
A. Since O is a highly electronegative element, it shows only ─2 oxidation state. However,
in OF2, F is more electronegative due to which O has +2 oxidation state.
44. Oxygen does not exhibit higher oxidation states (+4, +6) like the heavier members of its
group.
A. Due to the absence of d-orbitals in the valence shell of oxygen.
45. Stability of +6 state decreases and that of +4 state increases down Group 16.
A. Due to inert pair effect.
46. Acidic character of the hydrides of Group 16 increases down the group.
A. Due to decrease in H-E bond enthalpy. (E  Group 16 element)
47. Thermal stability of H2E decreases down the group.
A. H-E bond strength decreases.
48. Reducing character increases down the group.
A. H-E bond strength decreases.
49. Water has exceptionally high m.p. and b.p.
A. Due to intermolecular H-bonds, owing to the high electronegativity and small size of the
O atom.
50. Reducing property of dioxides reduces from SO2 to TeO2.
A. Stability of +6 oxidation state decreases from S to Te.
51. Stability of halides of Group 16 decreases in the order F > Cl > Br > I.
A. Stability of halides decrease with a decrease in electronegativity of the halogen.
52. SF6 is very stable, chemically inert and resistant to hydrolysis.
A. The S atom in SF6 is sterically protected by 6 F atoms.
53. SF4 and SeF6 are more easily hydrolysed.
A. They are less sterically protected. In SF4 steric protection is less due to lesser no. of F
atoms and in SeF6 steric protection is less due to larger size of Se.
54. Water is a liquid, while H2S is a gas at room temperature.
A. Due to the presence of intermolecular H-bonds in water.
55. Though the formation of oxides with elements is exothermic, some external heating is
required to initiate the reaction.
A. Bond dissociation enthalpy of O=O is very high.
56. The 2 O-O bond lengths in the ozone molecule are identical.
A. Ozone is a resonance hybrid of 2 canonical structures.
57. O3 is a powerful oxidizing agent.
A. O3  O2 + O
The above reaction is highly exothermic (H negative and S positive, thus, G
negative). The nascent O liberated makes O3 a powerful oxidizing agent.
58. At high temperatures, sulphur exists as S2. Both S2 and O2 are paramagnetic in nature.
A. Both S2 and O2 have unpaired electrons in the antibonding π* orbitals. Hence, they are
paramagnetic in nature.
59. SO2 decolourises acidified potassium permanganate solution (KMnO4).
A. SO2 is a reducing agent and reduces KMnO4 to colourless Mn2+ in an acidic medium.
5SO2 + 2MnO4- + 2H2O  5SO42- + 4H+ + 2Mn2+
[pink colour]
[colourless]
60. The 2 S-O bonds in SO2 are of the same length.
A. It is a resonance hybrid of 2 canonical forms.
61. Low temperature and high pressure are favourable conditions for the maximum yield of
SO3 from SO2.
A. 2SO2 + O2
2SO3 ;
H = - 196.6 kJ/mol
The above reaction is exothermic and reversible. The forward reaction leads to a decrease
in volume. Therefore, applying Le Chatlier’s Principle, low temperature and high
pressure will increase the yield of SO3.
62. In the manufacture of H2SO4 by the contact process, SO3 is absorbed in conc. H2SO4 to
form oleum, but SO3 is not directly absorbed in H2O to form H2SO4.
A. When SO3 is directly absorbed in H2O, a corrosive mist of H2SO4 is formed, which is
dangerous.
63. Ka2 <<Ka1 for the reaction of H2SO4 in water.
A. Ionisation of H2SO4 into H3O+ and HSO4- is highly favoured.
H2SO4 (aq) + H2O (l)  H3O+ (aq) + HSO4- (aq) ;
Ka1 very large
+
2However, ionization of HSO4 into H3O and SO4 is less favoured.
HSO4- (aq) + H2O (l)  H3O+ (aq) + SO42- (aq) ;
Ka2 small
64. Sugar is charred and turns black when conc. H2SO4 is added.
A. Conc. H2SO4 is a strong dehydrating agent. It removes elements of water from sugar to
form carbon.
C12H22O11
12C + 11H2O
65. Electron gain enthalpy for O  O- and O  O2- are -141 kJ/mol and 702 kJ/mol,
respectively. Yet oxides have O2- ions in them and not O- ions.
A. In O2-, O has a stable octet configuration. Moreover, the lattice enthalpy of the resulting
ionic compound will be higher if the charge on the constituent ions is greater.
66. SO2 is an air pollutant.
A. It dissolves in rainwater and causes acid rain since SO2 is an acidic oxide.
67. Oxygen and chlorine have nearly the same electronegativity, but oxygen forms H-bonds
while Cl does not.
A. O atom is smaller in size than Cl atom. Small size of an electronegative atom favours Hbonding.
68. H2SO4 is very low in volatility.
A. Due to intermolecular H-bonds.
69. Conc. H2SO4 is a good oxidizing agent.
A. It oxidises metals and non-metals and itself gets reduced to SO2.

Group 17 (F, Cl, Br, I, At): ns2 np5
70. Halogens have the smallest atomic radii in their respective periods.
A. Due to maximum effective nuclear charge on the outer electrons.
71. Halogens have maximum negative electron gain enthalpy in their corresponding periods.
A. The atoms of these elements have only one electron less than the stable noble gas
configuration. Also, the effective nuclear charge is the maximum in them. Thus, they
readily accept one electron.
72. Electron gain enthalpy of Group 17 becomes less negative down the group.
A. Due to an increase in the size of successive atoms.
73. F has lower negative electron gain enthalpy than Cl.
A. Due to the small size of the F atom, there is strong inter-electron repulsion in its small 2p
orbitals. Thus, the incoming electron does not experience much attraction.
74. F-F bond enthalpy is less than Cl-Cl bond enthalpy.
A. Bond enthalpy generally decreases with an increase in the size of an atom. However, F-F
bond enthalpy is less because of strong inter-electron repulsion among the lone pairs in
the F2 molecule, where they are much closer to each other due to the small size of the F
atom.
75. Although electron gain enthalpy of F is less negative as compared to Cl, F2 is a stronger
oxidizing agent than Cl2.
A. Oxidising power of F2 depends upon the following factors:
i)
electron gain enthalpy of F
ii)
enthalpy of dissociation of the F-F bond
iii)
hydration enthalpy of FThough F has low negative electron gain enthalpy, the F-F bond enthalpy is low and the
hydration enthalpy of F- is high.
76. F2 shows anomalous behavior.
A. Due to its very small size, high electronegativity, low F-F bond dissociation enthalpy and
the non-availability of d-orbitals in its valence shell.
77. Most of the reactions of halogens are exothermic.
A. Due to the small and strong bond formed by it with other elements.
78. F forms only one oxoacid HOF, while other halogens form many.
A. Due to its high electronegativity, small size and absence of d-orbitals.
79. HF is a liquid at room temperature while HCl is a gas.
A. Intermolecular H-bonds exists between HF molecules, due to the small size and high
electronegativity of the F atom.
80. Acidic strength of HX increases from HF to HI.
A. H-X bond dissociation enthalpy decreases down the group from F to I with increase in
size of the halogen atom.. Hence the H-F bond is the strongest and HF is the weakest
acid. H─I bond is the weakest so HI is the strongest acid
81. Stability of HX decreases from HF to HI.
A. H-X bond dissociation enthalpy decreases from H-F to H-I.
82. F exhibits only -1 oxidation state whereas the other halogens exhibit +1, +3, +5 and +7
oxidation states as well.
A. F is the most electronegative element. Thus, it shows only -1 oxidation state. F does not
have d-orbitals in its valence shell, and so cannot show oxidation state more than 1.
83. OF2 and O2F2 are essentially fluorides and not oxides.
A. Because F is more electronegative than O.
84. Stability of oxides formed is in the order I > Cl > Br.
A. Due to a combination of kinetic and thermodynamic factors.
85. Ionic character of halides of metals in oxidation state +1 is in the order:
MF > MCl > MBr > MI
A. Electronegativity decreases from F to I.
86. SnCl4, PbCl4, SbCl5 and UF6 are respectively more covalent than SnCl2, PbCl2, SbCl3 and
UF4.
A. The halides of elements in higher oxidation states are more covalent than those in lower
oxidation states. For example, SnCl4 is more covalent (Sn in +4 oxidation state) than
SnCl2 (Sn in +2 state).
87. I-Cl is more reactive than I2.
A. In general, inter-halogen compounds X-X’ are more reactive than halogens X2 since the
X-X’ bond is weaker the X-X bond.
88. Cl2 is a good bleaching agent.
A. This is because of its oxidizing action.
Cl2 + H2O  2HCl + O
Coloured substance + O  Colourless substance
89. All halogens are coloured.
A. The absorption of radiation in the visible region results in the excitation of outer electrons
to higher energy levels. By absorbing different quanta of radiation, halogens display
different colours.
90. When HCl reacts with finely powdered iron, ferrous chloride FeCl2 is formed, and not
ferric chloride FeCl3.
A. Fe + 2HCl  FeCl2 + H2
The H2 gas formed prevents the oxidation of FeCl2 to FeCl3.
91. Inter-halogen compound XX’3 (e.g. BrF3) has a bent ‘T’ shape.
A. Due to the presence of 3 bond pairs and 2 lone pairs of electrons, the molecule becomes
bent ‘T’ in shape to reduce the lp-bp repulsion.
92. IF7 (XX’7) is pentagonal bipyramidal in shape.
A. Due to the presence of 7 bond pairs of electrons.
93. IF5 is square pyramidal in shape.
A. Due to the presence of 5 bond pair and 1 lone pair of electrons on the central I atom.
94. There are no inter-halogen compounds with F as the central atom.
A. Due to the very small size, high electronegativity and absence of d-orbitals in the F atom.
95. Acidic character of oxoacids is in the order:
HOCl < HOClO < HOClO2 < HOClO3
A. Acidic character increases with increase in oxidation state of the halogen atom, also the
corresponding conjugate base is resonance stabilized.

Group 18 (He, Ne, Ar, Kr, Xe, Rn): ns2 np6
96. Group 18 elements are called noble gases.
A. They have very low reactivity due to the stable octet configuration in their valence shells.
97. They have very low b.p. and are liquefied only at low temperatures.
A. They are monoatomic and the interatomic forces are only weak dispersion forces.
98. It has been difficult to study the chemistry of Radon.
A. It is a radioactive element with a very short half-life. So it makes the study of radon
difficult.
99. Noble gases have comparatively large atomic sizes.
A. Noble gases exist in monoatomic form with very weak dispersion forces between the
atoms. So, the atomic radius measured in their case is the van der Waal’s radius, which is
the distance between the nuclei of 2 neighboring atoms not bonded to each other. Hence,
the large size.
100. No true compounds of He, Ne and Ar are yet known
A. Due to their very high ionization enthalpies.
101. Compound of Xe are mainly with F and O.
A. Because F and O are the most electronegative elements.
102. Hydrolysis of XeF6 is not a redox reaction.
A. Oxidation state of Xe is the same in XeF6 and the products of its hydrolysis, i.e. XeOF4
.
and XeO2F2.
103. He is used in diving apparatus.
A. It is used as a diluent for oxygen in the diving apparatus because of its low solubility in
.
the blood. This is to minimize the painful effects accompanying the decompression of
.
deep-sea divers.
104. Among the noble gases, only Xe is known to form many compounds with O and F.
A. Due to its lower I.E. as compared to the smaller elements of its group.( Rn is the largest
.
atom in the group. However, it is radioactive.)
105. Neil Bartlett could prepare the first inert gas compound Xe+PtF6- similar to O2+PtF6-.
A. Because the 1st ionization enthalpies of molecular oxygen O2 & Xe are nearly the same.
106. Inert gases have large positive values of electron gain enthalpy.
A. Due to their stable octet configuration, they do not have the tendency to gain electrons.
 Arrange The Following In The Increasing
Order Of The Property Indicated:
1. N, P, As, Sb, Bi
a)
Atomic and ionic radius
b)
Ionisation enthalpy
c)
Electronegativity
A. a) N < P < As < Sb < Bi
b) Bi< Sb < As < P < N
c) Bi = Sb < As < P < N
2. NH3, PH3, AsH3, SbH3, BiH3
a)
Thermal stability
b)
Bond strength or bond dissociation enthalpy
c)
Base strength
d)
Reducing character
e)
Boiling point
f)
Bond angle
A. a) BiH3 < SbH3 < AsH3< PH3 < NH3
b) BiH3 < SbH3 < AsH3< PH3 < NH3
c) BiH3 < SbH3 < AsH3 < PH3 < NH3
d) NH3 < PH3 < AsH3 < SbH3 < BiH3
e) PH3 < AsH3 < NH3 < SbH3 < BiH3
f) BiH3 < SbH3 < AsH3 < PH3 < NH3
3. O, S, Se, Te, Po
a)
Stability of +6 state
b)
Stability of +4 state
c)
Electronegativity
d)
Negative electron gain enthalpy
A. a) Po < Te < Se < S
b) S < Se < Te < Po
c) Po < Te < Se < S < O
d) O < Po < Te < Se < S
4. H2O, H2S, H2Se, H2Te
a)
Acidic character
b)
Bond dissociation enthalpy
c)
Thermal stability
d)
Reducing property
e)
Bond angle
A. a) H2O < H2S < H2Se < H2Te
b) H2Te < H2Se < H2S < H2O
c) H2Te < H2Se < H2S < H2O
d) H2O < H2S < H2Se < H2Te
e) H2Te < H2Se < H2S < H2O
5. F2, Cl2, Br2, I2
a) Oxidizing power
b) Bond dissociation enthalpy
c) Bond length
A. a) I2 < Br2 < Cl2 < F2
b) I2 < F2 < Br2 < Cl2
c) F2 < Cl2 < Br2 < I2
6. F, Cl , Br, I
a) Atomic and ionic radius
b) Negative electron gain enthalpy
c) Electronegativity
d) Enthalpy of hydration of X- ions
A. a) F < Cl < Br < I
b) I < Br < F < Cl
c) I < Br < Cl < F
d) I < Br < Cl < F
7. HF, HCl, HBr, HI
a) Boiling point
b) Enthalpy of dissociation
c) Acid strength
d) Ka values
e) pKa values
f) Bond length
g) Stability
A. a) HCl < HBr < HI < HF
b) HI < HBr < HCl < HF
c) HF < HCl < HBr < HI
d) HF < HCl < HBr < HI (higher the Ka value, stronger the acid)
e) HI < HBr < HCl < HF (lower the pKa value, stronger the acid)
f) HF < HCl < HBr < HI
g) HI < HBr < HCl < HF
8. MF, MCl, MBr, MI (ionic character)
A. MI < MBr < MCl < MF
9. HOCl, HOClO, HOClO2, HOClO3 (acid strength)
A. HOCl < HOClO < HOClO2 < HOClO3
10. N, P, As, Sb, Bi (stability of +3 and +5 oxidation states)
A. N < P < As < Sb < Bi (+3 state)
Bi < Sb < As < P < N (+5 state)
11. N, O, S, F (electronegativity)
A. S < N < O < F
12. He, Ne, Ar, Kr, Xe, Rn
a) Atomic radius
b) Ionization enthalpy
A. a) He < Ne < Ar < Kr < Xe < Rn
b) Rn < Xe < Kr < Ar < Ne < He
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