Page 1
DAIL Y PRACTICE
PROBLEMS
CHEMIS TR Y
SOLUTIONS
1. (d)
2. (d) The additional 1t -molecular electron in 02-is added to
the less stable antibonding orbital. Hence, it is easy to
remove that electron from 0 2 - ion. Thus I.E. (l.P.) of
o 2 - is lowest.
3. (a) Continuous increase as no. of shells increases down
the group.
4. (b)
5. (b) Closed shell (Ne), half filled (P) and completely filled
con figuration (Mg) are the cause of higher value of
I.E.
6. (d)
7. (b) Cu, Ag and Au are coinage metals. They belong to group
1B (d-block) of periodic table.
8. (d) In fourth period filling up of 3d orbital becomes
energetically favourable before the 4p orbital is filled.
9. (b)
K + ? K 2+ +e -. Since e- is to be removed from stable
configuration.
10. (a) For isoelectronic species ionic radii decreases as the
charge on ion decreases. Further on moving down in
a group ionic radii increases. Hence the correct order
is
o2- < N3- < SP.< p3-
11. (c) On moving down a group atomic radii increases.
12. (a) Incoming electrons occupies the smaller n = 2 shell,
also negative charge on oxygen co -) is another fa ctor
due to which incoming electron fe el repulsion.
13. (c)
Hence electron repulsion outweigh the stability gained
by achieving noble gas configuration.
14. (b) It is electronic configuration of alkali metal. Hence it
wiLl form basic oxide.
15. (b) The right sequence ofl.E 1 of Li <B <Be< C.
16. (c) AI ( 3s 2 p1) and Mg (3s2 ). Lower energy is required to
remove 3p1 electron than 3s1 electron (penetrating
effect is s > p > d > f) . Secondly Mg has stable
electron ic con figuration than AI
17. (c) Ireprese ntsLi, Ti represen tsK
TJT represents Br, IV represents I
V represents He
18. (c)
So, amongst these, IT represents most reactive metal
and V represents least reactive non-metal.
19. (c) Generally, the ionization enthalpies or energy increases
from left to right in a period and decreases from top to
bottom in a group. Several fa ctor such as atomic radius,
nuclear charge, shielding effect are responsible for
change of ionization enthalpies.
Here, Ist ionization enthalpy of A and B is greater than
group I (Li 520 kJmol -1 to Cs374 kJmol -1), which means
element A and B belong to group -2 and all three given
ionization enthalpy values are less for element B means
B wi II come below A.
20. (b) For dissolution, Hydration energy> Lattice energy.
BaS04 is sparingly soluble in water because its
hydration energy is lesser than the lattice energy and
thus ions are not separated from each other.
21. (c) The anomalous behaviour of first member of a group
in the s- and p-block element is due to their small size,
large charge/radius ratio and high electronegativity.
22. (c) In 3d6 4s2 the differe ntiating electron enters d sub
shell, Hence it represents transition metal, Fe.
23. (b) In the isoelectronic species, all isoelectronic anions
belong to the same period and cations to the next
period.
24. (b) On passing from left to right in a period acidic character
of the normal oxides of the elements increases with
increase in electroneg ativity.
25. (a) The electronic configuration clearly suggests that
it is a d-block element (having configuration
(n -1) d
1- 1o
ns 0 - 2 ) which starts from III B and goes
till 11 B. Hence with d3 configuration it would be
classified in the group.
26. (b) Correct order of increasing basic strength is
N? > PH3 > AsH 3 > Sb? > BiH 3
27. (c) Electron affinity of 9F is less than that of 17Cl
28. (c) ns2 np 1 is the electronic configuration oflll A period.
Alp3 is amphoteric oxide
29. (d) pS+ has more effective nuclear charge and smaller size than
p3+_
30. (a) For isoelectron ic species, size of anion increases as
negative charge increases. Thus the correct order is
N3- > o 2- > F
(1.71) (1.40) (1.36)
31. (a) Hydrogen bonding increases the boiling point.
Hydrogen bonds are fo rmed in compounds having F
or 0 or N with hydrogen. S, Se, Te cannot undergo
hydrogen bond fo rmation because of their larger size
and lower electronegativity values.
Page 2
DAIL Y PRACTICE
PROBLEMS
CHEMIS TR Y
SOLUTIONS
1. (d)
2. (d) The additional 1t -molecular electron in 02-is added to
the less stable antibonding orbital. Hence, it is easy to
remove that electron from 0 2 - ion. Thus I.E. (l.P.) of
o 2 - is lowest.
3. (a) Continuous increase as no. of shells increases down
the group.
4. (b)
5. (b) Closed shell (Ne), half filled (P) and completely filled
con figuration (Mg) are the cause of higher value of
I.E.
6. (d)
7. (b) Cu, Ag and Au are coinage metals. They belong to group
1B (d-block) of periodic table.
8. (d) In fourth period filling up of 3d orbital becomes
energetically favourable before the 4p orbital is filled.
9. (b)
K + ? K 2+ +e -. Since e- is to be removed from stable
configuration.
10. (a) For isoelectronic species ionic radii decreases as the
charge on ion decreases. Further on moving down in
a group ionic radii increases. Hence the correct order
is
o2- < N3- < SP.< p3-
11. (c) On moving down a group atomic radii increases.
12. (a) Incoming electrons occupies the smaller n = 2 shell,
also negative charge on oxygen co -) is another fa ctor
due to which incoming electron fe el repulsion.
13. (c)
Hence electron repulsion outweigh the stability gained
by achieving noble gas configuration.
14. (b) It is electronic configuration of alkali metal. Hence it
wiLl form basic oxide.
15. (b) The right sequence ofl.E 1 of Li <B <Be< C.
16. (c) AI ( 3s 2 p1) and Mg (3s2 ). Lower energy is required to
remove 3p1 electron than 3s1 electron (penetrating
effect is s > p > d > f) . Secondly Mg has stable
electron ic con figuration than AI
17. (c) Ireprese ntsLi, Ti represen tsK
TJT represents Br, IV represents I
V represents He
18. (c)
So, amongst these, IT represents most reactive metal
and V represents least reactive non-metal.
19. (c) Generally, the ionization enthalpies or energy increases
from left to right in a period and decreases from top to
bottom in a group. Several fa ctor such as atomic radius,
nuclear charge, shielding effect are responsible for
change of ionization enthalpies.
Here, Ist ionization enthalpy of A and B is greater than
group I (Li 520 kJmol -1 to Cs374 kJmol -1), which means
element A and B belong to group -2 and all three given
ionization enthalpy values are less for element B means
B wi II come below A.
20. (b) For dissolution, Hydration energy> Lattice energy.
BaS04 is sparingly soluble in water because its
hydration energy is lesser than the lattice energy and
thus ions are not separated from each other.
21. (c) The anomalous behaviour of first member of a group
in the s- and p-block element is due to their small size,
large charge/radius ratio and high electronegativity.
22. (c) In 3d6 4s2 the differe ntiating electron enters d sub
shell, Hence it represents transition metal, Fe.
23. (b) In the isoelectronic species, all isoelectronic anions
belong to the same period and cations to the next
period.
24. (b) On passing from left to right in a period acidic character
of the normal oxides of the elements increases with
increase in electroneg ativity.
25. (a) The electronic configuration clearly suggests that
it is a d-block element (having configuration
(n -1) d
1- 1o
ns 0 - 2 ) which starts from III B and goes
till 11 B. Hence with d3 configuration it would be
classified in the group.
26. (b) Correct order of increasing basic strength is
N? > PH3 > AsH 3 > Sb? > BiH 3
27. (c) Electron affinity of 9F is less than that of 17Cl
28. (c) ns2 np 1 is the electronic configuration oflll A period.
Alp3 is amphoteric oxide
29. (d) pS+ has more effective nuclear charge and smaller size than
p3+_
30. (a) For isoelectron ic species, size of anion increases as
negative charge increases. Thus the correct order is
N3- > o 2- > F
(1.71) (1.40) (1.36)
31. (a) Hydrogen bonding increases the boiling point.
Hydrogen bonds are fo rmed in compounds having F
or 0 or N with hydrogen. S, Se, Te cannot undergo
hydrogen bond fo rmation because of their larger size
and lower electronegativity values.
DPP/CC03--------------------------------------? S-9 I
2
attracted toward each other through weak van der 32. (b) A. Li+<AJ 2 +<M g +<K+
d.. I The cation wi th the greater positive charge will Waal 's force hence van der Waal ra u are very arge.
have a smaller radius beca\Jse of the greater 38. ( c) According to Dobereneir 's triad the atomic mass ofBr
attraction of the electrons to the nucleus. Anion will be average of the atomic masses ofCI & I
with the greater negative charge will have the
larger radius.
I
Positive charge ex: . .
di IOUIC ra US
Negative charge ex: ionic radius
B. Greater positive charge, increases effective nuclear
charge in case ofisoelectronic speci es. While for
same group elements effective nuclear charge
decreases do·wn the groups.
C. Cl > F > Br >I
electron affinity ofCI is highest in halogen fa mily.
D. F>CI>Br>T
electroneg ativi tyoffluorine (F) is higher than Cl,
Brandl.
33. (a) Because AI is amphoteric in nature so it dissolve in
both acid and base.
34. (b) Pb2 + (5dJo 6s2 ), has the most stable +2 oxidation state
because here the d-orbital is completely tilled and is
more stable than Fe 2 + (3d>). Again Ag + ( 4d10) is more
stable as here again the d-orbital is completely filled
and Ag 2 + is not easily obtained. Pb2 + is more stable
compared to Sn2 + ( 4d10 5s 2 ) because of its large size.
35. (b)
36. (a) Iodine with lower atomic weight than that oftellurimn
(Group VI) was placed in Group VII along with fluorine,
chlorine, bromine because of similarities in properties.
37. (c) Covalent radius is radius of an atom in its bound state
i.e., in fluorine it is half of distance between two
covalently bonded fluorine atoms; van der Waal radii
is one-h alf of the distance between the nuclei of two
identical non-bonded isolated atoms. These atoms are
= 35.5+127 = 81 .25
2
39. (b) ·: For Na ? Na + + e- IE1 = 5.1 eV
:. For Na+ +e - ?Na EF= -5.1 eV
(because the reaction is reverse)
40. (b) N has half filled atomic orbital, which is more stable.
1
41. (d) Number of moles = -- 35. 5
42. (a)
Given, 1 eV = 23.06 kcal moJ- 1
3.7 eV = 3.7 x 23.06 kcal moi -1
i.e. I mole realease energy
= 3.7 x 23.06 kcal
:. Energy released
=
- 1-x 3.7 x 23.06 kcal = 2.4 kcal
35.5
43. (a) o - ion exerts a fo rce of repulsion on the incoming
electron. The energy is required to overcome it.
44. (d) Oxides ofEka-Aluminium = Gap 3
Oxides ofEka-Silicon = Si02
Melting point ofEka-Aiuminium = Low (302 K)
Melting point ofEka-Si Iicon = High ( 1231 K)
45. (b) In case ofhalogens covalent radius is considered this
bond is fo rmed by ove rlapping of electron clouds;
while noble gases remain monoatomic, in this case
only way to obtain radius is through van der Waal
radii.
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