S-block Elements Practice Questions - DPP for NEET

 Page 1


1. (d)    Alkali metals are highly electropositive  and thus highly reducing
.Therefore reduction , double decomposition and  displacement 
methods for their extraction are not suitable .Only electrolytic
methods are useful for  their extraction.
2. (a) A reducing agent is a substance which can loose electron and
hence a reducing agent should have low ionisation energy. Now
since ionisation energy decreases from Li to Cs, the reducing
property should increase from Li to Cs. The only exception to this
is lithium. This is because the net process of converting an atom to
an ion takes place in 3 steps.
(i)  M(s) ? M(g) ?H = Sublimation energy
(ii)  M(g) ? M
+
(g) + e
–
?H = Ionisation energy
(iii)  M
+
(g)+H
2
O ? M
+
 (aq) ?H = Hydration energy
The large amount of energy liberated in hydration of Li (because of its
small size) makes the overall ?H negative. This accounts for the
higher oxidation potential of lithium i.e., its high reducing power.
3. (a) According to Fajan rules, ionic character increases with increase in
size of the cation and decrease in size of the anion. Thus, CsF has
higher ionic character than NaCl and hence bond in CsF is stronger
than in NaCl.
4. (b) BaCO
3
 forms a yellow ppt of barium chromate. BaCO
3
 forms a
white precipitate of BaSO
4
. BaCl
2
 is soluble in water.
5. (a) The weaker the base, the less stable is its carbonate. Since LiOH is
the weakest base, hence Li
2
CO
3
 has the lowest thermal stability.
6. (b) As outermost electronic configuration of alkali metals is ns
1
 and also
their size are largest in their respective periods so their 1st I.E will
be lowest among the given options. As second electron is to be
removed from complete shell or noble gas core, so the IInd I.E.
must be highest among the given options. So, option (b) is correct
Page 2


1. (d)    Alkali metals are highly electropositive  and thus highly reducing
.Therefore reduction , double decomposition and  displacement 
methods for their extraction are not suitable .Only electrolytic
methods are useful for  their extraction.
2. (a) A reducing agent is a substance which can loose electron and
hence a reducing agent should have low ionisation energy. Now
since ionisation energy decreases from Li to Cs, the reducing
property should increase from Li to Cs. The only exception to this
is lithium. This is because the net process of converting an atom to
an ion takes place in 3 steps.
(i)  M(s) ? M(g) ?H = Sublimation energy
(ii)  M(g) ? M
+
(g) + e
–
?H = Ionisation energy
(iii)  M
+
(g)+H
2
O ? M
+
 (aq) ?H = Hydration energy
The large amount of energy liberated in hydration of Li (because of its
small size) makes the overall ?H negative. This accounts for the
higher oxidation potential of lithium i.e., its high reducing power.
3. (a) According to Fajan rules, ionic character increases with increase in
size of the cation and decrease in size of the anion. Thus, CsF has
higher ionic character than NaCl and hence bond in CsF is stronger
than in NaCl.
4. (b) BaCO
3
 forms a yellow ppt of barium chromate. BaCO
3
 forms a
white precipitate of BaSO
4
. BaCl
2
 is soluble in water.
5. (a) The weaker the base, the less stable is its carbonate. Since LiOH is
the weakest base, hence Li
2
CO
3
 has the lowest thermal stability.
6. (b) As outermost electronic configuration of alkali metals is ns
1
 and also
their size are largest in their respective periods so their 1st I.E will
be lowest among the given options. As second electron is to be
removed from complete shell or noble gas core, so the IInd I.E.
must be highest among the given options. So, option (b) is correct
choice.
7. (a) As we move down the group, the lattice energies of carbonates
remain approximately the same. However the hydration energies
of the metal cation decreases from Be
2+
 to Ba
2+
,
 
hence the
solubilities of carbonates of the alkaline earth metal decrease down
the group mainly due to decreasing hydration energies of the
cations  from Be
2+ 
to Ba
2+
.
8. (a) 4KO
2
 + 2 CO
2
 ? 2 K
2
CO
3  
+ 3O
2
 .
KO
2
 is used as an oxidising agent. It is used as air purifier in space
capsules. Submarines and breathing masks as it produces oxygen
and remove carbon dioxide.
9. (d) Halides of group I and II impart characteristic colour to the flame
due to low IE of the central atom. However, ionization energy of
Be and Mg atoms is high due to their small size, hence they can't
be excited to higher levels by Bunsen burner flame. Thus, their
halides don't impart colour to flame
10. (c) According to Fajan's rule,
Size of cation  Ionic character.
Recall that size of metal (ion) increases while going down the group,
and decreases on crossing the periods from left to right. Thus Rb
+
(group I) is largest and Be
2+
 (group II)  is smallest in size. Hence
RbCl has greatest, and BeCl
2
 has lowest ionic character.
11. (d) Na
2
O
2
 is peroxide of sodium not super oxide. The formula of
sodium superoxide is NaO
2
.
12. (c) Ca and CaH
2
 both react with H
2
O to form H
2
 gas,
whereas K gives H
2
 while KO
2
 gives O
2
 and H
2
O
2
2K + 2 H
2
O  2KOH + H
2
Similarly, Na gives H
2
 while Na
2
 O
2
 gives H
2
O
2
Page 3


1. (d)    Alkali metals are highly electropositive  and thus highly reducing
.Therefore reduction , double decomposition and  displacement 
methods for their extraction are not suitable .Only electrolytic
methods are useful for  their extraction.
2. (a) A reducing agent is a substance which can loose electron and
hence a reducing agent should have low ionisation energy. Now
since ionisation energy decreases from Li to Cs, the reducing
property should increase from Li to Cs. The only exception to this
is lithium. This is because the net process of converting an atom to
an ion takes place in 3 steps.
(i)  M(s) ? M(g) ?H = Sublimation energy
(ii)  M(g) ? M
+
(g) + e
–
?H = Ionisation energy
(iii)  M
+
(g)+H
2
O ? M
+
 (aq) ?H = Hydration energy
The large amount of energy liberated in hydration of Li (because of its
small size) makes the overall ?H negative. This accounts for the
higher oxidation potential of lithium i.e., its high reducing power.
3. (a) According to Fajan rules, ionic character increases with increase in
size of the cation and decrease in size of the anion. Thus, CsF has
higher ionic character than NaCl and hence bond in CsF is stronger
than in NaCl.
4. (b) BaCO
3
 forms a yellow ppt of barium chromate. BaCO
3
 forms a
white precipitate of BaSO
4
. BaCl
2
 is soluble in water.
5. (a) The weaker the base, the less stable is its carbonate. Since LiOH is
the weakest base, hence Li
2
CO
3
 has the lowest thermal stability.
6. (b) As outermost electronic configuration of alkali metals is ns
1
 and also
their size are largest in their respective periods so their 1st I.E will
be lowest among the given options. As second electron is to be
removed from complete shell or noble gas core, so the IInd I.E.
must be highest among the given options. So, option (b) is correct
choice.
7. (a) As we move down the group, the lattice energies of carbonates
remain approximately the same. However the hydration energies
of the metal cation decreases from Be
2+
 to Ba
2+
,
 
hence the
solubilities of carbonates of the alkaline earth metal decrease down
the group mainly due to decreasing hydration energies of the
cations  from Be
2+ 
to Ba
2+
.
8. (a) 4KO
2
 + 2 CO
2
 ? 2 K
2
CO
3  
+ 3O
2
 .
KO
2
 is used as an oxidising agent. It is used as air purifier in space
capsules. Submarines and breathing masks as it produces oxygen
and remove carbon dioxide.
9. (d) Halides of group I and II impart characteristic colour to the flame
due to low IE of the central atom. However, ionization energy of
Be and Mg atoms is high due to their small size, hence they can't
be excited to higher levels by Bunsen burner flame. Thus, their
halides don't impart colour to flame
10. (c) According to Fajan's rule,
Size of cation  Ionic character.
Recall that size of metal (ion) increases while going down the group,
and decreases on crossing the periods from left to right. Thus Rb
+
(group I) is largest and Be
2+
 (group II)  is smallest in size. Hence
RbCl has greatest, and BeCl
2
 has lowest ionic character.
11. (d) Na
2
O
2
 is peroxide of sodium not super oxide. The formula of
sodium superoxide is NaO
2
.
12. (c) Ca and CaH
2
 both react with H
2
O to form H
2
 gas,
whereas K gives H
2
 while KO
2
 gives O
2
 and H
2
O
2
2K + 2 H
2
O  2KOH + H
2
Similarly, Na gives H
2
 while Na
2
 O
2
 gives H
2
O
2
Likewise Ba gives H
2
 while BaO
2
 gives H
2
O
2
Ba + 2 H
2
O  Ba (OH)
2
 + H
2
BaO
2
 + 2H
2
O  Ba (OH)
2
 + H
2 
O
2
13. (d) 2NaOH  
  
  
14. (b)
 
 Correct choice : (b)
15. (d) In Castner Kellner cell, sodium amalgam is formed at mercury
cathode.
16. (b)
17. (d) K and Mg, both form oxides
Mg form nitride also 
K does not form nitride.
18. (c) The atom becomes larger on descending the group, so the bonds
Page 4


1. (d)    Alkali metals are highly electropositive  and thus highly reducing
.Therefore reduction , double decomposition and  displacement 
methods for their extraction are not suitable .Only electrolytic
methods are useful for  their extraction.
2. (a) A reducing agent is a substance which can loose electron and
hence a reducing agent should have low ionisation energy. Now
since ionisation energy decreases from Li to Cs, the reducing
property should increase from Li to Cs. The only exception to this
is lithium. This is because the net process of converting an atom to
an ion takes place in 3 steps.
(i)  M(s) ? M(g) ?H = Sublimation energy
(ii)  M(g) ? M
+
(g) + e
–
?H = Ionisation energy
(iii)  M
+
(g)+H
2
O ? M
+
 (aq) ?H = Hydration energy
The large amount of energy liberated in hydration of Li (because of its
small size) makes the overall ?H negative. This accounts for the
higher oxidation potential of lithium i.e., its high reducing power.
3. (a) According to Fajan rules, ionic character increases with increase in
size of the cation and decrease in size of the anion. Thus, CsF has
higher ionic character than NaCl and hence bond in CsF is stronger
than in NaCl.
4. (b) BaCO
3
 forms a yellow ppt of barium chromate. BaCO
3
 forms a
white precipitate of BaSO
4
. BaCl
2
 is soluble in water.
5. (a) The weaker the base, the less stable is its carbonate. Since LiOH is
the weakest base, hence Li
2
CO
3
 has the lowest thermal stability.
6. (b) As outermost electronic configuration of alkali metals is ns
1
 and also
their size are largest in their respective periods so their 1st I.E will
be lowest among the given options. As second electron is to be
removed from complete shell or noble gas core, so the IInd I.E.
must be highest among the given options. So, option (b) is correct
choice.
7. (a) As we move down the group, the lattice energies of carbonates
remain approximately the same. However the hydration energies
of the metal cation decreases from Be
2+
 to Ba
2+
,
 
hence the
solubilities of carbonates of the alkaline earth metal decrease down
the group mainly due to decreasing hydration energies of the
cations  from Be
2+ 
to Ba
2+
.
8. (a) 4KO
2
 + 2 CO
2
 ? 2 K
2
CO
3  
+ 3O
2
 .
KO
2
 is used as an oxidising agent. It is used as air purifier in space
capsules. Submarines and breathing masks as it produces oxygen
and remove carbon dioxide.
9. (d) Halides of group I and II impart characteristic colour to the flame
due to low IE of the central atom. However, ionization energy of
Be and Mg atoms is high due to their small size, hence they can't
be excited to higher levels by Bunsen burner flame. Thus, their
halides don't impart colour to flame
10. (c) According to Fajan's rule,
Size of cation  Ionic character.
Recall that size of metal (ion) increases while going down the group,
and decreases on crossing the periods from left to right. Thus Rb
+
(group I) is largest and Be
2+
 (group II)  is smallest in size. Hence
RbCl has greatest, and BeCl
2
 has lowest ionic character.
11. (d) Na
2
O
2
 is peroxide of sodium not super oxide. The formula of
sodium superoxide is NaO
2
.
12. (c) Ca and CaH
2
 both react with H
2
O to form H
2
 gas,
whereas K gives H
2
 while KO
2
 gives O
2
 and H
2
O
2
2K + 2 H
2
O  2KOH + H
2
Similarly, Na gives H
2
 while Na
2
 O
2
 gives H
2
O
2
Likewise Ba gives H
2
 while BaO
2
 gives H
2
O
2
Ba + 2 H
2
O  Ba (OH)
2
 + H
2
BaO
2
 + 2H
2
O  Ba (OH)
2
 + H
2 
O
2
13. (d) 2NaOH  
  
  
14. (b)
 
 Correct choice : (b)
15. (d) In Castner Kellner cell, sodium amalgam is formed at mercury
cathode.
16. (b)
17. (d) K and Mg, both form oxides
Mg form nitride also 
K does not form nitride.
18. (c) The atom becomes larger on descending the group, so the bonds
becomes weaker (metallic bond), the cohesive force/energy
decreases and accordingly melting point also decreases.
19. (d) Lesser the lattice energy, more will be the solubility in H
2
O
.
20. (b)
All alkali metals decompose water with the evolution of hydrogen.
Be + 2H
2
O  No reaction
Ca, Sr, Ba and Ra decompose cold water readily with evolution of
hydrogen. Mg decomposes boiling water but Be is not attacked by
water even at high temperatures as its oxidation potential is lower
than the other members.
21. (c) Monovalent sodium and potassium ions and divalent magnesium
and calcium ions are found in large proportions in biological
fluids.
22. (b) Alkali metals readily lose electron to give monovalent M
+
 ion.
Hence they are never found in free state in nature.
23. (d) In alkaline earth metals, ionic size increases down the group. The
lattice energy remains constant because sulphate  ion is so large, so
that small change in cationic size does not make any difference.
On moving down the group the degree of hydration of metal ions
decreases very much leading to decrease in solubility.
? 
24. (a) The alkali metals dissolve in liquid ammonia without evolution of
hydrogen. The metal loses electrons and combine with ammonia
molecule.
M ?   M
+ 
(in liquid ammonia) + e
–
 (ammoniated)
M + (x + y) NH
3 
?[M(NH
3
)
x
]
+ 
+   e
–
(NH
3
)
y
Solvated electron
It is ammoniated electron which is responsible for colour.
Page 5


1. (d)    Alkali metals are highly electropositive  and thus highly reducing
.Therefore reduction , double decomposition and  displacement 
methods for their extraction are not suitable .Only electrolytic
methods are useful for  their extraction.
2. (a) A reducing agent is a substance which can loose electron and
hence a reducing agent should have low ionisation energy. Now
since ionisation energy decreases from Li to Cs, the reducing
property should increase from Li to Cs. The only exception to this
is lithium. This is because the net process of converting an atom to
an ion takes place in 3 steps.
(i)  M(s) ? M(g) ?H = Sublimation energy
(ii)  M(g) ? M
+
(g) + e
–
?H = Ionisation energy
(iii)  M
+
(g)+H
2
O ? M
+
 (aq) ?H = Hydration energy
The large amount of energy liberated in hydration of Li (because of its
small size) makes the overall ?H negative. This accounts for the
higher oxidation potential of lithium i.e., its high reducing power.
3. (a) According to Fajan rules, ionic character increases with increase in
size of the cation and decrease in size of the anion. Thus, CsF has
higher ionic character than NaCl and hence bond in CsF is stronger
than in NaCl.
4. (b) BaCO
3
 forms a yellow ppt of barium chromate. BaCO
3
 forms a
white precipitate of BaSO
4
. BaCl
2
 is soluble in water.
5. (a) The weaker the base, the less stable is its carbonate. Since LiOH is
the weakest base, hence Li
2
CO
3
 has the lowest thermal stability.
6. (b) As outermost electronic configuration of alkali metals is ns
1
 and also
their size are largest in their respective periods so their 1st I.E will
be lowest among the given options. As second electron is to be
removed from complete shell or noble gas core, so the IInd I.E.
must be highest among the given options. So, option (b) is correct
choice.
7. (a) As we move down the group, the lattice energies of carbonates
remain approximately the same. However the hydration energies
of the metal cation decreases from Be
2+
 to Ba
2+
,
 
hence the
solubilities of carbonates of the alkaline earth metal decrease down
the group mainly due to decreasing hydration energies of the
cations  from Be
2+ 
to Ba
2+
.
8. (a) 4KO
2
 + 2 CO
2
 ? 2 K
2
CO
3  
+ 3O
2
 .
KO
2
 is used as an oxidising agent. It is used as air purifier in space
capsules. Submarines and breathing masks as it produces oxygen
and remove carbon dioxide.
9. (d) Halides of group I and II impart characteristic colour to the flame
due to low IE of the central atom. However, ionization energy of
Be and Mg atoms is high due to their small size, hence they can't
be excited to higher levels by Bunsen burner flame. Thus, their
halides don't impart colour to flame
10. (c) According to Fajan's rule,
Size of cation  Ionic character.
Recall that size of metal (ion) increases while going down the group,
and decreases on crossing the periods from left to right. Thus Rb
+
(group I) is largest and Be
2+
 (group II)  is smallest in size. Hence
RbCl has greatest, and BeCl
2
 has lowest ionic character.
11. (d) Na
2
O
2
 is peroxide of sodium not super oxide. The formula of
sodium superoxide is NaO
2
.
12. (c) Ca and CaH
2
 both react with H
2
O to form H
2
 gas,
whereas K gives H
2
 while KO
2
 gives O
2
 and H
2
O
2
2K + 2 H
2
O  2KOH + H
2
Similarly, Na gives H
2
 while Na
2
 O
2
 gives H
2
O
2
Likewise Ba gives H
2
 while BaO
2
 gives H
2
O
2
Ba + 2 H
2
O  Ba (OH)
2
 + H
2
BaO
2
 + 2H
2
O  Ba (OH)
2
 + H
2 
O
2
13. (d) 2NaOH  
  
  
14. (b)
 
 Correct choice : (b)
15. (d) In Castner Kellner cell, sodium amalgam is formed at mercury
cathode.
16. (b)
17. (d) K and Mg, both form oxides
Mg form nitride also 
K does not form nitride.
18. (c) The atom becomes larger on descending the group, so the bonds
becomes weaker (metallic bond), the cohesive force/energy
decreases and accordingly melting point also decreases.
19. (d) Lesser the lattice energy, more will be the solubility in H
2
O
.
20. (b)
All alkali metals decompose water with the evolution of hydrogen.
Be + 2H
2
O  No reaction
Ca, Sr, Ba and Ra decompose cold water readily with evolution of
hydrogen. Mg decomposes boiling water but Be is not attacked by
water even at high temperatures as its oxidation potential is lower
than the other members.
21. (c) Monovalent sodium and potassium ions and divalent magnesium
and calcium ions are found in large proportions in biological
fluids.
22. (b) Alkali metals readily lose electron to give monovalent M
+
 ion.
Hence they are never found in free state in nature.
23. (d) In alkaline earth metals, ionic size increases down the group. The
lattice energy remains constant because sulphate  ion is so large, so
that small change in cationic size does not make any difference.
On moving down the group the degree of hydration of metal ions
decreases very much leading to decrease in solubility.
? 
24. (a) The alkali metals dissolve in liquid ammonia without evolution of
hydrogen. The metal loses electrons and combine with ammonia
molecule.
M ?   M
+ 
(in liquid ammonia) + e
–
 (ammoniated)
M + (x + y) NH
3 
?[M(NH
3
)
x
]
+ 
+   e
–
(NH
3
)
y
Solvated electron
It is ammoniated electron which is responsible for colour.
25. (c) Mg is more reducing in nature than carbon