Classification of Elements and Periodicity in Properties MCQ Questions & Answers in Inorganic Chemistry | Chemistry
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131.
Which of the following elements will have highest second ionisation enthalpy?
A
$$1{s^2}2{s^2}2{p^6}3{s^2}$$
B
$$1{s^2}2{s^2}2{p^6}3{s^1}$$
C
$$1{s^2}2{s^2}2{p^6}3{s^2}3{p^2}$$
D
$$1{s^2}2{s^2}2{p^6}3{s^2}3{p^3}$$
Answer :
$$1{s^2}2{s^2}2{p^6}3{s^1}$$
After losing one electron, the atom will get a stable configuration and its second ionisation energy will be very high. Hence, option (B) is correct.
132.
Which has most stable $$ + 2$$ oxidation state :
A
$$Sn$$
B
$$Pb$$
C
$$Fe\,$$
D
$$Ag$$
Answer :
$$Pb$$
TIPS/Formulae :
(i) Ion having half filled or full filled orbital have extra stability.
(ii) Larger the size of cation more will be its stability $$P{b^{2 + }}\left( {5{d^{10}}6{s^2}} \right),$$ has the most stable $$ + 2$$ oxidation state because here the $$d$$-orbital is completely filled and is more stable than $$F{e^{2 + }}\left( {3{d^6}} \right).$$ Again $$A{g^ + }\left( {4{d^{10}}} \right)$$ is more stable as here again the $$d$$-orbital is completely filled and $$A{g^{2 + }}$$ is not easily obtained. $$P{b^{2 + }}$$ is more stable compared to $$S{n^{2 + }}\left( {4{d^{10}}5{s^2}} \right)$$ because of its large size.
133.
Which of the following statements is wrong ?
A
van der Waal’s radius of iodine is more than its covalent radius
B
All isoelectronic ions belong to same period of the periodic table
C
$$I.E{._1}$$ of $$N$$ is higher than that of $$O$$ while $$I.E{._2}$$ of $$O$$ is higher than that of $$N$$
D
The electron gain enthalpy of $$N$$ is almost zero while that of $$P$$ is $$74.3\,kJ\,mo{l^{ - 1}}$$
Answer :
All isoelectronic ions belong to same period of the periodic table
In the isoelectronic species, all isoelectronic anions belong to the same period and cations to the next period.
134.
Which is correct increasing order of their tendency of the given elements to form $${M^{3 - }}$$ ion?
A
$$Bi > Sb > As > P > N$$
B
$$Bi < Sb < As < P < N$$
C
$$N < P < Sb < Bi < As$$
D
$$Bi > Sb \sim N \sim P > As$$
Answer :
$$Bi < Sb < As < P < N$$
On moving down the group, the stability of - 3 oxidation state decreases. This is due to the following reasons :
(i) On descending a group the size of the atom or ion increases. As a result, attraction of the nucleus per newly added electron decreases. (ii) A large anion cannot fit easily into lattice of a small cation. (iii) As the negative charge on the ion increases, it becomes more and more susceptible to polarisation.
135.
Among $$A{l_2}{O_3},Si{O_2},{P_2}{O_3}\,{\text{and}}\,S{O_2}$$ the correct order of acid strength is
As the size increases the basic nature of oxides changes to acidic nature i.e., acidic nature increases.
$$\mathop {S{O_2} > {P_2}{O_3}}\limits_{Acidic} > \mathop {Si{O_2}}\limits_{Weak\,acidic} > \mathop {A{l_2}{O_3}}\limits_{Amphoteric} $$
$$\,S{O_2}\,{\text{and}}\,{P_2}{O_3}$$ are acidic as their corresponding acids $${H_2}S{O_3}\,{\text{and}}\,{H_3}P{O_3}\,$$ are strong acids.
136.
If the atomic number of an element is 33, it will be placed in the periodic table in the
A
first group
B
third group
C
fifth group
D
seventh group
Answer :
fifth group
The electronic configuration of element with atomic number 33 is $$1{s^2},2{s^2}2{p^6},3{s^2}3{p^6},4{s^2},3{d^{10}},4{p^3}.$$
As, its last shell have five electrons and hence, its group is 10 + 5 = 15th or $$V A.$$
139.
In general, the properties that decrease and increase down a group in the periodic table, respectively, are:
A
atomic radius and electronegativity.
B
electron gain enthalpy and electronegativity.
C
electronegativity and atomic radius.
D
electronegativity and electron gain enthalpy.
Answer :
electronegativity and atomic radius.
Generally, electronegativity decreases down the group as the size increases. This can also be formulated as:
$${\text{Electronegativity}}\, \propto \frac{1}{{{\text{size}}}}$$
140.
Which ionisation potential $$(IP)$$ in the following equations involves the greatest amount of energy ?
A
$$Na \to N{a^ + } + {e^ - }$$
B
$${K^ + } \to {K^{2 + }} + {e^ - }$$
C
$${C^{2 + }} \to {C^{3 + }} + {e^ - }$$
D
$$C{a^ + } \to C{a^{2 + }} + {e^ - }$$
Answer :
$${K^ + } \to {K^{2 + }} + {e^ - }$$
$${K^ + } \to {K^2} + {e^ - }.$$ Since $${e^ - }$$ is to be removed from stable configuration.