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.

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.

In the isoelectronic species, all isoelectronic anions belong to the same period and cations to the next period.

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.

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.

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 = 15^th or $$V A.$$

Fully filled electronic configuration.

Effective nuclear charge (i,e. $$Z/e$$ ratio) decreases from
$${F^ - }$$ to $${N^{3 - }}$$  hence the radii follows the order :
$${F^ - } < {O^{2 - }} < {N^{3 - }}.\,Z/e\,\,{\text{for}}$$      $${F^ - } = \frac{9}{{10}} = 0.9,{\text{for}}\,\,{O^{2 - }} = \frac{8}{{10}} = .8,$$       $${\text{for}}\,\,{N^{3 - }} = \frac{7}{{10}} = 0.7$$

Generally, electronegativity decreases down the group as the size increases. This can also be formulated as:
$${\text{Electronegativity}}\, \propto \frac{1}{{{\text{size}}}}$$

$${K^ + } \to {K^2} + {e^ - }.$$    Since $${e^ - }$$  is to be removed from stable configuration.