Question

$\begin{array}{cc}\text{List I}& \text{List II}\\ \text{(A) N, F, O, Li}& \text{(P) Increasing order of |EA|}\\ \text{(B) F, N, Si, Mg, Rb}& \text{(Q) Increasing order of}{\text{IE}}_1\\ \text{(C) K, Na, Al, Mg}& {\text{(R) Increasing order of IE}}_2\\ \text{(D) N, B, O, Cl}& \text{(S) Incresing order of radius}\end{array}$
Which of the following is the only CORRECT combination?

• A. A-S; B-R; C-P; D-Q
• B. A-S; B-Q; C-Q; D-P
• C. A-R; B-S; C-Q; D-P
• D. A-P; B-Q; C-R; D-S

Answer 1

The correct option is C A-R; B-S; C-Q; D-P

(A) Ionisation energy increases from left to right in a period due to increase in effective nuclear charge which bound the outermost electrons strongly. However, due to noble gas configuration of $(L{i}^{+}:1s^2)$ , it has the highest second ionization energy followed by $(O^{+}:2s^{2}2p^3)$ which has half filled electronic configration. Hence, $I{E}_2$ for the given elements follows the order: $N<F<O<Li$

(B) On moving along the group from top to bottom atomic radius increases and while moving left to right in a preriod it decreases. Hence the radius follows the order: $F<N<Si<Mg<Rb$

(C) Ionisation energy decreases down the group due to increase in number of shells and increases as we move from left to right. However due to filled electronic configuration of $Mg$, its IE will be greater than $Al$.
$K<Na<Al<Mg$

(D) $EA$ refers the amount of energy released when an electron is added to a neutral atom. Electron affinity increases as we move from left to right along a period. However, due to half filled configuration, $N$ has the least EA. So correct order of electron affinity will be: $N<B<O<Cl$