Question

Which of the following order(s) for electron affinity is(are) correct?

(a)$S>O<Se$

(b)$CI>F$

(c)$S>O$

(d)$O>S$

(e)$N>P$

(f)$C>N$

• A. $a,b,c,e$
• B. $a,b,c,f$
• C. $b,c,d,e$
• D. $b,c,f$

Answer 1

The correct option is B $a,b,c,f$

In a period, with the increase in atomic number, electron affinity becomes more negative due to the increase in the effective nuclear charge. In a group, it becomes less negative on moving from top to bottom.

N has positive electron affinity due to the extra stability of exactly half-filled p orbitals.
In case of $F^{-}$ ion and $O^{6-}$ ion, due to small size, the inter-electronic repulsion is higher. Hence, less energy is evolved in the formation of the ion.

Due to this, the orders (b) CI > F and (c) S > O is correct.

The orders a, b, c, and f are correct whereas the orders d and e are incorrect.

Hence, the correct option is $\text{B}$