Question

Arrange the elements with the following electronic configurations in the increasing order of electron affinity

(i) $1s^{2}2s^{2}2p^3$ (ii) $1s^{2}2s^{2}2p^4$ (iii) $1s^{2}2s^{2}2p^5$ (iv) $1s^{2}2s^{2}2p^{6}3s^{2}3p^4$

• A. i < iv < ii< iii
• B. iv < iii < ii < i
• C. i < ii < iv < iii
• D. ii < iii < i < iv

Answer 1

The correct option is C

i < ii < iv < iii

The elements are

(i) N (ii) O (iii) F (iv) S

Now, we can arrange them in the order of electron affinity. Nitrogen has a negative electron affinity - meaning, it would "take" energy to put in one more electron into the already exactly half-filled $2p^3$ orbital. We can consider the electron affinity to be 0 as per the recent convention. Flourine is the most electronegative element and from the given elements, it will have the highest EA value. Sulphur has a higher electron affinity than oxygen. This is an anomaly.

N < O < S < F