Question

The electronic configurations of four elements are given below. Arrange these elements in the correct order of magnitudes (without sign) of their electron affinities:
$i.2s^{2}2p^5$ $ii.3s^{2}3p^5$
$iii.2s^{2}2p^4$ $iv.3s^{2}3p^4$
Select the correct answer using the codes given below:

• A. $\left(i\right)<\left(ii\right)<\left(iii\right)<\left(iv\right)$
• B. $\left(ii\right)>\left(i\right)>\left(iv\right)>\left(iii\right)$
• C. $\left(i\right)<\left(iii\right)<\left(iv\right)<\left(ii\right)$
• D. $\left(iii\right)>\left(iv\right)>\left(ii\right)>\left(i\right)$

Answer 1

The correct option is B $\left(ii\right)>\left(i\right)>\left(iv\right)>\left(iii\right)$

$i.2s^{2}2p^5\to 2s^{2}2p^6$
Addition of an electron leads to a stable configuration . So the electron affinity will be high.
$ii.3s^{2}3p^5\to 3s^{2}3p^6$
Electron affinity is higher than (i) because the electron repulsions would be higher in (i) (owing to its small size).
$iii.2s^{2}2p^4\to 2s^{2}2p^5$
Lesser energy will be released than (i) and (ii) because of its electronic configuration (The configuration is not as stable as $i$).
$iv.3s^{2}3p^4\to 3s^{2}3p^5$
It releases more energy than (iii) and less energy than (ii) (The configuration is not as stable as $ii$).