The increasing order of electron affinity of the electronic configurations of element is:
(I) $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{5}$
(II) $1 s^{2} 2 s^{2} 2 p^{3}$
(III) $1 s^{2} 2 s^{2} 2 p^{5}$
(IV) $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{1}$

I I I > I V > I I I > I

No worries! We‘ve got your back. Try BYJU‘S free classes today!

I I I > I V > I I > I

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

I > I V > I I I > I I

No worries! We‘ve got your back. Try BYJU‘S free classes today!

I V > I I I I > I I

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $I I I > I V > I I > I$
The affinity of electron depends on the outer shell configuration of an element. From given configurations. $I I I$ will have highest electron affinity as $2 p$ is more closer to the nucleus and it is deficient of $1$ electron to complete its octet This will be followed by $I V$ where $3 s$ orbital requires $1$ electron. $I I$ will have less electron affinity than $I V$ because half-filled p orbital is stable. $I$ will have the least electron affinity as $3 p$ is comparatively farther from nucleus. Hence, the required order is:
$I I I > I V > I I > I$

Suggest Corrections

Similar questions

I : 1 s^{2} 2 S^{2}

I I : 1 s^{2} 2 s^{2} 2 p^{6}

I I : 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2}

I V : 1 s^{2} 2 s^{2} 2 p^{3}

V : 1 s^{2} 2 s^{2} 2 p^{5}

I . E .

P \to 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{3} Q \to 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{1} R \to 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{10} 4 s^{2} 4 p^{3} S \to 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{10} 4 s^{2} 4 p^{1}

Join BYJU'S Learning Program