Question

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> The electron gain enthalpy (in $kJ/mol$) of fluorine, chlorine, bromine and iodine, respectively, are :

JEE MAIN - 2020

• A. $-333,-349,-325$ and $-296$
• B. $-333,-325,-349$ and $-296$
• C. $-349,-333,-325$ and $-296$
• D. $-296,-325,-333$ and $-349$

Answer 1

The correct option is A $-333,-349,-325$ and $-296$

The electron gain enthalpy of $F$ is less negative than that of $Cl$ because of its small size. The additional $e^{\ominus }$ in case of $F$ will be added to $2p$-orbital while in case of $Cl$ it will go in the $3p$-orbital.

As a $2p$-orbital occupies smaller region of space than $3p$-orbital, it will be more compact and $e^{\ominus }-e^{\ominus }$ repulsion would be more. Hence $e^{\ominus }$ gain enthalpy of $F$ is less than $Cl$.

On going down the group from $Cl$ to $I,$ due to decrease in size, $e^{\ominus }$ gain enthalpy decreases.

So order of $e^{\ominus }$ gain enthalpy,
$Cl>F>Br>I$
Correct option is (a).