Its value is negative and the ease with which an atom accepts electrons its electron gain enthalpy becomes more negative that is, its value increases.
With an increase in nuclear charge, the electron gain enthalpy also increases.
Due to the extra stability of nitrogen (it has half-filled valence shell $[ He ] \,1 s ^2 \,2 p ^3$, which provides extra stability to it), it is very difficult to add an electron to it.
Therefore, it has almost zero electron gain enthalpy.
In general, on moving down the group, the electron gain enthalpy decreases due to an increase in the atomic size, but on moving from $F$ to $Cl$ the electron gain enthalpy increases.
Due to the very small size of $F$ atom and absence of $3^{\text {rd }}$ shell, the incoming electron experiences repulsion due to the valence electrons and its electron gain enthalpy is less than $Cl$. After $Cl$ the electron gain enthalpy follows the general order and decreases on moving down to $Br$ and $I$.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$2H_2(g) + O_2(g) \to 2H_2O(l)$ ; $\Delta _fH^o_{298}(H_2O(l)) = -285.5\, kJ/mol$
What is $\Delta S^o_{298}$ for the given fuel cell reaction ?
Given $: O_2(g) + 4H^+(aq) + 4e^- \to 2H_2O(l)$ $E^o = 1.23\, V$
$PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)$
If total pressure at equilibrium of the reaction mixture is $P$ and degree of dissociation of $PCl_5$ is $x,$ the partial pressure of $PCl_3$ will be