pBlock ElementsII 1 Question 56
6. When the first electron gain enthalpy $\left(\Delta_{e_{g}} H\right)$ of oxygen is $-141 \mathrm{~kJ} / \mathrm{mol}$, its second electron gain enthalpy is
(2019 Main, 9 Jan II)
(a) a positive value
(b) a more negative value than the first
(c) almost the same as that of the first
(d) negative, but less negative than the first
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Solution:
- As given, the first electron gain enthalpy of oxygen can be shown as,
$$ \mathrm{O}(g)+e^{-} \longrightarrow \mathrm{O}^{-}(g) \Delta e_{g} H_{1}=-141 \mathrm{~kJ} / \mathrm{mol} $$
The expression of second electron gain enthalpy of oxygen will be,
$$ \mathrm{O}^{-}(g)+e^{-} \longrightarrow \mathrm{O}^{2-}(g) \quad \Delta e_{g} H_{2}=+ \text { ve } $$
$\Delta e_{g} H_{2}$ of oxygen is positive, i.e. endothermic, because a strong electrostatic repulsion will be observed between highy negative $\mathrm{O}^{-}$ and the incoming electron $\left(e^{-}\right)$. A very high amount of energy will be consumed (endothermic) by the system to overcome the electrostatic repulsion.
