The first ionisation energy of oxygen is less | vedclass

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(C) The electronic configuration of nitrogen $(N)$ is $1s^2 2s^2 2p^3$,which is a stable half-filled $p$-orbital configuration.Oxygen $(O)$ has the co

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Greater inter-electron repulsion between two electrons in the same $p$-orbital counter balances the increase in effective nuclear charge on moving from nitrogen to oxygen

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(C) The electronic configuration of nitrogen $(N)$ is $1s^2 2s^2 2p^3$,which is a stable half-filled $p$-orbital configuration.Oxygen $(O)$ has the configuration $1s^2 2s^2 2p^4$.Although oxygen has a higher effective nuclear charge than nitrogen,the $2p^4$ configuration of oxygen experiences greater inter-electron repulsion between two electrons in the same $p$-orbital.This repulsion makes it easier to remove an electron from oxygen compared to the stable half-filled configuration of nitrogen,thus resulting in a lower first ionisation energy for oxygen.

$A_{(g)} \to A^{+}_{(g)} + e^-, \ A_1$	$B_{(g)} \to B^{+}_{(g)} + e^-, \ B_1$
$B^{+}_{(g)} \to B^{2+}_{(g)} + e^-, \ B_2$	$C_{(g)} \to C^{+}_{(g)} + e^-, \ C_1$
$C^{+}_{(g)} \to C^{2+}_{(g)} + e^-, \ C_2$	$C^{2+}_{(g)} \to C^{3+}_{(g)} + e^-, \ C_3$

The first ionisation energy of oxygen is less than that of nitrogen. Which of the following is the correct reason for this observation?

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