The first ionisation energy of Na,B,N and O a | vedclass

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(B) $Na$ is an alkali metal and has the lowest ionisation energy due to its large atomic size. $B$,$N$,and $O$ are non-metals belonging to the same pe

Vedclass · Accepted Answer

$Na < B < O < N$

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(B) $Na$ is an alkali metal and has the lowest ionisation energy due to its large atomic size. $B$,$N$,and $O$ are non-metals belonging to the same period ($2^{nd}$ period). Generally,ionisation energy increases from left to right across a period due to an increase in effective nuclear charge. However,$N$ $(1s^2 2s^2 2p^3)$ has a stable half-filled $p$-orbital configuration,which makes its ionisation energy higher than that of $O$ $(1s^2 2s^2 2p^4)$. Therefore,the correct order is $Na < B < O < N$.

$(IP)_1$	$7.1 \ eV$
$(IP)_2$	$10.3 \ eV$
$(IP)_3$	$36.2 \ eV$
$(IP)_4$	$39.1 \ eV$

The first ionisation energy of $Na$,$B$,$N$ and $O$ atoms follows the order

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