If energy absorbed for conversion of M_{(g)} | vedclass

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(C) The energy absorbed for the conversion of $M_{(g)}$ to $M_{(g)}^{+}$ is the first ionization energy $(IE_1)$.For the elements $Li, Be, B, C, N$ (p

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$a < c < b < d < e$

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(C) The energy absorbed for the conversion of $M_{(g)}$ to $M_{(g)}^{+}$ is the first ionization energy $(IE_1)$.For the elements $Li, Be, B, C, N$ (period $2$),the general trend of $IE_1$ increases across the period due to increasing effective nuclear charge.However,there are exceptions due to stable electronic configurations:$Li (1s^2 2s^1) = a$$Be (1s^2 2s^2) = b$ (fully filled $s$-orbital,higher than $B$)$B (1s^2 2s^2 2p^1) = c$$C (1s^2 2s^2 2p^2) = d$$N (1s^2 2s^2 2p^3) = e$ (half-filled $p$-orbital,higher than $O$)The order is $Li < B < Be < C < N$,which corresponds to $a < c < b < d < e$.

$B$	$801$
$Al$	$577$
$Ga$	$579$
$In$	$558$
$Tl$	$589$

If energy absorbed for conversion of $M_{(g)}$ into $M_{(g)}^{+}$ for $Li, Be, B, C$ and $N$ is $a, b, c, d$ and $e$ (in $k\,cal/mol$) respectively,then the correct order will be:

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The first ionization enthalpy values (in $kJ\,mol^{-1}$) of group $13$ elements are :

$B$ $801$

$Al$ $577$

$Ga$ $579$

$In$ $558$

$Tl$ $589$

How would you explain this deviation from the general trend?

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