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(A) The first ionization enthalpy generally increases across a period from left to right due to an increase in effective nuclear charge.However,there

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$Li < B < Be < C < N$

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(A) The first ionization enthalpy generally increases across a period from left to right due to an increase in effective nuclear charge.However,there are exceptions due to electronic configuration stability:$1$. $Be$ $(1s^2 2s^2)$ has a fully filled $2s$ orbital,making it more stable than $B$ $(1s^2 2s^2 2p^1)$,so $Be > B$.$2$. $N$ $(1s^2 2s^2 2p^3)$ has a half-filled $2p$ subshell,which is more stable than $C$ $(1s^2 2s^2 2p^2)$,so $N > C$.The correct increasing order is $Li 						Element			First ionization enthalpy $(\Delta_i H / kJ \ mol^{-1})$							$Li$			$520$							$Be$			$899$							$B$			$801$							$C$			$1086$							$N$			$1402$

Arrange the following elements in increasing order of first ionization enthalpy:
$Li, Be, B, C, N$
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The correct order of first ionization enthalpy values of the following elements is:
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In which of the following electronic configurations will there be a large difference between the second and third ionization energy?

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The correct option among the following is:

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Arrange the following elements in increasing order of first ionization enthalpy:$Li, Be, B, C, N$Choose the correct answer from the options given below: