The ionization energy of boron is less than t | vedclass

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(C) The electronic configuration of $Be$ $(Z=4)$ is $1s^2 2s^2$,which represents a fully filled $2s$ subshell,providing extra stability.The electronic

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the outermost electron in boron occupies a $2p$ orbital

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(C) The electronic configuration of $Be$ $(Z=4)$ is $1s^2 2s^2$,which represents a fully filled $2s$ subshell,providing extra stability.The electronic configuration of $B$ $(Z=5)$ is $1s^2 2s^2 2p^1$.In $B$,the outermost electron is in the $2p$ orbital,which is shielded by the $2s^2$ electrons and is further from the nucleus than the $2s$ electrons of $Be$.Therefore,it is easier to remove the $2p$ electron from $B$ than the $2s$ electron from $Be$,resulting in a lower ionization energy for $B$ compared to $Be$.

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

The ionization energy of boron is less than that of beryllium because

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