The least stable ion is

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(B) The electronic configuration of $Li$ $(Z=3)$ is $1s^2 \, 2s^1$. Adding one electron forms $Li^-$ $(1s^2 \, 2s^2)$,which is stable due to a fully f

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$Be^{-}$

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(B) The electronic configuration of $Li$ $(Z=3)$ is $1s^2 \, 2s^1$. Adding one electron forms $Li^-$ $(1s^2 \, 2s^2)$,which is stable due to a fully filled $2s$ subshell.The electronic configuration of $Be$ $(Z=4)$ is $1s^2 \, 2s^2$. Adding one electron forms $Be^-$ $(1s^2 \, 2s^2 \, 2p^1)$. This process disrupts the stable,fully filled $2s^2$ configuration of $Be$,making $Be^-$ the least stable ion.The electronic configuration of $B$ $(Z=5)$ is $1s^2 \, 2s^2 \, 2p^1$. Adding one electron forms $B^-$ $(1s^2 \, 2s^2 \, 2p^2)$,which is relatively stable.The electronic configuration of $C$ $(Z=6)$ is $1s^2 \, 2s^2 \, 2p^2$. Adding one electron forms $C^-$ $(1s^2 \, 2s^2 \, 2p^3)$,which is stable due to the half-filled $p$-subshell.

Similar Questions

When a chloride ion is formed from an isolated gaseous chlorine atom,$3.8 \ eV$ of energy is released. This value is equal to the:

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The first electron affinity of $C, N$ and $O$ will be of the order

The formation of the $O^{2-}$ ion is first exothermic and then endothermic,as shown by the following reaction steps:
$O_{(g)} + e^- \to O^-_{(g)}; \Delta H^o = -142 \ kJ \ mol^{-1}$
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This is due to:

The correct order of electron affinity of $B, C, N, O$ is

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