The second ionisation energies of Li, Be, B a | vedclass

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(B) The second ionisation energy is the energy required to remove an electron from a $M^{+}$ cation in the gaseous state: $X^{+}_{(g)} \rightarrow X^{

Vedclass · Accepted Answer

$Li > B > C > Be$

Vedclass · Answer

(B) The second ionisation energy is the energy required to remove an electron from a $M^{+}$ cation in the gaseous state: $X^{+}_{(g)} \rightarrow X^{2+}_{(g)} + e^{-}$.For the given elements,the electronic configurations of the $M^{+}$ ions are:$Li^{+} (1s^2)$ - Stable noble gas configuration.$Be^{+} (1s^2 2s^1)$$B^{+} (1s^2 2s^2)$$C^{+} (1s^2 2s^2 2p^1)$$Li$ has the highest second ionisation energy because the electron is removed from a stable $1s^2$ noble gas core.Comparing $Be^{+}, B^{+},$ and $C^{+}$,the removal of the second electron from $B^{+}$ involves breaking a stable $2s^2$ configuration,making its $IE_2$ higher than $C^{+}$.Thus,the order is $Li > B > C > Be$.Therefore,the correct option is $B$.

The second ionisation energies of $Li, Be, B$ and $C$ are in the order

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