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(C) The first ionisation energy $(IE_1)$ is the energy required to remove the most loosely bound electron from an isolated gaseous atom.Alkali metals

Vedclass · Accepted Answer

$1s^2, 2s^2, 2p^6, 3s^1$

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(C) The first ionisation energy $(IE_1)$ is the energy required to remove the most loosely bound electron from an isolated gaseous atom.Alkali metals have the lowest first ionisation energy in their respective periods because they have a stable noble gas core plus one valence electron in the $ns^1$ orbital.Comparing the options:$A$: $1s^2, 2s^2, 2p^5$ (Fluorine,a halogen,has high $IE_1$)$B$: $1s^2, 2s^2, 2p^6, 3s^2, 3p^1$ (Aluminium,group $13$)$C$: $1s^2, 2s^2, 2p^6, 3s^1$ (Sodium,an alkali metal,group $1$)$D$: $1s^2, 2s^2, 2p^6$ (Neon,a noble gas,has very high $IE_1$)Therefore,the configuration $1s^2, 2s^2, 2p^6, 3s^1$ corresponds to an alkali metal,which has the minimum first ionisation energy among the given choices.

Which electronic configuration shows the minimum first ionisation energy?

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