The first ionization potentials left( eV righ | vedclass

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(D) The electronic configuration of $Be$ $(Z=4)$ is $1s^2 2s^2$.The electronic configuration of $B$ $(Z=5)$ is $1s^2 2s^2 2p^1$.$Be$ has a fully fille

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$9.32 \ eV, 8.29 \ eV$

Vedclass · Answer

(D) The electronic configuration of $Be$ $(Z=4)$ is $1s^2 2s^2$.The electronic configuration of $B$ $(Z=5)$ is $1s^2 2s^2 2p^1$.$Be$ has a fully filled $2s$ orbital,which is more stable than the $2p$ orbital of $B$.Therefore,more energy is required to remove an electron from $Be$ compared to $B$.The first ionization potential of $Be$ is approximately $9.32 \ eV$ and that of $B$ is $8.29 \ eV$.Thus,the correct option is $D$.

The first ionization potentials $\left( eV \right)$ of $Be$ and $B$ respectively are:

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