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(B) The ionization energy of an anion depends on the effective nuclear charge and the inter-electronic repulsions. As we move down the group from $O^{

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$S^{-}$

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(B) The ionization energy of an anion depends on the effective nuclear charge and the inter-electronic repulsions. As we move down the group from $O^{-}$ to $Te^{-}$,the size of the atom increases,which decreases the ionization energy. However,for $O^{-}$,the electron density is concentrated in a smaller $2p$ orbital,leading to significant inter-electronic repulsion. This makes the removal of an electron from $O^{-}$ easier compared to $S^{-}$. Thus,$S^{-}$ has a higher ionization energy than $O^{-}$. Therefore,the trend for ionization energy among these species is $S^{-} > Se^{-} > Te^{-} > O^{-}$.

Which of the following atomic species has maximum ionisation energy?

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