Question

With which of the following configuration an atom has the lowest ionization enthalpy?

• A. $1s^{2}2s^{2}2p^6$
• B. $1s^{2}2s^{2}2p^5$
• C. $1s^{2}2s^{2}2p^3$
• D. $1s^{2}2s^{2}2p^{6}3s^1$

Answer 1

The correct option is D

$1s^{2}2s^{2}2p^{6}3s^1$

In general, most atoms tend to attain stable octet configurations (or sometimes a fully-filled valence s-orbital configuration like $He$). $1s^2{s}^{2}2p^6$ already has the valence octet and thus it will difficult to remove a valence electron from such an atom. Fluorine has $1s^{2}2s^{2}2p^5$ configuration with 7 valence electrons. It would readily gain an electron than lose one. Oxygen has $1s^{2}2s^{2}2p^4$ configuration and would prefer to gain 2 electrons to get to the stable octet configuration. Nitrogen, with its $1s^{2}2s^{2}2p^3$ configuration, has an exactly half-filled $2p$ valence orbital. This also contributes to extra stability (fully filled and exactly half-filled orbitals).

The element with the $1s^{2}2s^{2}2p^{6}3s^1$ electronic configuration is eager to lose the $3s^1$ electron and get to the stable ${12}^2{s}^{2}2p^6$ octet.