Which one will have the highest 2nd ionisation energy?

The correct option is 1s² 2s² 2p⁶ 3s¹.

To understand why this electronic configuration results in the highest second ionization energy, let us break down the concepts step-by-step:

1. Understanding Second Ionization Energy:
Ionization energy is the energy required to remove an electron from an atom or ion. The first ionization energy involves removing the first electron from a neutral gaseous atom. The second ionization energy (${\mathrm{IE}}_2$) is the energy required to remove a second electron from a singly charged cation:
$M^{+}\to M^{2+}+e^{-}$

2. Analyzing the Cation Configurations:
Let us determine the electronic configurations of the univalent cations ($M^{+}$) formed after removing the first electron from each option:

• For 1s² 2s² 2p⁶ 3s¹ (Sodium, Na):
Removing the first electron (from the 3s orbital) yields the stable noble gas configuration:
$M^{+}:1s^{2}2s^{2}2p^6$ (Neon configuration, which has a fully filled shell and is highly stable).

• For 1s² 2s² 2p⁴ (Oxygen, O):
Removing the first electron (from the 2p orbital) yields:
$M^{+}:1s^{2}2s^{2}2p^3$ (Half-filled p-orbital configuration).

• For 1s² 2s² 2p⁶ (Neon, Ne):
Removing the first electron (from the 2p orbital) yields:
$M^{+}:1s^{2}2s^{2}2p^5$

• For 1s² 2s² 2p⁶ 3s² (Magnesium, Mg):
Removing the first electron (from the 3s orbital) yields:
$M^{+}:1s^{2}2s^{2}2p^{6}3s^1$

3. Comparing the Second Ionization Energies:
The second electron is removed from the $M^{+}$ cations described above. The cation formed from 1s² 2s² 2p⁶ 3s¹ has a noble gas core configuration ($1s^{2}2s^{2}2p^6$). This represents a highly stable closed-shell octet with a lower principal quantum number (n = 2 vs. n = 3) and a high effective nuclear charge. Breaking this extremely stable configuration requires an exceptionally large amount of energy compared to removing an electron from any of the other configurations. Therefore, it possesses the highest second ionization energy.