Sodium is not found in the +2 oxidation state because of its ?

Sodium (Na) belongs to Group 1 (alkali metals) of the periodic table and has an atomic number of 11 with the electronic configuration 1s² 2s² 2p⁶ 3s¹. The outermost electron in the 3s orbital is loosely bound and can be easily removed, leading to the formation of a Na⁺ ion with a configuration of 1s² 2s² 2p⁶, which is identical to the noble gas neon (Ne). This configuration is very stable because all the shells are completely filled.

When we talk about the possibility of sodium forming a +2 oxidation state, it would require the removal of one more electron from the stable, completely filled 2p orbital. The energy required for this process is called the second ionization potential (I₂), and for sodium, it is extremely high (approximately 4560 kJ/mol). This is because the second electron must be removed from a stable noble gas core, which resists further ionization.

Due to this high second ionization potential, it is energetically unfavorable for sodium to lose a second electron. As a result, sodium remains stable in the +1 oxidation state, forming ionic compounds like NaCl, Na₂O, and NaOH, where it exists as Na⁺ ions.

This behavior is common to all alkali metals (Li, Na, K, Rb, Cs, Fr). They all form compounds with a +1 oxidation state because their first ionization energies are low, but the second ionization energies are prohibitively high.

Thus, sodium is not found in the +2 oxidation state because removing a second electron from a noble gas configuration demands too much energy, making the process thermodynamically impossible under normal conditions.