Question

Sulphur and the rest of the elements of group 16 are less electronegative than oxygen. Therefore, their atoms cannot accept electrons easily. They can acquire the $n{s}^{2}n{p}^6$ configuration by sharing two electrons with the atoms of the other elements and thus, exhibit a $+2$ oxidation state in their compounds. In addition to this, their atoms have vacant $d$-orbitals in their valence shell. Electrons can be promoted here from $s$ and $p$-orbitals of the same shell, hence they can show $+4$ and $+6$ oxidation states.
Like sulphur, oxygen does not show $+4$ and $+6$ oxidation states. The reason is that oxygen:

• A. Is a gas, while sulphur is a solid
• B. Has a higher ionization enthalpy when compared to sulphur
• C. Has a higher electron affinity when compared to sulphur
• D. Has no $d$-orbitals in its valence shell

Answer 1

The correct option is D Has no $d$-orbitals in its valence shell

Sulphur shows +4 and +6 oxidation states due to the presence of vacant $3d$ orbitals to which electrons can be promoted from $3s$ and $3p$ filled orbitals.

Since oxygen atom does not have $2d$ orbitals, no electron can be promoted.