Question

Which of the following statements are correct for the $S{O}_4^{2-}$ ion?

• A. It is tetrahedral.
• B. All the $S-O$ bond length are equal, and shorter than expected.
• C. It contains four $\sigma$-bonds between the $S$ and the $O$ atoms, two $\pi$-bonds delocalized over the $S$ and the four $O$ atoms, and all the $S-O$ bonds have a bond order of $1.5$.
• D. Oxidation state of sulphur is $+6$ and all oxygen in $2$.

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A It is tetrahedral.
B All the $S-O$ bond length are equal, and shorter than expected.
C It contains four $\sigma$-bonds between the $S$ and the $O$ atoms, two $\pi$-bonds delocalized over the $S$ and the four $O$ atoms, and all the $S-O$ bonds have a bond order of $1.5$.
D Oxidation state of sulphur is $+6$ and all oxygen in $2$.

All of the following statements are correct for the $S{O}_4^{2-}$ ion.

(A) $S$ atom has $4$ bonding domains and zero lone pairs of electrons. Hence it is $s{p}^3$ hybridized with tetrahedral geometry and bond angles of ${109.5}^o$
(B) All the $S-O$ bond length are equal, and shorter than expected. This is due to delocalization of electrons (resonance).

(C) It contains four σ-bonds between the $S$ and the $O$ atoms, two π-bonds delocalized over the $S$ and the four $O$ atoms, and all the $S-O$ bonds have a bond order of $1.5$. Thus each $S-O$ bond has partial double bond character.

(D) Oxidation state of sulphur is $+6$ and all oxygen atoms have $-2$ oxidation states.

$+6+4(-2)=-2$