Question

Which of the following statements are correct for the ${\text{SO}}_4^{2 - }$ ion?
A.It is tetrahedral
B.All the ${\text{S}} - {\text{O}}$ bond length are equal and shorter than we expected
C.It contains four $\sigma $bonds between the ${\text{S}}$ and the ${\text{O}}$ atoms, two $\pi $ bonds delocalized over the ${\text{S}}$ and the four ${\text{O}}$ atoms, and all the ${\text{S}} - {\text{O}}$ bonds have a bond order of $1.5$
D.Oxidation state of Sulphur is $ + 6$ and all oxygen is $2$

Answer 1

Hint:To answer this question, you must be familiar with writing hybridisation and drawing the structure of a compound. Sulphur has six electrons in its valence shell and also has vacant $3d$ orbitals. The hybridization of sulphur is $s{p^3}$.

Complete step by step solution:
The sulphate anion has a central Sulphur atom bonded to four oxygen atoms.
Sulphur has atomic number 16. It has six electrons in its valence shell. The electronic configuration of Sulphur is: $S:\left[ {Ne} \right]3{s^2}3{p^4}$.
It excites two of its paired electrons to the vacant d orbital to increase its covalency and four unpaired electrons in the $3s$ and $3p$ orbitals. Each of these four electrons are used to form a single bond with each oxygen atom. The remaining two excited electrons will be used to form $\pi $ bonds.
Thus, the hybridisation of Sulphur will be $s{p^3}$. Thus, the four oxygen molecules are attached to the Sulphur atom in a tetrahedral arrangement.
We know that oxygen has a valency of 2. One of the two valencies of each oxygen atom is satisfied by a single covalent bond with the Sulphur atom. Sulphur has two more excited electrons present in the $3d$ orbital which form a double bond with one oxygen atom each. Now the remaining two oxygen atoms carry a negative charge in order to satisfy their valency.
The $\pi $ bonds and the negative charges are delocalized over the four oxygen atoms thus providing each oxygen atom a partial negative charge and a partial double bond character to the ${\text{S}} - {\text{O}}$ bonds.
Sulphur forms six bonds with more electronegative oxygen atoms and thus has an oxidation state of $ + 6$ and each oxygen atom has oxidation state of $ - 2$.
Thus, we can conclude that all the given statements are correct.

The correct options are A, B, C and D.

Note:
Sulphur and oxygen form a $p\pi - d\pi $ multiple bond. The double bond is known because the pi bond formed is between the $2p$ orbital of oxygen atom and the $3d$ orbital of the Sulphur atom.