Question

The hybridization of S atom in $SO_4^{2 - }$ is
A) $sp$
B) $s{p^2}$
C) $s{p^3}$
D) $s{p^3}d$

Answer 1

We know that the arrangement of hybridized orbitals around the central atom can give rise to some specific geometries which if known for a molecule can be used to deduce the hybridization of the central atom.

Complete answer:

We know that the concept of hybridization was given by Pauling as per whom these are hybrid orbitals which take part in bonding and not the atomic orbitals. Depending on the atomic orbitals that are being used in hybridization, we can have different types of hybridization. Let’s have a brief look at some:
sp hybridization: in this case, one s orbital and one p orbital undergo hybridization to give two sp hybrid orbitals which are used for bonding. Here, we have linear geometry. The examples include, $BeC{l_2}$ molecule.
$s{p^2}$ hybridization: in this case, one s orbital and two p orbitals undergo hybridization to give three $s{p^2}$ hybrid orbitals which are used for bonding. Here, we have trigonal planar geometry. The examples include, $BC{l_3}$ molecule.
$s{p^3}$ hybridization: in this case, one $s$ orbital and three $p$ orbitals undergo hybridization to give four $s{p^3}$ hybrid orbitals which are used for bonding. Here, we have tetrahedral geometry. The examples include, $C{H_4}$ molecule.

Now, let’s have a look at the given $SO_4^{2 - }$ ion. Here, the central atom is S to which four O atoms are attached. We can draw the structure of ion as follows:

As it is evident that the ion has a tetrahedral geometry in which there are four bond pairs and no lone pairs. So, four hybrid orbitals are used which means the hybridization of the central atom is $s{p^3}$.

Hence, the correct option is C.

Note: We have to be careful about the total number of valence electrons used in the bonding as well as the geometry. Pi bonds do not contribute to hybridisation only we need to count the bond pairs or lone pairs around the central atom.