Question

Explain hybridization of central atom in $PO_4^{3 - }$

Answer 1

Hint : The hybridization of the central atom is determined by finding the overall number of electrons that is involved in a chemical bonding. In the given question, the hybridization of $PO_4^{3 - }$

Complete Step By Step Answer:
In the given molecule, the phosphorus atom has valence of $5$ electrons. The hybridisation is found using the presence of sigma bonds and also the number of lone pairs. The presence of the sigma bond is $4$ and the presence of the lone pair is $0$ . This indicates that the steric number (steric number is the number of atoms, or lone pairs around a central atom. It is determines molecular geometry) is $4$ and hence the hybridisation is $sp{}^3$ and the shape of this molecule is a tetrahedral and there are two electrons are forming double bond and there is presence of $3$ electrons that forms $3$ sigma bonds.

Additional Information:
Hybridization is basically the idea of the atomic orbital’s fuse and forms new hybridized orbitals which in turn influences both molecular geometry as well as bonding properties. It is also known to be an expansion of VBT or valence bond theory. There are three types of hybridisation and they are; $sp,sp{}^2,sp{}^3$ .
$sp{}^3$ Hybridisation is the mixing character of one $2s$ orbital and three $2p$ orbital to create $4$ hybrid orbitals with the same characteristics. For an atom to be $sp{}^3$ Hybridisation it has to an s orbital and three p orbital

Note :
The orbital hybridisation is mixing of atomic orbitals transforming into new hybrid orbital which are suitable for pairing of electrons hence forming chemical bonds in VBT. The number of monovalent atoms which surrounds the central atom should be taken into account.