Question

What is the hybridization of the central atom in each of the following?

(a) \({\rm{Be}}{{\rm{H}}_{\rm{2}}}\)

(b) \({\rm{S}}{{\rm{F}}_{\rm{6}}}\)

(c) \({\rm{PO}}_{\rm{4}}^{{\rm{3 - }}}\)

(d) \({\rm{PC}}{{\rm{l}}_{\rm{5}}}\)

Short answer

Answer

(a) The hybridization is sp.

(b) The hybridization is \({\rm{s}}{{\rm{p}}^{\rm{3}}}{\rm{\;}}{{\rm{d}}^{\rm{2}}}\).

(c) The hybridization is \({\rm{s}}{{\rm{p}}^{\rm{3}}}\).

(d) The hybridization is \({\rm{s}}{{\rm{p}}^{\rm{3}}}{\rm{\;d}}\).

Step-by-step solution

Definition of Concept

Hybridization: When two or more different pure orbitals with comparable energy are mixed together to produce an equivalent amount of impure orbitals with equal energy and definite geometry, this is referred to as hybrid orbitals, and the phenomenon is referred to as hybridization.

Find the hybridization of the central atom

For\({\rm{Be}}{{\rm{H}}_{\rm{2}}}\)

The central atom is Be, which has two valence electrons, and the two hydrogen atoms each have one electron.

To become stable, it must share two electrons with the beryllium central atom. As a result, the number of electron pairs involved in bond formation is two.

Now,

The number of central atom bond pair electrons =2

The number of electrons with a single lone pair =0

Total electron valence =2

Hybridization =sp

Therefore, the required hybridization is sp.

Find the hybridization of the central atom

For\({\rm{S}}{{\rm{F}}_{\rm{6}}}\)

The central atom is sulphur, which has a valence electron of 6, and there are six fluorine atoms, each of which has one valence electron, so in order to become stable, it will share each electron of sulphur with each fluorine atom, and thus all six electrons will participate in bond formation.

Therefore,

The number of central atom bond pair electrons =6

The number of electrons with a single lone pair =0

Total electron valence =6

Hybridization =\({\rm{s}}{{\rm{p}}^{\rm{3}}}{\rm{\;}}{{\rm{d}}^{\rm{2}}}\)

Therefore, the required hybridization is \({\rm{s}}{{\rm{p}}^{\rm{3}}}{\rm{\;}}{{\rm{d}}^{\rm{2}}}\).

Find the hybridization of the central atom

For\({\rm{P}}{{\rm{O}}_{\rm{4}}}^{{\rm{3 - }}}\)

Because the central atom is phosphorus, which has 5 valence electrons, and there are four oxygen atoms, each of which has 6 valence electrons, each phosphorus and oxygen atom has shared two pairs of electrons, which are referred to as bond pair electrons, but there are two pairs of electrons that did not participate in bond formation and are referred to as lone pairs of electron.

Phosphorus has two bond pair electrons.

The number of electrons in a lone pair =2

The total number of valence electrons in the system is =4.

Hybridization =\({\rm{s}}{{\rm{p}}^{\rm{3}}}\)

Therefore, the required hybridization is \({\rm{s}}{{\rm{p}}^{\rm{3}}}\).

Find the hybridization of the central atom

For\({\rm{PC}}{{\rm{l}}_{\rm{5}}}\)

The central atom, phosphorus, has five valence electrons, and each chlorine atom shares one pair of electrons with the phosphorus atom, so there are five valence electrons involved in bond formation.

Phosphorus has five bond pair electrons.

The number of electrons in a lone pair =0

Valence electrons total =5

Hybridization =\({\rm{s}}{{\rm{p}}^{\rm{3}}}{\rm{\;d}}\)

Therefore, the required hybridization is\({\rm{s}}{{\rm{p}}^{\rm{3}}}{\rm{\;d}}\).