Amongst $$SF_4, XeF_4, CF_4$$ and $$H_2O$$, the number of species with two lone pairs of electrons is ______

We need to find the number of species with exactly two lone pairs of electrons on the central atom among $$SF_4$$, $$XeF_4$$, $$CF_4$$, and $$H_2O$$.

Sulfur has 6 valence electrons.

In $$SF_4$$, sulfur forms 4 bonds with fluorine atoms.

Lone pairs on S = $$\frac{6 - 4}{2} = 1$$ lone pair.

The geometry of $$SF_4$$ is see-saw (sp$$^3$$d hybridization) and it has 1 lone pair.

Xenon has 8 valence electrons.

In $$XeF_4$$, xenon forms 4 bonds with fluorine atoms.

Lone pairs on Xe = $$\frac{8 - 4}{2} = 2$$ lone pairs.

The geometry of $$XeF_4$$ is square planar (sp$$^3$$d$$^2$$ hybridization) and it has 2 lone pairs.

Carbon has 4 valence electrons.

In $$CF_4$$, carbon forms 4 bonds with fluorine atoms.

Lone pairs on C = $$\frac{4 - 4}{2} = 0$$ lone pairs.

The geometry of $$CF_4$$ is tetrahedral (sp$$^3$$ hybridization) and it has 0 lone pairs.

Oxygen has 6 valence electrons.

In $$H_2O$$, oxygen forms 2 bonds with hydrogen atoms.

Lone pairs on O = $$\frac{6 - 2}{2} = 2$$ lone pairs.

The geometry of $$H_2O$$ is bent (sp$$^3$$ hybridization) and it has 2 lone pairs.

The species with two lone pairs are $$XeF_4$$ and $$H_2O$$.

Therefore, the number of species with two lone pairs is 2.