In the hydrides of group $15$ and group $16\mathrm{}$ (except $\mathrm{N}{\mathrm{H}}_3$ and ${\mathrm{H}}_2\mathrm{O}$ ) the energy difference between $3\mathrm{s}$ and $3\mathrm{p}$ orbital’s is quite high. Hybridization increases the energy of $3\mathrm{s}$ orbital so much that lone pair rather prefers to occupy unhybridized s orbital. For example, in $\mathrm{P}{\mathrm{H}}_3$ , $600\mathrm{}\mathrm{k}\mathrm{J}\mathrm{}\mathrm{m}\mathrm{o}{\mathrm{l}}^{-1}$ of energy is required to hybridize the central atom. So, to avoid such energy demanding hybridization P forms bonds with unhybridized p orbitals leaving the lone pair in the spherical s orbital which leads to a bond angle close to $90°$ .