$\mathrm{C}(\mathrm{OH}{)}_4$  is unstable because a carbon atom cannot hold more than one $-\mathrm{OH}$ groups but $\mathrm{Si}(\mathrm{OH}{)}_4$ is a stable compound because :

Triagonal bipyramidal geometry is present in both PCl₅ and PBr₅, which is an irregular and unstable structure.

As a result, PCl₅ separates into two more sturdy octahedral and tetrahedral structures.

The formula for PCl₅ in solid form is $\left[{\mathrm{PCl}}_4{]}^{+}\right[{\mathrm{PCl}}_6{]}^{-}$.

In $[{\mathrm{PCl}}_4{]}^{+}$ and $[{\mathrm{PCl}}_6{]}^{-}$. P is hybridised in the sp³ and sp³ d² states, respectively.

The solid form of PBr₅ is $[{\mathrm{PBr}}_4{]}^{+}$Br^-, and the hybridization state of P in $[{\mathrm{PBr}}_4{]}^{+}$ is sp³ hybridized.

Hence the correct option is A and D.