Question

The molecules like $Br{F}_5$ and $XeO{F}_4$ are square pyramidal in shape. What is the type of hybridisation shown in these molecules?

• A. ${\text{dsp}}^3$
• B. ${\text{dsp}}^2$
• C. ${\text{sp}}^{3}d$
• D. ${\text{sp}}^3{d}^2$

Answer 1

Answer: (D)

${\text{sp}}^3{d}^2$

The electron configuration of the Bromine atom is represented as;

$1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}4s^{2}4p^5$

During the hybridization one 4s, three 4p and two 4d orbitals take part in the process giving rise to $s{p}^3{d}^2$ hybrid orbitals. 5 sigma bonds ate also formed between the central atom of bromine with fluorine atoms. Anyone of the hybrid orbitals also contains the lone pairs.

In case of $XeO{F}_4$ the central atom will use its s-orbital, its p-orbitals and as many of its d-orbitals as needed to mix together to make the hybrid orbitals. So, for $XeO{F}_4$, Xe will need its s orbital, all three of its p-orbitals, and 2 of its d-orbitals, and its hybridization state will be $s{p}^3{d}^2$, or $d^{2}s{p}^3$

The answer is option D.