Question

$s{p^3}{d^2}$ hybridisation is not displayed by:
(A) ${\left[ {Cr{F_6}} \right]^{3 - }}$
(B) $S{F_6}$
(C) $P{F_5}$
(D) $Br{F_5}$

Answer 1

Here we should know that $1s,3p,2d$orbitals undergo intermixing to form the six identical orbitals. These are further directed towards the octahedron. $s{p^3}{d^2}$ hybridisation here is not displayed by the $P{F_5}$ this is so because the central atom of P atoms have five sets of electron pairs that are completely busy with the bonding part.

Complete answer: Firstly we will talk about the concept of hybridisation here that is the concept of mixing atomic orbitals into new hybrid orbitals with different shapes and so on which is further suitable for pairing of electrons to form the chemical bonds in the valence bond theory. When we further talk about the $s{p^3}{d^2}$or the ${d^2}s{p^3}$ are the hybridisation for the octahedral geometry.in octahedron, the bond are formed parallel to the three axis. Hence, we can say that the is $d{x^2} - d{y^2}$ and $d{z^2}$will be further used to form the hybrid orbitals.Here in the question we should basically see the last and the second last option because rest others are having the six atoms attached to the central atom but in these two options only five atoms are attached to the central metal atom. $s{p^3}{d^2}$ hybridisation here is not displayed by the $P{F_5}$ this is so because the central atom of P atoms have five sets of electron pairs that are completely busy with the bonding part and further here the hybridisation we are talking about corresponds to electron pairs.

Hence the correct option is (C).

Note: Hence, we can conclude from here that the option C is correct. We should know that the hybrid orbitals are further useful in expansion of molecular geometry and the atomic bonding properties and are symmetrically disposed in space.