Question

Among the following, the pair in which the two species are not isostructural, is

• A. $Si{F}_4$ and $S{F}_4$
• B. $I{O}_3^{-}$ and $Xe{O}_3$
• C. $B{H}_4^{-}$ and $N{H}_4^{+}$
• D. $P{F}_6^{-}$ and $S{F}_6$

Answer 1

Answer: (A)

$Si{F}_4$ and $S{F}_4$ are not isostructural because $Si{F}_4$ is tetrahedral due to $s{p}^3$-hybridisation of $Si$.
${}_{14}Si=1s^2,2s^2,2p^6,3s^{2}3p^2$ (In ground state)

${}_{14}Si=1s^2,2s^{2}2p^6,3s^{1}3p^3$ (In excited state)
Hence, four equivalent sp ${}^3$-hybrid orbitals are obtained and they are overlapped by four p-orbitals of four fluorine atoms on their axes. Thus it shows following structure

While $S{F}_4$ is not tetrahedral but it is distorted tetrahedral because in it S is sp ${}^{3}d$ hybrid.
${}_{16}S=1s^2,2s^{2}2p^6,3s^{2}3p_x^{2}3p_y^{1}3p_z^1$
(In ground state)
$=1s^2,2s^{2}2p^6,3s^{2}3p_x^{1}3p_y^{1}3p_z^{1}3d_{xy}^1$
$s{p}^{3}d$-hybridisation
(In first excitation state)

Hence, five $s{p}^{3}d$ hybrid orbitals are obtained. One orbital is already paired and rest four are overlapped with four $p$-orbitals of four fluorine atoms on their axis in trigonal bipyramidal form.
This structure is distorted from trigonal bi-pyramidal to tetrahedral due to involvement of repulsion between lone pair and bond pair.