Question

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

• A. $ Si{{F}_{4}} $ and $ S{{F}_{4}} $
• B. $ IO_{3}^{-} $ and $ Xe{{O}_{3}} $
• C. $ BH_{4}^{-} $ and $ NH_{4}^{+} $
• D. $ PF_{6}^{-} $ and $ S{{F}_{6}} $

Answer 1

Answer: (A)

$SiF _{4}$ and $SF _{4}$ are not isostructural because $SiF _{4}$ is tetrahedral due to $s p^{3}$-hybridisation of $Si$.
${ }_{14} Si =1 s^{2}, 2 s^{2}, 2 p^{6}, 3 s ^{2} 3 p ^{2}$ (In ground state)

${ }_{14} Si =1 s^{2}, 2 s^{2} 2 p^{6}, 3 s^{1} 3 p^{3}$ (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 $SF _{4}$ is not tetrahedral but it is distorted tetrahedral because in it S is sp $^{3} d$ hybrid.
${ }_{16} S =1 s^{2}, 2 s^{2} 2 p^{6}, 3 s^{2} 3 p_{x}^{2} 3 p_{y}^{1} 3 p_{z}^{1}$
(In ground state)
$=1 s^{2}, 2 s^{2} 2 p^{6}, 3 s^{2} 3 p_{x}^{1} 3 p_{y}^{1} 3 p_{z}^{1} 3 d_{x y}^{1}$
$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.