Question

In which of the following molecules/ions are all the bonds not equal?

• A. $S{F}_4$
• B. $Si{F}_4$
• C. $Xe{F}_4$
• D. $B{F}_4^{-}$

Answer 1

The correct option is A $S{F}_4$

In $S{F}_4$, sulphur ungergoes $s{p}^{3}d$ hybridization. It contains two axial and two equatorial bonds and one position is occupied by a lone pair of electrons. So here the bond angles are ${90}^0$ and ${120}^o$. Hence the length of axial bonds are different from equitorial bonds.

$Si{F}_4$ is $s{p}^3$ hybridised and all the four bonds are of equal length.

$Xe{F}_4$ is $s{p}^3{d}^2$ hybridised. this its geometry is octahedral and shape is square planner. Here also all the four bonds are of equal length.
$B{F}_4^{-}$s is $s{p}^3$ hybridised. So all the bond length of $B{F}_4^{-}$ are also equal.