Question

Which of the following possess two lone pair of electrons on the central atom and square planar in shape?

(I) $S{F}_4$ (II) $Xe{O}_4$ (III) $Xe{F}_4$ (IV) $IC{l}_4^{-}$

• A. I, II
• B. II, IV
• C. III, IV
• D. All

Answer 1

The correct option is C

III, IV

Firstly, calculate the value of electron pairs and geometry.

$S{F}_4\Rightarrow \frac{(V+L+A-C)}{2}=\frac{(6+4)}{2}=5\Rightarrow$ [Trigonal Bipyramidal]

$Xe{O}_4\Rightarrow \frac{8}{2}=4\Rightarrow \left[AVSEPR\right]\Rightarrow$ Tetrahedral

$Xe{F}_4\Rightarrow \frac{(8+4)}{2}=6\Rightarrow$ Octahedral

$IC{l}_4^{-}\Rightarrow \frac{(7+4-1)}{2}=6\Rightarrow$ Octahedral

$\Rightarrow S{F}_4\Rightarrow$ ep = 5 $\Rightarrow$ 1 lone pair

$\Rightarrow Xe{O}_4\Rightarrow$ ep = 4 $\Rightarrow$ 0 lone pair

$\Rightarrow Xe{F}_4\Rightarrow$ ep = 6 $\Rightarrow$ 2 lone pairs

$\Rightarrow IC{l}_4^{-}\Rightarrow$ ep = 6 $\Rightarrow$ 2 lone pairs

And structure of $Xe{F}_4$ and $IC{l}_4^{-}$

Therefore, both $Xe{F}_4$ and $IC{l}_4^{-}$ possess two lone pairs and has square planar shape.