Question

Match the following :

	List I		List II
A	$\left[ CrF _{6}\right]^{3-}$	I	$s p^{3} d^{2}$, square planar
B	$XeF _{4}$	II	$s p^{3} d$, square planar
C	$PCl _{5}$	III	$s p^{3} d^{2}$, square pyramid
D	$BrF _{5}$	IV	$s p^{3} d$, trigonal bipyramidal
		V	$s p^{3} d^{2}$, octahedral

• A. A - III , B - I , C - IV ,D - V
• B. A - III , B - I , C - II ,D - V
• C. A - V , B - I , C - IV ,D - III
• D. A - V , B - II , C - IV ,D - III

Answer 1

Answer: (C)

(A) $\left[ CrF _{6}\right]^{3-}$

Oxidation state of $Cr =+3$

Electronic configuration of $Cr ^{+3}=[ Ar ] 3 d^{3} 4 s^{0}$



So, hybridisation of $\left[ CrF _{6}\right]^{3-} 3 d=d^{2} s p^{3}$

Shape of molecule is octahedral.

(B) $X e F _{4}$

Number of $e^{-}$ pair (Bond pair + lone pair)

$(n)=\frac{\text {Number of valence $e^{-}+$ Number of univalent atom } \pm \text { charge }}{2}$

So, for $XeF_{4}(n)=\frac{8+4}{2}=\frac{12}{2}=6$

$\Rightarrow (4$ bond pair $+2$ lone pair $)$

It means $XeF _{4}$ hybridisation $s p^{3} d^{2}$ and square planar shape.

(C) $P C l _{5}$

Number of $e^{-}$ pair in $PCl _{5}(n)=\frac{5+5}{2}=\frac{10}{2}=5$

$\Rightarrow (5$ bond pair $)$

It means $PCl _{5}$ hybridisation $s p^{3} d$ and trigonal bipyramidal shapes.

(D) $BrF _{5}$

Number of $e^{-}$ pair in $BrF_{5}(n)=\frac{7+5}{2}=\frac{12}{2}=6$

$ \Rightarrow (5$ bond pair $+ 1$ lone pair $)$

It means $BrF _{5}$ hybridisation $s p^{3} d^{2}$ and square pyramidal shape.