Question

Match the following :

	List I		List II
A	${\left[Cr{F}_6\right]}^{3-}$	I	$s{p}^3{d}^2$, square planar
B	$Xe{F}_4$	II	$s{p}^{3}d$, square planar
C	$PC{l}_5$	III	$s{p}^3{d}^2$, square pyramid
D	$Br{F}_5$	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) ${\left[Cr{F}_6\right]}^{3-}$

Oxidation state of $Cr=+3$

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



So, hybridisation of ${\left[Cr{F}_6\right]}^{3-}3d=d^{2}s{p}^3$

Shape of molecule is octahedral.

(B) $Xe{F}_4$

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 $Xe{F}_4(n)=\frac{8+4}{2}=\frac{12}{2}=6$

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

It means $Xe{F}_4$ hybridisation $s{p}^3{d}^2$ and square planar shape.

(C) $PC{l}_5$

Number of $e^{-}$ pair in $PC{l}_5(n)=\frac{5+5}{2}=\frac{10}{2}=5$

$\Rightarrow (5$ bond pair $)$

It means $PC{l}_5$ hybridisation $s{p}^{3}d$ and trigonal bipyramidal shapes.

(D) $Br{F}_5$

Number of $e^{-}$ pair in $Br{F}_5(n)=\frac{7+5}{2}=\frac{12}{2}=6$

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

It means $Br{F}_5$ hybridisation $s{p}^3{d}^2$ and square pyramidal shape.