Question
The group of molecules having identical shape is :
A.
$$PC{l_5},I{F_5},Xe{O_2}{F_2}$$
B.
$$B{F_3},PC{l_3},Xe{O_3}$$
C.
$$S{F_4},Xe{F_4},CC{l_4}$$
D.
$$Cl{F_4},XeO{F_2},XeF_3^ + $$
Answer :
$$Cl{F_4},XeO{F_2},XeF_3^ + $$
Solution :
$$\eqalign{
& Cl{F_3} \to {\text{Hybridisation}} \cr
& = 3 + \frac{1}{2}\left[ {7 - 3} \right] = 5\left( {s{p^3}d} \right) \cr
& XeO{F_2} \to {\text{Hybridisation}} \cr
& = 3 + \frac{1}{2}\left[ {8 - 4} \right] = 5\left( {s{p^3}d} \right) \cr
& XeF_3^ + \to {\text{Hybridisation}} \cr
& = 3 + \frac{1}{2}\left[ {8 - 3 - 1} \right] \cr
& = 5\left( {s{p^3}d} \right) \cr} $$
All molecules have $$s{p^3}d$$ hybridization and 2 lone pairs. Hence all have identical ( $$T$$ - shape).