Question

Which of the following has $s{p}^3$ hybridization?

• A. $Xe{O}_3$
• B. $BC{l}_3$
• C. $Xe{F}_4$
• D. $BB{r}_3$

Answer 1

Answer: (A)

$Xe{O}_3$

To calculate the hybridization, we have:

$s=\frac{\text{valence shell electron of central metal + number of mono atomic atoms}}{2}$

In $Xe{O}_3$, we have, $s=\frac{8+0}{2}=4$

if $s=4$, we have hybridization as $s{p}^3$

In $BC{l}_3$, we have, $s=\frac{3+3}{2}=3$

if $s=3$, we have hybridization as $s{p}^2$

In $Xe{F}_4$, we have $s=\frac{8+4}{2}=6$

if $s=6$, we have hybridization as $s{p}^3{d}^2$

In $BB{r}_3$, we have, $s=3$ and we have hybridization as $s{p}^2$.

Hence, option A is correct.