Question

Which of the following have state of hybridization of the central atom not same as in the others?

• A. B in $B{F}_3$
• B. O in $H_3{O}^{+}$
• C. N in $N{H}_3$
• D. P in $PC{l}_3$

Answer 1

Answer: (A)

B in $B{F}_3$

In $B{F}_3$, the ground state electronic configuration of Boron is $1s^{2}2s^{2}2P{x}^1$. excited state electronic configuration

is $1s^{2}2s^{1}2P{x}^{1}2P{y}^1$

One electron from one ${}^{\prime }{s}^{\prime }$ orbital and two electrons from ${}^{\prime }{p}^{\prime }$ orbital overlaps with ${}^{\prime }{p}^{\prime }$ orbital electrons of Flourine to form three $s{p}^2$ hybridised bonds. So, $B{F}_3$ has $s{p}^2$ hybridisation.

Flourine $\left(a\right)$

But in $H_3{O}^{+},N{H}_3$ and $PC{l}_3$, the hybridisation involves one electron from ${}^{\prime }{s}^{\prime }$ orbital, $3$ electrons for ${}^{\prime }{p}^{\prime }$ orbitals in central orbital forming $s{p}^3$ hybridisation.