Question

In which of the following molecules / ions $B{F}_3,N{O}_2^{-},N{H}_2^{-}$ and $H_{2}O$, the central atom is $s{p}^2$ hybridized?

• A. $B{F}_3$ and $N{O}_2{}^{-}$
• B. $N{O}_2{}^{-}$ and $N{H}_2$
• C. $N{H}_2{}^{-}$ and $H_{2}O$
• D. $N{O}_2{}^{-}$ and $H_{2}O$

Answer 1

Answer: (A)

$B{F}_3$ and $N{O}_2{}^{-}$

To find the hybridization of central atoms:

For a molecules,

$\frac{Totalno.ofvalenceelectronsofallatoms}{8}$

i.e. for ${BF}_3$

It is $\frac{24}{8}=3$

This obtained quotient can be called as co ordination number.

Co ordination number Hybridization type

$2$ $sp$

$3$ ${sp}^2$

$4$ ${sp}^3$

$5$ ${sp}^{3}d$

$6$ ${sp}^3{d}^2$

For Ions,

Co ordination number $=\frac{1}{2}(v+x-c+a)$

where, $v=$ no. of valence electrons in central atom

$x=$ no. of mono valent electrons around central atom

$c=$ $+ve$ charge on Cation

$a=$ $-ve$ charge on Anion

For ${NO}_2^{-}$

It is $\frac{1}{2}(5+0-0+1)=3$

Hence, According to the given table ${BF}_3$ and ${NO}_2^{-}$ have ${sp}^2$ hybridization.