Question

The state of hybridisation of central atom in dimer of $B{H}_3$ and $Be{H}_2$?

• A. $s{p}^2,s{p}^2$
• B. $s{p}^3,s{p}^2$
• C. $s{p}^3,s{p}^3$
• D. $s{p}^2,sp$

Answer 1

The correct option is D $s{p}^2,sp$

In $B{H}_3$

We know that,

Boron has three valence electron, so it is supposed to make $3$ bond in a molecules with hybridization$s{p}^2$ as only $S$ and two $p$ are used in hybridization because last $p$ orbital vacant.

In $Be{H}_3$

Hybridization of $Be{H}_2$ is $‘s{p}^{\prime }$

You can proceed in this way

$Be$ Has $2$ electrons in their valence shell and $Hhas1$

So, $2+2=4$

$\frac{4}{2}=2$ (For hydrogen only we divide by 2 otherwise by 8)

$2=sp$ $-Hybridisation$

Hence,

Required solution is $B{H}_3=s{p}^2,Be{H}_2=sp$