What is the hybridisation of \(\mathrm{Be}\) in \(\mathrm{BeF}_2\) molecule?
What is the hybridisation of \(\mathrm{Be}\) in \(\mathrm{BeF}_2\) molecule?
\(d s p^2\)
\(s p^2 d\)
\(\mathrm{sp}\)
\(s p^3\)
Solution
Hybridisation state of \(\mathrm{Be}\) in \(\mathrm{BeF}_2\) is \(s p\).
\(\begin{aligned}
& \mathrm{F}-\mathrm{Be}-\mathrm{F} \\
& \mathrm{H}=\frac{1}{2}(V+M-C+A) \\
&=\frac{1}{2}(2+2-0+0)=2(s p)
\end{aligned}\)
where,
\(V=\) number of valence electrons of \(\mathrm{Be}=2\)
\(H=\) number of monovalent \(\mathrm{F}\) atom \(=2\)
\(C=\) number of cationic charge \(=\mathbf{0}\)
\(A=\) number of anionic charge \(=0\)
So, the hybridisation of \(\mathrm{Be}\) in \(\mathrm{BeF}_2\) molecule is \(s p\).