Question

In $TeC{l}_4$, the central atom tellurium involves:

• A. sp${}^3$-hybridisation
• B. sp${}^3$d -hybridisation
• C. sp${}^3$d${}^2$-hybridisation
• D. dsp${}^2$-hybridisation

Answer 1

Answer: (B)

sp${}^3$d -hybridisation

Tellurium is in group 6, so it has six valence electrons. Chlorine is in group 7 so it has 7 valence electrons.

Since $TeC{l}_4$ has four bond pairs and one unbounded pair, its geometry is based on the trigonal bipyramidal structure.

But since there are only four bond pairs, the molecule takes a see-saw shape and the unbonded electrons take the place of a bonded element.

For trigonal bipyramidal structures, the hybridization is $s{p}^{3}d$.

Hence, option $B$ is correct.