Question

In $\ce{TeCl4}$ the central atom tellurium involves

• A. $sp^3$ hybridization
• B. $sp^3d$ hybridization
• C. $sp^3d^2$ hybridization
• D. $dsp^2$ hybridization

Answer 1

Answer: (B)

Hybridisation $= \frac{1}{2}$ [Number of valence electrons of central atom + no. of monovalent atoms attached to it + negative charge if any - positive charge if any]
$= \frac{1}{2} \left[6+4+0-0\right] =5=sp^{3}d$
$\left[\because Te \left(52\right)=\left[Kr\right]4d^{10}5s^{2}5p^{4}\right]$