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] $