Question:

In ${TeCl4}$ the central atom tellurium involves

Updated On: Apr 26, 2024
  • $sp^3$ hybridization
  • $sp^3d$ hybridization
  • $sp^3d^2$ hybridization
  • $dsp^2$ hybridization
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The Correct Option is B

Solution and Explanation

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] $
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