Q11 / 120 · VITEEE 2007 · Chemistry

In $\mathrm{TeCl}_{4}$ the central atom tellurium involves

  • A
    $\mathrm{sp}^{3}$ hybridization
  • Correct answer
    B
    sp ${ }^{3}$ d hybridization
  • C
    $\mathrm{sp}^{3} \mathrm{~d}^{2}$ hybridization
  • D
    ds$p^{ 2}$ hybridization

Solution

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] $\begin{array}{l} =\frac{1}{2}[6+4+0-0]=5=\mathrm{sp}^{3} \mathrm{~d} \\ {\left[\because \mathrm{Te}(52)=[\mathrm{Kr}] 4 \mathrm{~d}^{10} 5 \mathrm{~s}^{2} 5 \mathrm{p}^{4}\right]} \end{array}$

Hence, the correct answer is option: 'B'.

Relevant Questions