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In $\mathrm{TeCl}_{4}$ the central atom tellurium involves
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Verified Answer
The correct answer is:
sp ${ }^{3}$ d hybridization
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]
$\begin{array}{l}
=\frac{1}{2}[6+4+0-0]=5=\mathrm{sp}^{3} \mathrm{~d} \\
{\left[\because \mathrm{Te}(52)=[\mathrm{Kr}] 4 \mathrm{~d}^{10} 5 \mathrm{~s}^{2} 5 \mathrm{p}^{4}\right]}
\end{array}$
$\begin{array}{l}
=\frac{1}{2}[6+4+0-0]=5=\mathrm{sp}^{3} \mathrm{~d} \\
{\left[\because \mathrm{Te}(52)=[\mathrm{Kr}] 4 \mathrm{~d}^{10} 5 \mathrm{~s}^{2} 5 \mathrm{p}^{4}\right]}
\end{array}$
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