MCQ
In $\ce{TeCl_4},$ the central atom tellurium involves:
- A$sp^3$ hybridisation
- ✓$\ce{sp^3d}$ hybridisation
- C$\ce{dsp^2}$ hybridisation
- D$\ce{sp^3d^2}$ hybridisation
✓
Answer
Correct option: B.
$\ce{sp^3d}$ hybridisation
Number of hybrid orbitals $=\frac{1}{2} ($no. of electrons in valence shell of atom $+$ no. of monovalent atoms $-$ charge no cation $+$ charge on the anion$.)$
Number of hybride orbitals $=\frac{1}{2}(6+4+0+0) = 5$
Hence$, \ce{TeCl_4}$ shows $\ce{sp^3d}$ hybridisation.
Number of hybride orbitals $=\frac{1}{2}(6+4+0+0) = 5$
Hence$, \ce{TeCl_4}$ shows $\ce{sp^3d}$ hybridisation.
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