Question
In $\ce{TeCl4},$ the central atom tellurium involves:
✓
Answer
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$, \ce{TeCl4}$ shows $sp^3d$ hybridisation.
Number of hybride orbitals $=\frac{1}{2}(6+4+0+0) = 5$
Hence$, \ce{TeCl4}$ shows $sp^3d$ hybridisation.
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