MCQ
In TeCl4, the central atom tellurium involves:
- Asp3 hybridisation
- Bsp3d hybridisation
- Cdsp2 hybridisation
- Dsp3d2 hybridisation
Explanation:
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, TeCl4 shows sp3d hybridisation.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$(i)$ $\left[ FeF _{6}\right]^{3-}$
$(ii)$ $\left[ Co \left( NH _{3}\right)_{6}\right]^{3+}$
$(iii)$ $\left[ NiCl _{4}\right]^{2-}$
$(iv)$ $\left[ Cu \left( NH _{3}\right)_{4}\right]^{2+}$