Chemical Bonding — Easy JEE chemistry MCQ
The hybridization of the central atom in $\text{XeF}_4$ is:
- A. sp³
- B. sp³d
- C. sp³d²
- D. dsp²
Solution
In XeF₄:
- Xe is the central atom
- Number of valence electrons in Xe = 8
- Number of F atoms = 4 (each contributes 1 electron for bonding)
Total electron pairs around Xe:
- Bonding pairs = 4 (with 4 F atoms)
- Lone pairs = (8 - 4)/2 = 2
Total electron pairs = 4 + 2 = 6
For 6 electron pairs, the hybridization is **sp³d²**
The geometry is octahedral with 2 lone pairs in axial positions, giving a square planar molecular shape.