Answer: sp 3 d 2 (one 3s, three 3p, two 3d orbitals).
- A sp<sup>3</sup>d<sup>2</sup> (one 3s, three 3p, two 3d orbitals)
- B sp<sup>3</sup>d, giving a trigonal bipyramidal geometry instead
- C sp<sup>3</sup>, giving a simple tetrahedral geometry instead
- D sp<sup>2</sup>, giving a trigonal planar geometry instead
Correct answer: A. sp<sup>3</sup>d<sup>2</sup> (one 3s, three 3p, two 3d orbitals)
Explanation: SF<sub>6</sub> has 6 bonding pairs; S uses one 3s + three 3p + two 3d = sp<sup>3</sup>d<sup>2</sup>, giving octahedral geometry.
Xenon tetrafluoride (XeF4) has six electron domains around xenon — four Xe–F bonds plus two lone pairs. The lone pairs occupy opposite axial positions, leaving the four fluorine atoms in a square-planar shape (Xe–F ≈ 194 pm). Image: ChemSim, Public Domain, via Wikimedia Commons.
Concept context
Covers the nitrogen family (Group 15), oxygen family (Group 16), halogens (Group 17), and noble gases (Group 18). One of the highest-weightage inorganic chapters in NEET and JEE - includes preparation and properties of HNO₃, H₂SO₄, interhalogens, and xenon compounds.