Q.Determine the hybridisation of the central sulphur atom in $\text{SF}_6$ and relate it to its octahedral shape.

Hybridization is the mixing of valence orbitals belonging to the SAME atom, recast into an equal number of new, mutually EQUIVALENT hybrid orbitals — introduced because plain (unhybridized) orbital overlap cannot explain how, e.g., beryllium forms two bonds, boron three, or carbon four, given their ground-state electron configurations, nor why all four C–H bonds of methane come out experimentally identical. It proceeds in two steps: first, formation of the excited state, where a paired electron in a lower orbital (e.g. 2s) is promoted into an empty, only-slightly-higher-energy orbital (e.g. 2p) so the atom gains as many half-filled orbitals as its bonding valency requires; second, mixing and recasting, where these now-comparable-energy orbitals combine into new orbitals identical to each other in shape and energy, oriented in space for minimum mutual repulsion. Hybridization requires the mixing orbitals to belong to the same atom and to have nearly the same energy (2s/2p qualify; 3s/2p do not). Three types of s-p hybrid orbital follow directly from how many 's' and 'p' orbitals combine: $sp^3$ (one s + three p → four hybrids, 109°28' apart, tetrahedral — e.g. CH4, and NH3/H2O with one/two hybrid orbitals holding lone pairs instead of bonds, which compresses their bond angles); $sp^2$ (one s + two p → three hybrids, 120° apart, trigonal planar, with one unhybridized p orbit …