Q.Write the hybridisation present in the following compounds.

Simple atomic-orbital overlap explains diatomic bonding well, but fails for polyatomic molecules — methane's four C-H bonds are experimentally IDENTICAL, even though carbon's unhybridised 2s and 2p orbitals differ in energy and shape and offer only two unpaired electrons, not four. Linus Pauling's solution was hybridisation: the mixing of atomic orbitals on the SAME atom, of comparable energy, into an EQUAL NUMBER of new, equivalent hybrid orbitals of identical energy, oriented with maximum symmetry to minimise electron-electron repulsion.

sp (linear, e.g. BeCl₂). Be's ground state 2s² is promoted to 2s¹2p¹; the one 2s and one 2p orbital mix into two collinear sp orbitals (180° apart), each overlapping a Cl 3p orbital.

sp² (trigonal planar, e.g. BF₃). B's ground state 2s²2p¹ is promoted to 2s¹2pₓ¹2p_y¹; 2s plus two 2p orbitals mix into three coplanar sp² orbitals (120° apart), each overlapping an F 2p orbital.

sp³ (tetrahedral, e.g. CH₄). C's ground state 2s²2pₓ¹2p_y¹ is promoted to 2s¹2pₓ¹2p_y¹2p_z¹; all four valence orbitals mix into four sp³ orbitals (109°28' apart), each overlapping an H 1s orbital.

sp³d (trigonal bipyramidal, e.g. PCl₅). P's ground state 3s²3pₓ¹3p_y¹3p_z¹ promotes one 3s electron into a vacant 3d_z² orbital; five orbitals (one 3s, three 3p, one 3d) mix into five sp³d orbitals, each overlapping a Cl 3p orbital.

sp³d² (octahedral, e.g. SF₆). S's ground state promotes one 3s and one 3p electron into two vacant 3d orbitals (3d_z², 3d_(x²−y²)); six orbitals mix into six sp³d² orbitals, each overlapping an F 2p orbital. …