sp³d³ Hybridization

This involves the mixing of one s three p and three d-orbitals forming seven sp³d³ hybrid orbitals having pentagonal bipyramidal geometry. The geometry of IF₇ molecule can be explained on the basis of sp³d³ Hybridization.

Geometry of IF

₇ molecule

The outer electronic configuration of iodine atom is 5s² 5p⁵. To make seven bonds, one s and two p orbitals are promoted to the higher vacant 5d orbitals as shown in fig.1.29 (b).

fig 1.29 (a) - Formation of IF₇ molecule involving sp³d³-Hybridization

fig 1.29 (b) - Pentagonal bipyramidal geometry of IF₇ molecule

These seven orbitals are then hybridized to give seven sp³d³ hybrid orbitals. Each of these sp³d³ hybrid orbitals overlaps with 2p orbitals of fluorine to form IF₇ molecule having pentagonal bipyramidal geometry.

In this geometry, all the bond angles are not equal. Five F- atoms are directed towards the vertices of a regular pentagon making an angle of 72^o. The other two F-atoms are at right angles (90^o) to the plane. Due to different bond angles, the bonds are different in length. The axial bonds are larger than equatorial bonds.