Past exam of the chemistry course of the year 1 of the natural sciences course of the University of Cambridge / 2022 / Paper 1 / 6 / a / i / Solution

In molecular boron trichloride, boron uses three approximately $s{p}^2$ hybrids to form three equivalent B–Cl sigma bonds in a trigonal plane. Its perpendicular empty $2p$ orbital can accept electron density from the compatible combination of filled chlorine $3p$ lone pairs. The resulting pi donation is delocalized over the three B–Cl connections.

In the elementary model, one net pi bond is distributed over three equivalent sigma bonds, giving formal average B–Cl bond order $1+1/3=4/3$. The corresponding resonance contributors each put a B=Cl connection in a different position, with formal negative boron and positive donating chlorine. They do not mean one experimentally distinct double bond and two single bonds. The $4/3$ value is the idealized sigma-plus-delocalized-pi bookkeeping value; a computed bond index need not equal that exact rational number.