The Foundations of Chemistry

BF 3 and BCl 3 are gases at room
temperature. Liquid BBr 3 and solid
BI 3 are shown here.

292 CHAPTER 7: Chemical Bonding

Solution

Step 1: The skeleton is Cl Be Cl

see limitation type A

g

Step 2: N 2    8 (for Cl) 1   4 (for Be) 20 eneeded

Step 3: A 2    7 (for Cl) 1   2 (for Be) 16 eavailable

Step 3: SNA 20 e 16 e 4 eshared

Step 3: ClSBeSCl

Step 4:

Calculation of formal charges shows that

for Be, FC 2 (20)0 and for Cl, FC 7 (16) 0

In BeCl 2 , the chlorine atoms achieve the argon configuration, [Ar], and the beryllium

atom has a share of only four electrons. Compounds such as BeCl 2 , in which the central

atom shares fewer than 8 e, are sometimes referred to as electron deficientcompounds.

This “deficiency” refers only to satisfying the octet rule for the central atom. The term

does not imply that there are fewer electrons than there are protons in the nuclei, as in

the case of a cation, because the molecule is neutral.

A Lewis formula can be written for BeCl 2 that doessatisfy the octet rule (see structure

shown in the margin). Let us evaluate the formal charges for that formula:

for Be, FC 2 (40)2 and for Cl, FC 7 (24) 1

We have said that the most favorable structure for a molecule is one in which the formal

charge on each atom is zero, if possible. In case some atoms did have nonzero formal

charges, we would expect that the more electronegative atoms (Cl) would be the ones with

lowest formal charge. Thus, we prefer the Lewis structure shown in Example 7-5 over

the one in the margin.

One might expect a similar situation for compounds of the other IIA metals, Mg, Ca,

Sr, Ba, and Ra. These elements, however, have lower ionization energiesand larger radii

than Be, so they usually form ions by losing two electrons.

EXAMPLE 7-6 Limitations of the Octet Rule

Write the Lewis formula for boron trichloride, BCl 3 , a covalent compound.

Plan

This covalent compound of boron is an example of limitation type B. As we follow the steps

in writing the Lewis formula, we use six electronsas the number neededby boron in step 2. Steps

3 and 4 should show only three pairs of electrons for boron.

Solution

Cl

Step 1: The skeleton is Cl B Cl

Cl Be Cl or Cl Be Cl

See the Saunders Interactive
General Chemistry CD-ROM,
Screen 9.7, Electron-Deficient
Compounds.

Cl Be Cl