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Chemical Bonding 135
Don’t Forget!
Don’t Forget!
For CH 4O, we would have:
1 C 4 H 1 O
N 8 + 4 (2) =8 + 8 = 24
A 4 + 4 (1) =4 + 6 = 14
S = N-A = 24-14 = 10 bonds = S/2 = 10/2 = 5
The result, 5, indicates that there are five bonds present in this compound. You
should begin by placing one bond (pair of electrons) between each pair of atoms.
Use a dash to represent a bond. If you still have bonds remaining after placing a dash
between each pair of atoms, it means you have double or triple bonds. Next, distribute
the remaining available electrons around the atoms so that each atom has its
full octet, 8 electrons (either bonding or nonbonding, shared or not). Hydrogen is an
exception; it only gets 2. Figure 9-2 shows the Lewis structural formula of CH 4O.
In the N A S rule, only the valence electrons are counted. This rule assumes
that all atoms, except hydrogen, are going to obey the octet rule.
H
H C
O H
H
Figure 9-2 Lewis structural formula of CH4O It is also possible to write a Lewis structure for polyatomic anions or cations. The
N A = S rule can be used, but for an anion extra electrons equal to the magnitude
of the negative charge must be added to the electrons available. If the ion is
a cation, you will need to subtract number of electrons equal to the charge.
Metals nonmetals l ionic bonds
Nonmetals nonmetals l covalent bonds