Let us see how this relationship applies to some species whose Lewis formulas we have
already shown.
For F 2 ,
N 2 8 (for two F atoms) 16 eneeded
A 2 7 (for two F atoms) 14 eavailable
SNA 16 14 2 eshared
The Lewis formula for F 2 shows 14 valence electrons total, with 2 eshared in a single
bond.
For HF,
N 1 2 (for one H atom) 1 8 (for one F atom) 10 eneeded
A 1 1 (for one H atom) 1 7 (for one F atom) 8 eavailable
SNA 10 8 2 eshared
The Lewis formula for HF shows 8 valence electrons total, with 2 eshared in a single
bond.
For H 2 O,
N 2 2 (for two H atoms) 1 8 (for one O atom) 12 eneeded
A 2 1 (for two H atoms) 1 6 (for one O atom) 8 eavailable
SNA 12 8 4 eshared
The Lewis formula for H 2 O shows 8 valence electrons total, with a total of 4 eshared,
2 ein each single bond.
For CO 2 ,
N 1 8 (for one C atom) 2 8 (for two O atoms) 24 eneeded
A 1 4 (for one C atom) 2 6 (for two O atoms) 16 eavailable
SNA 24 16 8 eshared
The Lewis formula for CO 2 shows 16 valence electrons total, with a total of 8 eshared,
4 ein each double bond.
For NH 4 ,
N 1 8 (for one N atom) 4 2 (for four H atoms) 16 eneeded
A 1 5 (for one N atom) 4 1 (for four H atoms) 1 (for 1charge) 8 eavailable
SNA 16 8 8 eshared
The Lewis formula for NH 4 shows 8 valence electrons total, with all 8 eshared, 2 e
in each single bond.
The following general steps describe the use of the SNA relationship in
constructing dot formulas for molecules and polyatomic ions.
Writing Lewis Formulas
1.Select a reasonable (symmetrical) “skeleton” for the molecule or polyatomic ion.
a.The least electronegative elementis usually the central element, except that H is
never the central element, because it forms only one bond. The least elec-
282 CHAPTER 7: Chemical Bonding
F F
H F
H O
H
O C O
The 1ionic charge is due to a
deficiencyof one erelative to the
neutral atoms.
H
H
H N H