Sometimes the steps in 2.5 give more than one valid Lewis diagram for the same formula. Two questions follow. If the diagrams are equally good, which one is real? And if they are not equally good, how do you pick the best? Resonance answers the first; formal charge answers the second.
One formula, several diagrams
Draw the nitrate ion, NO₃⁻. It has 5 + 3 × 6 + 1 = 24 valence electrons. N goes in the center with three O atoms around it. After the outer atoms get their lone pairs, N has only six electrons, so one O lone pair becomes an N=O double bond. But which O? Any of the three works. That gives three diagrams that place the atoms identically and differ only in where the double bond is drawn.
Diagrams like these are resonance structures. They are linked with a double-headed arrow (↔). A rule worth memorizing: resonance structures never move atoms; they only move electrons (lone pairs and the extra pairs of multiple bonds).
What resonance means
A single diagram of nitrate predicts one short N=O bond and two longer N–O bonds. Measurements show something else: all three N–O bonds are the same length, between a typical N–O single bond and an N=O double bond. No single diagram describes that.
The real ion is the resonance hybrid, one structure that is a blend of all the valid diagrams. The extra bonding pair is not on one O; it is spread over all three N–O bonds. Electrons spread over several atoms like this are delocalized (you met delocalized electrons in metals in 2.1; here the spreading is over a few atoms).
The ion does not switch between the diagrams. A mule is not a horse on some days and a donkey on others; it is always a mule, a blend of both. Each diagram is just our best attempt to draw it with lines and dots.
Bond order in a hybrid
The bond order is the number of bonding pairs between two atoms: 1 for a single bond, 2 for a double, 3 for a triple. In a hybrid the bond order can be a fraction. Count the bonding pairs between the central atom and the atoms that share the resonance, and divide by the number of those bonds.
Worked example. Find the bond order of each N–O bond in nitrate, NO₃⁻.
Each resonance structure has one double bond (2 pairs) and two single bonds (1 pair each): 2 + 1 + 1 = 4 bonding pairs.
These 4 pairs are shared equally by 3 bonds: bond order = 4 ÷ 3 = 1.33 (4/3).
So each N–O bond is between a single bond (1) and a double bond (2): longer and weaker than N=O, shorter and stronger than N–O. Ozone, O₃, works the same way: 3 pairs over 2 bonds, bond order 1.5.
Formal charge
When the diagrams for one formula are not equivalent, you need a way to compare them. Formal charge is a bookkeeping number for each atom in a diagram. It compares the valence electrons the atom brings with the electrons the diagram assigns to it: all its lone-pair electrons, plus half of each bond (a bond is shared, so each atom gets half).
formal charge = valence electrons − lone-pair electrons − ½ (bonding electrons)
Two checks: the formal charges in a diagram add up to the overall charge of the species (zero for a molecule), and formal charge is not the real charge on the atom. It is a tool for choosing diagrams.
Worked example. Find the formal charges in this diagram of the thiocyanate ion, SCN⁻: S=C=N, with two lone pairs on S and two on N.
S (6 valence): 6 − 4 (two lone pairs) − ½(4) = 0.
C (4 valence): 4 − 0 − ½(8) = 0.
N (5 valence): 5 − 4 − ½(4) = −1.
Check: 0 + 0 + (−1) = −1, the ion's charge.
Choosing the best diagram
Among valid diagrams (octets kept for period 2 atoms, correct electron count), the best one:
- has formal charges as close to zero as possible, and
- puts any negative formal charge on the more electronegative atom.
The reasoning is physical: separating charge costs energy, and a more electronegative atom attracts electrons more strongly, so it holds extra electron density at lower energy. The best diagram contributes most to the hybrid; poorer ones contribute less.
Worked example. Which diagram of SCN⁻ is best? Electronegativity: N 3.0, S 2.6, C 2.6.
A: S=C=N (two lone pairs on each end): S 0, C 0, N −1.
B: S–C≡N (three lone pairs on S, one on N): S 6 − 6 − 1 = −1, C 0, N 5 − 2 − 3 = 0.
C: S≡C–N (one lone pair on S, three on N): S 6 − 2 − 3 = +1, C 0, N 5 − 6 − 1 = −2.
C has the largest formal charges, so it is worst. A and B both have just one −1. A puts it on N, the more electronegative atom, so A is the best diagram; B also contributes to the hybrid.
Formal charge can sometimes favor a diagram with an expanded octet (for example, sulfate drawn with two S=O bonds has smaller formal charges). Both styles of sulfate diagram appear in textbooks, and the exam accepts a diagram with octets and correct formal charges; if a question asks you to use formal charge to choose, follow the rule above.
Common mistakes
- "The molecule flips between structures." It has one hybrid structure.
- Moving atoms. Resonance moves electrons only.
- Counting a whole bond for one atom in formal charge. Each atom gets half the bonding electrons.
- Bond order 1.5 for carbonate. Count the pairs: 4 over 3 bonds, 4/3.