Multiple Bonds and Polyatomic Ions

After single bonds and terminal lone pairs are placed, an ordinary octet-obeying center such as C, N or O that still has fewer than eight electrons usually needs a multiple bond. Treat known electron-deficient or radical exceptions separately. Bring in one lone pair from an outer atom for each pair of electrons the centre is missing: short by two means one double bond, short by four means either two double bonds or one triple bond.

Promoting a lone pair into a bond lets that pair count around two atoms rather than only one. A terminal O with three lone pairs and one single bond already has eight surrounding electrons. When one lone pair becomes a second bond, O has two lone pairs and a double bond and still has eight, while the central atom gains two electrons in its surrounding count. No electrons are created. For ordinary second-period centers, stop once the center reaches eight rather than adding extra multiple bonds to eliminate a charge.

Bond order is the number of shared pairs between two atoms: 1 for a single bond, 2 for a double, 3 for a triple. Higher bond order pulls the nuclei closer, so the bond gets shorter and stronger. The C-C bond in ethane is longer and weaker than the C=C in ethene, which is longer and weaker again than the triple bond in ethyne.

Compare bond orders within a similar atom pair and environment: C–C, C=C and C≡C are a useful family. You cannot infer that every triple bond is shorter than every double bond between any elements, because atomic size also changes bond length. Likewise a double bond is not exactly twice as strong as a single bond; it adds a π interaction to a σ interaction rather than duplicating an identical bond. Resonance can make several equivalent bonds have an intermediate average order.

For an anion, add one electron per unit of negative charge before drawing. For a cation, remove one electron per unit of positive charge. Nitrate NO₃⁻ has 5 + 3(6) + 1 = 24 electrons, while ammonium NH₄⁺ has 5 + 4 - 1 = 8.

For nitrate, first draw three N–O single bonds and complete the oxygen octets: six bonding electrons plus eighteen nonbonding electrons equals the required twenty-four, but N has only six surrounding electrons. Convert one oxygen lone pair into an N=O bond. The final count is eight bonding electrons plus sixteen lone-pair electrons. N now has an octet. Its formal charge is +1, each single-bonded O is −1 and the double-bonded O is 0, totaling −1.

Two checks catch nearly every error. First, the electrons drawn must equal the adjusted total. Second, the sum of the formal charges on all atoms must equal the overall charge on the ion, which is zero for a neutral molecule.

This nitrate drawing is one of three equivalent contributors because the double bond can occupy any of the three N–O positions. A single contributor does not imply that the actual ion has one permanently special oxygen. Brackets with an outside − charge describe the entire ion; individual formal charges inside belong to the chosen contributor. A correct sum alone is insufficient if nitrogen has been drawn with ten electrons.

A polyatomic ion has covalent bonding within it and an overall net charge. In Ca(NO₃)₂, two nitrate ions balance one Ca²⁺ ion; the parentheses count whole nitrate groups, not additional N–O bonds. When drawing nitrate itself, use its twenty-four-electron budget and put the completed structure in brackets with − outside.

For CN⁻, 4 + 5 + 1 = 10 electrons give a C≡N triple bond and one lone pair on each atom. Formal charges are −1 on C and 0 on N. The ion's overall charge must be shown even though a valid local structure and octets have already been achieved. Hydrogen never receives a double bond in the ordinary structures considered here, and terminal fluorine normally retains a single bond and three lone pairs.