Formal Charge and Resonance Problem Set

Introduction

Formal charge problems test both arithmetic and chemical interpretation. A correct electron total can still produce a poor Lewis drawing, and a pleasing drawing can hide the wrong net charge. Work systematically: count electrons, distribute them, calculate every formal charge, and only then compare resonance contributors.

Core explanation

For an atom in a Lewis representation, formal charge FC = V − N − B/2, where V is the free atom's valence-electron count, N is nonbonding electrons shown on that atom and B is bonding electrons in its drawn bonds. Equivalently, subtract the number of lone electrons and one electron per bond line from V. All atom formal charges must sum to the charge on the species. Formal charge is a bookkeeping allocation of shared electrons; it is not the measured physical charge at each atom.

Start with ammonium, NH₄⁺. The electron budget is 5 + 4(1) − 1 = 8. Four N–H single bonds use all eight electrons. Nitrogen has four bond lines and no lone electrons, so FC(N) = 5 − 0 − 4 = +1. Each hydrogen has one bond, so FC(H) = 1 − 1 = 0. The total +1 verifies the ion's charge. The geometry is tetrahedral, and the four N–H bonds are equivalent in this simple description.

For hydroxide, OH⁻, the budget is 6 + 1 + 1 = 8. One O–H bond uses two electrons and oxygen retains three lone pairs. FC(O) = 6 − 6 − 1 = −1; FC(H) = 0. A frequent error is to assign oxygen −1 merely because it is electronegative, without checking the pictured electrons. The formal charge follows the Lewis arrangement and formula.

For nitrate, NO₃⁻, the budget is 5 + 3(6) + 1 = 24. A common octet-respecting contributor has one N=O and two N–O single bonds. Nitrogen has four bond lines, giving FC(N) = +1. The double-bonded oxygen has two lone pairs and FC = 0; each single-bonded oxygen has three lone pairs and FC = −1. The sum is −1. There are three equivalent ways to choose the oxygen shown with the double bond. These are contributors to one delocalised ion, not three ions rapidly switching positions.

For carbonate, CO₃²⁻, the budget is 4 + 3(6) + 2 = 24. One C=O and two C–O bonds give carbon four bond lines and FC(C) = 0. The double-bonded oxygen has FC = 0; each single-bonded oxygen FC = −1, summing to −2. Again three symmetry-equivalent contributors account for equal C–O bonds in the physical ion. Do not draw three double bonds simultaneously while keeping carbon to an octet; that would change the electron count and central-atom valence.

For less symmetric alternatives, compare plausible contributors by octets and charge placement. Smaller unnecessary charge separation often favours a contributor, and negative formal charge is generally more plausible on a more electronegative atom, but these are useful criteria rather than infallible laws. Measured structure and electronic theory ultimately decide which Lewis model represents the species best.

Step-by-step reasoning

1. Total valence electrons, adding one per negative charge and subtracting one per positive charge. 2. Connect atoms, complete suitable terminal octets, then assess the centre. 3. Calculate each formal charge explicitly and sum them. 4. Move only electron pairs, not nuclei, to construct alternate resonance contributors. 5. Use symmetry and formal-charge criteria to describe the actual delocalised bonding.

Visual explanation

Draw three nitrate triangles side by side. Put the N=O double line on a different oxygen in each. Under each, write N +1, two O −1 and one O 0; then draw one central resonance hybrid with three equal N–O bonds.

Real-world analogy

Three maps can emphasise different routes through the same city; the city itself does not jump between maps. Resonance drawings are representations that together describe one electron distribution, not frames of a mechanical animation.

Real-world example

Nitrate in water and carbonate in minerals are identified as polyatomic ions, whose equivalent oxygen positions influence their measured bond lengths. Formal-charge work helps distinguish the internal covalent structure from the ions' overall charge.

Why?

Why does a nitrate contributor put +1 on nitrogen even though the entire ion is −1? Formal charges must be summed across atoms: +1 on N and −1 on each of two single-bonded oxygens give −1 overall.

Common misconception

“The double bond in nitrate stays on one specific oxygen.” A single contributor labels it that way, but equivalent contributors and experimental symmetry support equivalent N–O bonding in the ion.

Worked example

Check carbonate in one contributor. Carbon has V = 4, N = 0 and four bond lines: FC = 0. Double-bonded oxygen has V = 6, N = 4 and two bond lines: FC = 0. Each single-bonded oxygen has V = 6, N = 6 and one bond: FC = −1. The total is −2, matching CO₃²⁻. Its three contributors place the double line at three equivalent sites.

Quick check

1. What must the formal charges in NH₄⁺ sum to? Answer: +1, the net charge of the ion.

Exam focus

Show the total electron budget and individual charge arithmetic. Verify the charge sum and central-atom octet. Write resonance arrows between valid contributors, never between structures with atoms in different positions.

Advanced insight

Formal charges can be useful even when they look unlike computed partial charges. The scheme assigns half the electrons in each bond to each atom by convention; physical electron density need not divide equally. Its value lies in disciplined comparison of Lewis representations.

Summary

Formal charge is an electron-bookkeeping test for Lewis structures. Ammonium has +1 on N; hydroxide has −1 on O; nitrate and carbonate have sets of equivalent resonance contributors. Charge totals and electron budgets guard against attractive but impossible drawings.

Practice questions

1. How many valence electrons belong in the Lewis budget for NO₃⁻? Answer: 24. 2. What is nitrogen's formal charge in NH₄⁺? Answer: +1. 3. In a carbonate contributor, what are the single-bonded oxygens' formal charges? Answer: −1 each. 4. May resonance change which oxygen nucleus occupies a position? Answer: No. Resonance changes electron placement while the atomic skeleton stays fixed.