Choosing the Best Lewis Structure with Formal Charge

First require the correct connectivity, total valence electrons and overall charge. Then require ordinary second-period octets, apart from a justified radical or electron-deficient exception. A structure with ten electrons on N is invalid even if a formal-charge calculation looks attractive. Formal-charge preferences cannot rescue an invalid electron budget or forbidden octet expansion.

Among valid drawings, prefer fewer and smaller formal-charge magnitudes when other factors are comparable. All neutral structures already have a net charge of zero, so “closest to zero” must refer to individual charges and charge separation, not to their sum. Next, a negative charge is generally better accommodated on a more electronegative atom. These are introductory chemical heuristics, not exact calculated resonance weights.

Formal-charge and electronegativity rules are introductory ranking heuristics, not quantitative resonance-weight measurements. In SCN⁻, S=C=N has formal charges (0,0,−1), while S–C≡N has (−1,0,0). The stated electronegativity tiebreaker selects the first because N is more electronegative than S; this alone does not establish the actual weights of the contributors. A third valid octet drawing S≡C–N has (+1,0,−2), placing +1 on sulfur, not nitrogen. The real ion is a resonance hybrid.

To compare N₂O contributors, keep N–N–O connectivity and use sixteen valence electrons in each. The triple/single form has terminal N = 0, central N = +1, O = −1. The double/double form has terminal N = −1, central N = +1, O = 0. Both satisfy octets and have equal charge magnitudes; placing the negative formal charge on O favors the first under the electronegativity criterion. Reversing the connectivity to N–O–N is not merely another resonance contributor.

Ions must carry charge somewhere, so the goal is the best placement, not zero everywhere. In nitrate, the negative charges sit on oxygen atoms rather than on nitrogen, which is exactly what the second rule demands.

Nitrate illustrates why zero everywhere is impossible for an anion. Its required charge is −1, and the standard octet contributors each have +1 on N and two −1 oxygens. Drawing three N=O bonds would appear to change the charge distribution but gives nitrogen twelve surrounding electrons, which fails the validity check. Show all equivalent valid contributors rather than selecting one oxygen as the permanent location of the negative charge.

A useful written answer names the favored drawing, lists the distinguishing formal charges and states the criterion. For OCN⁻: “Both octet-valid contributors have overall −1 charge, but the O–C≡N contributor places the negative formal charge on O rather than N. With equal charge magnitudes, oxygen's greater electronegativity favors that contributor in this model.”

Check that the same criterion really distinguishes the candidates. Equivalent carbonate contributors have equal formal-charge patterns related by symmetry, so no oxygen's version is preferred over the others. For non-equivalent contributors involving different elements or bond types, electronegativity alone may not establish actual physical weights. Report the conclusion under the stated heuristic without claiming that a minor contributor disappears from the description.