Three 16-Electron Linear Triatomics: Resonance, Formal Charge, and Measured Bond Lengths
Three linear triatomic species with 16 valence electrons are examined: CO₂, N₃⁻, and SCN⁻. Measured bond lengths are: both C–O bonds in CO₂ = 116 pm; both N–N bonds in N₃⁻ = 116 pm; in SCN⁻, S–C = 168 pm and C–N = 118 pm (typical single-bond values for comparison: S–C ≈ 181 pm, C–N ≈ 147 pm; C–N triple bond ≈ 116 pm). A student argues that N₃⁻ must have two unequal N–N bonds because its two resonance structures place the triple bond on different sides, and draws SCN⁻ as S=C=N⁻ with the negative formal charge on sulfur, calling that the best structure because "the charge belongs on the bigger atom."
1. Draw the relevant resonance structures for N₃⁻ and SCN⁻, assign formal charges to every atom, and identify the major contributor for each species. Then evaluate the student's two claims.
Read the worked answer
N₃⁻ (16 e⁻): the dominant structure is [N=N=N]⁻, each terminal N carrying two lone pairs and a double bond, giving formal charges (−1, +1, −1). A second type, N≡N–N with three lone pairs on the terminal N, gives (0, +1, −2); the −2 charge makes it a minor contributor. The three resonance forms are related by symmetry, so the hybrid has identical electron density on both ends and therefore equal N–N bond lengths. The student's claim is wrong: resonance structures are not separately observable states that alternate; the molecule is a single averaged structure, and here the two "sided" forms are degenerate anyway.
SCN⁻ (16 e⁻): two competing structures are [S=C=N]⁻ with formal charges (0, 0, −1) and [S–C≡N]⁻ with charges (−1, 0, 0). The better structure places the negative charge on the more electronegative atom, nitrogen, so [S–C≡N]⁻ dominates. The student's "bigger atom holds the charge" rule is not a valid criterion - electronegativity, not size, decides.
2. Use the measured lengths to argue for a bond order greater than 2 for the N–N bonds in N₃⁻ and greater than 2 for the C–N bond in SCN⁻. Explain why only within-species comparisons (measured vs. expected single/double/triple for the same pair of atoms) are valid.
Read the worked answer
N₃⁻: a typical N=N double bond is about 125 pm, and a triple bond about 110 pm. The measured 116 pm lies between them, so the hybrid has bond order greater than 2 (approaching 3), reflecting a contribution from the N≡N–N form; that also explains the shortening relative to a pure double bond, while the two bonds remain equal by symmetry.
SCN⁻: 118 pm matches almost exactly the C–N triple bond (about 116 pm), so the C–N bond is essentially triple. The S–C distance of 168 pm lies between the single-bond value (181 pm) and a double bond (about 155 pm), as expected when a minor S=C character is mixed in with the dominant S–C≡N⁻ form.
Across different pairs of atoms, bond length is not a clean bond-order meter - 116 pm appears in CO₂ (C–O, order 2), in N₃⁻ (N–N, order >2), and near SCN⁻'s C–N (order 3) simply because atomic radii differ. The valid test compares a measured length with the tabulated single, double, and triple lengths for that same pair of atoms.
3. Predict the net dipole moment of each of the three species (zero or nonzero) and justify each in terms of geometry and charge distribution.
Read the worked answer
CO₂: two identical C=O dipoles pointing in opposite directions along a linear axis cancel exactly. Net dipole = 0; the molecule is nonpolar.
N₃⁻: symmetric, with both terminal nitrogens chemically equivalent (identical formal charges and identical bond lengths). Any bond dipoles cancel along the axis, and the −1 charge is shared equally by both ends. Net dipole = 0.
SCN⁻: the two ends are different atoms with different electronegativities and different charge distribution (the hybrid puts more negative character at nitrogen). The two bond dipoles do not cancel, so there is a net dipole along the molecular axis: SCN⁻ is polar. Since all three have net charge, ion–dipole forces dominate in water regardless, but only SCN⁻ has a permanent dipole that can orient in an applied field, and its nitrogen end is the better hydrogen-bond acceptor.