The hydrogen cyanide lewis structure places carbon in the middle, giving the arrangement \( \mathrm{H-C \equiv N} \) with a single bond to hydrogen, a triple bond between carbon and nitrogen, and one lone pair on nitrogen.
Step 1: Count total valence electrons
Count valence electrons from each atom and add them:
\[ \text{Valence e}^- = 1(\text{H}) + 4(\text{C}) + 5(\text{N}) = 10 \]
The final hydrogen cyanide lewis structure must contain exactly 10 valence electrons (5 electron pairs) distributed among bonds and lone pairs.
Step 2: Choose the skeletal (connectivity) structure
Hydrogen forms only one bond and is almost always terminal, so carbon is placed in the center: \( \mathrm{H-C-N} \).
Common connectivity error: writing \( \mathrm{H-N-C} \) does not match typical bonding preferences for these atoms because hydrogen cannot be a central atom and carbon commonly serves as the central atom when present with H and a more electronegative atom like N.
Step 3: Add single bonds and subtract electrons used
Start with single bonds H–C and C–N. Each single bond uses 2 electrons:
\[ 2 \text{ bonds} \times 2 \, e^- = 4 \, e^- \quad \Rightarrow \quad 10 - 4 = 6 \, e^- \text{ remaining} \]
Step 4: Complete octets and adjust with multiple bonds
Place the remaining 6 electrons as lone pairs to satisfy octets on the non-hydrogen atoms. If all 6 were placed as lone pairs on nitrogen, carbon would still lack a full octet. The fix is to convert lone pairs on nitrogen into bonding pairs between C and N until carbon has an octet.
Converting two lone pairs on N into bonding pairs creates a triple bond between C and N, giving: \( \mathrm{H-C \equiv N} \), and leaving one lone pair on N.
Step 5: Verify the electron count
Check that the structure uses exactly 10 valence electrons:
\[ \underbrace{2}_{\mathrm{H-C}} + \underbrace{6}_{\mathrm{C \equiv N}} + \underbrace{2}_{\text{lone pair on N}} = 10 \, e^- \]
Step 6: Compute formal charges
Formal charge is computed as:
\[ \mathrm{FC} = (\text{valence e}^-) - (\text{nonbonding e}^-) - \frac{1}{2}(\text{bonding e}^-) \]
| Atom | Valence \(e^-\) | Nonbonding \(e^-\) | Bonding \(e^-\) | \(\mathrm{FC}\) |
|---|---|---|---|---|
| H | 1 | 0 | 2 (one single bond) | \(1 - 0 - \tfrac{1}{2}\cdot 2 = 0\) |
| C | 4 | 0 | 8 (one single + one triple) | \(4 - 0 - \tfrac{1}{2}\cdot 8 = 0\) |
| N | 5 | 2 (one lone pair) | 6 (triple bond) | \(5 - 2 - \tfrac{1}{2}\cdot 6 = 0\) |
Zero formal charge on every atom indicates that \( \mathrm{H-C \equiv N} \) is the dominant Lewis structure.
Step 7: Molecular geometry and hybridization
In VSEPR terms, carbon has two electron domains (the H–C bond and the C≡N bond; a multiple bond counts as one domain for geometry), so the electron-domain geometry is linear and the molecular shape is linear: \( \angle \mathrm{H-C-N} \approx 180^\circ \).
The bonding is consistent with sp hybridization at carbon (and at nitrogen): one sp orbital forms the H–C \( \sigma \) bond, the other sp orbital forms the C–N \( \sigma \) bond, and two unhybridized p orbitals on each atom overlap to make the two \( \pi \) bonds in the triple bond.
Quick checklist for the hydrogen cyanide lewis structure
- Total valence electrons: \(10\).
- Connectivity: H is terminal; carbon is central.
- Bonding: H–C single bond; C≡N triple bond.
- Lone pairs: one lone pair on N; none on C or H.
- Formal charges: all atoms \(0\).
- Shape: linear (\(\approx 180^\circ\)).