The carbon atom in the cyanate ion (NCO⁻) is sp hybridized. Carbon sits at the center of this linear ion, forming two sigma bonds (one to nitrogen and one to oxygen) with no lone pairs, giving it a steric number of 2. A steric number of 2 always corresponds to sp hybridization and a bond angle of 180°.
Why Carbon Is sp Hybridized
Hybridization depends on how many groups of electrons surround an atom, counting only sigma bonds and lone pairs. In NCO⁻, the central carbon forms one sigma bond with nitrogen and one sigma bond with oxygen. It has no lone pairs. That gives carbon a steric number of 2, which maps directly to sp hybridization.
With sp hybridization, two of carbon’s orbitals (one s and one p) mix to form two sp hybrid orbitals pointing in opposite directions, 180° apart. The remaining two unhybridized p orbitals sit perpendicular to the molecular axis and participate in pi bonding with nitrogen and oxygen. This is why NCO⁻ is perfectly linear, with a measured bond angle of 180°.
Resonance Structures and Bonding
NCO⁻ has three resonance structures, and they help explain what the pi bonds look like:
- [N≡C–O]⁻: A triple bond between carbon and nitrogen, a single bond to oxygen, and the negative charge on oxygen. This is the most important contributor because it places the negative charge on the most electronegative atom and minimizes charge separation.
- [N=C=O]⁻: Two double bonds with a negative charge on nitrogen and a positive charge on… actually, this version places a negative charge on oxygen and positive on nitrogen, with charge separation.
- [N⁻–C=O] or similar: The least favorable structure, with a positive charge on carbon and a negative charge on nitrogen.
The dominant resonance structure features a triple bond between carbon and nitrogen and a single bond to oxygen. Orbital analysis confirms this: the carbon-nitrogen bond behaves like a true triple bond (one sigma plus two pi bonds), while oxygen lone pairs align parallel to the pi system. The C–N bond length in NCO⁻ is about 1.19 Å, shorter than a typical single bond, and the C–O bond is about 1.22 Å. X-ray data from sodium cyanate crystals show similar values: 1.13 Å for C–N and 1.21 Å for C–O.
Regardless of which resonance structure you emphasize, the carbon always forms two sigma bonds and zero lone pairs. The pi bonds (whether arranged as a triple bond on one side or double bonds on both sides) use the unhybridized p orbitals. So the hybridization stays sp across all resonance contributors.
How to Determine This Yourself
If you want to find the hybridization of any atom in a molecule, the process is straightforward. Count the number of sigma bonds the atom forms (each connection to a neighboring atom counts as one sigma bond, regardless of bond order). Then count any lone pairs on that atom. Add those two numbers together to get the steric number.
For carbon in NCO⁻: two sigma bonds plus zero lone pairs equals a steric number of 2. A steric number of 2 means sp. A steric number of 3 would mean sp², and 4 would mean sp³. The multiple bonds (double or triple) don’t change the steric number because they involve pi bonds, which use unhybridized p orbitals rather than hybrid orbitals.
Comparison With Similar Molecules
NCO⁻ is isoelectronic with carbon dioxide (CO₂), meaning they have the same number of electrons and the same overall shape. In CO₂, the carbon is also sp hybridized, forming two sigma bonds with no lone pairs, producing a linear molecule. Acetylene (C₂H₂) and hydrogen cyanide (HCN) follow the same pattern: sp-hybridized carbon atoms with linear geometry.
The key theme is that sp hybridization produces linear geometry with 180° bond angles. Whenever you see a carbon atom bonded to exactly two other atoms with no lone pairs, you can confidently assign sp hybridization, whether the bonds are single, double, or triple. The cyanate ion fits this pattern cleanly.