To draw the Lewis structure for the cyanide ion (CN–), we need to follow these steps:
- Count the total valence electrons: Carbon (C) has 4 valence electrons and nitrogen (N) has 5 valence electrons. The cyanide ion carries a negative charge, which adds 1 additional electron. Therefore, the total number of valence electrons is:
- 4 (from C) + 5 (from N) + 1 (for the negative charge) = 10 valence electrons
- Arrange the atoms: Place carbon (C) at the center, as it is less electronegative than nitrogen (N). Thus, the basic arrangement will be C — N.
- Add bonds between atoms: Form a triple bond between C and N to satisfy the bonding requirements. This uses 6 of our 10 valence electrons:
- Each bond represents 2 electrons, so for the triple bond: 3 bonds x 2 = 6 electrons.
- Distribute the remaining electrons: After using 6 electrons in bonds, we have 4 electrons left. Place these electrons as lone pairs:
- Since there are already 6 electrons involved in the triple bond, there are no lone pairs left on carbon or nitrogen in the cyanide ion.
- Calculate the formal charges: The formal charge can be calculated using the formula:
Formal Charge = Valence Electrons – (Non-Bonding Electrons + 0.5 * Bonding Electrons)
For Carbon:
4 (valence) – (0 lone pairs + 0.5 * 6 bonding) = + 1
For Nitrogen:
5 (valence) – (0 lone pairs + 0.5 * 6 bonding) = -1
The final Lewis structure for the cyanide ion looks like this:
:N≡C:
This structure represents a carbon atom triple bonded to a nitrogen atom. The negative charge is typically represented on the nitrogen atom, indicating that it has one extra electron, and the formal charge of carbon is +1, while nitrogen has a formal charge of -1. Thus, the overall charge of the cyanide ion is -1, consistent with our calculations.