To draw the Lewis dot structure for the cyanide ion (CN⁻), follow these steps:
- Determine the total number of valence electrons: Carbon has 4 valence electrons, and nitrogen has 5. Since the cyanide ion has a negative charge, add one extra electron. This gives a total of 4 + 5 + 1 = 10 valence electrons.
- Draw the skeletal structure: Carbon and nitrogen are connected by a triple bond. Place the carbon atom first, followed by the nitrogen atom.
- Distribute the electrons: Place the remaining electrons around the atoms to satisfy the octet rule. Carbon will have a lone pair, and nitrogen will have a lone pair as well.
a. Hybridization State:
- Carbon in CN⁻ is sp hybridized. This is because it forms one sigma bond with nitrogen and has two lone pairs.
- Nitrogen in CN⁻ is also sp hybridized. It forms one sigma bond with carbon and has two lone pairs.
b. Sigma and Pi Bonds:
- The triple bond between carbon and nitrogen consists of one sigma bond and two pi bonds. The sigma bond is formed by the overlap of sp hybrid orbitals from carbon and nitrogen. The two pi bonds are formed by the side-by-side overlap of unhybridized p orbitals.
c. Non-Bonding Electrons:
- Both carbon and nitrogen have lone pairs of electrons. Carbon has one lone pair, and nitrogen has one lone pair. These are non-bonding electrons.