What is the bond-order of CN?

[Note: Write the answer after multiplying it with 4]

The bond order of the cyanide ion (CN) can be determined through its molecular orbital configuration. To find the bond order, we first need to understand the number of bonding and antibonding electrons in the molecule. The bond order is calculated using the formula: Bond Order = (Number of bonding electrons - Number of antibonding electrons) / 2 For CN, we can start by determining its electron configuration.

Step-by-Step Calculation of Bond Order

Cyanide (CN) consists of a carbon atom and a nitrogen atom. The total number of valence electrons in CN is:

• Carbon (C) has 4 valence electrons.
• Nitrogen (N) has 5 valence electrons.

Adding these together gives us a total of 9 valence electrons for CN.

Molecular Orbital Diagram

In the molecular orbital theory, these 9 electrons will fill the molecular orbitals in the following order:

• σ(1s) - 2 electrons
• σ*(1s) - 2 electrons
• σ(2s) - 2 electrons
• σ*(2s) - 0 electrons
• σ(2p) - 2 electrons
• π(2p) - 2 electrons
• π*(2p) - 0 electrons

Now, let's count the bonding and antibonding electrons:

• Bonding electrons: 2 (σ(1s)) + 2 (σ(2s)) + 2 (σ(2p)) + 2 (π(2p)) = 8
• Antibonding electrons: 2 (σ*(1s)) + 0 (σ*(2s)) + 0 (π*(2p)) = 2

Calculating Bond Order

Now we can plug these values into the bond order formula:

Bond Order = (8 - 2) / 2 = 6 / 2 = 3

However, since you requested the bond order multiplied by 4, we perform that calculation:

Bond Order x 4 = 3 x 4 = 12

Final Thoughts

The bond order of CN, when multiplied by 4, results in a value of 12. This indicates a strong triple bond between the carbon and nitrogen atoms, which is consistent with the known stability and reactivity of the cyanide ion in various chemical contexts.