Consider the following species :

CN⁺, CN^–, NO and CN

Which one of these will have the highest bond order?

Concept:

• Bond order is calculated as half of the number of bonding electrons minus the number of anti-bonding electrons.
• Hence, the mathematical formula is written as, 
• Bond order = (bonding orbital - antibonding orbital) /2
• The value of bond order determines the nature of the bond.

Explanation:

→ NO : (σ1s)², (σ*1s)²,(σ2s)², (σ*2s)2, (σ2p_z)², (π2p_x)² =(π2py)2,(π*2p_x)¹ = (π*2p_y)⁰

\(BO = \frac{{10 - 5}}{2} = 2.5\)

→ CN^- : (σ1s)2, (σ*1s)2, (σ2s)2, (σ*2s)2, (π2px)2 = (π2p_y)2 (π2p_z)2

\(BO = \frac{{10 - 4}}{2} =3\)

→ CN : (σ1s)², (σ∗1s)², (σ2s)²,(σ∗ 2s)², (π2p_x)² = (π2p_y)²,(σ2p_z)¹

\(BO = \frac{{9 - 4}}{2} =2.5\)

→ CN⁺ : (σ1s)², (σ∗1s)², (σ2s)²,(σ∗ 2s)², (π2p_x)² = (π2p_y) ²

\(BO = \frac{{8 - 4}}{2} =2\)

Therefore, CN–  has the highest bond order.

Hence, the correct option for the given question is, 2) CN–