Question

The bond order of dioxygen is \(m\). The bond order values of \(N_2^{+}\) and \(C_2^{2-}\) are respectively

• A. \(\dfrac{5m}{4},\ \dfrac{3m}{2}\)
• B. \(\dfrac{3m}{2},\ \dfrac{5m}{4}\)
• C. \(\dfrac{m}{2},\ \dfrac{m}{3}\)
• D. \(\dfrac{2m}{3},\ \dfrac{m}{2}\)

Hint:

Bond order is calculated using \[ \text{B.O.}=\frac{N_b-N_a}{2}. \] Removing an electron from a bonding orbital decreases bond order by \(0.5\), while adding an electron to a bonding orbital increases bond order by \(0.5\).

Answer 1

The Correct Option is A

Step 1: Determine bond orders via MO electron counting.
O₂: 16 e⁻, B.O. = (10–6)/2 = 2 = m. N₂⁺: remove 1 bonding e⁻ from N₂ (B.O.=3) → B.O. = 2.5 = 5/2 = 5m/4. C₂²⁻: add 2 e⁻ to C₂ (B.O.=2) into π bonding → B.O. = 3 = 3m/2.

Step 2: Final Answer:
B.O.(N₂⁺) = 5m/4, B.O.(C₂²⁻) = 3m/2 (option 1).