Question

The set of species having only fractional bond order values is

• A. \(C_2^{2-},\,N_2,\,O_2^{2-}\)
• B. \(O_2^{+},\,O_2^{-},\,N_2^{+}\)
• C. \(O_2^{2+},\,O_2,\,C_2^{2-}\)
• D. \(Li_2,\,H_2^{+},\,C_2\)

Hint:

Species formed by adding or removing one electron from a stable molecule often show fractional bond order because the electron enters or leaves a molecular orbital singly.

Answer 1

The Correct Option is B

Step 1: Recall the formula for bond order.
According to molecular orbital theory, \[ \text{Bond Order} = \frac{N_b-N_a}{2} \] where \[ N_b=\text{number of bonding electrons} \] and \[ N_a=\text{number of antibonding electrons} \] Fractional bond order occurs when the difference \[ N_b-N_a \] is odd.

Step 2: Check option (1).
For \[ C_2^{2-} \] bond order is \[ 3 \] For \[ N_2 \] bond order is \[ 3 \] For \[ O_2^{2-} \] bond order is \[ 1 \] All are integral values.
Hence, option (1) is incorrect.

Step 3: Check option (2).
For \[ O_2^{+} \] bond order is \[ 2.5 \] For \[ O_2^{-} \] bond order is \[ 1.5 \] For \[ N_2^{+} \] bond order is \[ 2.5 \] All have fractional bond orders.
Hence, option (2) is correct.

Step 4: Check options (3) and (4).
For \[ O_2^{2+} \] bond order is \[ 3 \] For \[ O_2 \] bond order is \[ 2 \] For \[ C_2^{2-} \] bond order is \[ 3 \] Thus, option (3) contains only integral bond orders.
Similarly, \[ Li_2 \] has bond order \[ 1 \] \[ H_2^{+} \] has bond order \[ 0.5 \] \[ C_2 \] has bond order \[ 2 \] Since all are not fractional, option (4) is also incorrect.

Step 5: Final conclusion.
Therefore, the correct set containing only fractional bond order values is \[ \boxed{O_2^{+},\,O_2^{-},\,N_2^{+}} \]