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 bond order formula and condition for fractional values.
\[ \text{Bond Order} = \frac{N_b - N_a}{2} \] A fractional bond order results when $(N_b - N_a)$ is an odd number.
Step 2: Evaluate bond order of $O_2^+$ (15 electrons).
MO filling ends at $(\pi^*2p)^1$. $N_b = 10$, $N_a = 5$. Bond order = $(10-5)/2 = 2.5$. Fractional.
Step 3: Evaluate bond order of $O_2^-$ (17 electrons).
One extra electron in $\pi^*$: $(\pi^*2p)^3$. $N_b = 10$, $N_a = 7$. Bond order = $(10-7)/2 = 1.5$. Fractional.
Step 4: Evaluate bond order of $N_2^+$ (13 electrons).
Configuration: $(\sigma1s)^2(\sigma^*1s)^2(\sigma2s)^2(\sigma^*2s)^2(\pi2p)^4(\sigma2p)^1$. $N_b = 9$, $N_a = 4$. Bond order = $(9-4)/2 = 2.5$. Fractional.
Step 5: Confirm all three species have fractional bond orders.
$O_2^+$: 2.5, $O_2^-$: 1.5, $N_2^+$: 2.5. All are fractional. This is the only option with exclusively fractional values for all three species.
Step 6: State the final answer.
\[ \boxed{O_2^+,\ O_2^-,\ N_2^+} \]