The bond order and magnetic property of acetylide ion are same as that of

🧠 Find which species has the same bond order and magnetic property as the acetylide ion $\mathrm{C_2^{2-}}$.

🗺️ MO analysis of $\mathrm{C_2^{2-}}$ (14 electrons):

\sigma_{1s}^2\sigma^*_{1s}^2\sigma_{2s}^2\sigma^*_{2s}^2\pi_{2p}^4\sigma_{2p}^2

Bond order = $(10-4)/2 = \mathbf{3}$; all electrons paired → diamagnetic

Check each option:

| Species | Total e⁻ | Bond Order | Unpaired e⁻ | Magnetic | |---------|-----------|-----------|------------|---------| | $\mathrm{O_2^+}$ | 15 | 2.5 | 1 | Paramagnetic | | $\mathrm{N_2^+}$ | 13 | 2.5 | 1 | Paramagnetic | | $\mathrm{NO^+}$ | 14 | 3 | 0 | Diamagnetic ✓ | | $\mathrm{O_2^-}$ | 17 | 1.5 | 1 | Paramagnetic |

⚡ $\mathrm{NO^+}$ is isoelectronic with $\mathrm{C_2^{2-}}$ (both 14 electrons) → same MO configuration → BO = 3 and diamagnetic.

$\boxed{\text{Answer: } \mathrm{NO^+}}$