We want to find which of the given diatomic species (among $\mathrm{O}_2^+$, $\mathrm{O}_2^-$, $\mathrm{N}_2^+$, and $\mathrm{NO}^+$) has the same bond order and the same magnetic property (dia- or paramagnetic) as the acetylide ion, $\mathrm{C}_2^{2-}$.

1. Electron Count and MO Considerations

(a) Acetylide Ion, $\mathbf{C_2^{2-}}$

A neutral $\mathrm{C}_2$ molecule would have $2 \times 6 = 12$ electrons.

In $\mathrm{C}_2^{2-}$, there are 2 extra electrons, giving a total of $12 + 2 = 14$ electrons.

(b) Matching Total Electrons Among the Options

Let us see how many electrons are in each species:

$\mathrm{O}_2$ has $2 \times 8 = 16$ electrons

$\mathrm{O}_2^+$ has $16 - 1 = 15$ electrons

$\mathrm{O}_2^-$ has $16 + 1 = 17$ electrons

$\mathrm{N}_2$ has $2 \times 7 = 14$ electrons

$\mathrm{N}_2^+$ has $14 - 1 = 13$ electrons

$\mathrm{NO}$ has $7 + 8 = 15$ electrons

$\mathrm{NO}^+$ has $15 - 1 = 14$ electrons

Hence the only diatomic species in the list that has 14 electrons (like $\mathrm{C}_2^{2-}$) is $\mathrm{NO}^+$.

2. Bond Order and Magnetic Property

(a) $\mathrm{C}_2^{2-}$ (14 Electrons)

From molecular orbital (MO) theory for 14-electron diatomics (similar to $\mathrm{N}_2$ with 14e), we generally get:

Bond order = 3

All electrons paired in MOs → diamagnetic.

(b) $\mathrm{NO}^+$ (also 14 Electrons)

Similarly, $\mathrm{NO}^+$ also has 14 electrons and, by the same reasoning in MO theory, has:

Bond order = 3

Diamagnetic (no unpaired electrons).

3. Conclusion

Among the given choices:

$\mathrm{O}_2^+$ has 15 electrons (not 14)

$\mathrm{O}_2^-$ has 17 electrons

$\mathrm{N}_2^+$ has 13 electrons

$\mathrm{NO}^+$ has 14 electrons

Therefore, the species with the same bond order (3) and the same magnetic property (diamagnetic) as $\mathrm{C}_2^{2-}$ is:

$ \boxed{\mathrm{NO}^{+} \text{ (Option D).}} $