JEE Main 2023 — Chemical Bonding And Molecular Structure Question with Solution

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:

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