Bond length is inversely related to bond order: a higher bond order means a stronger, shorter bond. To compare these nitrogen species, we first calculate bond orders using MO theory. Neutral N_2 has 14 electrons: \sigma(1s)^2 \sigma^(1s)^2 \sigma(2s)^2 \sigma^(2s)^2 \pi(2p)^4 \sigma(2p_z)^2, giving bond order = (10 - 4)/2 = 3. N_2^+ has 13 electrons (one removed from \sigma(2p_z)), giving bond order = (9 - 4)/2 = 2.5. N_2^- has 15 electrons (one added to \pi^(2p)), giving bond order = (10 - 5)/2 = 2.5. N_2^{2+} has 12 electrons (two removed from the bonding \sigma(2p_z) orbital), giving bond order = (8 - 4)/2 = 2. Therefore the order of bond orders is: N_2 (3) > N_2^+ = N_2^- (2.5) > N_2^{2+} (2). Since bond length decreases as bond order increases, N_2 has the shortest bond length. Option N_2^+ is wrong because removing an electron from the bonding \sigma(2p_z) orbital reduces bond order from 3 to 2.5. Option N_2^- adds an electron to an antibonding \pi^ orbital, also reducing bond order to 2.5. Option N_2^{2+} loses two bonding electrons, further reducing bond order to 2. This is a fundamental JEE MO theory application from NCERT Class 11. Verification: the experimental N≡N bond length is 110 pm, the shortest among all nitrogen species listed, consistent with a bond order of 3.