To determine the bond order for each nitrogen species, we need to write their molecular orbital (MO) configurations and apply the bond order formula.

Step 1: Determine the total number of electrons for each species.

• ${\mathrm{N}}_2$: $2\times 7=14$ electrons
• ${\mathrm{N}}_2^{-}$: $14+1=15$ electrons
• ${\mathrm{N}}_2^{2-}$: $14+2=16$ electrons

Step 2: Write the MO configuration for each species.

For ${\mathrm{N}}_2$ (14 electrons):
${\sigma }_{1s}^2{\sigma }_{1s}^{\ast 2}{\sigma }_{2s}^2{\sigma }_{2s}^{\ast 2}{\pi }_{2p_x}^2{\pi }_{2p_y}^2{\sigma }_{2p_z}^2$

For ${\mathrm{N}}_2^{-}$(15 electrons):
${\sigma }_{1s}^2{\sigma }_{1s}^{\ast 2}{\sigma }_{2s}^2{\sigma }_{2s}^{\ast 2}{\pi }_{2p_x}^2{\pi }_{2p_y}^2{\sigma }_{2p_z}^2{\pi }_{2p_x}^{\ast 1}$

For ${\mathrm{N}}_2^{2-}$ (16 electrons):
${\sigma }_{1s}^2{\sigma }_{1s}^{\ast 2}{\sigma }_{2s}^2{\sigma }_{2s}^{\ast 2}{\pi }_{2p_x}^2{\pi }_{2p_y}^2{\sigma }_{2p_z}^2{\pi }_{2p_x}^{\ast 1}{\pi }_{2p_y}^{\ast 1}$

Step 3: Calculate the bond order for each species.

• ${\mathrm{N}}_2$: Bond Order $=\frac{1}{2}(10-4)=3$
• ${\mathrm{N}}_2^{-}$: Bond Order $=\frac{1}{2}(10-5)=2.5$
• ${\mathrm{N}}_2^{2-}$: Bond Order $=\frac{1}{2}(10-6)=2$

Step 4: Rank the species in terms of increasing bond order.

The bond orders are: ${\mathrm{N}}_2^{2-}(2)<{\mathrm{N}}_2^{-}(2.5)<{\mathrm{N}}_2(3)$

Option Analysis:

• A) ${\mathrm{N}}_2^{2-}<{\mathrm{N}}_2^{-}<{\mathrm{N}}_2$: This order (2 < 2.5 < 3) is correct.
• B) ${\mathrm{N}}_2<{\mathrm{N}}_2^{2-}<{\mathrm{N}}_2^{-}$: This order (3 < 2 < 2.5) is incorrect.
• C) ${\mathrm{N}}_2^{-}<{\mathrm{N}}_2^{2-}<{\mathrm{N}}_2$: This order (2.5 < 2 < 3) is incorrect.
• D) ${\mathrm{N}}_2^{-}<{\mathrm{N}}_2<{\mathrm{N}}_2^{2-}$: This order (2.5 < 3 < 2) is incorrect.

Correct Answer: (A)