According to MO theory which of the following lists ranks the nitrogen species in terms of increasing bond order ?

Key Idea Bond order $=\frac{\mathrm{N}_{\mathrm{b}}-\mathrm{N}_{\mathrm{a}}}{2}$=2Nb​−Na​​ where, $\mathrm{N}_{\mathrm{b}}=$Nb​= number of electrons in bonding $M O$MO $\mathrm{N}_{\mathrm{a}}=$Na​= number of electrons in anti bonding $M O$MO $$\begin{aligned} & \mathrm{N}_2(7+7=14)=\sigma 1 \mathrm{~s}^2, \dot{*} 1 \mathrm{~s}^2, \sigma 2 \mathrm{~s}^2, \stackrel{\dot{*}}{\sigma} 2 \mathrm{~s}^2, \pi 2 \mathrm{p}_{\mathrm{x}}^2 \\ & \approx \pi 2 \mathrm{p}_{\mathrm{y}}^2, \sigma 2 \mathrm{p}_{\mathrm{z}}^2 \end{aligned}$$​N2​(7+7=14)=σ1 s2,∗˙1 s2,σ2 s2,σ∗˙2 s2,π2px2​≈π2py2​,σ2pz2​​ $$\begin{aligned} & \mathrm{BO}= \frac{10-4}{2}=3 \\ & \mathrm{~N}_2(7+7+1=15) \\ &= \sigma 1 \mathrm{~s}^2, \dot{*} 1 \mathrm{~s}^2, \sigma 2 \mathrm{~s}^2, \\ & \stackrel{*}{\sigma} 2 \mathrm{~s}^2, \sigma 2 \mathrm{p}_z^2, \pi 2 \mathrm{p}_{\mathrm{x}}^2 \approx \pi 2 \mathrm{p}_{\mathrm{y}}^2, \pi 2 \mathrm{p}_{\mathrm{x}}^1 \end{aligned}$$​BO=210−4​=3 N2​(7+7+1=15)=σ1 s2,∗˙1 s2,σ2 s2,σ∗2 s2,σ2pz2​,π2px2​≈π2py2​,π2px1​​ $$\begin{aligned} & \mathrm{BO}=\frac{10-5}{2}=2.5 \\ & \mathrm{~N}_2^{2-}(7+7+2=16) \end{aligned}$$​BO=210−5​=2.5 N22−​(7+7+2=16)​ $$\begin{aligned} & =\sigma 1 \mathrm{~s}^2, \star 1 \mathrm{~s}^2, \sigma 2 \mathrm{~s}^2, \star 2 \mathrm{~s}^2, \sigma 2 \mathrm{p}_{\mathrm{z}}^2, \pi 2 \mathrm{p}_{\mathrm{z}}^2 \\ & \approx \pi 2 \mathrm{p}_{\mathrm{y}}^2, \pi 2 \mathrm{p}_{\mathrm{x}}^1 \approx \pi 2 \mathrm{p}_{\mathrm{y}}^1 \\ & \end{aligned}$$​=σ1 s2,⋆1 s2,σ2 s2,⋆2 s2,σ2pz2​,π2pz2​≈π2py2​,π2px1​≈π2py1​​ $$\mathrm{BO}=\frac{10-6}{2}=2$$BO=210−6​=2 Hence, the increasing order of B.O is, $$\mathrm{N}_2^{2-} < \mathrm{N}_2^{-} < \mathrm{N}_2$$N22−​<N2−​<N2​