Which of the following has the …

NO
Number of electrons = 15
Electronic configuration = {tex} \sigma \mathrm{1s}^{2}, \sigma^{*} \operatorname{1s}^{2}, \sigma 2 s^{2}, \sigma^{*} 2 s^{2}, \sigma 2 p_{z}^{2}, {/tex}{tex} \pi 2 p_{x}^{2}=\pi 2 p_{y}^{2}, \pi^{*} 2 p_{x}^{1} {/tex}
Bond order = {tex}\frac {1} {2}{/tex}(10 - 5) = 2.5
Magnetic property = Paramagnetic.

NO⁺
Number of electron = 14
Electronic configuration = {tex} \sigma \mathrm{1s}^{2}, \sigma^{*} \mathrm{1s}^{2}, \sigma 2 s^{2}, \sigma^{*} 2 s^{2}, {/tex}{tex} \sigma 2 p_{z}^{2}, \pi 2 \rho_{x}^{2}=\pi 2 p_{y}^{2} {/tex}
Bond order = {tex}\frac {1} {2}{/tex}(10 - 4) = 3
Magnetic property = Diamagnetic.

{tex} \mathrm{N} O^{2+} {/tex}
Number of electrons = 13
Electronic configuration = {tex} \sigma \mathrm{1s}^{2}, \sigma^{*} \mathrm{1s}^{2}, \sigma 2 s^{2}, \sigma^{*} 2 s^{2}, {/tex}{tex} \sigma 2 p_{z}^{2}, \pi 2 \rho_{x}^{2}=\pi 2 p_{y}^{1} {/tex}
Bond order = {tex}\frac {1} {2}{/tex} (9 - 4) = 2.5
Magnetic property = Paramagnetic

NO^-
Number of electrons = 16
Electronic configuration = {tex} \sigma \mathrm{1s}^{2}, \sigma^{*} \mathrm{1s}^{2}, \sigma 2 s^{2}, \sigma^{*} 2 s^{2}, {/tex}{tex} \pi 2 p_{x}^{2}=\pi 2 p_{y}^{2}, \pi^{*} 2 p_{x}^{2} {/tex}
Bond order = {tex}\frac {1} {2}{/tex}(10 - 6) = 2
Magnetic property = Diamagnetic

Since {tex} \mathrm{NO}^{+} {/tex} has the highest bond order and therefore, it has the shortest bond length.