Question

Select correct statement(s) :

• A. Bond length of $N{O}^{+}>NO$
• B. Bond order of $N{O}^{+}>NO$
• C. Bond energy of $NO>N{O}^{+}$
• D. $NO$ is paramagnetic but $N{O}^{+}$ is diamagnetic

Answer 1

Answer: (B), (D)

The correct options are
B Bond order of $N{O}^{+}>NO$
D $NO$ is paramagnetic but $N{O}^{+}$ is diamagnetic

Electronic configuration of $NO=\sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2{\sigma }^{\ast }2s^2\pi 2p_x^2\pi 2p_y^2\sigma 2p_z^2{\pi }^{\ast }2p_x^1$
$B.O.ofNO=\frac{10-5}{2}=2.5$
As one unpaired electron is there in ${\pi }^{\ast }2p_x^1$orbital. It is paramagnetic.
Electronic configuration of $N{O}^{+}=\sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2{\sigma }^{\ast }2s^2\pi 2p_x^2\pi 2p_y^2\sigma 2p_z^2$
$N{O}^{+}=\frac{10-4}{2}=3.0$
No unpaired electron is there. Hence, diamagnetic.