Question

What is the correct order of bond strength for $NO,N{O}^{+}$ and $N{O}^{-}$ ?

• A. $NO>N{O}^{+}>N{O}^{-}$
• B. $N{O}^{+}>NO>N{O}^{-}$
• C. $N{O}^{-}>NO>N{O}^{+}$
• D. $N{O}^{+}>N{O}^{-}>NO$

Answer 1

Answer: (B)

Electronic configuration of $NO,N{O}^{+}$ and $N{O}^{-}:$
NO : $KK\sigma 2s^2{\sigma }^{\ast }2s^2\sigma 2p_z^2\pi 2p_x^2\pi 2p_y^2{\pi }^{\ast }2p_x^1$
$N{O}^{+}:KK\sigma 2s^2{\sigma }^{\ast }2s^2\sigma 2p_z^2\pi 2p_x^2\pi 2p_y^2$
$N{O}^{-}:KK\sigma 2s^2{\sigma }^{\ast }2s^2\sigma 2p_z^2\pi 2p_x^2\pi 2p_y^2{\pi }^{\ast }2p_x^1$
$={\pi }^{\ast }2p_y^1$
Bond order $=\frac{1}{2}(N_b-N_a)$
B.O. of NO $=\frac{1}{2}(8-3)=2.5$
B.O. of $N{O}^{+}=\frac{1}{2}(8-2)=3$
B.O. of $N{O}^{-}=\frac{1}{2}(8-4)=2$
Bond strength $\propto$ Bond order Thus order of bond strength is :
$N{O}^{+}>NO>N{O}^{-}$