Question

Question 20

Which of the following options represents the correct bond order?

(a) $O_2^{-}>O_2>O_2^{+}$

(b) $O_2^{-}<O_2<O_2^{+}$

(c) $O_2^{-}>O_2<O_2^{+}$

(d) $O_2^{-}<O_2>O_2^{+}$

Answer 1

Answer: (b) $O_2^{-}<O_2<O_2^{+}$

Electronic configuration of O_2(16 electrons)

= $\sigma 1s^2,{\sigma }^{\ast }1s^2,\sigma 2s^2,{\sigma }^{\ast }2s^2,\sigma 2p_z^2,\pi 2p_x^2\approx \pi 2p_y^2,{\pi }^{\ast }2p_x^1\approx {\pi }^{\ast }2p_y^1$

Bond order = $\frac{1}{2}(N_b-N_a)=\frac{1}{2}(10-6)=2$

Electronic configuration of $O_2^{+}$ (15 electrons)

= $\sigma 1s^2,{\sigma }^{\ast }1s^2,\sigma 2s^2,{\sigma }^{\ast }2s^2,\sigma 2p_z^2,\pi 2p_x^2\approx \pi 2p_y^2,{\pi }^{\ast }2p_x^1\approx {\pi }^{\ast }2p_y^0$

Bond order = $\frac{1}{2}(N_b-N_a)=\frac{1}{2}(10-5)=2.5$

Electronic configuration of $O-2^{-}$ (17 electrons)

=$\sigma 1s^2,{\sigma }^{\ast }1s^2,\sigma 2s^2,{\sigma }^{\ast }2s^2,\sigma 2p_z^2,\pi 2p_x^2\approx \pi 2p_y^2,{\pi }^{\ast }2p_x^2\approx {\pi }^{\ast }2p_y^1$

Bond order = $\frac{1}{2}(N_b-N_a)=\frac{1}{2}(10-7)=1.5$

Thus, the order of bond order is $O_2^{-}<O_2<O_2^{+}$