Neet Solved Paper 2015 Question 9
Question: Which of the following options represents the correct bond order?
Options:
A) $ O_2^{-}>O _2>O_2^{+} $
B) $ O_2^{-}<O _2<O_2^{+} $
C) $ O_2^{-}>O _2<O_2^{+} $
D) $ O_2^{-}\lt O _2 \gt O_2^{+} $
Show Answer
Answer:
Correct Answer: B
Solution:
Bond order of $ O_2^{-} $
$O_2^{-}=\sigma 1s^{2},\overset{*}{\mathop{\sigma }},1s^{2},\sigma 2s^{2}$ $\overset{*}{\mathop{\sigma }},2s^{2}\sigma {2p_z}^{2}(\pi {2p_x}^{2}=\pi 2p_y^{2}) ({{\pi }^*}{2p_x}^2$ =${{\pi }^*}{2p_y}^1) $
Bond order $ =\frac{\text{number of electrons in BMO}-\text{number of elections ABMO}}{2} $
$ =\frac{10-7}{2}=\frac{3}{2}=1.5 $
$ O_2^{+}=\sigma 1s^{2},\overset{*}{\mathop{\sigma }},1s^{2},\sigma 2s^{2},$ $\overset{*}{\mathop{\sigma }},2s^{2}\sigma 2p_z^{2} $ $ (\pi 2p_x^{2}=\pi {2p_y}^{2})({{\pi }^{*}}2p_x^{1}$ =${{\pi }^{*}}{2p_y}^{0}) $
$ BO=\frac{10-5}{2}=\frac{5}{2}=2.5 $
$ O _2=\sigma 1s^{2}\overset{*}{\mathop{\sigma }},1s^{2},\sigma 2s^{2},$ $\overset{*}{\mathop{\sigma }},2s^{2}\sigma 2p_z^{2}(\pi 2p_x^{2}=\pi 2p_y^{2}) $ $ (\overset{*}{\mathop{\pi }},{2p_x}^{1}$ =$\overset{*}{\mathop{\pi }},{2p_y}^{1}) $
$ BO=\frac{10-6}{2}=\frac{4}{2}=2 $ So, the correct sequence is $ O_2^{-}<O _2<O_2^{+} $
