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^{+} $$
