- Exam
- NEET
- Year
- 2015
- Subject
- Chemistry
- Chapter
- Chemical Bonding and Molecular Structure
NEET 2015 · Chemistry · Chemical Bonding and Molecular Structure
Decreasing order of stability of O _ 2 , O _ 2 ^ - , O _ 2 ^ + and...
Decreasing order of stability of and is:
Options
- A
- B
- C
- D
Correct answer
A.
Step-by-step solution
B . O = \dfrac{{N}_{b} - {N}_{a}}{2}{O}_{2}^{-} ⇒ \sigma1{s}^{2} < \overset{*}{\sigma} 1{s}^{2} < \sigma2{s}^{2} < \overset{*}{\sigma} 2{s}^{2} < \sigma2{p}_{z}^{2} < \pi2{p}_{x}^{2} = \pi2{p}_{y}^{2} < \overset{*}{\pi} 2{p}_{x}^{2} = \overset{*}{\pi} 2{p}_{y}^{1}{O}_{2}^{-} = 1.5 ⇒ \dfrac{10 - 7}{2} = 1.5{O}_{2} ⇒ \sigma1{s}^{2} < \overset{*}{\sigma} 1{s}^{2} < \sigma2{s}^{2} < \overset{*}{\sigma} 2{s}^{2} < \sigma2{p}_{z}^{2} < \pi2{p}_{x}^{2} = \pi2{p}_{y}^{2} < \overset{*}{\pi} 2{p}_{x}^{1} = \overset{*}{\pi} 2{p}_{y}^{1}{O}_{2} = 2.0 ⇒ \dfrac{10 – 6}{2} = 2{O}_{2}^{+} ⇒ \sigma1{s}^{2} < \overset{*}{\sigma} 1{s}^{2} < \sigma2{s}^{2} < \overset{*}{\sigma} 2{s}^{2} < \sigma2{p}_{z}^{2} < \pi2{p}_{x}^{2} < \pi2{p}_{y}^{2} < \overset{*}{\pi} 2{p}_{x}^{1} < \overset{*}{\pi} 2{p}_{y}^{0}{O}_{2}^{+} = 2.5 ⇒ \dfrac{10 - 5}{2} = 2.5{O}_{2}^{2 -} ⇒ \sigma1{s}^{2} < \overset{*}{\sigma}1{s}^{2} < \sigma2{s}^{2} < \overset{*}{\sigma}2{s}^{2} < \sigma2{p}_{z}^{2} < \pi2{p}_{x}^{2} = \pi2{p}_{y}^{2} < \overset{*}{\pi}2{p}_{x}^{2} = \overset{*}{\pi}2{p}_{y}^{2}{O}_{2}^{2 +} ⇒ B . O = \dfrac{10 - 4}{2} = 3.0{O}_{2}^{-} < {O}_{2}^{+} < {O}_{2}^{2 +}{O}_{2} ⟶ 16{e}^{-}{O}_{2}^{-} ⟶ 17{e}^{-}{O}_{2}^{+} ⟶ 15{e}^{-}{O}_{2}^{2 -} ⟶ 18{e}^{-}\begin{cases} & {O}_{2}^{+} & {O}_{2} \\ Bondorder & 2.5 & 2\end{cases}
Stability order \left[{O}_{2}^{+} > {O}_{2} > {O}_{2}^{-} > {O}_{2}^{2 -}\right]