Q55 / 180 · NEET 2015 (Phase 2) · Chemistry

Decreasing order of stability of $O_{2}, O_{2}^{-}, O_{2}^{+}$ and $O_{2}^{2 -}$ is:

Correct answer
A
$O_{2}^{+} > O_{2} > O_{2}^{-} > O_{2}^{2 -}$
B
$O_{2}^{2 -} > O_{2}^{-} > O_{2} > O_{2}^{+}$
C
$O_{2} > O_{2}^{+} > O_{2}^{2 -} > O_{2}^{-}$
D
$O_{2}^{-} > O_{2}^{2 -} > O_{2}^{+} > O_{2}$

Solution

$B . O = \frac{N_{b} - N_{a}}{2}$

$O_{2}^{-} \Rightarrow σ 1 s^{2} < \overset{*}{σ} 1 s^{2} < σ 2 s^{2} < \overset{*}{σ} 2 s^{2} < σ 2 p_{z}^{2} < π 2 p_{x}^{2} = π 2 p_{y}^{2} < \overset{*}{π} 2 p_{x}^{2} = \overset{*}{π} 2 p_{y}^{1}$

B.O for $O_{2}^{-} = 1.5 \Rightarrow \frac{10 - 7}{2} = 1.5$

$O_{2} \Rightarrow σ 1 s^{2} < \overset{*}{σ} 1 s^{2} < σ 2 s^{2} < \overset{*}{σ} 2 s^{2} < σ 2 p_{z}^{2} < π 2 p_{x}^{2} = π 2 p_{y}^{2} < \overset{*}{π} 2 p_{x}^{1} = \overset{*}{π} 2 p_{y}^{1}$

B.O for $O_{2} = 2.0 \Rightarrow \frac{10 - 6}{2} = 2$

$O_{2}^{+} \Rightarrow σ 1 s^{2} < \overset{*}{σ} 1 s^{2} < σ 2 s^{2} < \overset{*}{σ} 2 s^{2} < σ 2 p_{z}^{2} < π 2 p_{x}^{2} < π 2 p_{y}^{2} < \overset{*}{π} 2 p_{x}^{1} < \overset{*}{π} 2 p_{y}^{0}$

B.O for $O_{2}^{+} = 2.5 \Rightarrow \frac{10 - 5}{2} = 2.5$

$O_{2}^{^{2 -}} \Rightarrow σ 1 s^{2} < \overset{*}{σ} 1 s^{2} < σ 2 s^{2} < \overset{*}{σ} 2 s^{2} < σ 2 p_{z}^{2} < π 2 p_{x}^{2} = π 2 p_{y}^{2} < \overset{*}{π} 2 p_{x}^{2} = \overset{*}{π} 2 p_{y}^{2}$

$O_{2}^{2 +} \Rightarrow B . O = \frac{10 - 4}{2} = 3.0$

$O_{2}^{-} < O_{2}^{+} < O_{2}^{2 +}$

O_{2} ⟶ 16 e^{-}

O_{2}^{-} ⟶ 17 e^{-}

O_{2}^{+} ⟶ 15 e^{-}

O_{2}^{2 -} ⟶ 18 e^{-}

\begin{matrix} O_{2}^{+} & O_{2} \\ B o n d o r d e r & 2.5 & 2 \end{matrix} \begin{matrix} O_{2}^{-} & O_{2}^{2 -} \\ 1.5 & 1 \end{matrix}

S t a b i l i t y α B o n d o r d e r

Stability order

[O_{2}^{+} > O_{2} > O_{2}^{-} > O_{2}^{2 -}]

Hence, the correct answer is option: 'A'.

Decreasing order of stability of O2, O2-, O2+ and O22- is:

Solution

Relevant Questions

Decreasing order of stability of $O_{2}, O_{2}^{-}, O_{2}^{+}$ and $O_{2}^{2 -}$ is: