KCET · Chemistry · Chemical Bonding and Molecular Structure
The increasing order of bond order of \( \mathrm{O}_{2}, \mathrm{O}_{2}^{+}, \mathrm{O}_{2}^{-} \), and \( \mathrm{O}_{2}^{--} \)is
- A \( \mathrm{O}_{2}^{+}, \mathrm{O}_{2}, \mathrm{O}_{2}^{-}, \mathrm{O}_{2}^{2-} \)
- B \( \mathrm{O}_{2}^{2-}, \mathrm{O}_{2}^{-}, \mathrm{O}_{2}^{+}, \mathrm{O}_{2} \)
- C \( \mathrm{O}_{2}, \mathrm{O}_{2}^{+}, \mathrm{O}_{2}^{-}, \mathrm{O}_{2}^{2-} \)
- D \( \mathrm{O}_{2}{ }^{2-}, \mathrm{O}_{2}{ }^{-}, \mathrm{O}_{2}, \mathrm{O}_{2}{ }^{+} \)
Answer & Solution
Correct Answer
(A) \( \mathrm{O}_{2}^{+}, \mathrm{O}_{2}, \mathrm{O}_{2}^{-}, \mathrm{O}_{2}^{2-} \)
Step-by-step Solution
Detailed explanation
Electronic configuration of \( \mathrm{O}_{2} \) :
\( \sigma 1 s^{2} \sigma^{*} 1 s^{2} \sigma 2 s^{2} \sigma^{*} 2 s^{2} \sigma 2 p_{z}{ }^{2} \Pi 2 p_{x}{ }^{2}=\Pi 2 p_{y}{ }^{2} \Pi^{*} 2 p_{x}{ }^{1}=\Pi{ }^{*} 2 p_{y}{ }^{1} \) Bond order \( =\frac{\left[\text { No. of bonding } e^{-} s-\text { No. of anti-bonding } e^{-} s\right]}{2} \)
\[
\begin{array}{l}
O_{2}=\frac{1}{2}(10-6)=2 \\
O_{2}^{-}=\frac{1}{2}(10-7)=1.5 \\
O_{2}^{2-}=\frac{1}{2}(10-8)=1 \\
O_{2}^{+}=\frac{1}{2}(10-5)=\frac{5}{2}=2.5
\end{array}
\]
Therefore, the correct decreasing order of bond order is \( \mathrm{O}_{2}{ }^{+}>\mathrm{O}_{2}>\mathrm{O}_{2}{ }^{-}>\mathrm{O}_{2}{ }^{2-} \).
\( \sigma 1 s^{2} \sigma^{*} 1 s^{2} \sigma 2 s^{2} \sigma^{*} 2 s^{2} \sigma 2 p_{z}{ }^{2} \Pi 2 p_{x}{ }^{2}=\Pi 2 p_{y}{ }^{2} \Pi^{*} 2 p_{x}{ }^{1}=\Pi{ }^{*} 2 p_{y}{ }^{1} \) Bond order \( =\frac{\left[\text { No. of bonding } e^{-} s-\text { No. of anti-bonding } e^{-} s\right]}{2} \)
\[
\begin{array}{l}
O_{2}=\frac{1}{2}(10-6)=2 \\
O_{2}^{-}=\frac{1}{2}(10-7)=1.5 \\
O_{2}^{2-}=\frac{1}{2}(10-8)=1 \\
O_{2}^{+}=\frac{1}{2}(10-5)=\frac{5}{2}=2.5
\end{array}
\]
Therefore, the correct decreasing order of bond order is \( \mathrm{O}_{2}{ }^{+}>\mathrm{O}_{2}>\mathrm{O}_{2}{ }^{-}>\mathrm{O}_{2}{ }^{2-} \).
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