Q49 / 180 · KCET 2016 · Chemistry
The increasing order of bond order of \( \mathrm{O}_{2}, \mathrm{O}_{2}^{+}, \mathrm{O}_{2}^{-} \), and \( \mathrm{O}_{2}^{--} \)is
- Correct answerA\( \mathrm{O}_{2}^{+}, \mathrm{O}_{2}, \mathrm{O}_{2}^{-}, \mathrm{O}_{2}^{2-} \)
- 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\( \mathrm{O}_{2}{ }^{2-}, \mathrm{O}_{2}{ }^{-}, \mathrm{O}_{2}, \mathrm{O}_{2}{ }^{+} \)
Solution
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 \( \mathrm{O}_{2}{ }^{+}>\mathrm{O}_{2}>\mathrm{O}_{2}{ }^{-}>\mathrm{O}_{2}{ }^{2-} \).
Hence, the correct answer is option: 'A'.