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KCET · Chemistry · Chemical Bonding and Molecular Structure

When \(\mathrm{O}_{2}\) is converted into \(\mathrm{O}_{2}^{+}\)

  1. A both paramagnetic character and bond order increase
  2. B bond order decreases
  3. C paramagnetic character increases
  4. D paramagnetic character decreases and the bond order increases
Verified Solution

Answer & Solution

Correct Answer

(D) paramagnetic character decreases and the bond order increases

Step-by-step Solution

Detailed explanation

(i) \(\mathrm{O}_{2}-(\sigma 1 s)^{2}(\sigma * 1 s)^{2}(\sigma 2 s)^{2}\left(\sigma^{*} 2 s\right)^{2}\)
\[
\begin{aligned}
&\left(\sigma 2 p_{z}\right)^{2}\left(\pi 2 p_{x}\right)^{2}\left(\pi 2 p_{y}\right)^{2}\left(\pi * 2 p_{x}\right)^{1} \\
&\left(\pi^{*} 2 p_{y}\right)^{1}
\end{aligned}
\]
Bond order \(=\frac{N_{b}-N_{a}}{2}\)
\[
=\frac{8-4}{2}
\]
\[
=2
\]
\(\mathrm{O}_{2}\) molecule having 2 unpaired electron.
(ii) \(\mathrm{O}_{2}^{+}-(\sigma 1 s)^{2}\left(\sigma^{*} 1 s\right)^{2}(\sigma 2 s)^{2}\left(\sigma^{*} 2 s\right)^{2}\)
\[
\begin{aligned}
&\left(\sigma 2 p_{z}\right)^{2}\left(\pi 2 p_{x}\right)^{2}\left(\pi 2 p_{y}\right)^{2}\left(\pi^{*} 2 p_{x}\right)^{1} \\
&\left(\pi^{*} 2 p_{y}\right)^{0}
\end{aligned}
\]
Bond order \(=\frac{8-3}{2}\)
\[
=2.5
\]
\(\mathrm{O}_{2}^{+}\)ion having only 1 unpaired electron.
Hence, when \(\mathrm{O}_{2}\) is converted into \(\mathrm{O}_{2}^{+}\) paramagnetic character decrease and the bond order increases.