KCET · Chemistry · Solid State
Which of the following is diamagnetic?
- A \(\mathrm{H}_{2}^{+}\)
- B \(\mathrm{He}_{2}^{+}\)
- C \(\mathrm{O}_{2}\)
- D \(\mathrm{N}_{2}\)
Answer & Solution
Correct Answer
(D) \(\mathrm{N}_{2}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{MO}\) configuration of \(\mathrm{H}_{2}^{+}: \sigma 1 s^{1}\)
\(\mathrm{MO}\) configuration of \(\mathrm{He}_{2}^{+}: \sigma 1 s^{2}, \star^{*} s^{1}\)
\(\mathrm{MO}\) configuration of \(\mathrm{N}_{2}\) :
\(\sigma 1 s^{2}\), ซั \(1 s^{2}, \sigma 2 s^{2}, \stackrel{*}{\sigma} 2 s^{2}, \pi 2 p_{y}^{2}=\pi 2 p_{z}^{2}, \sigma 2 p_{x}^{2}\)
\(\mathrm{MO}\) configuration of \(\mathrm{O}_{2}\) :
\(\sigma 1 s^{2}, \sigma^{*} 1 s^{2}, \sigma 2 s^{2}, \sigma^{*} 2 s^{2}, \sigma 2 p_{x}^{2}, \pi 2 p_{y}^{2}\)
\(=\pi 2 p_{z}^{2}, \pi 2 p_{y}^{1}=\pi_{\pi}^{*} 2 p_{z}^{1}\)
Hence, \(\mathrm{O}_{2}\) is paramagnetic as it possesses two unpaired electrons.
\(\mathrm{MO}\) configuration of \(\mathrm{He}_{2}^{+}: \sigma 1 s^{2}, \star^{*} s^{1}\)
\(\mathrm{MO}\) configuration of \(\mathrm{N}_{2}\) :
\(\sigma 1 s^{2}\), ซั \(1 s^{2}, \sigma 2 s^{2}, \stackrel{*}{\sigma} 2 s^{2}, \pi 2 p_{y}^{2}=\pi 2 p_{z}^{2}, \sigma 2 p_{x}^{2}\)
\(\mathrm{MO}\) configuration of \(\mathrm{O}_{2}\) :
\(\sigma 1 s^{2}, \sigma^{*} 1 s^{2}, \sigma 2 s^{2}, \sigma^{*} 2 s^{2}, \sigma 2 p_{x}^{2}, \pi 2 p_{y}^{2}\)
\(=\pi 2 p_{z}^{2}, \pi 2 p_{y}^{1}=\pi_{\pi}^{*} 2 p_{z}^{1}\)
Hence, \(\mathrm{O}_{2}\) is paramagnetic as it possesses two unpaired electrons.
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