Question

Which of the following species is not paramagnetic?

• A. $\text{C}\text{O}$
• B. $\text{O}_{2}$
• C. $\text{B}_{2}$
• D. $\text{N}\text{O}$

Answer 1

Answer: (A)

CO has no unpaired electron while other have unpaired electrons so, paramagnetic in nature.(Apply M.O.T)
$\text{Molecular orbital configuration}\Rightarrow $
$\text{CO }:\left(\text{like N}\right)_{2} \, ;\text{ σ}1s^{2} < \left(\text{σ}\right)^{\star}\left(\text{1S}\right)^{2} < \text{σ}2\left(\text{S}\right)^{2} < \left(\sigma \right)^{\star}2\left(\text{S}\right)^{2} < \text{\pi }2\text{P}_{\text{x}}^{2}\text{= \pi }2\text{P}_{\text{y}}^{2} < \text{σ}2\text{P}_{\text{z}}^{2} \, \, \left(\right.\text{No unpaired electrons)}$
$\left(\text{O}\right)_{2}:\text{σ}1\left(\text{S}\right)^{2} < \left(\text{σ}\right)^{\star}\left(\text{1S}\right)^{2} < \text{σ}2\left(\text{S}\right)^{2} < \left(\text{σ}\right)^{\star}2\left(\text{S}\right)^{2} < \text{σ}2\text{P}_{\text{z}}^{2} < \text{\pi }2\text{P}_{\text{x}}^{2}=\text{\pi }2\text{P}_{\text{y}}^{2} < \left(\text{\pi }\right)^{\star}2\text{P}_{\text{x}}^{\text{1}}\text{= }\left(\pi \right)^{\star}2\text{P}_{\text{y}}^{\text{1}} \, \, \left(\right. 2 \text{unpaired electrons} \left.\right)$
$\left(\text{B}\right)_{2}:\text{σ}1\left(\text{S}\right)^{2} < \left(\text{σ}\right)^{\star}1\left(\text{S}\right)^{2} < \text{σ}2\left(\text{S}\right)^{2} < \left(\text{σ}\right)^{\star}2\left(\text{S}\right)^{2} < \text{\pi }2\text{P}_{\text{x}}^{1}=\text{\pi }2\text{P}_{\text{y}}^{\text{1}} \, (2\text{unpaired electrons}\left.\right)$
$\text{NO }:\text{σ}1\left(\text{S}\right)^{2} < \left(\text{σ}\right)^{\star}1\left(\text{S}\right)^{2} < \text{σ}2\left(\text{S}\right)^{2} < \left(\text{σ}\right)^{\star}2\left(\text{S}\right)^{2} < \text{σ}2\text{P}_{\text{z}}^{2} < \text{\pi }2\text{P}_{\text{x}}^{2}=\text{\pi }2\text{P}_{\text{y}}^{2} < \left(\text{\pi }\right)^{\star}2\text{P}_{\text{y}}^{\text{1}}=\left(\text{ \pi }\right)^{\star}2\text{P}_{\text{x}}^{0}\left(\right. 1 \text{unpaired electrons} \left.\right)$