Paramagnetic Species Generation in Specific Chemical Reactions

To determine which reagent pairs result in the formation of paramagnetic species, let us analyze each reaction individually:

1. Option (A): $\text{Na}$ with excess liquid $\text{NH}_3$ When an alkali metal such as sodium dissolves in excess liquid ammonia, it forms a dilute solution containing solvated cations and ammoniated electrons: $\text{Na} + (x+y)\text{NH}_3 \rightarrow [\text{Na}(\text{NH}_3)_x]^+ + [e(\text{NH}_3)_y]^-$ The ammoniated electrons are unpaired, which imparts a deep blue color and strong paramagnetic character to the solution. Therefore, option (A) yields a paramagnetic species.

2. Option (B): $\text{K}$ with excess $\text{O}_2$ Potassium burns in excess oxygen to form potassium superoxide: $\text{K} + \text{O}_2 \rightarrow \text{KO}_2$ The superoxide ion is $\text{O}_2^-$, containing 17 valence electrons. Its molecular orbital configuration is: \sigma_{1s}^2\ \sigma^*_{1s}^2\ \sigma_{2s}^2\ \sigma^*_{2s}^2\ \sigma_{2p_z}^2\ (\pi_{2p_x}^2 = \pi_{2p_y}^2)\ (\pi^*_{2p_x}^2 = \pi^*_{2p_y}^1) The presence of one unpaired electron in the degenerate $\pi^*$ antibonding molecular orbital makes $\text{KO}_2$ paramagnetic. Therefore, option (B) yields a paramagnetic species.

3. Option (C): $\text{Cu}$ with dilute $\text{HNO}_3$ The reaction of copper with dilute nitric acid is represented by: $3\text{Cu} + 8\text{HNO}_3\ (\text{dil.}) \rightarrow 3\text{Cu}(\text{NO}_3)_2 + 2\text{NO} \uparrow + 4\text{H}_2\text{O}$ This reaction forms two distinct paramagnetic species:

   • $\text{Cu}^{2+}$ in $\text{Cu}(\text{NO}_3)_2$, having a $d^9$ electronic configuration ($[\text{Ar}]\,3d^9$) with one unpaired electron.
   • Nitric oxide ($\text{NO}$), an odd-electron molecule with 11 valence electrons, possessing one unpaired electron in its $\pi^*_{2p}$ orbital. Therefore, option (C) yields paramagnetic species.

4. Option (D): $\text{O}_2$ with $2$-ethylanthraquinol The industrial auto-oxidation of $2$-ethylanthraquinol by oxygen produces $2$-ethylanthraquinone and hydrogen peroxide: $\text{2-ethylanthraquinol} + \text{O}_2 \xrightarrow{\text{air}} \text{2-ethylanthraquinone} + \text{H}_2\text{O}_2$ Both products, $2$-ethylanthraquinone and $\text{H}_2\text{O}_2$, possess fully paired electron shells and are diamagnetic. Therefore, option (D) does not yield paramagnetic species.

Hence, the correct options are (A), (B), and (C).