Question

The pair(s) of reagents that yield paramagnetic species is/are
(JEE Advanced, 2014)

• A. Na and excess of $N{H}_3$
• B. K and excess of $O_2$
• C. Cu and dilute $HN{O}_3$
• D. $O_2$ and 2-ethylanthraquinol

Answer 1

Answer: (A), (B), (C)

The correct options are
A Na and excess of $N{H}_3$
B K and excess of $O_2$
C Cu and dilute $HN{O}_3$

Dilute solution sof Na in liquid ammonia is paramagnetic due to ammoniated $e^{-1}s$
$N{a}^{\oplus }+(x+y)N{H}_3\to N{a}^{+}(N{H}_3{)}_x+e^{-}(N{H}_3{)}_y$
$K+\underset{\left(excess\right)}{O_2}\to \underset{Paramagnetic}{K{O}_2}$
$O_{\overline{2}}$ is paramagnetic due to unpaired electron in antibonding orbital
$3Cu+\underset{diluted}{8HN{O}_3}\to 3Cu(N{O}_3{)}_2+2NO+4H_{2}O$
NO is paramagnetic due to unpaired electrons on "N".