Q. The pair(s) of reagents that yield paramagnetic species is/are
The s-Block Elements
Solution:
Reaction of alkali metals with ammonia depends upon the physical state of ammonia whether it is in gaseous state or liquid state. If ammonia is considered as a gas then reaction will be
(a) $Na + \underset{(Excess)}{NH_3} \rightarrow NaNH_2 + \frac{1}{2} H_2 $
$ (NaNH_2 + 1/2 H_2$ are diamagnetic )
If ammonia is considered as a liquid then reaction will be
$ M + (x + y) NH_3 \rightarrow [M(NH_3)_x]^+ + [e(NH_3)_y ]^- $
$\bullet$ Ammoniated electron
$\bullet$Blue colour
$\bullet$ Paramagnetic
$\bullet$ Very strong reducing agent
(b) $K + \underset{\text{(Excess)}}{O_2} \rightarrow \underset{\text{Potassium superoxide paramagnetic}}{ KO_2(K^+ , O_2^-)} $
(c) $3Cu + 8HNO_3 \rightarrow \underset{\text{Paramagnetic}}{ 3Cu(NO_3)_2} + \underset{\text{Paramagnetic}}{2NO} + 4H_2O $
Hence, option (a), (b) and (c) are correct choices.
