QJEEAdv14Ch5c47: JEE Advanced 2014 Chemistry - Class 12 JEEAdv14Ch5c47 Solution: The pair(s) of reagents that yield paramagnetic species is/are

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) $Na + \mathop {N{H_3}}\limits_{(Excess)} \to NaN{H_2} + {1 \over 2}{H_2}$

(NaNH₂ + 1/2H₂ are diamagnetic) If ammonia is considered as a liquid then reaction will be

$M + (x + y)N{H_3} \to {[M{(N{H_3})_x}]^ + } + {[e{(N{H_3})_y}]^ - }$

$\bullet$ ammoniated electron

$\bullet$ blue colour

$\bullet$ paramagnetic

$\bullet$ very strong reducing agent

(b) $K + \mathop {{O_2}}\limits_{(Excess)} \to \mathop {K{O_2}({K^ + },O_2^ - )}\limits_{Potassium\,\sup eroxide\,Paramagnetic}$

(c) $3Cu + 8HN{O_3} \to \mathop {3Cu{{(N{O_3})}_2}}\limits_{Paramagnetic} + \mathop {2NO}\limits_{Paramagnetic} + 4{H_2}O$

(d)
Hence, option (a), (b) and (c) are correct choices.