sblock Elements 1 Question 21
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39. The pair(s) of reagents that yield paramagnetic species is/are
======= ####39. The pair(s) of reagents that yield paramagnetic species is/are
3e0f7ab6f6a50373c3f2dbda6ca2533482a77bed (a) $\mathrm{Na}$ and excess of $\mathrm{NH}_{3}$
(b) $\mathrm{K}$ and excess of $\mathrm{O}_{2}$
(c) $\mathrm{Cu}$ and dilute $\mathrm{HNO}_{3}$
(d) $\mathrm{O}_{2}$ and 2-ethylanthraquinol
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Solution:
- PLAN Paramagnetic character of species can be easily explained on the basis of presence of unpaired electrons, i.e. compounds containing unpaired electron(s) is/are paramagnetic.
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) $\mathrm{Na}+\underset{\text { (Excess) }}{\mathrm{NH}{3}} \longrightarrow \mathrm{NaNH}{2}+\frac{1}{2} \mathrm{H}_{2}$
( $\mathrm{NaNH}{2}+1 / 2 \mathrm{H}{2}$ are diamagnetic)
If ammonia is considered as a liquid then reaction will be
$$ M+(x+y) \mathrm{NH}{3} \longrightarrow\left[M\left(\mathrm{NH}{3}\right){x}\right]^{+}+\left[e\left(\mathrm{NH}{3}\right)_{y}\right] $$
(b) $\mathrm{K}+\underset{\text { (Excess) }}{\mathrm{O}{2}} \longrightarrow \underset{\text { Potassium superoxide }}{\mathrm{KO}{2}\left(\mathrm{~K}^{+}, \mathrm{O}_{2}^{-}\right)}$ paramagnetic
(c) $3 \mathrm{Cu}+8 \mathrm{HNO}{3} \longrightarrow \underset{\text { Paramagnetic }}{3 \mathrm{Cu}\left(\mathrm{NO}{3}\right){2}}+\underset{\text { Paramagnetic }}{2 \mathrm{NO}}+4 \mathrm{H}{2} \mathrm{O}$
(d)
Hence, option (a), (b) and (c) are correct choices.
