sblock Elements - Result Question 29
39. The pair(s) of reagents that yield paramagnetic species is/are
(a) $Na$ and excess of $NH _3$
(b) $K$ and excess of $O _2$
(c) $Cu$ and dilute $HNO _3$
(d) $O _2$ and 2-ethylanthraquinol
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Answer:
Correct Answer: 39. (a,b,c)
Solution:
- 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) $Na+\underset{\text { (Excess) }}{NH _3} \longrightarrow NaNH _2+\dfrac{1}{2} H _2$
( $NaNH _2+1 / 2 H _2$ are diamagnetic)
If ammonia is considered as a liquid then reaction will be
$$ M+(x+y) NH _3 \longrightarrow\left[M\left(NH _3\right) _x\right]^{+}+\left[e\left(NH _3\right) _y\right] $$
(b) $K+\underset{\text { (Excess) }}{O _2} \longrightarrow \underset{\text { Potassium superoxide paramagnetic }}{KO _2\left(K^{+}, O _2^{-}\right)}$
(c) $3 Cu+8 HNO _3 \longrightarrow \underset{\text { Paramagnetic }}{3 Cu\left(NO _3\right) _2}+\underset{\text { Paramagnetic }}{2 NO}+4 H _2 O$
(d)
Hence, option (a), (b) and (c) are correct choices.
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