The pair(s) of reagents that yield paramagnetic species is/are

Question

The pair(s) of reagents that yield paramagnetic species is/are (A) Na and excess of NH3  (B) K and excess of O2  (C) Cu and dilute HNO3  (D)  O2 and 2-ethylanthraquinol

Tardigrade Admin · Accepted Answer

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)

N a + (E x cess) N H_{3} \to N a N H_{2} + \frac{1}{2} H_{2}

(N a N H_{2} + 1/2 H_{2}

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}]^{-}

∙

Ammoniated electron

∙

Blue colour

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Paramagnetic

∙

Very strong reducing agent
(b)

K + (Excess) O_{2} \to Potassium superoxide paramagnetic K O_{2} (K^{+}, O_{2}^{-})

(c)

3 C u + 8 H N O_{3} \to Paramagnetic 3 C u (N O_{3})_{2} + Paramagnetic 2 NO + 4 H_{2} O

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

Q. The pair(s) of reagents that yield paramagnetic species is/are

$N a$ and excess of $N H_{3}$

$K$ and excess of $O_{2}$

$C u$ and dilute $H N O_{3}$

$O_{2}$ and $2$ -ethylanthraquinol

Q. The pair(s) of reagents that yield paramagnetic species is/are

Na and excess of NH3​

K and excess of O2​

Cu and dilute HNO3​

O2​ and 2-ethylanthraquinol

$N a$ and excess of $N H_{3}$

$K$ and excess of $O_{2}$

$C u$ and dilute $H N O_{3}$

$O_{2}$ and $2$ -ethylanthraquinol