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

N a

and excess of

N H_{3}

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K

and excess of

O_{2}

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C u

and dilute

H N O_{3}

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O_{2}

and 2 - ethylanthraquinol

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Solution

The correct options are
A $N a$ and excess of $N H_{3}$
B $K$ and excess of $O_{2}$
C $C u$ and dilute $H N O_{3}$
(A) $N a$ and excess of $N H_{3}$ yields solvated electrons which are unpaired electrons. Hence, the species is paramagnetic.

(B) $K$ and excess of $O_{2}$ forms superoxide $K O_{2}$ which contains paramagnetic superoxide ion $O_{2}^{-}$ with one unpaired electron.

(C) $C u$ reacts with dilute $H N O_{3}$ to form cupric nitrate with 1 unpaired electron in $C u^{2 +}$ and also in $N O$ . This results in paramagnetism.

$3 C u + 8 H N O_{3} (d i l) \to 3 C u (N O_{3})_{2} + 4 H_{2} O + 2 N O$

(D) Refer image.

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