The pair(s) of reagents that yield paramagnetic species is/are
(JEE Advanced, 2014)

Na and excess of

N H_{3}

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K and excess of

O_{2}

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Cu and dilute

H N O_{3}

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O_{2}

and 2-ethylanthraquinol

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Solution

The correct option is C Cu and dilute $H N O_{3}$
Dilute solution sof Na in liquid ammonia is paramagnetic due to ammoniated $e^{- 1} s$
$N a^{\oplus} + (x + y) N H_{3} \to N a^{+} (N H_{3})_{x} + e^{-} (N H_{3})_{y}$
$K + O_{2} (e x c e s s) \to K O_{2} P a r a m a g n e t i c$
$O_{¯ 2}$ is paramagnetic due to unpaired electron in antibonding orbital
$3 C u + 8 H N O_{3} d i l u t e d \to 3 C u (N O_{3})_{2} + 2 N O + 4 H_{2} O$
NO is paramagnetic due to unpaired electrons on "N".

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