JEE Advanced 2014 · Paper 1 · Chemistry Q30

ChemistryMultiple correct+3Question 30 of 60

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

Na and excess of

\ce{NH3}

K and excess of

\ce{O2}

Cu and dilute

\ce{HNO3}

\ce{O2}

and 2-ethylanthraquinol

Answer and worked solution

Official answer: A, B, C

(A) Sodium dissolves in liquid ammonia to give ammoniated electrons, $\ce{e-(NH3)_x}$ , which are unpaired. Paramagnetic.
(B) Potassium burns in excess oxygen to the superoxide $\ce{KO2}$ ; $\ce{O2-}$ has one unpaired electron. Paramagnetic.
(C) $\ce{3Cu + 8HNO3 -> 3Cu(NO3)2 + 2NO + 4H2O}$ . NO has an odd electron and $\ce{Cu^{2+}}$ ( $d^9$ ) is also paramagnetic.
(D) This is the industrial route to $\ce{H2O2}$ : 2-ethylanthraquinol + $\ce{O2}$ gives the anthraquinone and $\ce{H2O2}$ , both diamagnetic.

(A), (B), (C)

Topics:Alkali metals Group 15 Water and hydrogen peroxide

JEE Advanced 2014 · Paper 1, Chemistry question 30