s-Block Elements — JEE Advanced · 2014 · 25 May 2014 · Paper 1 Question with Solution

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

(A)Na and excess of $NH_{3}$ ✓ Correct
(B)K and excess of $O_{2}$ ✓ Correct
(C)Cu and dilute $HNO_{3}$ ✓ Correct
(D) $O_{2}$ and 2-ethylanthraquinol

Answer: (A), (B), (C)

Solution

Paramagnetic species have unpaired electrons:

(A) **Na + excess

NH_{3}

**: Sodium dissolves in liquid ammonia to form solvated electrons, which are paramagnetic:

Na \to Na^{+} + e_{(am)}^{-}

. Correct.

(B) **K + excess

O_{2}

**: Potassium reacts with excess oxygen to form potassium superoxide,

KO_{2}

, which contains the superoxide ion

O_{2}^{-}

with one unpaired electron (paramagnetic). Correct.

(C) **Cu + dilute

HNO_{3}

**: Produces

Cu (NO_{3})_{2}

and NO gas. NO has one unpaired electron in its π* orbital, making it paramagnetic. Correct.

(D) **

O_{2}

+ 2-ethylanthraquinol**: This is a redox reaction producing anthraquinone and water/hydrogen peroxide. The products are diamagnetic. Incorrect.

Correct answers: (A), (B), (C).

JEE Advanced · 2014 · 25 May 2014 · Paper 1 · One or more correct options

LIST-I	LIST-II
(I) $H_{2} O_{2}$	(P) $Mg (HCO_{3})_{2} + Ca (OH)_{2} \to$

LIST-I

LIST-II

(I)

H_{2} O_{2}

(P)

Mg (HCO_{3})_{2} + Ca (OH)_{2} \to

JEE Advanced · 2014 · 25 May 2014 · Paper 1 · s-Block Elements

Solution

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