JEE Advanced 2019 · Paper 2 · Chemistry Q3

ChemistryMultiple correct+4 / -1Question 21 of 54

Consider the following reactions (unbalanced)

$\ce{Zn}$ + hot conc. $\ce{H2SO4} \to$ G + R + X

$\ce{Zn}$ + conc. $\ce{NaOH} \to$ T + Q

G + $\ce{H2S}$ + $\ce{NH4OH} \to$ Z (a precipitate) + X + Y

Choose the correct option(s)

Z is dirty white in colour

The oxidation state of Zn in T is +1

R is a V-shaped molecule

Bond order of Q is 1 in its ground state

Answer and worked solution

Official answer: A, C, D

Identify the species.

$\ce{Zn + 2H2SO4 -> ZnSO4 + SO2 + 2H2O}$ , so G is $\ce{ZnSO4}$ , R is $\ce{SO2}$ and X is $\ce{H2O}$ .
$\ce{Zn + 2NaOH -> Na2ZnO2 + H2}$ , so T is $\ce{Na2ZnO2}$ and Q is $\ce{H2}$ .
$\ce{ZnSO4 + H2S + 2NH4OH -> ZnS v + (NH4)2SO4 + 2H2O}$ , so Z is ZnS and Y is $\ce{(NH4)2SO4}$ .

The statements.

(A), (C), (D)

Topics:Group 16 Transition element properties VSEPR theory

JEE Advanced 2019 · Paper 2, Chemistry question 3