JEE Advanced 2020 · Paper 2 · Chemistry Q10

ChemistryMultiple correct+4 / -2Question 28 of 54

Choose the correct statement(s) among the following.

\ce{SnCl2.2H2O}

is a reducing agent.

\ce{SnO2}

reacts with KOH to form

\ce{K2[Sn(OH)6]}

A solution of

\ce{PbCl2}

in HCl contains

\ce{Pb^{2+}}

and

\ce{Cl-}

ions.

The reaction of

\ce{Pb3O4}

with hot dilute nitric acid to give

\ce{PbO2}

is a redox reaction.

Answer and worked solution

Official answer: A, B

(A) Sn(II) is readily oxidised to Sn(IV), so $\ce{SnCl2.2H2O}$ is a reducing agent. True.
(B) $\ce{SnO2}$ is amphoteric: $\ce{SnO2 + 2KOH + 2H2O -> K2[Sn(OH)6]}$ . True.
(C) In HCl, $\ce{PbCl2}$ dissolves as the chloro complex $\ce{[PbCl4]^{2-}}$ , not as free $\ce{Pb^{2+}}$ and $\ce{Cl-}$ . False.
(D) $\ce{Pb3O4}$ behaves as $\ce{2PbO.PbO2}$ . Nitric acid dissolves the PbO part and leaves $\ce{PbO2}$ : $\ce{Pb3O4 + 4HNO3 -> 2Pb(NO3)2 + PbO2 + 2H2O}$ . No oxidation state changes, so it is not a redox reaction. False.

(A), (B)

Topics:Group 14 Oxidation number

JEE Advanced 2020 · Paper 2, Chemistry question 10