JEE Advance - Chemistry (2020 - Paper 2 Offline - No. 10)
Choose the correct statement(s) among the following :
SnCl2 . 2H2O is a reducing agent.
SnO2 reacts with KOH to form K2[Sn(OH)6].
A solution of PbCl2 in HCl contains Pb2+ and Cl$$-$$ ions.
The reaction of Pb3O4 with hot dilute nitric acid to give PbO2 is a redox reaction.
Explanation
Sn2+ of stannous chloride dihydrate (SnCl2 . 2H2O) tends to convert into Sn4+.
Hence, statement (a) is correct.
(b) SnO2 reacts with KOH and gives K2SnO3 . 3H2O or K2[Sn(OH)6] because it is amphoteric in nature.
$$Sn{O_2} + KOH\buildrel {} \over \longrightarrow {K_2}Sn{O_3} + {H_2}O$$
or $${K_2}[Sn{(OH)_6}]$$
Hence, statement (b) is correct.
(c) In conc. HCl, PbCl2 exists as chloroplumbous acid, H2[PbCl4]
$$\mathop {PbC{l_2}}\limits^{II} + 2HCl\buildrel {} \over \longrightarrow \mathop {{H_2}[PbC{l_4}]}\limits^{II} $$
Hence, statement (c) is incorrect.
(d) Pb3O4 is a mixture of $$(2\mathop {PbO}\limits^{ + 2} + {\mathop {PbO}\limits^{ + 4} _2})$$
$$P{b_3}{O_4} + 4HN{O_3}\buildrel {} \over \longrightarrow \mathop {2Pb{{(N{O_3})}_2}}\limits^{ + 2} + \mathop {Pb{O_2}}\limits^{ + 4} + 2{H_2}O$$
It is not a redox reaction. Thus, the statement (d) is incorrect.
Hence, statement (a) is correct.
(b) SnO2 reacts with KOH and gives K2SnO3 . 3H2O or K2[Sn(OH)6] because it is amphoteric in nature.
$$Sn{O_2} + KOH\buildrel {} \over \longrightarrow {K_2}Sn{O_3} + {H_2}O$$
or $${K_2}[Sn{(OH)_6}]$$
Hence, statement (b) is correct.
(c) In conc. HCl, PbCl2 exists as chloroplumbous acid, H2[PbCl4]
$$\mathop {PbC{l_2}}\limits^{II} + 2HCl\buildrel {} \over \longrightarrow \mathop {{H_2}[PbC{l_4}]}\limits^{II} $$
Hence, statement (c) is incorrect.
(d) Pb3O4 is a mixture of $$(2\mathop {PbO}\limits^{ + 2} + {\mathop {PbO}\limits^{ + 4} _2})$$
$$P{b_3}{O_4} + 4HN{O_3}\buildrel {} \over \longrightarrow \mathop {2Pb{{(N{O_3})}_2}}\limits^{ + 2} + \mathop {Pb{O_2}}\limits^{ + 4} + 2{H_2}O$$
It is not a redox reaction. Thus, the statement (d) is incorrect.
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