1

JEE Advanced 2020 Paper 2 Offline

MCQ (More than One Correct Answer)
Choose the correct statement(s) among the following :
A
SnCl2 . 2H2O is a reducing agent.
B
SnO2 reacts with KOH to form K2[Sn(OH)6].
C
A solution of PbCl2 in HCl contains Pb2+ and Cl$$-$$ ions.
D
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.
2

JEE Advanced 2015 Paper 1 Offline

MCQ (More than One Correct Answer)

Fe3+ is reduced to Fe2+ by using

A
H2O2 in presence of NaOH.
B
Na2O2 in water.
C
H2O2 in presence of H2SO4.
D
Na2O2 in presence of H2SO4.

Explanation

Fe3+ is reduced to Fe2+ by H2O2/NaOH and Na2O2/H2O.

$$2F{e^{3 + }} + {H_2}{O_2} + O{H^ - } \to 2F{e^{2 + }} + 2{H_2}O + {O_2}$$

$$N{a_2}{O_2} + {H_2}O \to {H_2}{O_2} + NaOH$$

3

JEE Advanced 2014 Paper 1 Offline

MCQ (More than One Correct Answer)
For the reaction,

$${I^ - } + ClO_3^ - + {H_2}S{O_4} \to C{l^ - } + HSO_4^ - + {I_2}$$

the correct statement(s) in the balanced equation is/are
A
stoichiometric coefficient of HSO$$_4^ - $$ is 6
B
iodide is oxidised
C
sulphur is reduced
D
H2O is one of the products

Explanation

Oxidation half-reaction,

$$2{I^ - } \to {I_2} + 2{e^ - }$$ .... (i)

Here, I$$-$$ is converted into I2. Oxidation number of I is increasing from $$-$$1 to 0 hence, this is a type of oxidation reaction.

Reduction half-reaction

$$6{H^ + } + ClO_3^ - + 6{e^ - } \to C{l^ - } + 3{H_2}O$$ ..... (ii)

Here, H2O releases as a product. Hence, option (d) is correct.

Multiplying equation (i) by 3 and adding in equation (ii)


Stoichiometric coefficient of $$HSO_4^ - $$ is 6.

Hence, option (a), (b) and (d) are correct.

Joint Entrance Examination

JEE Main JEE Advanced WB JEE

Graduate Aptitude Test in Engineering

GATE CSE GATE ECE GATE EE GATE ME GATE CE GATE PI GATE IN

Medical

NEET

CBSE

Class 12