1

JEE Advanced 2021 Paper 2 Online

MCQ (More than One Correct Answer)
The correct statement(s) related to oxoacids of phosphorous is(are)
A
Upon heating, H3PO3 undergoes disproportionation reaction to produce H3PO4 and PH3.
B
While H3PO3 can act as reducing agent, H3PO4 cannot.
C
H3PO3 is a monobasic acid.
D
The H atom of P-H bond in H3PO3 is not ionizable in water.

Explanation

(a) H3PO3 gives H3PO4 and PH3 on heating. This reaction is known as disproportionation reaction.

$$4{H_3}P{O_3}\buildrel \Delta \over \longrightarrow 3{H_3}P{O_4} + P{H_3}$$

(b) P in H3PO4 is in its highest oxidation state i.e. + 5 hence, it cannot act as reducing agent. But P in H3PO3 is in oxidation state, + 3 hence, it can act as reducing agent.

(c) H3PO3 contains two $$-$$OH groups, hence it is a dibasic acid.


(d) Hydrogen attached to P, does not ionise in water. Therefore, options (a), (b) and (d) are correct.
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JEE Advanced 2019 Paper 2 Offline

MCQ (More than One Correct Answer)
Consider the following reactions (unbalanced).

$$Zn + Hot\,conc.\,{H_2}S{O_4}\mathrel{\mathop{\kern0pt\longrightarrow} \limits_{}} G + R + X$$

$$Zn + conc.\,NaOH\mathrel{\mathop{\kern0pt\longrightarrow} \limits_{}} T + Q$$

$$G + {H_2}S + N{H_4}OH\mathrel{\mathop{\kern0pt\longrightarrow} \limits_{}} Z\,(a\,precipitate) + X + Y$$

Choose the correct option(s).
A
The oxidation state of Zn in T is +1
B
R is a V-shaped molecule
C
Bond order of Q is 1 in its ground state
D
Z is dirty white in colour

Explanation

When Zn react with hot conc. H2SO4 then SO2 is released and ZnSO4 is obtained.



R(SO2) molecule is V-shaped



Thus, option (b) is correct.

When Zn is react with conc. NaOH then H2 gas is evolved and Na2ZnO2 is obtained.

$$Zn + 2NaOH(conc.)\mathrel{\mathop{\kern0pt\longrightarrow} \limits_{}} \mathop {N{a_2}Zn{O_2}}\limits_{(T)} + \mathop {{H_2}}\limits_{(Q)} \uparrow $$

In ground state, H$$-$$H (Q)

(bond order = 1)

Thus, option (c) is correct.

The oxidation state of Zn in T(Na2ZnO2) is +2

Thus, option (a) is incorrect.

$$\mathop {ZnS{O_4}}\limits_{(G)} + {H_2}S + N{H_4}OH\mathrel{\mathop{\kern0pt\longrightarrow} \limits_{}} \mathop {ZnS}\limits_{(Z)} \downarrow + \mathop {2{H_2}O}\limits_{(X)} + \mathop {{{(N{H_4})}_2}S{O_4}}\limits_{(Y)} $$

ZnS (Z) compound is dirty white coloured.

Thus, option (d) is correct.
3

JEE Advanced 2019 Paper 1 Offline

MCQ (More than One Correct Answer)
A tin chloride Q undergoes the following reactions (not balanced)

$$Q + C{l^ - } \to X$$

$$Q + M{e_3}N \to Y$$

$$Q + CuC{l_2} \to Z + CuCl$$

X is a monoanion having pyramidal geometry. Both Y and Z are neutral compounds.

Choose the correct option(s).
A
There is a coordinate bond in Y
B
The central atom in Z has one lone pair of electrons.
C
The oxidation state of the central atom in Z is +2
D
The central atom in X is sp3 hybridised

Explanation

Sn can exist in +2 or +4 oxidation state. So, Q in the given reactions can be SnCl2 or SnCl4. Since X is monoanion having trigonal pyramidal geometry (sp3 with one lone pair as per VSEPR), so Q is SnCl2, and the first reaction is

The second reaction is Stephen’s reaction, where nitrile is converted to aldehyde via formation of iminium salt.


There is a coordinate bond in the Cl2Sn.N(CH3)3 in between nitrogen and Sn metal.

Z is oxidised product and oxidation state of Sn is +4 in Z compound. Structure of SnCl4 (Z) is
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JEE Advanced 2018 Paper 1 Offline

MCQ (More than One Correct Answer)
Based on the compounds of group $$15$$ elements, the correct statement(s) is (are)
A
$$B{i_2}{O_5}$$ is more basic than $${N_2}{O_5}$$
B
$$N{F_3}$$ is more covalent than $$Bi{F_3}$$
C
$$P{H_3}$$ boils at lower temperature than $$N{H_3}$$
D
The $$N-N$$ single bond is stronger than the $$P-P$$ single bond

Explanation

Option (A) : Correct. The basicity of oxides usually increases on descending a group. Therefore, Bi2O5 is more basic than N2O5.

Option (B) : Correct. Covalent nature of a molecule depends on the electronegativity difference between bonded atoms.

Option (C) : Correct. Boiling point of NH3 is more than that of PH3 due to hydrogen bonding.

Option (D) : Incorrect. P$$-$$P single bond is stronger than N$$-$$N single bond. This is due to the fact that N is small in size, due to smaller size of atoms lone pair of repulsion will be more.

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