JEE Advanced 2023 · Paper 1 · Chemistry Q14

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ChemistryList match+3 / -1Question 48 of 51

Match the reactions (in the given stoichiometry of the reactants) in List-I with one of their products given in List-II and choose the correct option.

List-I

P
$\ce{P2O3 + 3H2O ->}$
Q
$\ce{P4 + 3NaOH + 3H2O ->}$
R
$\ce{PCl5 + CH3COOH ->}$
S
$\ce{H3PO2 + 2H2O + 4AgNO3 ->}$

List-II

1
$\ce{P(O)(OCH3)Cl2}$
2
$\ce{H3PO3}$
3
$\ce{PH3}$
4
$\ce{POCl3}$
5
$\ce{H3PO4}$

P → 2; Q → 2; R → 5; S → 3

P → 3; Q → 5; R → 4; S → 2

P → 5; Q → 2; R → 1; S → 3

P → 2; Q → 3; R → 4; S → 5

Answer and worked solution

Official answer: D

(P) $\ce{P2O3 + 3H2O -> 2H3PO3}$ → 2.

(Q) $\ce{P4 + 3NaOH + 3H2O -> PH3 + 3NaH2PO2}$ → 3.

(R) $\ce{PCl5 + CH3COOH -> CH3COCl + POCl3 + HCl}$ → 4.

(S) $\ce{H3PO2 + 2H2O + 4AgNO3 -> 4Ag + H3PO4 + 4HNO3}$ → 5. Hypophosphorous acid is a strong reducing agent.

P→2, Q→3, R→4, S→5.

Topics:Group 15 Oxoacids

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JEE Advanced 2023 · Paper 1, Chemistry question 14