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JEE Main 202624 JanuaryEvening Shift
Exam
JEE Main
Year
2026
Date
24 January
Shift
Evening Shift
Subject
Chemistry
Chapter
Coordination Compounds

JEE Main 2026 · 24 January · Evening Shift · Chemistry · Coordination Compounds

A chromium complex with a formula CrCl_ 3 · 6 H_ 2 O has a spin...

A chromium complex with a formula CrCl3⋅6H2O\mathrm{CrCl}_{3} \cdot 6 \mathrm{H}_{2} \mathrm{O} has a spin only magnetic moment value of 3.87 BM and its solution conductivity corresponds to 1:21: 2 electrolyte. 2.75 g of the complex solution was initially passed through a cation exchanger. The solution obtained after the process was reacted with excess of AgNO3\mathrm{AgNO}_{3}. The amount of AgCl formed in the above process is ____\_\_\_\_ g. (Nearest integer)
[Given: Molar mass in gmol−1Cr:52;Cl:35.5,Ag:108,O:16,H:1\mathrm{g} \mathrm{mol}^{-1} \mathrm{Cr}: 52 ; \mathrm{Cl}: 35.5, \mathrm{Ag}: 108, \mathrm{O}: 16, \mathrm{H}: 1]

Correct answer

0

Step-by-step solution

The complex CrCl3⋅6H2O\text{CrCl}_3 \cdot 6\text{H}_2\text{O} with magnetic moment 3.87 BM indicates 3.87=n(n+2)3.87 = \sqrt{n(n+2)} giving n = 3 unpaired electrons, confirming Cr³⁺.
A 1:2 electrolyte means structure is [Cr(H2O)5Cl]2+Cl2−[\text{Cr(H}_2\text{O})_5\text{Cl}]^{2+}\text{Cl}_2^-.
Molar mass = 52 + 5(18) + 35.5 + 6(18) = 285.5 g/mol.
Moles of complex = 2.75/285.5 = 0.00963 mol.
After cation exchange, 2 Cl⁻ remain per formula unit (from the 1:2 electrolyte), giving 0.00963 × 2 = 0.01926 mol of free Cl⁻.
Reacting with excess AgNO₃: mass of AgCl = 0.01926 × 143.5 = 2.76 g ≈ 3 g (nearest integer).
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