JEE Main 2026 — Jan 24, Shift 2 — Chemistry solution

JEE Main2026Jan 24, Shift 2Chemistry

A chromium complex with a formula $CrCl_{3} \cdot 6 H_{2} O$ has a spin only magnetic moment value of 3.87 BM and its solution conductivity corresponds to 1 : 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 $AgNO_{3}$ . The amount of AgCl formed in the above process is ________ g. (Nearest integer)

[Given : Molar mass in g $mol^{- 1}$ Cr : 52; Cl : 35.5, Ag : 108, O : 16, H : 1]

Your answer

Solution

$μ = 3.87$ BM corresponds to 3 unpaired electrons $\Rightarrow$ $Cr^{3 +}$ ( $d^{3}$ ). 1:2 electrolyte means two ionisable groups outside the coordination sphere.

The complex is therefore $[Cr (H_{2} O)_{5} Cl] Cl_{2} \cdot H_{2} O$ . Molar mass $= 52 + 5 (18) + 35.5 + 2 (35.5) + 18 = 266.5$ g/mol.

Moles of complex $= \frac{2.75}{266.5} \approx 0.01032$ .

Each formula unit releases 2 ionisable $Cl^{-}$ outside the sphere (the cation exchanger removes the cationic complex but the free $Cl^{-}$ remain available for precipitation):

Moles of $AgCl = 2 \times 0.01032 \approx 0.02063$ .

Mass of AgCl $= 0.02063 \times 143.5 \approx 2.96$ g $\approx 3$ g.

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