JEE Main 2026 — Jan 24, Shift 1 — Chemistry solution

JEE Main2026Jan 24, Shift 1Chemistry

Consider a mixture 'X' which is made by dissolving 0.4 mol of $[C o (N H_{3})_{5} S O_{4}] B r$ and 0.4 mol of $[C o (N H_{3})_{5} B r] S O_{4}$ in water to make 4 L of solution. When 2 L of mixture 'X' is allowed to react with excess of $A g N O_{3}$ , it forms precipitate 'Y'. The rest 2 L of mixture 'X' reacts with excess $B a C l_{2}$ to form precipitate 'Z'. Which of the following statements is CORRECT.

A
0.2 mol of 'Z' is formed
B
'Y' is $B a S O_{4}$ and 'Z' is $A g B r$
C
0.4 mol of 'Z' is formed
D
0.1 mol of 'Y' is formed

Solution

In 4 L of solution, mixture X contains 0.4 mol of each complex. In 2 L, there is 0.2 mol of $[C o (N H_{3})_{5} S O_{4}] B r$ and 0.2 mol of $[C o (N H_{3})_{5} B r] S O_{4}$ .

Only ions in the outer (ionisable) sphere react:

With excess $A g N O_{3}$ : only the ionisable $B r^{-}$ from $[C o (N H_{3})_{5} S O_{4}] B r$ precipitates as AgBr. So Y = AgBr, moles = 0.2.

With excess $B a C l_{2}$ : only the ionisable $S O_{4}^{2 -}$ from $[C o (N H_{3})_{5} B r] S O_{4}$ precipitates as $B a S O_{4}$ . So Z = $B a S O_{4}$ , moles = 0.2.

$∴$ 0.2 mol of Z is formed. Option (1) is correct.

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