Q.A complex has a molecular formula MSO₄Cl.6H₂O. The aqueous solution of it gives white precipitate with Barium chloride solution and no precipitate is obtained when it is treated with silver nitrate solution. If the secondary valence of the metal is six, which one of the following correctly represents the complex?
Step 1. A white precipitate with BaCl₂ confirms free (ionisable) SO₄²⁻ is present -- so sulfate sits OUTSIDE the coordination sphere as the counter ion.
Step 2. No precipitate with AgNO₃ confirms there is NO free Cl⁻ -- so the single chloride must be bound INSIDE the coordination sphere as a ligand.
Step 3. The secondary valence (coordination number) is given as six, and the coordination sphere must contain the one Cl⁻ plus water molecules to reach six donor atoms: 5 H₂O + 1 Cl = 6 donor groups, giving the coordination entity [M(H₂O)₅Cl].
Step 4. The formula MSO₄Cl.6H₂O has 6 total water molecules; 5 are used inside the coordination sphere, leaving 1 as free lattice water, written after the sulfate.
Step 5. Putting it together: [M(H₂O)₅Cl]SO₄.H₂O, matching option (c).
(c) [M(H₂O)₅Cl]SO₄.H₂O
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