The coordination complex that gives white precipitate of molar mass 1…

The correct answer is Option (3) → [Co(NH₃)₅(SO₄)]Cl

In a coordination compound, the ligands inside the square bracket are bound to the metal and do not ionise. Only the ions outside the bracket (the counter ions) are free in solution and can be precipitated.

A white precipitate with AgNO₃ is silver chloride. Its molar mass is:

    \( \displaystyle M(\ce{AgCl})=108+35.5=143.5\ \mathrm{g\,mol^{-1}} \)

So the complex must have a chloride ion outside the coordination sphere. \( \ce{[Co(NH3)5(SO4)]Cl} \) ionises to give one free \( \ce{Cl^{-}} \) ion, which forms 143.5 g of AgCl for each mole of complex:

    \( \displaystyle \ce{[Co(NH3)5(SO4)]Cl -> [Co(NH3)5(SO4)]^+ + Cl^-} \)

    \( \displaystyle \ce{Cl^- + AgNO3 -> AgCl v + NO3^-} \)

Why the other options are incorrect

• Option (1) [Co(NH₃)₅(SO₄)]Br — Incorrect. The free \( \ce{Br^{-}} \) ion gives pale yellow AgBr (molar mass 188 g mol⁻¹).

• Option (2) [Co(NH₃)₅(SO₄)]I — Incorrect. The free \( \ce{I^{-}} \) ion gives yellow AgI (molar mass 235 g mol⁻¹).

• Option (4) [Co(NH₃)₅Cl]SO₄ — Incorrect. The chlorine is a ligand inside the bracket and does not ionise, so no AgCl forms. The free \( \ce{SO4^{2-}} \) ion gives a white precipitate with BaCl₂ instead.

NCERT (2026-27 edition):
Werner’s Theory of Coordination Compounds – …