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The complex $[Cr(NH_3)_6]Cl_3$ on reaction with excess $AgNO_3$ produces:
A1 mole AgCl per mole of complex
B2 moles AgCl per mole of complex
C3 moles AgCl per mole of complex
DNo precipitate
Answer & Solution
Correct answer: C. 3 moles AgCl per mole of complex
All three Cl⁻ are outside the coordination sphere (counter-ions, ionizable). All three precipitate with Ag⁺ as AgCl. This is Werner's classic argument for determining how many Cl are coordinated.
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