Total number of moles of AgCl precipitated on addition of excess of AgNO₃ to one mole each of the following complexes [Co(NH₃)₄Cl₂]Cl,[Ni(H₂O)₆]Cl₂,[Pt(NH₃)₂Cl₂] and [Pd(NH₃)₄]Cl₂ is ___________.

Solution

Chemistry ma’am · 3 steps

Only the chlorides outside the square bracket are free ions, so only they meet Ag⁺ and precipitate.

1. Ionisable chloride. In water, a complex splits into the complex ion and the counter ions outside the bracket. Chlorides inside the bracket stay bonded to the metal, so excess AgNO₃ cannot pull them out as AgCl.

   Complex	Ions in water	Free Cl⁻ per mole
   [Co(NH₃)₄Cl₂]Cl	[Co(NH₃)₄Cl₂]⁺ + Cl⁻	1
   [Ni(H₂O)₆]Cl₂	[Ni(H₂O)₆]²⁺ + 2Cl⁻	2
   [Pt(NH₃)₂Cl₂]	neutral, does not ionise	0
   [Pd(NH₃)₄]Cl₂	[Pd(NH₃)₄]²⁺ + 2Cl⁻	2

2. Add them up. Each free Cl⁻ gives one AgCl, so the total is 1 + 2 + 0 + 2 = 5 mol.
3. So. The total amount of AgCl precipitated is 5 mol.

The common slip: 9. Counting every chlorine in each formula gives 3 + 2 + 2 + 2 = 9. The four chlorides bonded to Co and Pt are ligands and do not precipitate.

The idea to keep. Count chlorides outside the square bracket only. Ligand chlorides are held by the metal and never precipitate with AgNO₃.

Answer from the NTA final answer key. Solution by PrepWiser.