Number of ions produced on ionisation of Cr(NH3)4Cl3 reacting with AgNO3 to give 1 mole of AgCl, will be
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 2
\( \ce{AgNO3} \) precipitates only the chloride ions that are outside the coordination sphere (the ionisable ones). One mole of AgCl per mole of complex means one \( \ce{Cl^{-}} \) is ionisable and the other two are bonded to Cr:
\( \ce{Cr(NH3)4Cl3} \) = \( \ce{[Cr(NH3)4Cl2]Cl} \)
\( \ce{[Cr(NH3)4Cl2]Cl} \) → \( \ce{[Cr(NH3)4Cl2]^{+}} \) + \( \ce{Cl^{-}} \)
So the compound gives 2 ions in solution (Cr stays at coordination number 6: 4 \( \ce{NH3} \) + 2 Cl).