A metal complex with a formula MCl4.3NH3 is involved in sp3d2 hybridisation. It upon reaction with e — JEE Mains Chemistry Past Papers Chemistry Question

💡 Solution & Explanation

# Solution: MCl4·3NH3 Complex with AgNO3 **Step 1: Determine the coordination geometry** sp³d² hybridization indicates octahedral geometry with 6 ligands around the metal center. **Step 2: Identify the ligands in the complex** - Total ligands = 4 Cl⁻ + 3 NH₃ = 7 species - Since only 6 can coordinate (octahedral), one Cl⁻ must be outside the coordination sphere as a counter ion - Structure: [MCl₃(NH₃)₃]⁺·Cl⁻·3NH₃ **Step 3: Determine chloride ions available for precipitation** - Inside coordination sphere: 3 Cl⁻ (coordinated, not precipitable) - Outside coordination sphere: 1 Cl⁻ (free, can precipitate) - Total free Cl⁻ = 1 mole per mole of complex **Step 4: Calculate moles of AgCl precipitate** AgNO₃ reacts with free Cl⁻: Ag⁺ + Cl⁻ → AgCl↓ Only 1 mole of Cl⁻ (the counter ion) reacts with excess AgNO₃. **Step 5: Calculate moles of AgNO₃ consumed** 1 mole of AgNO₃ is needed to precipitate 1 mole of AgCl. Therefore, the answer is **1 mole of AgCl** (or **1 mole of AgNO₃ is consumed**).