JEE Main 2017 · 2 Apr · Shift 1 · Chemistry Q50

← JEE Main 2017 · 2 Apr · Shift 1

ChemistrySingle correct+4 / -1Question 50 of 90

On treatment of 100 mL of 0.1 M solution of $\ce{CoCl3.6H2O}$ with excess $\ce{AgNO3}$ ; $1.2\times10^{22}$ ions are precipitated. The complex is :

\ce{[Co(H2O)6]Cl3}

\ce{[Co(H2O)5Cl]Cl2.H2O}

\ce{[Co(H2O)4Cl2]Cl.2H2O}

\ce{[Co(H2O)3Cl3].3H2O}

Answer and worked solution

Official answer: C

The solution contains $0.1\times0.1=0.01$ mol of complex, and $1.2\times10^{22}$ ions is $0.02$ mol.

The key's answer, $\ce{[Co(H2O)4Cl2]Cl.2H2O}$ , counts the precipitated ions as both $\ce{Ag+}$ and $\ce{Cl-}$ : 0.02 mol of ions is 0.01 mol of AgCl, i.e. one ionisable chloride per formula unit.

If "ions" meant chloride ions only, there would be two per formula unit, which would be $\ce{[Co(H2O)5Cl]Cl2.H2O}$ . The answer follows the official key.

Topics:Isomerism in complexes Mole concept

← Previous question Next question →

JEE Main 2017 · 2 Apr · Shift 1, Chemistry question 50