Question

50 ml of 0.2 M solution of a compound with empirical formula $CoC{l}_3.4N{H}_3$ on treatment with excess of $AgN{O}_3\left(aq\right)$ yields 1.435 g of AgCl. Ammonia is not removed by treatment with concentrated $H_{2}S{O}_4$. The formula of the compound is:

• A. $CoC{l}_3{\left(N{H}_3\right)}_4$
• B. $\left[CoC{l}_2{\left(N{H}_3\right)}_4\right]Cl$
• C. $\left[Co{\left(N{H}_3\right)}_4\right]C{l}_3$
• D. $\left[CoC{l}_3\left(N{H}_3\right)\right]N{H}_3$

Answer 1

Answer: (B)

$\left[CoC{l}_2{\left(N{H}_3\right)}_4\right]Cl$

No. of moles of given complex $=50\times 0.2=10$ millimole and no. of millimoles of AgCl formed $=1.435/143.5)=0.01$ mole $=10$ millimole. As no.of moles of AgCl formed $=$ no. of moles of chloride ion given by complex. So, no. of moles of chloride ion given by complex $=10$ millimole. That means every mole of complex gives one mole of chloride ion. So, compound is $\left[CoC{l}_2\right(N{H}_3{)}_4]Cl$.