Question

One pink solid (X) has the formula $CoC{l}_3.5N{H}_3.H_{2}O.A$ solution of this salt (X) in water is also pink and precipitates 3 mole AgCl on treatment with silver nitrate solution. When the pink solid (X) is heated, it loses $1mole{H}_{2}O$ to yield a purple solid (Y) with the same molar ratio of $N{H}_3:Cl:Co$. The purple solid releases two of its chlorides rapidly on dissolution and after treatment with $AgN{O}_3$ solution. The coordination number of cobalt in complexes (X) and (Y) is six. The IUPAC name of pink (X) and purple (Y) complexes respectively are:

• A. Tetraamminechloridoaquacobalt(III) chloride and pentaamminechloridocobalt(III) chloride.
• B. Pentaamminechloridocobalt(III) chloride and pentaammineaquacobalt(III) chloride.
• C. Pentaammineaquacobalt(III) chloride and pentaamminechloridocobalt(III) chloride.
• D. Pentaammineaquacobaltate(III) chloride and pentaamminechloridocobaltate(III) chloride.

Answer 1

The correct option is C Pentaammineaquacobalt(III) chloride and pentaamminechloridocobalt(III) chloride.

From the data given in the question, it can be concluded that pink solid X contains three chloride ions outside the coordination sphere. Hence, treatment with $AgN{O}_3$ gives three moles of AgCl. Pink solid X also contains one water molecule inside its coordination sphere. Hence the formula of pink solid is $\left[Co\right(N{H}_3{)}_5{H}_{2}O]C{l}_3$.Its IUPAC name is pentaammineaquacobalt(III) chloride.
The purple solid Y contains two chloride ions outside its coordination sphere. When X is heated, it loses one water molecule to give Y. So the structure of Y is $\left[Co\right(N{H}_3{)}_{5}Cl]C{l}_2$. Its IUPAC name is pentaamminechloridocobalt(III) chloride.