Question

Which of the following is the most likely structure of $CrC{l}_3\cdot 6H_{2}O,$ if 1/3 of total chlorine of the compound is precipitated by adding $AgN{O}_3$ to its aqueous solution:

• A. $CrC{l}_3\cdot 6H_{2}O$
• B. $\left[Cr\right(H_{2}O{)}_{3}C{l}_3\left]\right(H_{2}O{)}_3$
• C. $\left[CrC{l}_2\right(H_{2}O{)}_4]Cl\cdot 2H_{2}O$
• D. $[CrCl\cdot (H_{2}O{)}_5]C{l}_2\cdot H_{2}O$

Answer 1

The correct option is C $\left[CrC{l}_2\right(H_{2}O{)}_4]Cl\cdot 2H_{2}O$

According to the question, in $CrC{l}_3\cdot 6H_{2}O,1/3$ of total chlorine of the compound is precipitated by adding $AgN{O}_3$ to its aqueous solution.
$\because$ Total number of Cl in compound is 3. Hence precipitated Cl = 1
$\therefore$ Structure of compound is $\left[CrC{l}_2\right(H_{2}O{)}_4]Cl\cdot 2H_{2}O$
$CrC{l}_2\left(H_{2}O{)}_4\right]Cl.2H_{2}O\stackrel{AgN{O}_3}{\to }\left[CrC{l}_2\right(H_{2}O{)}_4{]}^{+}+AgCl$
Therefore, the correct option is (c).