Question

$A,B$ and $C$ are three complexes of chromium(III) with the empirical formula: $H_{12}{O}_{6}C{l}_{3}Cr$. All the three complexes have water and chloride ions as ligands. Complex $A$ does not react with concentrated $H_{2}S{O}_4$, whereas complexes $B$ and $C$ lose $6.75\%$ and $13.5\%$ of their original weight, respectively, on treatment with concentrated $H_{2}S{O}_4$. If $C$ is $\left[Cr\right(H_{2}O{)}_{x}C{l}_y]Cl\cdot z{H}_{2}O,$ find the x, y and z.

• A. $x=4,y=2,z=2$
• B. $x=3,y=2,z=3$
• C. $x=2,y=2,z=4$
• D. None of these

Answer 1

The correct option is A $x=4,y=2,z=2$

$A$ does not lose water with conc.$H_{2}S{O}_4,$ indicating that $H_{2}O$ are in coordinate sphere.
Thus, $A$ is $\left[Cr\right(H_{2}O{)}_6]C{l}_3$
Molar mass of the complex$=52+108+106.5$$=266.5g$ $mo{l}^{-1}$
After loss with conc.$H_{2}S{O}_4$, molecular weight of $\left(B\right)=\frac{266.5\times 93.25}{100}$ $=149.18$
Net loss$=266.5-249.18$
$=17.32g=1mol$ $H_{2}O$
Thus, one water molecule is lost per molecule of the complex. Thus, complex $\left(B\right)$ is $\left[Cr\right(H_{2}O{)}_{6}C{l}_2]Cl\cdot H_{2}O$
After loss with conc.$H_{2}S{O}_4$, molar mass of $\left(C\right)$ is $=\frac{266.5\times 86.5}{100}$ $=230.25g$ $mo{l}^{-1}$
Net loss$=35.98g$
Number of water molecules lost$=\frac{35.98}{18}=2$. Thus, two $H_{2}O$ molecules are as hydrate molecules.
Thus, $\left(C\right)$ is $\left[Cr\right(H_{2}O{)}_{4}C{l}_2]Cl\cdot 2H_{2}O$. Hence, option A is correct.