Question

A, B and C are three complexes of chromium (III) with the empirical formula: ${{\rm{H}}_{{\rm{12}}}}{{\rm{O}}_{\rm{6}}}{\rm{C}}{{\rm{l}}_{\rm{3}}}{\rm{Cr}}$. All the three complexes have water and chloride ions as ligands. Complex A does not react with concentrated H2SO4, whereas complexes B and C lose 6.75% and 13.5% of their original weight, respectively, on treatment with concentrated ${{\rm{H}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}$. If C is $\left[ {{\rm{Cr}}{{\left( {{{\rm{H}}_{\rm{2}}}{\rm{O}}} \right)}_x}{\rm{C}}{{\rm{l}}_y}} \right]{\rm{Cl}}{\rm{.}}z{{\rm{H}}_2}{\rm{O}}$, 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

Hint: Hint: We know that a compound in which metal atoms bound to a number of neutral molecules or anions is termed as coordination compound or complex compound. A coordination entity constitutes a central metal atom bond to a fixed number of molecules or ions. For example, $\left[ {{\rm{CoC}}{{\rm{l}}_{\rm{3}}}{{\left( {{\rm{N}}{{\rm{H}}_{\rm{3}}}} \right)}_3}} \right]$, $\left[ {{\rm{Ni}}{{\left( {{\rm{CO}}} \right)}_{\rm{4}}}} \right]$ etc.

Complete step by step answer:
Here, three complexes of chromium (III) namely A, B and C are given. The empirical formula of the complexes is ${{\rm{H}}_{{\rm{12}}}}{{\rm{O}}_{\rm{6}}}{\rm{C}}{{\rm{l}}_{\rm{3}}}{\rm{Cr}}$ and water and chlorine are ligands in all the compounds. Among all the components, only water undergoes reaction with ${{\rm{H}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}$.


Let’s first identify the structure of complex A. Given that in all the complex oxidation state of chromium is III. Also A does not undergoes reaction with ${{\rm{H}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}$, that means, all the water molecules are in the coordination sphere. So, structure of complex A is $\left[ {{\rm{Cr}}{{\left( {{{\rm{H}}_{\rm{2}}}{\rm{O}}} \right)}_6}} \right]{\rm{C}}{{\rm{l}}_{\rm{3}}}$.

Now, come to the complex B. Here also the oxidation state of chromium ion is III. And when it reacts with ${{\rm{H}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}$, it loses 6.75% of its original weight.

To know the number of water molecules reacted with ${{\rm{H}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}$, we have to calculate the empirical mass of the complex.

${{\rm{H}}_{{\rm{12}}}}{{\rm{O}}_{\rm{6}}}{\rm{C}}{{\rm{l}}_{\rm{3}}}{\rm{Cr}} = 12 \times 1 + 6 \times 16 + 3 \times 35.5 + 52 = 266.5\,{\rm{g}}\,{\rm{mo}}{{\rm{l}}^{ - 1}}$


${\rm{Weight}}\,{\rm{loss by}}\,{\rm{complex B = 6}}{\rm{.75\% of 266}}{\rm{.5}}$

$ \Rightarrow {\rm{Weight}}\,{\rm{loss}} = \dfrac{{6.75}}{{100}} \times 266.5 = 18\,{\rm{g}}$

18 g weight loss indicates that one molecule of water is outside the coordination sphere. So, the complex B is $\left[ {{\rm{Cr}}{{\left( {{{\rm{H}}_{\rm{2}}}{\rm{O}}} \right)}_5}{\rm{Cl}}} \right]{\rm{C}}{{\rm{l}}_2}{\rm{.}}{{\rm{H}}_2}{\rm{O}}$.

Now, come to the complex C. Here also the oxidation state of chromium is also III. And when it reacts with ${{\rm{H}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}$, it loses 13.5% of its original weight.

Now, we have to calculate the number of water molecules reacted with ${{\rm{H}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}$ using the empirical mass of the complex, that is, $266.5\,{\rm{g}}\,{\rm{mo}}{{\rm{l}}^{ - 1}}$

${\rm{Weight}}\,{\rm{loss by}}\,{\rm{complex C}} = {\rm{13}}{\rm{.5\% of 266}}{\rm{.5}}$

$ \Rightarrow {\rm{Weight}}\,{\rm{loss}} = \dfrac{{13.5}}{{100}} \times 266.5 = 36\,{\rm{g}}$

36 g weight loss indicates that, two molecules of water $\left( {\dfrac{{36\,{\rm{g}}}}{{18\,{\rm{g}}}} = 2\,{\rm{mol}}} \right)$
 are outside the coordination sphere. So, the structure of complex C is $\left[ {{\rm{Cr}}{{\left( {{{\rm{H}}_{\rm{2}}}{\rm{O}}} \right)}_4}{\rm{C}}{{\rm{l}}_2}} \right]{\rm{Cl}}{\rm{.2}}{{\rm{H}}_2}{\rm{O}}$.

So, if we compare complex C with $\left[ {{\rm{Cr}}{{\left( {{{\rm{H}}_{\rm{2}}}{\rm{O}}} \right)}_x}{\rm{C}}{{\rm{l}}_y}} \right]{\rm{Cl}}{\rm{.}}z{{\rm{H}}_2}{\rm{O}}$, value of x, y and z is 4, 2 and 2 respectively.

So, the correct answer is Option A .

Note:
Isomerism is the phenomenon in which more than one compound possesses the same chemical formula but their arrangement in the compound is different. Coordination compounds also show isomerism. Ionisation isomerism is isomerism in which a counter ion in a complex salt is a potential ligand and can displace a ligand which can then become the counter ion. For example, $\left[ {{\rm{Co}}{{\left( {{\rm{N}}{{\rm{H}}_{\rm{3}}}} \right)}_5}{\rm{S}}{{\rm{O}}_{\rm{4}}}} \right]{\rm{Br}}$ and $\left[ {{\rm{Co}}{{\left( {{\rm{N}}{{\rm{H}}_{\rm{3}}}} \right)}_5}{\rm{Br}}} \right]{\rm{S}}{{\rm{O}}_{\rm{4}}}$.