Chapter 19, Problem 84AE

Interpretation Introduction

Interpretation: The structure of three salts formed needs to be determined if 0.27 g of octahedral complex of Cr reacts with strong dehydrating agent and lost mass of 0.036 g. Reaction of 270 mg of second salt with same dehydrating agent show mass loss of 18 mg. The 3^rd salt did not lose any mass while treating with dehydrating agent.

  ${\text{Co(en)}}_{\text{3}}^{\text{3+}}{\text{+ e}}^{\text{-}}\stackrel{}{\to }{\text{Co(en)}}_{\text{3}}^{\text{2+}}$

Given:

  $\begin{array}{l}{\text{Co}}_{}^{\text{3+}}{\text{+ e}}^{\text{-}}\stackrel{}{\to }{\text{Co}}_{}^{\text{2+}}\text{ ξ°=1}\text{.82V}\\ {\text{Co(en)}}_{\text{3}}^{\text{3+}}{\text{ = K}}_{\text{f}}\text{ = 2}{\text{.0×10}}^{\text{47}}\\ {\text{ Co(en)}}_{\text{3}}^{\text{2+}}{\text{= K}}_{\text{f}}\text{ = 1}{\text{.5×10}}^{\text{12}}\end{array}$

Concept Introduction:

Crystal field theory is the theory given to explain the bonding in the coordination complexes. As ligand approaches towards the metal ion, the d-orbital of metal ion divide according to the energy of metal ion. On the basis of energy and degeneracy, the d-orbital can be classified as ${\text{t}}_{\text{2g}}$ and ${\text{e}}_{\text{g}}$ orbitals. The electrons will fill according to their energy levels.

In octahedral complex, the ${\text{t}}_{\text{2g}}$ orbitals are lower energy orbitals in d-orbital splitting whereas ${\text{e}}_{\text{g}}$ orbitals are higher energy orbitals.

Answer to Problem 84AE

  

Explanation of Solution

Given information:

Addition of excess of aqueous silver nitrate to 100.0 mL portion of 0.100 M solution of each salt forms silver chloride,

• 1^st solution= 1430 mg AgCl
• 2^nd solution = 2870 mg AgCl
• 3^rd solution = 4300 mg AgCl

Two salts are green and one is violet here.

Compound 1:

  ${\text{Moles CrCl}}_{\text{3}}{\text{.6H}}_{\text{2}}\text{O = 0}\text{.27 g × }\frac{\text{1 mole}}{\text{266}\text{.5 g }}\text{=1}{\text{.0×10}}^{\text{-3}}{\text{moles CrCl}}_{\text{3}}{\text{.6H}}_{\text{2}}\text{O}$

Calculate moles of water of hydration:

  ${\text{Moles of H}}_{\text{2}}\text{O = 0}\text{.036 g × }\frac{\text{1 mole}}{\text{18}\text{.0 g }}\text{= 0}{\text{.002 moles H}}_{\text{2}}\text{O}$

  $\frac{{\text{Moles of H}}_{\text{2}}\text{O }}{\text{mole compound}}\text{ = }\frac{\text{0}\text{.002 moles}}{\text{0}\text{.001 moles }}\text{= 2}\text{.0 }$

With 2 moles of water; the formula must be ${\text{[Cr(H}}_{\text{2}}{\text{O)}}_{\text{4}}{\text{Cl}}_{\text{2}}\text{]Cl}{\text{.2H}}_{\text{2}}\text{O}$ .

Mass of AgCl = 1430 mg

Check the mass of Cl in ${\text{[Cr(H}}_{\text{2}}{\text{O)}}_{\text{4}}{\text{Cl}}_{\text{2}}\text{]Cl}{\text{.2H}}_{\text{2}}\text{O}$ :

  $\text{Moles of AgCl =0}\text{.100 L × }\frac{\text{0}{\text{.100 moles [Cr(H}}_{\text{2}}{\text{O)}}_{\text{4}}{\text{Cl}}_{\text{2}}\text{]Cl }}{\text{1 L }}\text{× }\frac{{\text{ 1 mole Cl}}^{\text{- }}}{{\text{1 mole [Cr(H}}_{\text{2}}{\text{O)}}_{\text{4}}{\text{Cl}}_{\text{2}}\text{]Cl}}\times \frac{\text{ 1 mole AgCl }}{\text{1 mole Cl-}}\times \frac{\text{ 143}\text{.4 g }}{\text{1 mole }}\text{= 1}\text{.43 g =1430 mg AgCl}$ Compound 2:

  ${\text{Moles CrCl}}_{\text{3}}{\text{.6H}}_{\text{2}}\text{O = 0}\text{.27 g × }\frac{\text{1 mole}}{\text{266}\text{.5 g }}\text{=1}{\text{.0×10}}^{\text{-3}}{\text{moles CrCl}}_{\text{3}}{\text{.6H}}_{\text{2}}\text{O}$

Calculate moles of water of hydration:

  ${\text{Moles of H}}_{\text{2}}\text{O = 0}\text{.018 g × }\frac{\text{1 mole}}{\text{18}\text{.0 g }}\text{= 0}{\text{.001 moles H}}_{\text{2}}\text{O}$

  $\frac{{\text{Moles of H}}_{\text{2}}\text{O }}{\text{mole compound}}\text{ = }\frac{\text{0}\text{.001 moles}}{\text{0}\text{.001 moles }}\text{= 1}\text{.0 }$

With 1 moles of water; the formula must be ${\text{[Cr(H}}_{\text{2}}{\text{O)}}_5{\text{Cl]Cl}}_{\text{2}}{\text{.H}}_{\text{2}}\text{O}$ .

Mass of AgCl = 2870 mg

Check the mass of Cl in ${\text{[Cr(H}}_{\text{2}}{\text{O)}}_5{\text{Cl]Cl}}_{\text{2}}{\text{.H}}_{\text{2}}\text{O}$ :

  $\text{Moles of AgCl =0}{\text{.0200 moles Cl}}^{\text{-}}\text{ × }\frac{\text{1 mole AgCl }}{{\text{1 mole Cl}}^{\text{-}}}\times \frac{\text{ 143}\text{.4 g AgCl }}{\text{1 mole AgCl }}\text{= 2}\text{.87 g = 2870 mg AgCl}$

Compound 3:

  ${\text{Moles CrCl}}_{\text{3}}{\text{.6H}}_{\text{2}}\text{O = 0}\text{.27 g × }\frac{\text{1 mole}}{\text{266}\text{.5 g }}\text{=1}{\text{.0×10}}^{\text{-3}}{\text{moles CrCl}}_{\text{3}}{\text{.6H}}_{\text{2}}\text{O}$

Moles of water of hydration is 0 as no water lose is there. Thus the formula must be ${\text{[Cr(H}}_{\text{2}}{\text{O)}}_{\text{6}}{\text{]Cl}}_{\text{3}}$ .

Mass of AgCl = 4300 mg

Check the mass of Cl in ${\text{[Cr(H}}_{\text{2}}{\text{O)}}_{\text{6}}{\text{]Cl}}_{\text{3}}$ :

  $\text{Moles of AgCl =0}{\text{.0300 moles Cl}}^{\text{-}}\text{ × }\frac{\text{1 mole AgCl }}{{\text{1 mole Cl}}^{\text{-}}}\times \frac{\text{ 143}\text{.4 g AgCl }}{\text{1 mole AgCl }}\text{= 4}\text{.30 g = 4300 mg AgCl}$

Hence the structure of complexes should be:

  

Conclusion

Thus,