Chapter 20, Problem 59E

Interpretation Introduction

Interpretation: Three test tubes containing different complex ions are given. The color of light absorbed by each given ion is to be predicted. The identity of complex ion in each test tube is to be predicted.

Concept introduction: The capacity to absorb light by complex ion is directly proportional to crystal field splitting energy and energy absorbed is inversely proportional to the wavelength.

To determine: The color of light absorbed by each ion $\text{Cr}{\left({\text{H}}_2\text{O}\right)}_{\text{6}}{}^{3+},\text{Cr}{\left({\text{NH}}_3\right)}_{\text{6}}{}^{3-},\text{Cr}{\left({\text{H}}_2\text{O}\right)}_{\text{4}}{\text{Cl}}_{\text{2}}{}^{+}$ and the identity of complex ion in each test tube.

The order of increasing crystal field splitting energy for complex ions $\text{Cr}{\left({\text{H}}_2\text{O}\right)}_{\text{6}}{}^{3+},\text{Cr}{\left({\text{NH}}_3\right)}_{\text{6}}{}^{3-},\text{Cr}{\left({\text{H}}_2\text{O}\right)}_{\text{4}}{\text{Cl}}_{\text{2}}{}^{+}$ is,

$\text{Cr}{\left({\text{NH}}_3\right)}_{\text{6}}{}^{3-}>\text{Cr}{\left({\text{H}}_2\text{O}\right)}_{\text{6}}{}^{3+}>\text{Cr}{\left({\text{H}}_2\text{O}\right)}_{\text{4}}{\text{Cl}}_{\text{2}}{}^{+}$

Answer to Problem 59E

Answer

Light absorbed by $\text{Cr}{\left({\text{H}}_2\text{O}\right)}_{\text{4}}{\text{Cl}}_{\text{2}}{}^{+}$ is red. Light absorbed by $\text{Cr}{\left({\text{H}}_2\text{O}\right)}_{\text{6}}{}^{3+}$ is yellow-green. Light absorbed by $\text{Cr}{\left({\text{NH}}_3\right)}_{\text{6}}{}^{3-}$ is blue.

The violet color test tube contains $\text{Cr}{\left({\text{H}}_2\text{O}\right)}_{\text{6}}{}^{3+}$. The yellow color test tube contains $\text{Cr}{\left({\text{NH}}_3\right)}_{\text{6}}{}^{3-}$. The green color test tube contains $\text{Cr}{\left({\text{H}}_2\text{O}\right)}_{\text{4}}{\text{Cl}}_{\text{2}}{}^{+}$.

Explanation of Solution

Explanation

The spectrochemical series (crystal field splitting energy) of strong to weak ligand is,

${\text{CN}}^{-}>{\text{NO}}_{\text{2}}{}^{-}>\text{en}>{\text{NH}}_{\text{3}}>{\text{H}}_{\text{2}}\text{O}>{\text{OH}}^{-}>{\text{F}}^{-}>{\text{Cl}}^{-}>{\text{SCN}}^{-}>{\text{Br}}^{-}$

Hence, order of increasing crystal field splitting energy for complex ions $\text{Cr}{\left({\text{H}}_2\text{O}\right)}_{\text{6}}{}^{3+},\text{Cr}{\left({\text{NH}}_3\right)}_{\text{6}}{}^{3-},\text{Cr}{\left({\text{H}}_2\text{O}\right)}_{\text{4}}{\text{Cl}}_{\text{2}}{}^{+}$ is,

$\text{Cr}{\left({\text{NH}}_3\right)}_{\text{6}}{}^{3-}>\text{Cr}{\left({\text{H}}_2\text{O}\right)}_{\text{6}}{}^{3+}>\text{Cr}{\left({\text{H}}_2\text{O}\right)}_{\text{4}}{\text{Cl}}_{\text{2}}{}^{+}$

The energy absorbed by light is given by the expression,

$E=\frac{hc}{\lambda }$

Where,

• $E$ is the energy absorbed.
• $h$ is the Plank’s constant.
• $c$ is the velocity of light.

It is clear from the above equation that, energy absorbed is inversely proportional to the wavelength. It means that higher the energy, lower will be the wavelength.

Since, order of increasing crystal field splitting energy for complex ions $\text{Cr}{\left({\text{H}}_2\text{O}\right)}_{\text{6}}{}^{3+},\text{Cr}{\left({\text{NH}}_3\right)}_{\text{6}}{}^{3-},\text{Cr}{\left({\text{H}}_2\text{O}\right)}_{\text{4}}{\text{Cl}}_{\text{2}}{}^{+}$ is,

$\text{Cr}{\left({\text{NH}}_3\right)}_{\text{6}}{}^{3-}>\text{Cr}{\left({\text{H}}_2\text{O}\right)}_{\text{6}}{}^{3+}>\text{Cr}{\left({\text{H}}_2\text{O}\right)}_{\text{4}}{\text{Cl}}_{\text{2}}{}^{+}$

Therefore, order of increasing wavelength of light absorbed for complex ions $\text{Cr}{\left({\text{H}}_2\text{O}\right)}_{\text{6}}{}^{3+},\text{Cr}{\left({\text{NH}}_3\right)}_{\text{6}}{}^{3-},\text{Cr}{\left({\text{H}}_2\text{O}\right)}_{\text{4}}{\text{Cl}}_{\text{2}}{}^{+}$ is,

$\text{Cr}{\left({\text{H}}_2\text{O}\right)}_{\text{4}}{\text{Cl}}_{\text{2}}{}^{+}>\text{Cr}{\left({\text{H}}_2\text{O}\right)}_{\text{6}}{}^{3+}>\text{Cr}{\left({\text{NH}}_3\right)}_{\text{6}}{}^{3-}$

Hence, the complex ion $\text{Cr}{\left({\text{H}}_2\text{O}\right)}_{\text{4}}{\text{Cl}}_{\text{2}}{}^{+}$ will absorb the light at higher wavelength whereas complex ion $\text{Cr}{\left({\text{NH}}_3\right)}_{\text{6}}{}^{3-}$ will absorb the light at lower wavelength.

These complex ions absorb light in visible spectrum.

Refer to Table $\text{20-16}$.

The light absorbed by complex ions is,

Complex	Absorbed color	Observed color
$\text{Cr}{\left({\text{H}}_2\text{O}\right)}_{\text{4}}{\text{Cl}}_{\text{2}}{}^{+}$	Red	Green
$\text{Cr}{\left({\text{H}}_2\text{O}\right)}_{\text{6}}{}^{3+}$	Yellow-green	Violet
$\text{Cr}{\left({\text{NH}}_3\right)}_{\text{6}}{}^{3-}$	Blue	Yellow

The given test tubes are,

Figure 1

From the above table, it is clear that,

• The violet color test tube contains $\text{Cr}{\left({\text{H}}_2\text{O}\right)}_{\text{6}}{}^{3+}$.
• The yellow color test tube contains $\text{Cr}{\left({\text{NH}}_3\right)}_{\text{6}}{}^{3-}$.
• The green color test tube contains $\text{Cr}{\left({\text{H}}_2\text{O}\right)}_{\text{4}}{\text{Cl}}_{\text{2}}{}^{+}$.

Conclusion

Conclusion

The color of light absorbed by each ion $\text{Cr}{\left({\text{H}}_2\text{O}\right)}_{\text{6}}{}^{3+},\text{Cr}{\left({\text{NH}}_3\right)}_{\text{6}}{}^{3-},\text{Cr}{\left({\text{H}}_2\text{O}\right)}_{\text{4}}{\text{Cl}}_{\text{2}}{}^{+}$ and the identity of complex ion in each test tube is,

Complex	Absorbed color	Observed color
$\text{Cr}{\left({\text{H}}_2\text{O}\right)}_{\text{4}}{\text{Cl}}_{\text{2}}{}^{+}$	Red	Green
$\text{Cr}{\left({\text{H}}_2\text{O}\right)}_{\text{6}}{}^{3+}$	Yellow-green	Violet
$\text{Cr}{\left({\text{NH}}_3\right)}_{\text{6}}{}^{3-}$	Blue	Yellow