Chapter 23, Problem 23.120P

(a)

Interpretation Introduction

Interpretation:

The wavelength of maximum absorption and the crystal field splitting energy have to be calculated.

Concept introduction:

Crystal field splitting: The energy gap between the splitting of d-orbitals of the metal ion in presence of ligands is known as the crystal field splitting $\text{(Δ)}$.  The magnitude of $\text{(Δ)}$ is depends on the nature of metal ions and the ligands.

The fact that colour depends on the ligand allows us to create a spectrochemical series, which ranks the ability of a ligand to split $\text{d-orbital}$ energies.

The relative magnitude of $\Delta$ for a series of octahedral complexes of a given metal ion.  As $\Delta$ increases, higher energies (shorter wavelengths) of light must be absorbed to excite electrons.

The spectrochemical series is

  $I^{-}<C{l}^{-}<F^{-}<O{H}^{-}<H_{2}O<SC{N}^{-}<en<N{O}_2^{-}<C{N}^{-}<CO$

Explanation of Solution

$\begin{array}{l}{\left[{\text{Cr(H}}_{\text{2}}{\text{O)}}_{\text{6}}\right]}^{\text{3+}}\text{562}\text{nm}\\ \text{E=}\frac{\text{hc}}{\text{λ}}\\ \text{=}\frac{\text{(6}{\text{.626×10}}^{\text{-34}}\text{J}·\text{s)(3}{\text{.00×10}}^{\text{8}}\text{m/s)}}{\text{562}\text{nm}}\left(\frac{\text{1}\text{nm}}{{\text{10}}^{\text{-9}}\text{m}}\right)\left(\frac{\text{6}{\text{.022×10}}^{\text{23}}}{\text{mol}}\right)\left(\frac{\text{1}\text{}\text{kJ}}{{\text{10}}^{\text{3}}\text{J}}\right)\\ \text{=}\text{212}\text{.99878}\\ \text{=}\text{213}\text{kJ/mol}\end{array}$

$\begin{array}{l}{\left[{\text{Cr(CN)}}_{\text{6}}\right]}^{\text{3-}}381\text{nm}\\ \text{E=}\frac{\text{hc}}{\text{λ}}\\ \text{=}\frac{\text{(6}{\text{.626×10}}^{\text{-34}}\text{J}·\text{s)(3}{\text{.00×10}}^{\text{8}}\text{m/s)}}{381\text{nm}}\left(\frac{\text{1}\text{nm}}{{\text{10}}^{\text{-9}}\text{m}}\right)\left(\frac{\text{6}{\text{.022×10}}^{\text{23}}}{\text{mol}}\right)\left(\frac{\text{1}\text{}\text{kJ}}{{\text{10}}^{\text{3}}\text{J}}\right)\\ \text{=}\text{314}\text{.187}\\ \text{=}314\text{kJ/mol}\end{array}$

$\begin{array}{l}{\left[{\text{Cr(Cl)}}_{\text{6}}\right]}^{\text{3-}}735\text{nm}\\ \text{E=}\frac{\text{hc}}{\text{λ}}\\ \text{=}\frac{\text{(6}{\text{.626×10}}^{\text{-34}}\text{J}·\text{s)(3}{\text{.00×10}}^{\text{8}}\text{m/s)}}{735\text{nm}}\left(\frac{\text{1}\text{nm}}{{\text{10}}^{\text{-9}}\text{m}}\right)\left(\frac{\text{6}{\text{.022×10}}^{\text{23}}}{\text{mol}}\right)\left(\frac{\text{1}\text{}\text{kJ}}{{\text{10}}^{\text{3}}\text{J}}\right)\\ \text{=}162.864\\ \text{=}163\text{kJ/mol}\end{array}$

$\begin{array}{l}{\left[{\text{Cr(NH}}_{\text{3}}{\text{)}}_{\text{6}}\right]}^{\text{3+}}462\text{nm}\\ \text{E=}\frac{\text{hc}}{\text{λ}}\\ \text{=}\frac{\text{(6}{\text{.626×10}}^{\text{-34}}\text{J}·\text{s)(3}{\text{.00×10}}^{\text{8}}\text{m/s)}}{462\text{nm}}\left(\frac{\text{1}\text{nm}}{{\text{10}}^{\text{-9}}\text{m}}\right)\left(\frac{\text{6}{\text{.022×10}}^{\text{23}}}{\text{mol}}\right)\left(\frac{\text{1}\text{}\text{kJ}}{{\text{10}}^{\text{3}}\text{J}}\right)\\ \text{=}259.1024\\ \text{=}259\text{kJ/mol}\end{array}$

$\begin{array}{l}{\left[{\text{Ir(NH}}_{\text{3}}{\text{)}}_{\text{6}}\right]}^{\text{3+}}244\text{nm}\\ \text{E=}\frac{\text{hc}}{\text{λ}}\\ \text{=}\frac{\text{(6}{\text{.626×10}}^{\text{-34}}\text{J}·\text{s)(3}{\text{.00×10}}^{\text{8}}\text{m/s)}}{244\text{nm}}\left(\frac{\text{1}\text{nm}}{{\text{10}}^{\text{-9}}\text{m}}\right)\left(\frac{\text{6}{\text{.022×10}}^{\text{23}}}{\text{mol}}\right)\left(\frac{\text{1}\text{}\text{kJ}}{{\text{10}}^{\text{3}}\text{J}}\right)\\ \text{=}490.5955\\ \text{=}491\text{kJ/mol}\end{array}$

$\begin{array}{l}{\left[{\text{Fe(H}}_2{\text{O)}}_{\text{6}}\right]}^{\text{2+}}966\text{nm}\\ \text{E=}\frac{\text{hc}}{\text{λ}}\\ \text{=}\frac{\text{(6}{\text{.626×10}}^{\text{-34}}\text{J}·\text{s)(3}{\text{.00×10}}^{\text{8}}\text{m/s)}}{966\text{nm}}\left(\frac{\text{1}\text{nm}}{{\text{10}}^{\text{-9}}\text{m}}\right)\left(\frac{\text{6}{\text{.022×10}}^{\text{23}}}{\text{mol}}\right)\left(\frac{\text{1}\text{}\text{kJ}}{{\text{10}}^{\text{3}}\text{J}}\right)\\ \text{=}123.9185\\ \text{=}124\text{kJ/mol}\end{array}$

$\begin{array}{l}{\left[{\text{Fe(H}}_2{\text{O)}}_{\text{6}}\right]}^{\text{2+}}730\text{nm}\\ \text{E=}\frac{\text{hc}}{\text{λ}}\\ \text{=}\frac{\text{(6}{\text{.626×10}}^{\text{-34}}\text{J}·\text{s)(3}{\text{.00×10}}^{\text{8}}\text{m/s)}}{730\text{nm}}\left(\frac{\text{1}\text{nm}}{{\text{10}}^{\text{-9}}\text{m}}\right)\left(\frac{\text{6}{\text{.022×10}}^{\text{23}}}{\text{mol}}\right)\left(\frac{\text{1}\text{}\text{kJ}}{{\text{10}}^{\text{3}}\text{J}}\right)\\ \text{=}163.9798\\ \text{=}164\text{kJ/mol}\end{array}$

$\begin{array}{l}{\left[{\text{Co(H}}_2{\text{O)}}_{\text{6}}\right]}^{\text{2+}}405\text{nm}\\ \text{E=}\frac{\text{hc}}{\text{λ}}\\ \text{=}\frac{\text{(6}{\text{.626×10}}^{\text{-34}}\text{J}·\text{s)(3}{\text{.00×10}}^{\text{8}}\text{m/s)}}{405\text{nm}}\left(\frac{\text{1}\text{nm}}{{\text{10}}^{\text{-9}}\text{m}}\right)\left(\frac{\text{6}{\text{.022×10}}^{\text{23}}}{\text{mol}}\right)\left(\frac{\text{1}\text{}\text{kJ}}{{\text{10}}^{\text{3}}\text{J}}\right)\\ \text{=}296.5686\\ \text{=}296\text{kJ/mol}\end{array}$

$\begin{array}{l}{\left[{\text{Rh(NH}}_{\text{3}}{\text{)}}_{\text{6}}\right]}^{\text{3+}}295\text{nm}\\ \text{E=}\frac{\text{hc}}{\text{λ}}\\ \text{=}\frac{\text{(6}{\text{.626×10}}^{\text{-34}}\text{J}·\text{s)(3}{\text{.00×10}}^{\text{8}}\text{m/s)}}{295\text{nm}}\left(\frac{\text{1}\text{nm}}{{\text{10}}^{\text{-9}}\text{m}}\right)\left(\frac{\text{6}{\text{.022×10}}^{\text{23}}}{\text{mol}}\right)\left(\frac{\text{1}\text{}\text{kJ}}{{\text{10}}^{\text{3}}\text{J}}\right)\\ \text{=}405.7807\\ \text{=}406\text{kJ/mol}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The spectrochemical series for the ligand in the Chromium complex has to be written.

Concept introduction:

The fact that colour depends on the ligand allows us to create a spectrochemical series, which ranks the ability of a ligand to split $\text{d-orbital}$ energies.

The relative magnitude of $\Delta$ for a series of octahedral complexes of a given metal ion.  As $\Delta$ increases, higher energies (shorter wavelengths) of light must be absorbed to excite electrons.

The spectrochemical series is

  $I^{-}<C{l}^{-}<F^{-}<O{H}^{-}<H_{2}O<SC{N}^{-}<en<N{O}_2^{-}<C{N}^{-}<CO$

Explanation of Solution

Comparing the energies of the chromium complexes, we find the energy increases in the order:

${\text{Cl}}^{\text{–}}{\text{< H}}_{\text{2}}{\text{O < NH}}_{\text{3}}{\text{ < CN}}^{\text{–}}$

This is an abbreviated spectrochemical series.

(c)

Interpretation Introduction

Interpretation:

The Iron data is affect the oxidation state has to be explained.

Concept introduction:

Oxidation number: (Oxidation state) A number equal to the magnitude of the charge an atom would have if its shared electrons were transferred to the atom that attracts them more strongly.

The spectrochemical series is

  $I^{-}<C{l}^{-}<F^{-}<O{H}^{-}<H_{2}O<SC{N}^{-}<en<N{O}_2^{-}<C{N}^{-}<CO$

Explanation of Solution

The higher the oxidation number the greater the crystal field splitting $(\Delta ).$

  $\begin{array}{l}\left({\text{Fe}}^{\text{3+}}\text{ vs}{\text{. Fe}}^{\text{2+}}\right)\\ {\text{Fe}}^{\text{2+}}\text{=}{\text{3d}}^{\text{6}}{\text{4s}}^{\text{0}}\\ {\text{eg}}^{\text{2}}\text{and}{\text{t}}_{\text{2}}{\text{g}}^{\text{4}}\text{=-0}{\text{.4Δ}}_{\text{o}}\\ {\text{Fe}}^{\text{3+}}\text{=}{\text{3d}}^{\text{5}}{\text{4s}}^{\text{0}}\\ {\text{Fe}}^{\text{3+}}{\text{eg}}^{\text{2}}\text{and}{\text{t}}_{\text{2}}{\text{g}}^{\text{3}}{\text{=0Δ}}_{\text{o}}\end{array}$

(d)

Interpretation Introduction

Interpretation:

The Cobalt, Rhodium and Iron atom affect the periodic number has to be discussed.

Concept introduction:

The fact that colour depends on the ligand allows us to create a spectrochemical series, which ranks the ability of a ligand to split $\text{d-orbital}$ energies.

The relative magnitude of $\Delta$ for a series of octahedral complexes of a given metal ion.  As $\Delta$ increases, higher energies (shorter wavelengths) of light must be absorbed to excite electrons.

The spectrochemical series is

  $I^{-}<C{l}^{-}<F^{-}<O{H}^{-}<H_{2}O<SC{N}^{-}<en<N{O}_2^{-}<C{N}^{-}<CO$

Explanation of Solution

When moving down a column on the periodic table, in this case from $\text{Co to Rh to Ir,}$ the greater the energy required, the greater the crystal field splitting $(\Delta ).$