Chapter 9, Problem 9D.23E

(a)

Interpretation Introduction

Interpretation:

The ligand field splitting has to be estimated.

Concept Introduction:

The splitting energy of the complex can be calculated using the formula,

  ${\text{Δ}}_{\text{o}}\text{=}\frac{\text{hc}}{\text{λ}}$

Where

${\text{Δ}}_{\text{o}}$ is the splitting energy of the complex.

$\text{λ}$ is the wavelength.

h is the Planck’s constant $\left(\text{6}{\text{.626×10}}^{\text{-34}}\text{J}\text{s}\right)\text{.}$

c is the velocity of the light $\left({\text{3×10}}^{\text{8}}\text{m}{\text{s}}^{\text{-1}}\right).$

Answer to Problem 9D.23E

1. (i) The ligand field splitting of ${\left[\text{Cr}{\left(\text{Cl}\right)}_{\text{6}}\right]}^{\text{3-}}$ is $\text{161}\text{.69}\text{kJ}{\text{mol}}^{\text{-1}}.$
2. (ii) The ligand field splitting of ${\left[\text{Cr}{\left({\text{NH}}_{\text{3}}\right)}_{\text{6}}\right]}^{\text{3+}}$ is $\text{260}\text{.12}\text{kJ}{\text{mol}}^{\text{-1}}\text{.}$
3. (iii) The ligand field splitting of ${\left[\text{Cr}{\left({\text{H}}_{\text{2}}\text{O}\right)}_{\text{6}}\right]}^{\text{3+}}$ is $\text{208}\text{.09}\text{kJ}{\text{mol}}^{\text{-1}}\text{.}$

Explanation of Solution

(i).

The ligand field splitting of ${\left[\text{Cr}{\left(\text{Cl}\right)}_{\text{6}}\right]}^{\text{3-}}$ can be calculated as,

  ${\text{Δ}}_{\text{o}}\text{=}\frac{\text{hc}}{\text{λ}}$

  ${\text{Δ}}_{\text{o}}\text{=}\frac{\left(\text{6}{\text{.626×10}}^{\text{-34}}\text{J}\text{.}\text{s}\right)\left({\text{3×10}}^{\text{8}}\text{m}{\text{s}}^{\text{-1}}\right)}{\text{740}\text{nm×}\frac{\text{1}\text{m}}{{\text{10}}^{\text{9}}\text{nm}}}$

  ${\text{Δ}}_{\text{o}}\text{=}\frac{\text{1}{\text{.987×10}}^{\text{-25}}\text{J}}{\text{7}{\text{.4×10}}^{\text{-7}}}$

  ${\text{Δ}}_{\text{o}}\text{=0}{\text{.2685×10}}^{\text{-18}}\text{J}\text{.}$

The splitting energy in terms of kilojoulespermole is,

  ${\text{Δ}}_{\text{o}}\text{=0}{\text{.2685×10}}^{\text{-18}}\text{J×}\frac{\text{1kJ}}{\text{1000J}}\text{×}\frac{\text{6}{\text{.022×10}}^{\text{23}}}{\text{1}\text{mol}}$

  ${\text{Δ}}_{\text{o}}\text{=161}\text{.69}\text{kJ}{\text{mol}}^{\text{-1}}$

The ligand field splitting of ${\left[\text{Cr}{\left(\text{Cl}\right)}_{\text{6}}\right]}^{\text{3-}}$ is $\text{161}\text{.69}\text{kJ}{\text{mol}}^{\text{-1}}.$

(ii)

The ligand field splitting of ${\left[\text{Cr}{\left({\text{NH}}_{\text{3}}\right)}_{\text{6}}\right]}^{\text{3+}}$ can be calculated as,

  ${\text{Δ}}_{\text{o}}\text{=}\frac{\text{hc}}{\text{λ}}$

  ${\text{Δ}}_{\text{o}}\text{=}\frac{\left(\text{6}{\text{.626×10}}^{\text{-34}}\text{J}\text{.}\text{s}\right)\left({\text{3×10}}^{\text{8}}\text{m}{\text{s}}^{\text{-1}}\right)}{\text{460}\text{nm×}\frac{\text{1}\text{m}}{{\text{10}}^{\text{9}}\text{nm}}}$

  ${\text{Δ}}_{\text{o}}\text{=}\frac{\text{1}{\text{.987×10}}^{\text{-25}}\text{J}}{\text{4}{\text{.6×10}}^{\text{-7}}}$

  ${\text{Δ}}_{\text{o}}\text{=0}{\text{.4319×10}}^{\text{-18}}\text{J}\text{.}$

The splitting energy in terms of kilojoulespermole is,

  ${\text{Δ}}_{\text{o}}\text{=0}{\text{.4319×10}}^{\text{-18}}\text{J×}\frac{\text{1kJ}}{\text{1000J}}\text{×}\frac{\text{6}{\text{.022×10}}^{\text{23}}}{\text{1}\text{mol}}$

  ${\text{Δ}}_{\text{o}}\text{=260}\text{.12}\text{kJ}{\text{mol}}^{\text{-1}}$

The ligand field splitting of ${\left[\text{Cr}{\left({\text{NH}}_{\text{3}}\right)}_{\text{6}}\right]}^{\text{3+}}$ is $\text{260}\text{.12}\text{kJ}{\text{mol}}^{\text{-1}}\text{.}$

(iii)

The ligand field splitting of ${\left[\text{Cr}{\left({\text{H}}_{\text{2}}\text{O}\right)}_{\text{6}}\right]}^{\text{3+}}$ can be calculated as,

  ${\text{Δ}}_{\text{o}}\text{=}\frac{\text{hc}}{\text{λ}}$

  ${\text{Δ}}_{\text{o}}\text{=}\frac{\left(\text{6}{\text{.626×10}}^{\text{-34}}\text{J}\text{.}\text{s}\right)\left({\text{3×10}}^{\text{8}}\text{m}{\text{s}}^{\text{-1}}\right)}{575\text{nm×}\frac{\text{1}\text{m}}{{\text{10}}^{\text{9}}\text{nm}}}$

  ${\text{Δ}}_{\text{o}}\text{=}\frac{\text{1}{\text{.987×10}}^{\text{-25}}\text{J}}{\text{5}{\text{.6×10}}^{\text{-7}}}$

  ${\text{Δ}}_{\text{o}}\text{=0}{\text{.3455×10}}^{\text{-18}}\text{J}\text{.}$

The splitting energy in terms of kilojoulespermole is,

  ${\text{Δ}}_{\text{o}}\text{=0}{\text{.3455×10}}^{\text{-18}}\text{J×}\frac{\text{1kJ}}{\text{1000J}}\text{×}\frac{\text{6}{\text{.022×10}}^{\text{23}}}{\text{1}\text{mol}}$

  ${\text{Δ}}_{\text{o}}\text{=208}\text{.09}\text{kJ}{\text{mol}}^{\text{-1}}$

The ligand field splitting of ${\left[\text{Cr}{\left({\text{H}}_{\text{2}}\text{O}\right)}_{\text{6}}\right]}^{\text{3+}}$ is $\text{208}\text{.09}\text{kJ}{\text{mol}}^{\text{-1}}\text{.}$

(b)

Interpretation Introduction

Interpretation:

The ligands have to be arranged in the order of increasing field strength.

Answer to Problem 9D.23E

The order of ligands in the order of increasing field strength is,

  ${\left[\text{Cr}{\left(\text{Cl}\right)}_{\text{6}}\right]}^{\text{3-}}\text{<}{\left[\text{Cr}{\left({\text{OH}}_{\text{2}}\right)}_{\text{6}}\right]}^{\text{3+}}\text{<}{\left[\text{Cr}{\left({\text{NH}}_{\text{3}}\right)}_{\text{6}}\right]}^{\text{3+}}$

Explanation of Solution

The ligand field splitting energy of ${\left[\text{Cr}{\left(\text{Cl}\right)}_{\text{6}}\right]}^{\text{3-}}$ is $\text{161}\text{.69}\text{kJ}{\text{mol}}^{\text{-1}}$, ${\left[\text{Cr}{\left({\text{NH}}_{\text{3}}\right)}_{\text{6}}\right]}^{\text{3+}}$ is $\text{260}\text{.12}\text{kJ}{\text{mol}}^{\text{-1}}$ and ${\left[\text{Cr}{\left({\text{H}}_{\text{2}}\text{O}\right)}_{\text{6}}\right]}^{\text{3+}}$ is $\text{208}\text{.09}\text{kJ}{\text{mol}}^{\text{-1}}.$  Based on the value of splitting energy the increasing order of the ligands is,

  ${\left[\text{Cr}{\left(\text{Cl}\right)}_{\text{6}}\right]}^{\text{3-}}\text{<}{\left[\text{Cr}{\left({\text{OH}}_{\text{2}}\right)}_{\text{6}}\right]}^{\text{3+}}\text{<}{\left[\text{Cr}{\left({\text{NH}}_{\text{3}}\right)}_{\text{6}}\right]}^{\text{3+}}$