Chapter 22, Problem 22.80QP

(a)

Interpretation Introduction

Interpretation: To determine the magnetic property of the given compounds.

Concept Introduction:

Spectrochemical series: The list of ligands arranged in an ascending order of $\text{(Δ)}$(the splitting of d-orbitals in presence of various ligands).

$\begin{array}{l}\stackrel{{}^{{\text{I}}^{\text{-}}\text{<}{\text{Br}}^{\text{-}}\text{<}{\text{SCN}}^{\text{-}}\text{<}{\text{Cl}}^{\text{-}}\text{<}{\text{S}}^{\text{2-}}\text{<}{\text{F}}^{\text{-}}\text{<}{\text{OH}}^{\text{-}}\text{<}{\text{O}}^{\text{2-}}\text{<}{\text{H}}_{\text{2}}\text{O}\text{<}{\text{NCS}}^{\text{-}}\text{<}{\text{edta}}^{\text{4-}}\text{<}{\text{NH}}_{\text{3}}\text{< en}\text{<}{\text{NO}}^{\text{2-}}\text{<}{\text{CN}}^{\text{-}}\text{<}\text{CO}}}{\to }\\ {}_{{}^{{}^{\begin{array}{l}\text{weak-field}\text{increasing(Δ)}\text{strong-field}\\ \text{ligands}\text{ligands}\end{array}}}}\end{array}$

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.

To Identify: To determine the magnetic property of the given compounds.

Answer to Problem 22.80QP

The ${\text{Cu}}^{\text{3+}}$ion is unstable.

Explanation of Solution

Interpret the given information.

The ${\text{Cu}}^{\text{3+}}$ion is unstable because of the high charge density on the small cation and the coordination of the small and highly charged fluoride ligands is possible but in the biological compounds the ligands to extremely large to disperse the charge and get stabilized.

(b)

Interpretation Introduction

Interpretation: To determine the magnetic property of the given compounds.

Concept Introduction:

Spectrochemical series: The list of ligands arranged in an ascending order of $\text{(Δ)}$(the splitting of d-orbitals in presence of various ligands).

$\begin{array}{l}\stackrel{{}^{{\text{I}}^{\text{-}}\text{<}{\text{Br}}^{\text{-}}\text{<}{\text{SCN}}^{\text{-}}\text{<}{\text{Cl}}^{\text{-}}\text{<}{\text{S}}^{\text{2-}}\text{<}{\text{F}}^{\text{-}}\text{<}{\text{OH}}^{\text{-}}\text{<}{\text{O}}^{\text{2-}}\text{<}{\text{H}}_{\text{2}}\text{O}\text{<}{\text{NCS}}^{\text{-}}\text{<}{\text{edta}}^{\text{4-}}\text{<}{\text{NH}}_{\text{3}}\text{< en}\text{<}{\text{NO}}^{\text{2-}}\text{<}{\text{CN}}^{\text{-}}\text{<}\text{CO}}}{\to }\\ {}_{{}^{{}^{\begin{array}{l}\text{weak-field}\text{increasing(Δ)}\text{strong-field}\\ \text{ligands}\text{ligands}\end{array}}}}\end{array}$

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.

To Identify: To determine the magnetic property of the given compounds.

Answer to Problem 22.80QP

The systematic name of the given compound ${\text{K}}_3{\text{CuF}}_6$ is $\text{Potassium}\text{hexafluorocuprate(III)}$. Possesses octahedral geometry and paramagnetic.

Explanation of Solution

Interpret the given information.

The systematic name of the given compound ${\text{K}}_3{\text{CuF}}_6$ is $\text{Potassium}\text{hexafluorocuprate(III)}$.

The six ligands around the metal ion give octahedral geometry.

Magnetic property of the compound:

The high field complex ion possesses eight d-electrons and the two electrons are unpaired and exhibits paramagnetic.

(c)

Interpretation Introduction

Interpretation: To determine the magnetic property of the given compounds.

Concept Introduction:

Spectrochemical series: The list of ligands arranged in an ascending order of $\text{(Δ)}$(the splitting of d-orbitals in presence of various ligands).

$\begin{array}{l}\stackrel{{}^{{\text{I}}^{\text{-}}\text{<}{\text{Br}}^{\text{-}}\text{<}{\text{SCN}}^{\text{-}}\text{<}{\text{Cl}}^{\text{-}}\text{<}{\text{S}}^{\text{2-}}\text{<}{\text{F}}^{\text{-}}\text{<}{\text{OH}}^{\text{-}}\text{<}{\text{O}}^{\text{2-}}\text{<}{\text{H}}_{\text{2}}\text{O}\text{<}{\text{NCS}}^{\text{-}}\text{<}{\text{edta}}^{\text{4-}}\text{<}{\text{NH}}_{\text{3}}\text{< en}\text{<}{\text{NO}}^{\text{2-}}\text{<}{\text{CN}}^{\text{-}}\text{<}\text{CO}}}{\to }\\ {}_{{}^{{}^{\begin{array}{l}\text{weak-field}\text{increasing(Δ)}\text{strong-field}\\ \text{ligands}\text{ligands}\end{array}}}}\end{array}$

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.

To Identify: To determine the magnetic property of the given compounds.

Answer to Problem 22.80QP

The square-planar geometry will be diamagnetic in nature.

Explanation of Solution

Interpret the given information.

In complex ion of ${\text{Cu}}^{3+}$, the square-planar geometry is;

In complex ion of ${\text{Cu}}^{3+}$, the square-planar geometry possesses no unpaired electrons and the complex exhibit diamagnetic.