Chapter 23, Problem 23.90P

(a)

Interpretation Introduction

Interpretation:

The orbital splitting diagram given ${\left[MoC{l}_6\right]}^3{}^{-}$ complex has to be drawn using the spectrochemical series.

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.

In an octahedral complex, the six ligands approach the central metal atom lying at the symmetry along the Cartesian axes.  The orbitals lying along axes $d_{z^2}and{d}_{z^2-y^2}$ get repelled more strongly by the negative ligands and are raised in energy relative to the average energy in between the axes $d_{xy},d_{xz}and{d}_{zy}$

Splitting of five d-orbitals in an octahedral crystals field is as follows:

Figure 1

Explanation of Solution

Electron configuration of $Mo:\left[Kr\right]5s^{1}4d^5$

Charge on $Mo$: Each chloride ligand has a –1 charge for a total charge of –6, so $Mo$ has a +3 charge to make the overall complex charge equal to –3.

Electron configuration of $M{o}^{3+}:\left[Kr\right]4d^3$

Six ligands indicate an octahedral arrangement. Using Hund’s rule, fill the lower energy $t_{2}g$ orbitals first, filling empty orbitals before pairing electrons within an orbital.

(b)

Interpretation Introduction

Interpretation:

The orbital splitting diagram given ${\left[Ni{(H_{2}O)}_6\right]}^{2+}$ complex has to be drawn using the spectrochemical series.

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.

In an octahedral complex, the six ligands approach the central metal atom lying at the symmetry along the Cartesian axes.  The orbitals lying along axes $d_{z^2}and{d}_{z^2-y^2}$ get repelled more strongly by the negative ligands and are raised in energy relative to the average energy in between the axes $d_{xy},d_{xz}and{d}_{zy}$

Splitting of five d-orbitals in an octahedral crystals field is as follows:

Figure 1

Explanation of Solution

Electron configuration of $Ni:\left[Ar\right]4s^{2}3d^8$

Charge on $Ni$: The aqua ligands are neutral, so the charge on $Ni$ is +2.

Electron configuration of $N{i}^{2+}:\left[Ar\right]3d^8$

Six ligands indicate an octahedral arrangement.  Use Hund’s rule to fill the orbitals.

$H_{2}O$ is a weak-field ligand, so the splitting energy, $\Delta ,$ is not large enough to overcome the resistance to electron pairing.  One electron occupies each of the five d orbitals before pairing in the $t_{2}g$ orbitals, and the complex is called high-spin.

(c)

Interpretation Introduction

Interpretation:

The orbital splitting diagram given ${\left[Ni{(CN)}_4\right]}^2{}^{-}$ complex has to be drawn using the spectrochemical series.

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.

Splitting of five d-orbitals in a square planer crystals field is as follows:

Figure 1

Explanation of Solution

Electron configuration of $Ni:\left[Ar\right]4s^{2}3d^8$

Charge on $Ni$: Each cyanide ligand has a –1 charge for a total charge of –4, so $Ni$ has a +2 charge to make the overall complex charge equal to –2.

Electron configuration of $N{i}^{2+}:\left[Ar\right]3d^8$

The coordination number is 4 and the complex is square planar.

The complex is low-spin because $C{N}^{–}$ is a strong-field ligand.  Electrons pair in one set of orbitals before occupying orbitals of higher energy.