Chapter 19, Problem 41QAP

Give the number of unpaired electrons in octahedral complexes with strong-field ligands for
(a) Rh³+
(b) Mn³+
(c) Ag+
(d) Pt⁴+
(e) Au³+

Interpretation Introduction

(a)

Interpretation:

The number of unpaired electrons in octahedral complexes with strong field ligands for the metal ion ${\text{Rh}}^{3+}$ needs to be determined.

Concept introduction:

Coordination compounds are those in which a transition metal atom is bonded to a ligand which can be neutral, cation or anion. A transition metal cation has no outer s- electrons available for bonding, rather the inner d -electrons (in case of 3d transition metal elements) are available for making coordinate bonds with the ligands. Electrons are distributed in the five d- orbitals according to Hund’s rule which results in a maximum number of unpaired electrons. The abbreviated electronic configuration of an element depicts the electronic configuration of the elements by making use of noble gas configuration as they have fully-filled electron shells.

Answer to Problem 41QAP

There are zero unpaired electrons in Rh³⁺.

Explanation of Solution

Rhodium is a 4d transition metal element and its atomic number is 45.Its abbreviated electronic configuration can be written as [Kr] 4d⁸5s¹.

When it loses three electrons it leads to the formation of Rh³+ cation, and its abbreviated electronic configuration is written as [Kr] 4d⁶.

The distribution of electrons in the 4d orbitals when no ligand is present is given as follows:

In case of octahedral complexes, the distribution of electrons in the five d-orbitals takes place according to the crystal field theory, according to which in octahedral complexes as the ligand approaches the central metal atom, its d -orbitals get split into lower energy orbitals ${\text{(d}}_{\text{xy}}{\text{,d}}_{\text{yz}}{\text{,d}}_{\text{xz}}\text{)}$ which are known as ${\text{t}}_{\text{2g}}$ and the higher energy orbitals are ${\text{d}}_{x^2-y^2}{\text{ , d}}_{z^2}$ also known as ${\text{e}}_g$ orbitals.

Strong field ligands interact directly with the metal d-electrons and causes pairing of the electrons. The compound so formed is a low spin complex.

The distribution of electrons is shown below:

In case of low spin complexes, the unpaired electrons present in the higher energy gets paired with the electrons present in the t_2g orbitals. From the above electronic distribution, Rh³+ does not contain any unpaired electrons.

Interpretation Introduction

(b)

Interpretation:

The number of unpaired electrons in octahedral complexes with strong field ligands for the metal ion ${\text{Mn}}^{3+}$ needs to be determined.

Concept introduction:

Coordination compounds are those in which a transition metal atom is bonded to a ligand which can be neutral, cation or anion. A transition metal cation has no outer s- electrons available for bonding, rather the inner d -electrons (in case of 3d transition metal elements) are available for making coordinate bonds with the ligands. Electrons are distributed in the five d- orbitals according to Hund’s rule which results in a maximum number of unpaired electrons. The abbreviated electronic configuration of an element depicts the electronic configuration of the elements by making use of noble gas configuration as they have fully-filled electron shells.

Answer to Problem 41QAP

There are two unpaired electrons in Mn³⁺

Explanation of Solution

Manganese is a 3d transition metal element and its atomic number is 25.Its abbreviated electronic configuration can be written as [Ar] 3d⁵4s².

When it loses three electrons it leads to the formation of Mn³+ cation, and its abbreviated electronic configuration is written as [Ar] 3d⁴.

The distribution of electrons in the 3d orbitals when no ligand is present is given as follows:

In case of octahedral complexes, the distribution of electrons in the five d-orbitals takes place according to the crystal field theory. According to this, in octahedral complexes as the ligand approaches the central metal atom, its d -orbitals get split into lower energy orbitals ${\text{(d}}_{\text{xy}}{\text{,d}}_{\text{yz}}{\text{,d}}_{\text{xz}}\text{)}$ which are known as ${\text{t}}_{\text{2g}}$ and the higher energy orbitals are ${\text{d}}_{x^2-y^2}{\text{ , d}}_{z^2}$ also known as ${\text{e}}_g$ orbitals.

Strong field ligands interact directly with the metal d-electrons and causes pairing of the electrons. The compound so formed is a low spin complex.

The distribution of electrons is shown below:

In case of low spin complexes, the unpaired electrons present in the higher energy gets paired with the electrons present in the t_2g orbitals. From the above electronic distribution, Mn³+ does not contains two unpaired electrons.

Interpretation Introduction

(c)

Interpretation:

The number of unpaired electrons in octahedral complexes with strong field ligands for the metal ion ${\text{Ag}}^{+}$ needs to be determined.

Concept introduction:

Coordination compounds are those in which a transition metal atom is bonded toa ligand which can be neutral, cation or anion. A transition metal cation has no outer s- electrons available for bonding, rather the inner d -electrons (in case of 3d transition metal elements) are available for making coordinate bonds with the ligands. Electrons are distributed in the five d- orbitals according to Hund’s rule which results in a maximum number of unpaired electrons. The abbreviated electronic configuration of an element depicts the electronic configuration of the elements by making use of noble gas configuration as they have fully-filled electron shells.

Answer to Problem 41QAP

There are zero unpaired electrons in Ag⁺.

Explanation of Solution

Silver is a 4d transition metal element and its atomic number is 47.Its abbreviated electronic configuration can be written as [Kr] 4d¹⁰ 5s¹.

When it losesone electron it leads to the formation of Ag+ cation, and its abbreviated electronic configuration is written as [Kr] 4d¹⁰.

The distribution of electrons in the 4d orbitals when no ligand is present is given as follows:

In case of octahedral complexes, the distribution of electrons in the five d-orbitals takes place according to the crystal field theory according to which, in octahedral complexes, as the ligand approaches the central metal atom, its d -orbitals get split into lower energy orbitals ${\text{(d}}_{\text{xy}}{\text{,d}}_{\text{yz}}{\text{,d}}_{\text{xz}}\text{)}$ which are known as ${\text{t}}_{\text{2g}}$ and the higher energy orbitals are ${\text{d}}_{x^2-y^2}{\text{ , d}}_{z^2}$ also known as ${\text{e}}_g$ orbitals.

Strong field ligands interact directly with the metal d-electrons and causes pairing of the electrons. The compound so formed is a low spin complex.

The distribution of electrons is shown below:

The electrons remain paired in case of strong filed ligands also. From the above electronic distribution, Ag+ does not contain any unpaired electrons.

Interpretation Introduction

(d)

Interpretation:

The number of unpaired electrons in octahedral complexes with strong field ligands for the metal ion ${\text{Pt}}^{4+}$ needs to be determined.

Concept introduction:

Coordination compounds are those in which a transition metal atom is bonded toa ligand which can be neutral, cation or anion. A transition metal cation has no outer s- electrons available for bonding, rather the inner d -electrons (in case of 3d transition metal elements) are available for making coordinate bonds with the ligands. Electrons are distributed in the five d- orbitals according to Hund’s rule which results in a maximum number of unpaired electrons. The abbreviated electronic configuration of an element depicts the electronic configuration of the elements by making use of noble gas configuration as they have fully-filled electron shells.

Answer to Problem 41QAP

There are zero unpaired electrons in Pt⁴⁺.

Explanation of Solution

In case of transition metal cations, the electrons that are present beyond the noble gas are located in their inner d- orbitals (5d orbitals in case of 5d transition metal elements), this means that they have no outer s- electrons and the distribution of electrons is according to Hund’s rule which states that when orbitals of equal energy are available, then electrons enter singly in the respective orbitals, this gives rise to maximum number of unpaired electrons in transition metal cations.

Platinum is a 5d transition metal element and its atomic number is 78.Its abbreviated electronic configuration can be written as [Xe] 4f¹⁴5d⁹6s¹.

When it losesfour electrons it leads to the formation of Pt⁴+ cation, and its abbreviated electronic configuration is written as [Xe] 4f¹⁴ 5d⁶

The distribution of electrons in the 5d orbitals when no ligand is present is given as follows:

In case of octahedral complexes, the distribution of electrons in the five d-orbitals takes place according to the crystal field theory according to which in octahedral complexes as the ligand approaches the central metal atom its d -orbitals get split into lower energy orbitals ${\text{(d}}_{\text{xy}}{\text{,d}}_{\text{yz}}{\text{,d}}_{\text{xz}}\text{)}$ which are known as ${\text{t}}_{\text{2g}}$ and the higher energy orbitals are ${\text{d}}_{x^2-y^2}{\text{ , d}}_{z^2}$ also known as ${\text{e}}_g$ orbitals.

Strong field ligands interact directly with the metal d-electrons and causes pairing of the electrons. The compound so formed is a low spin complex.

The distribution of electrons is shown below:

In case of low spin complexes the unpaired electrons present in the higher energy gets paired with the electrons present in the t_2g orbitals. From the above electronic distribution, Pt⁴+ does not contain any unpaired electrons.

Interpretation Introduction

(e)

Interpretation:

The number of unpaired electrons in octahedral complexes with strong field ligands for the metal ion ${\text{Au}}^{3+}$ needs to be determined.

Concept introduction:

Coordination compounds are those in which a transition metal atom is bonded toa ligand which can be neutral, cation or anion. A transition metal cation has no outer s- electrons available for bonding, rather the inner d -electrons (in case of 3d transition metal elements) are available for making coordinate bonds with the ligands. Electrons are distributed in the five d- orbitals according to Hund’s rule which results in a maximum number of unpaired electrons. The abbreviated electronic configuration of an element depicts the electronic configuration of the elements by making use of noble gas configuration as they have fully-filled electron shells.

Answer to Problem 41QAP

There are zero unpaired electrons in Au³⁺.

Explanation of Solution

Gold is a 5d transition metal element and its atomic number is 79. Its abbreviated electronic configuration can be written as [Xe] 4f¹⁴ 5d¹⁰ 6s¹.

When it loses three electrons it leads to the formation of Au³+ cation, and its abbreviated electronic configuration is written as [Xe] 4f¹⁴ 5d⁸

The distribution of electrons in the 5d orbitals when no ligand is present is given as follows:

In case of octahedral complexes, the distribution of electrons in the five d-orbitals takes place according to the crystal field theory according to which in octahedral complexes as the ligand approaches the central metal atom its d -orbitals get split into lower energy orbitals ${\text{(d}}_{\text{xy}}{\text{,d}}_{\text{yz}}{\text{,d}}_{\text{xz}}\text{)}$ which are known as ${\text{t}}_{\text{2g}}$ and the higher energy orbitals are ${\text{d}}_{x^2-y^2}{\text{ , d}}_{z^2}$ also known as ${\text{e}}_g$ orbitals.

Strong field ligands interact directly with the metal d-electrons and causes pairing of the electrons. The compound so formed is a low spin complex.

The distribution of electrons is shown below:

In case of low spin complexes the unpaired electron present in the higher energy ${\text{d}}_{z^2}$ orbital gets paired with one of the the electrons present in the ${\text{d}}_{x^2-y^2}$ orbital.From the above electronic distribution, Au³+ does not contain any unpaired electrons.