Chapter 19, Problem 42QAP

Interpretation Introduction

(a)

Interpretation:

The number of unpaired electrons in octahedral complexes with weak field ligands in Rh³⁺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 42QAP

There are four unpaired electrons in Rh³⁺.

Explanation of Solution

In case of transition metal cations, the electrons that are present beyond the noble gas are in their inner d- orbitals (4d orbitals in case of 4d transition metal elements), this means that they have no outer s- electrons. 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.

Rhodium is a 4d transition metal element and its atomic number is 45. Its abbreviated electronic configuration can be written as [Kr] 4d⁸ 5s¹.

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:

$\begin{array}{ccccc}\upharpoonleft \downharpoonright & \upharpoonleft & \upharpoonleft & \upharpoonleft & \upharpoonleft \end{array}$

In case of octahedral complexes, the distribution of electrons in the five d-orbitals takes place as per 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.

Weak field ligands are unable to interact with the metal d-orbitals; therefore they are unable to cause pairing of the metal d-electrons. The compound so formed is a high spin complex which contains maximum number of unpaired electrons.

The distribution of electrons is shown below:

$\begin{array}{l}\begin{array}{cc}\upharpoonleft & \upharpoonleft \end{array}\\ {\text{ d}}_{x^2-y^2}{\text{ d}}_{z^2}\end{array}$

$\begin{array}{l}\begin{array}{ccc}\upharpoonleft \downharpoonright & \upharpoonleft & \upharpoonleft \end{array}\\ {\text{d}}_{\text{xy}}{\text{ d}}_{\text{yx}}{\text{ d}}_{\text{xz}}\end{array}$

Thus, high spin complex of Rh³⁺ contains three unpaired electrons when it is bonded to weak field ligands.

Interpretation Introduction

(b)

Interpretation:

The number of unpaired electrons in octahedral complexes with weak field ligands for Mn³⁺ 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 42QAP

There are four 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] 3d⁵ 4s².

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:

$\begin{array}{ccccc}\upharpoonleft & \upharpoonleft & \upharpoonleft & \upharpoonleft & \end{array}$

In case of octahedral complexes, the distribution of electrons in the five d-orbitals takes place as per 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.

Weak field ligands are unable to interact with the metal d-orbitals; therefore they are unable to cause pairing of the metal d-electrons. The compound so formed is a high spin complex which contains maximum number of unpaired electrons.

The distribution of electrons is shown below:

$\begin{array}{l}\begin{array}{cc}\upharpoonleft & \end{array}\\ {\text{ d}}_{x^2-y^2}{\text{ d}}_{z^2}\end{array}$

$\begin{array}{l}\begin{array}{ccc}\upharpoonleft & \upharpoonleft & \upharpoonleft \end{array}\\ {\text{d}}_{\text{xy}}{\text{ d}}_{\text{yx}}{\text{ d}}_{\text{xz}}\end{array}$

Thus, the high spin complex of Mn³⁺ contains four unpaired electrons.

Interpretation Introduction

(c)

Interpretation:

The number of unpaired electrons in octahedral complexes with weak field ligands for Ag⁺ 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 42QAP

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 loses one 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:

$\begin{array}{ccccc}\upharpoonleft \downharpoonright & \upharpoonleft \downharpoonright & \upharpoonleft \downharpoonright & \upharpoonleft \downharpoonright & \upharpoonleft \downharpoonright \end{array}$

In case of octahedral complexes, the distribution of electrons in the five d-orbitals takes place as per 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.

Weak field ligands are unable to interact with the metal d-orbitals; therefore, they are unable to cause pairing of the metal d-electrons. The compound so formed is a high spin complex which contains maximum number of unpaired electrons.

The distribution of electrons is shown below:

$\begin{array}{l}\begin{array}{cc}\upharpoonleft \downharpoonright & \upharpoonleft \downharpoonright \end{array}\\ {\text{ d}}_{x^2-y^2}{\text{ d}}_{z^2}\end{array}$

$\begin{array}{l}\begin{array}{ccc}\upharpoonleft \downharpoonright & \upharpoonleft \downharpoonright & \upharpoonleft \downharpoonright \end{array}\\ {\text{d}}_{\text{xy}}{\text{ d}}_{\text{yx}}{\text{ d}}_{\text{xz}}\end{array}$

The electrons remain paired in case of weak field 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 weak field ligands for Pt⁴⁺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 42QAP

There are four unpaired electrons in Pt⁴⁺.

Explanation of Solution

Platinum is a 5d transition metal element and its atomic number is 78. Its abbreviated electronic configuration can be written as [Xe] 4f¹⁴ 5d⁹ 6s¹.

When it loses four 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:

$\begin{array}{ccccc}\upharpoonleft \downharpoonright & \upharpoonleft & \upharpoonleft & \upharpoonleft & \upharpoonleft \end{array}$

In case of octahedral complexes, the distribution of electrons in the five d-orbitals takes place as per 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.

Weak field ligands are unable to interact with the metal d-orbitals; therefore, they are unable to cause pairing of the metal d-electrons. The compound so formed is a high spin complex which contains maximum number of unpaired electrons.

The distribution of electrons is shown below:

$\begin{array}{l}\begin{array}{cc}\upharpoonleft & \upharpoonleft \end{array}\\ {\text{ d}}_{x^2-y^2}{\text{ d}}_{z^2}\end{array}$

$\begin{array}{l}\begin{array}{ccc}\upharpoonleft \downharpoonright & \upharpoonleft & \upharpoonleft \end{array}\\ {\text{d}}_{\text{xy}}{\text{ d}}_{\text{yx}}{\text{ d}}_{\text{xz}}\end{array}$

Thus, the high spin complex of Pt⁴⁺ contains four unpaired electrons.

Interpretation Introduction

(e)

Interpretation:

The number of unpaired electrons in octahedral complexes with weak field ligands for Au³⁺ 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 42QAP

There are two 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:

$\begin{array}{ccccc}\upharpoonleft \downharpoonright & \upharpoonleft \downharpoonright & \upharpoonleft \downharpoonright & \upharpoonleft & \upharpoonleft \end{array}$

In case of octahedral complexes, the distribution of electrons in the five d-orbitals takes place as per 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.

Weak field ligands are unable to interact with the metal d-orbitals; therefore, they are unable to cause pairing of the metal d-electrons. The compound so formed is a high spin complex which contains maximum number of unpaired electrons.

The distribution of electrons is shown below:

$\begin{array}{l}\begin{array}{cc}\upharpoonleft & \upharpoonleft \end{array}\\ {\text{ d}}_{x^2-y^2}{\text{ d}}_{z^2}\end{array}$

$\begin{array}{l}\begin{array}{ccc}\upharpoonleft \downharpoonright & \upharpoonleft \downharpoonright & \upharpoonleft \downharpoonright \end{array}\\ {\text{d}}_{\text{xy}}{\text{ d}}_{\text{yx}}{\text{ d}}_{\text{xz}}\end{array}$

From the above electronic distribution, Au³⁺ contains two unpaired electrons.