Chapter 22, Problem 1PS

(a)

Interpretation Introduction

Interpretation:

The electronic configuration of ${\text{Cr}}^{\text{3+}}$ ion and the magnetic property of the ion have to be identified.

Concept introduction:

The electron configuration tells us in which orbitals the electrons for an element are located. Three rules are,

• Electrons fill orbitals staring with lowest n and moving upwards(Aufbau principle)
• No two electrons can fill one orbital with the same spin(Pauli Exclusion Principle)
• For orbitals within the same subshell, the electrons fill each orbital singly before any orbital gets a second electron (Hund’s rule).

Para magnetism refers to the magnetic state of an atom with one or more unpaired electrons.

Diamagnetism has no unpaired electrons (electrons are paired).

Answer to Problem 1PS

The electronic configuration of ${\text{Cr}}^{\text{3+}}$ is ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^3{\text{4s}}^0$ and ${\text{Cr}}^{\text{3+}}$ is paramagnetic in nature.

Explanation of Solution

The electronic configuration of chromium is ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{5}}{\text{4s}}^{\text{1}}$. The electronic configuration of ${\text{Cr}}^{\text{3+}}$ is ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^3{\text{4s}}^0$. Therefore there are three unpaired electrons are there so the ${\text{Cr}}^{\text{3+}}$ is paramagnetic in nature.

(b)

Interpretation Introduction

Interpretation:

The electronic configuration of ${\text{V}}^{\text{2+}}$ ion and the magnetic property of the ion has to be identified.

Concept introduction:

The electron configuration tells us in which orbitals the electrons for an element are located. Three rules are,

• Electrons fill orbitals staring with lowest n and moving upwards(Aufbau principle)
• No two electrons can fill one orbital with the same spin(Pauli Exclusion Principle)
• For orbitals within the same subshell, the electrons fill each orbital singly before any orbital gets a second electron (Hund’s rule).

Para magnetism refers to the magnetic state of an atom with one or more unpaired electrons.

Diamagnetism has no unpaired electrons (electrons are paired).

Answer to Problem 1PS

The electronic configuration of ${\text{V}}^{\text{2+}}$ is ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^3{\text{4s}}^0$ and ${\text{V}}^{\text{2+}}$ is paramagnetic in nature.

Explanation of Solution

The electronic configuration of vanadium is ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^3{\text{4s}}^2$. The electronic configuration of ${\text{V}}^{\text{2+}}$ is ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^3{\text{4s}}^0$. Therefore there are three unpaired electrons are there so the ${\text{V}}^{\text{2+}}$ is paramagnetic in nature.

(c)

Interpretation Introduction

Interpretation:

The electronic configuration of ${\text{Ni}}^{\text{2+}}$ ion and the magnetic property of the ion has be identified.

Concept introduction:

The electron configuration tells us in which orbitals the electrons for an element are located. Three rules are,

• Electrons fill orbitals staring with lowest n and moving upwards(Aufbau principle)
• No two electrons can fill one orbital with the same spin(Pauli Exclusion Principle)
• For orbitals within the same subshell, the electrons fill each orbital singly before any orbital gets a second electron (Hund’s rule).

Para magnetism refers to the magnetic state of an atom with one or more unpaired electrons.

Diamagnetism has no unpaired electrons (electrons are paired).

Answer to Problem 1PS

The electronic configuration of ${\text{Ni}}^{\text{2+}}$ is ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^8{\text{4s}}^0$ and ${\text{Ni}}^{\text{2+}}$ is paramagnetic in nature.

Explanation of Solution

The electronic configuration of nickel is ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^8{\text{4s}}^{\text{2}}$. The electronic configuration of ${\text{Ni}}^{\text{2+}}$ is ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^8{\text{4s}}^0$. Therefore there are two unpaired electrons are there so the ${\text{Ni}}^{\text{2+}}$ is paramagnetic in nature.

(d)

Interpretation Introduction

Interpretation:

The electronic configuration of ${\text{Cr}}^{\text{3+}}$ ion and the magnetic property of the ion has to be identified.

Concept introduction:

The electron configuration tells us in which orbitals the electrons for an element are located. Three rules are,

• Electrons fill orbitals staring with lowest n and moving upwards(Aufbau principle)
• No two electrons can fill one orbital with the same spin(Pauli Exclusion Principle)
• For orbitals within the same subshell, the electrons fill each orbital singly before any orbital gets a second electron (Hund’s rule).

Para magnetism refers to the magnetic state of an atom with one or more unpaired electrons.

Diamagnetism has no unpaired electrons (electrons are paired).

Answer to Problem 1PS

The electronic configuration of ${\text{Cu}}^{\text{+}}$ is ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{10}}{\text{4s}}^0$ and ${\text{Cu}}^{\text{+}}$ is diamagnetic in nature.

Explanation of Solution

The electronic configuration of copper is ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{10}}{\text{4s}}^{\text{1}}$. The electronic configuration of ${\text{Cu}}^{\text{+}}$ is ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{10}}{\text{4s}}^0$. Therefore there is no unpaired electrons are there so the ${\text{Cu}}^{\text{+}}$ is diamagnetic in nature.