Chapter 23, Problem 1E

By means of orbital diagrams, write electron configurations for the following transition element atom and ions (a) V ;(b) $C{r}^{2+}$ ; (c) $M{n}^{2+}$ ; (d) $F{e}^{2+}$ ; (e) $C{u}^{2+}$ ; (f) $N{I}^{2+}$ .

Interpretation Introduction

(a)

Interpretation:

To write the electronic configuration of V by using orbital diagram.

Concept introduction:

Orbital Diagram: It is pictorial representation of electrons in any atom. It represents shell number “n”, subshell “l” as well as spin of electrons. The orbitals present in subshell are represented by lines or boxes. In one orbital there can be maximum 2 electrons.

Subshell	Number of orbitals	Number of maximum electrons
s	1	2
p	3	6
d	5	10
f	7	14

The spin of electron is represented by arrows. The up arrow means 1/₂ spin and down arrow means - 1/₂ spin. There are some rules to fill electrons in these orbitals as:

1. Aufbau Principle:

It states that the electron goes in orbital which has lowest energy.

The series of orbital from lowest to highest energy:

$1s2s 2p3s3p4s3d 4p5s4d5p6s4f 5d 6p7s 5f 7p$

2. Pauli’s Exclusion Principle:

   It states that no two electron can have all the same quantum number, that’ why in an orbital, the spin or one electron is up and spin of second electrons is down.

3. Hund’s Rule:

   It states that each orbital in a subshell is single occupied with one electron before any one electron is occupied double and also the spin of all electrons

Electronic configuration:

It is notation which shows the distribution of electrons in orbitals of atoms, ion or molecule. The configuration lists the shell by principal quantum number, n= 1,2,3,4…. and subshell by azimuthal quantum number, l = s, p, d, f and electrons that can be present in it. With the help of orbital diagram electronic configuration can be written.

Answer to Problem 1E

The electronic configuration of V is ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{4s}}^{\text{2}}{\text{3d}}^{\text{3}}$

$\begin{array}{l}\begin{array}{c}\upharpoonleft \downharpoonright \end{array}\begin{array}{c}\upharpoonleft \downharpoonright \end{array}\begin{array}{ccc}\upharpoonleft \downharpoonright & \upharpoonleft \downharpoonright & \upharpoonleft \downharpoonright \end{array}\begin{array}{c}\upharpoonleft \downharpoonright \end{array}\begin{array}{ccc}\upharpoonleft \downharpoonright & \upharpoonleft \downharpoonright & \upharpoonleft \downharpoonright \end{array}\begin{array}{c}\upharpoonleft \downharpoonright \end{array}\begin{array}{ccccc}\upharpoonleft & \upharpoonleft & \upharpoonleft & & \end{array}\\ {\text{ 1s}}^2{\text{ 2s}}^2{\text{ 2p}}^6{\text{ 3s}}^2{\text{ 3p}}^6{\text{ 4s}}^2{\text{ 3d}}^3\end{array}$

Explanation of Solution

The atomic number of Vanadium, V is 23. V is neutral atom, hence the total number of electrons = 23

According to Aufbau principle, the lowest energy orbital is 1s which can have maximum 2 electrons. Next is 2s which has 2 electrons. Similarly, 2p has 6 electrons, 3s has 2 electrons 3p can have 6 electrons, 4s has 2 electrons. So far, total electrons added to orbitals are 20 electrons out of 23. Next comes 3d, it can have 10 electrons maximum but V has only 3 electrons left, which goes in 3d. The orbital diagram of V is:

$\begin{array}{l}\begin{array}{c}\upharpoonleft \downharpoonright \end{array}\begin{array}{c}\upharpoonleft \downharpoonright \end{array}\begin{array}{ccc}\upharpoonleft \downharpoonright & \upharpoonleft \downharpoonright & \upharpoonleft \downharpoonright \end{array}\begin{array}{c}\upharpoonleft \downharpoonright \end{array}\begin{array}{ccc}\upharpoonleft \downharpoonright & \upharpoonleft \downharpoonright & \upharpoonleft \downharpoonright \end{array}\begin{array}{c}\upharpoonleft \downharpoonright \end{array}\begin{array}{ccccc}\upharpoonleft & \upharpoonleft & \upharpoonleft & & \end{array}\\ {\text{ 1s}}^2{\text{ 2s}}^2{\text{ 2p}}^6{\text{ 3s}}^2{\text{ 3p}}^6{\text{ 4s}}^2{\text{ 3d}}^3\end{array}$

The electronic configuration of V is: ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{4s}}^{\text{2}}{\text{3d}}^{\text{3}}$

Interpretation Introduction

(b)

Interpretation:

To write the electronic configuration of ${\text{Cr}}^{2+}$ by using orbital diagram.

Concept introduction:

Orbital Diagram: It is pictorial representation of electrons in any atom. It represents shell number “n”, subshell “l” as well as spin of electrons. The orbitals present in subshell are represented by lines or boxes. In one orbital there can be maximum 2 electrons.

Subshell	Number of orbitals	Number of maximum electrons
s	1	2
p	3	6
d	5	10
f	7	14

The spin of electron is represented by arrows. The up arrow means 1/₂ spin and down arrow means - 1/₂ spin. There are some rules to fill electrons in these orbitals as:

1. Aufbau Principle:

   It states that the electron goes in orbital which has lowest energy.

The series of orbital from lowest to highest energy:

$1s2s 2p3s3p4s3d 4p5s4d5p6s4f 5d 6p7s 5f 7p$

2. Pauli’s Exclusion Principle:

   It states that no two electron can have all the same quantum number, that’ why in an orbital, the spin or one electron is up and spin of second electrons is down.

3. Hund’s Rule:

   It states that each orbital in a subshell is single occupied with one electron before any one electron is occupied double and also the spin of all electrons.

Electronic configuration:

It is notation which shows the distribution of electrons in orbitals of atoms, ion or molecule. The configuration lists the shell by principal quantum number, n= 1,2,3,4…. and subshell by azimuthal quantum number, l = s, p, d, f and electrons that can be present in it. With the help of orbital diagram electronic configuration can be written.

Answer to Problem 1E

The electronic configuration of Cr²+ is ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{4}}$.

$\begin{array}{l}\begin{array}{c}\upharpoonleft \downharpoonright \end{array}\begin{array}{c}\upharpoonleft \downharpoonright \end{array}\begin{array}{ccc}\upharpoonleft \downharpoonright & \upharpoonleft \downharpoonright & \upharpoonleft \downharpoonright \end{array}\begin{array}{c}\upharpoonleft \downharpoonright \end{array}\begin{array}{ccc}\upharpoonleft \downharpoonright & \upharpoonleft \downharpoonright & \upharpoonleft \downharpoonright \end{array}\begin{array}{ccccc}\upharpoonleft & \upharpoonleft & \upharpoonleft & \upharpoonleft & \end{array}\\ {\text{ 1s}}^2{\text{ 2s}}^2{\text{ 2p}}^6{\text{ 3s}}^2{\text{ 3p}}^6{\text{ 3d}}^4\end{array}$

Explanation of Solution

The atomic number of chromium ion, Cr is 24.

In the electronic configuration of Cr, there is an exception and the electronic configuration is written as: ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{4s}}^{\text{1}}{\text{3d}}^{\text{5}}$. Since the 3d⁵ is more stable configuration ( half filled configuration) than 3d⁴, one electrons from 4s² shifts to 3d which makes the configuration 4s¹ 3d⁵ instead of 4s² 3d⁴.

Here, Cr has +2 charge on it, which means 2 electrons have been lost so total electrons present in Cr²+ is 24 − 2 = 22 electrons. The electrons start losing from outermost orbital.

Here, outermost orbital is 4s so it loses one electron and one from 3d.

The orbital diagram is:

$\begin{array}{l}\begin{array}{c}\upharpoonleft \downharpoonright \end{array}\begin{array}{c}\upharpoonleft \downharpoonright \end{array}\begin{array}{ccc}\upharpoonleft \downharpoonright & \upharpoonleft \downharpoonright & \upharpoonleft \downharpoonright \end{array}\begin{array}{c}\upharpoonleft \downharpoonright \end{array}\begin{array}{ccc}\upharpoonleft \downharpoonright & \upharpoonleft \downharpoonright & \upharpoonleft \downharpoonright \end{array}\begin{array}{ccccc}\upharpoonleft & \upharpoonleft & \upharpoonleft & \upharpoonleft & \end{array}\\ {\text{ 1s}}^2{\text{ 2s}}^2{\text{ 2p}}^6{\text{ 3s}}^2{\text{ 3p}}^6{\text{ 3d}}^4\end{array}$

The electronic configuration of Cr²+ is ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{4}}$.

Interpretation Introduction

Interpretation:

To write the electronic configuration of ${\text{Mn}}^{2+}$ by using orbital diagram.

Concept introduction:

Orbital Diagram: It is pictorial representation of electrons in any atom. It represents shell number “n”, subshell “l” as well as spin of electrons. The orbitals present in subshell are represented by lines or boxes. In one orbital there can be maximum 2 electrons.

Subshell	Number of orbitals	Number of maximum electrons
s	1	2
p	3	6
d	5	10
f	7	14

The spin of electron is represented by arrows. The up arrow means 1/₂ spin and down arrow means - 1/₂ spin. There are some rules to fill electrons in these orbitals as:

1. Aufbau Principle:

   It states that the electron goes in orbital which has lowest energy.

The series of orbital from lowest to highest energy:

$1s2s 2p3s3p4s3d 4p5s4d5p6s4f 5d 6p7s 5f 7p$

2. Pauli’s Exclusion Principle:

   It states that no two electron can have all the same quantum number, that’ why in an orbital, the spin or one electron is up and spin of second electrons is down.

3. Hund’s Rule:

   It states that each orbital in a subshell is single occupied with one electron before any one electron is occupied double and also the spin of all electrons.

Electronic configuration:

It is notation which shows the distribution of electrons in orbitals of atoms, ion or molecule. The configuration lists the shell by principal quantum number, n= 1,2,3,4…. and subshell by azimuthal quantum number, l = s, p, d, f and electrons that can be present in it. With the help of orbital diagram electronic configuration can be written.

Answer to Problem 1E

The electronic configuration of Mn²+ is ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{5}}$

$\begin{array}{l}\begin{array}{c}\upharpoonleft \downharpoonright \end{array}\begin{array}{c}\upharpoonleft \downharpoonright \end{array}\begin{array}{ccc}\upharpoonleft \downharpoonright & \upharpoonleft \downharpoonright & \upharpoonleft \downharpoonright \end{array}\begin{array}{c}\upharpoonleft \downharpoonright \end{array}\begin{array}{ccc}\upharpoonleft \downharpoonright & \upharpoonleft \downharpoonright & \upharpoonleft \downharpoonright \end{array}\begin{array}{ccccc}\upharpoonleft & \upharpoonleft & \upharpoonleft & \upharpoonleft & \upharpoonleft \end{array}\\ {\text{ 1s}}^2{\text{ 2s}}^2{\text{ 2p}}^6{\text{ 3s}}^2{\text{ 3p}}^6{\text{ 3d}}^5\end{array}$

Explanation of Solution

The atomic number of Manganese, Mn is 25.

The electronic configuration of Mn is ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{4s}}^{\text{2}}{\text{3d}}^{\text{5}}\text{.}$

Here, Mn shows charge of +2 means 2 electrons are lost from outermost orbital, 4d.

Hence, total electrons present in Mn²+ are 23.

$\begin{array}{l}\begin{array}{c}\upharpoonleft \downharpoonright \end{array}\begin{array}{c}\upharpoonleft \downharpoonright \end{array}\begin{array}{ccc}\upharpoonleft \downharpoonright & \upharpoonleft \downharpoonright & \upharpoonleft \downharpoonright \end{array}\begin{array}{c}\upharpoonleft \downharpoonright \end{array}\begin{array}{ccc}\upharpoonleft \downharpoonright & \upharpoonleft \downharpoonright & \upharpoonleft \downharpoonright \end{array}\begin{array}{ccccc}\upharpoonleft & \upharpoonleft & \upharpoonleft & \upharpoonleft & \upharpoonleft \end{array}\\ {\text{ 1s}}^2{\text{ 2s}}^2{\text{ 2p}}^6{\text{ 3s}}^2{\text{ 3p}}^6{\text{ 3d}}^5\end{array}$

The electronic configuration of Mn²+ is ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{5}}$

Interpretation Introduction

(d)

Interpretation:

To write the electronic configuration of Fe²+ by using orbital diagram.

Concept introduction:

Orbital Diagram: It is pictorial representation of electrons in any atom. It represents shell number “n”, subshell “l” as well as spin of electrons. The orbitals present in subshell are represented by lines or boxes. In one orbital there can be maximum 2 electrons.

Subshell	Number of orbitals	Number of maximum electrons
s	1	2
p	3	6
d	5	10
f	7	14

The spin of electron is represented by arrows. The up arrow means 1/₂ spin and down arrow means - 1/₂ spin. There are some rules to fill electrons in these orbitals as:

1. Aufbau Principle:

   It states that the electron goes in orbital which has lowest energy.

The series of orbital from lowest to highest energy:

$1s2s 2p3s3p4s3d 4p5s4d5p6s4f 5d 6p7s 5f 7p$

2. Pauli’s Exclusion Principle:

   It states that no two electron can have all the same quantum number, that’ why in an orbital, the spin or one electron is up and spin of second electrons is down.

3. Hund’s Rule:

   It states that each orbital in a subshell is single occupied with one electron before any one electron is occupied double and also the spin of all electrons.

Electronic configuration:

It is notation which shows the distribution of electrons in orbitals of atoms, ion or molecule. The configuration lists the shell by principal quantum number, n= 1,2,3,4…. and subshell by azimuthal quantum number, l = s, p, d, f and electrons that can be present in it. With the help of orbital diagram electronic configuration can be written.

Answer to Problem 1E

The electronic configuration of Fe²+ is ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{6}}$.

$\begin{array}{l}\begin{array}{c}\upharpoonleft \downharpoonright \end{array}\begin{array}{c}\upharpoonleft \downharpoonright \end{array}\begin{array}{ccc}\upharpoonleft \downharpoonright & \upharpoonleft \downharpoonright & \upharpoonleft \downharpoonright \end{array}\begin{array}{c}\upharpoonleft \downharpoonright \end{array}\begin{array}{ccc}\upharpoonleft \downharpoonright & \upharpoonleft \downharpoonright & \upharpoonleft \downharpoonright \end{array}\begin{array}{ccccc}\upharpoonleft \downharpoonright & \upharpoonleft & \upharpoonleft & \upharpoonleft & \upharpoonleft \end{array}\\ {\text{ 1s}}^2{\text{ 2s}}^2{\text{ 2p}}^6{\text{ 3s}}^2{\text{ 3p}}^6{\text{ 3d}}^6\end{array}$

Explanation of Solution

The atomic number of Iron, Fe is 26.

The electronic configuration of Fe is ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{4s}}^{\text{2}}{\text{3d}}^{\text{6}}$.

Here, Fe shows charge of +2 means 2 electrons are lost from outermost orbital, 4d.

Hence, total electrons present in Fe²+ are 24.

The orbital diagram of Fe²+ is:

$\begin{array}{l}\begin{array}{c}\upharpoonleft \downharpoonright \end{array}\begin{array}{c}\upharpoonleft \downharpoonright \end{array}\begin{array}{ccc}\upharpoonleft \downharpoonright & \upharpoonleft \downharpoonright & \upharpoonleft \downharpoonright \end{array}\begin{array}{c}\upharpoonleft \downharpoonright \end{array}\begin{array}{ccc}\upharpoonleft \downharpoonright & \upharpoonleft \downharpoonright & \upharpoonleft \downharpoonright \end{array}\begin{array}{ccccc}\upharpoonleft \downharpoonright & \upharpoonleft & \upharpoonleft & \upharpoonleft & \upharpoonleft \end{array}\\ {\text{ 1s}}^2{\text{ 2s}}^2{\text{ 2p}}^6{\text{ 3s}}^2{\text{ 3p}}^6{\text{ 3d}}^6\end{array}$

The electronic configuration of Fe²+ is ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{6}}$.

Interpretation Introduction

(e)

Interpretation:

To write the Electronic configuration of Cu²+ by using orbital diagram.

Concept introduction:

Orbital Diagram: It is pictorial representation of electrons in any atom. It represents shell number “n”, subshell “l” as well as spin of electrons. The orbitals present in subshell are represented by lines or boxes. In one orbital there can be maximum 2 electrons.

Subshell	Number of orbitals	Number of maximum electrons
s	1	2
p	3	6
d	5	10
f	7	14

The spin of electron is represented by arrows. The up arrow means 1/₂ spin and down arrow means - 1/₂ spin. There are some rules to fill electrons in these orbitals as:

1. Aufbau Principle:

   It states that the electron goes in orbital which has lowest energy.

The series of orbital from lowest to highest energy:

$1s2s 2p3s3p4s3d 4p5s4d5p6s4f 5d 6p7s 5f 7p$

2. Pauli’s Exclusion Principle:

   It states that no two electron can have all the same quantum number, that’ why in an orbital, the spin or one electron is up and spin of second electrons is down.

3. Hund’s Rule:

   It states that each orbital in a subshell is single occupied with one electron before any one electron is occupied double and also the spin of all electrons.

Electronic configuration:

It is notation which shows the distribution of electrons in orbitals of atoms, ion or molecule. The configuration lists the shell by principal quantum number, n= 1,2,3,4…. and subshell by azimuthal quantum number, l = s, p, d, f and electrons that can be present in it. With the help of orbital diagram electronic configuration can be written.

Answer to Problem 1E

The electronic configuration of Cu²+ is ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{9}}$.

$\begin{array}{l}\begin{array}{c}\upharpoonleft \downharpoonright \end{array}\begin{array}{c}\upharpoonleft \downharpoonright \end{array}\begin{array}{ccc}\upharpoonleft \downharpoonright & \upharpoonleft \downharpoonright & \upharpoonleft \downharpoonright \end{array}\begin{array}{c}\upharpoonleft \downharpoonright \end{array}\begin{array}{ccc}\upharpoonleft \downharpoonright & \upharpoonleft \downharpoonright & \upharpoonleft \downharpoonright \end{array}\begin{array}{ccccc}\upharpoonleft \downharpoonright & \upharpoonleft \downharpoonright & \upharpoonleft \downharpoonright & \upharpoonleft \downharpoonright & \upharpoonleft \end{array}\\ {\text{ 1s}}^2{\text{ 2s}}^2{\text{ 2p}}^6{\text{ 3s}}^2{\text{ 3p}}^6{\text{ 3d}}^9\end{array}$

Explanation of Solution

The atomic number of Copper, Cu is 29.

The electronic configuration of Cu is 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹ 3d¹⁰. Cu shows exception in configuration as 4s¹ 3d¹⁰ instead 4s² 3d⁹ because 10 electron in 3d orbital is stable configuration so out of 2 electrons, one electron from 4s jumps to 3d leaving 1 electron in 4s and 10 electron in 3d.

Here, Cu shows charge of +2 means 1 electron from outermost 4s and one electron from 3d are lost. Hence, total electrons present in Cu²+ are 27.

$\begin{array}{l}\begin{array}{c}\upharpoonleft \downharpoonright \end{array}\begin{array}{c}\upharpoonleft \downharpoonright \end{array}\begin{array}{ccc}\upharpoonleft \downharpoonright & \upharpoonleft \downharpoonright & \upharpoonleft \downharpoonright \end{array}\begin{array}{c}\upharpoonleft \downharpoonright \end{array}\begin{array}{ccc}\upharpoonleft \downharpoonright & \upharpoonleft \downharpoonright & \upharpoonleft \downharpoonright \end{array}\begin{array}{ccccc}\upharpoonleft \downharpoonright & \upharpoonleft \downharpoonright & \upharpoonleft \downharpoonright & \upharpoonleft \downharpoonright & \upharpoonleft \end{array}\\ {\text{ 1s}}^2{\text{ 2s}}^2{\text{ 2p}}^6{\text{ 3s}}^2{\text{ 3p}}^6{\text{ 3d}}^9\end{array}$

The orbital diagram of Cu²+ is:

The electronic configuration of Cu²+ is ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{9}}$.

Interpretation Introduction

(f)

Interpretation:

To write the electronic configuration of Ni²+ by using orbital diagram.

Concept introduction:

Orbital Diagram: It is pictorial representation of electrons in any atom. It represents shell number “n”, subshell “l” as well as spin of electrons. The orbitals present in subshell are represented by lines or boxes. In one orbital there can be maximum 2 electrons.

Subshell	Number of orbitals	Number of maximum electrons
s	1	2
p	3	6
d	5	10
f	7	14

The spin of electron is represented by arrows. The up arrow means 1/₂ spin and down arrow means - 1/₂ spin. There are some rules to fill electrons in these orbitals as:

1. Aufbau Principle:

   It states that the electron goes in orbital which has lowest energy.

The series of orbital from lowest to highest energy:

$1s2s 2p3s3p4s3d 4p5s4d5p6s4f 5d 6p7s 5f 7p$

2. Pauli’s Exclusion Principle:

   It states that no two electron can have all the same quantum number, that’ why in an orbital, the spin or one electron is up and spin of second electrons is down.

3. Hund’s Rule:

   It states that each orbital in a subshell is single occupied with one electron before any one electron is occupied double and also the spin of all electrons.

Electronic configuration:

It is notation which shows the distribution of electrons in orbitals of atoms, ion or molecule. The configuration lists the shell by principal quantum number, n= 1,2,3,4…. and subshell by azimuthal quantum number, l = s, p, d, f and electrons that can be present in it. With the help of orbital diagram electronic configuration can be written.

Answer to Problem 1E

The electronic configuration of Ni²+ is ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{8}}$.

$\begin{array}{l}\begin{array}{c}\upharpoonleft \downharpoonright \end{array}\begin{array}{c}\upharpoonleft \downharpoonright \end{array}\begin{array}{ccc}\upharpoonleft \downharpoonright & \upharpoonleft \downharpoonright & \upharpoonleft \downharpoonright \end{array}\begin{array}{c}\upharpoonleft \downharpoonright \end{array}\begin{array}{ccc}\upharpoonleft \downharpoonright & \upharpoonleft \downharpoonright & \upharpoonleft \downharpoonright \end{array}\begin{array}{ccccc}\upharpoonleft \downharpoonright & \upharpoonleft \downharpoonright & \upharpoonleft \downharpoonright & \upharpoonleft & \upharpoonleft \end{array}\\ {\text{ 1s}}^2{\text{ 2s}}^2{\text{ 2p}}^6{\text{ 3s}}^2{\text{ 3p}}^6{\text{ 3d}}^8\end{array}$

Explanation of Solution

The atomic number of Nickel, Ni is 28.

The electronic configuration of Ni is ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{4s}}^{\text{2}}{\text{3d}}^{\text{8}}\text{.}$

Here, Ni shows charge of +2 means 2 electrons from outermost 4s are lost. Hence total electrons present in Ni²+ are 26.

The orbital diagram of Ni²+ is:

$\begin{array}{l}\begin{array}{c}\upharpoonleft \downharpoonright \end{array}\begin{array}{c}\upharpoonleft \downharpoonright \end{array}\begin{array}{ccc}\upharpoonleft \downharpoonright & \upharpoonleft \downharpoonright & \upharpoonleft \downharpoonright \end{array}\begin{array}{c}\upharpoonleft \downharpoonright \end{array}\begin{array}{ccc}\upharpoonleft \downharpoonright & \upharpoonleft \downharpoonright & \upharpoonleft \downharpoonright \end{array}\begin{array}{ccccc}\upharpoonleft \downharpoonright & \upharpoonleft \downharpoonright & \upharpoonleft \downharpoonright & \upharpoonleft & \upharpoonleft \end{array}\\ {\text{ 1s}}^2{\text{ 2s}}^2{\text{ 2p}}^6{\text{ 3s}}^2{\text{ 3p}}^6{\text{ 3d}}^8\end{array}$

The electronic configuration of Ni²+ is ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{8}}$.