Chapter 6, Problem 40QAP

Write the orbital diagram tor an atom of

(a) Na (b) O (c) Co

(d) Cl

Interpretation Introduction

(a)

Interpretation:

The orbital diagram for Na should be written.

Concept introduction:

The simplest method for describing the arrangement of electrons in an atom is by writing its electronic configuration. Since the set of four quantum numbers is used to describe the atomic orbitals in an atom, therefore by writing the electronic configuration one can get details of the number of electrons present in each sublevel. To show the distribution of electrons in the various orbitals, orbital diagrams are used. The filling of electrons in the atomic orbitals takes place according to the Aufbau principal which states that when an atom is present in its ground state, electrons are filled in order of increasing energy of the orbitals, which means that firstly lower energy orbitals are filled, and then filling of higher energy orbitals takes place.

Answer to Problem 40QAP

The orbital diagram for Na is:

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

Explanation of Solution

When the electronic configuration of an atom is written, it describes the number of electron present in each sublevel by the superscript. Atomic number of an element gives the total number of electrons present in an atom. Since the atomic number of sodium atom denoted by Na is 11, therefore its ground state electronic configuration is:

${\text{1s}}^{\text{2}}{\text{2s}}^2{\text{ 2p}}^{\text{6}}{\text{3s}}^1$

According to Hund’s Rule, when several orbitals having equal energy are available, then electrons are filled singly with parallel spins. No two electrons can have same spin in a given orbital. The most stable arrangement of electrons is the one in which two electrons present in two different orbitals have parallel spins. The total number of orbitals in a given sublevel are given by 2l+1, where l = 0,1,2,3 for s, p, d and f sublevels respectively. In case of Sodium, since the electrons are present in both the s and p-sublevel, therefore l=0 for s- sublevel,

This means 2l+1 = 2(0) +1= 1

This means that I orbital is present in each s- sublevel.

In case of p-sublevel the total number of orbitals are:

2l +1

2 (1) +1

2 +1=3.

The orbital diagram for its electronic configuration is shown below:

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

Interpretation Introduction

(b)

Interpretation:

The orbital diagram for O should be written.

Concept introduction:

The simplest method for describing the arrangement of electrons in an atom is by writing its electronic configuration. Since the set of four quantum numbers is used to describe the atomic orbitals in an atom, therefore by writing the electronic configuration one can get details of the number of electrons present in each sublevel. The filling of electrons in the atomic orbitals takes place according to the Aufbau principal which states that when an atom is present in its ground state, electrons are filled in order of increasing energy of the orbitals, which means that firstly lower energy orbitals are filled, and then filling of higher energy orbitals takes place.

To show the distribution of electrons in the various orbitals, orbital diagrams are used.

Answer to Problem 40QAP

The orbital diagram for O is:

$\begin{array}{l}1s\text{ 2s 2p }\\ \begin{array}{c}\upharpoonleft \downharpoonright \end{array}\begin{array}{c}\upharpoonleft \downharpoonright \end{array}\begin{array}{ccc}\upharpoonleft \downharpoonright & \upharpoonleft & \upharpoonleft \end{array}\end{array}$

Explanation of Solution

When the electronic configuration of an atom is written, it describes the number of electron present in each sublevel by the superscript. Atomic number of an element gives the total number of electrons present in an atom. Since the atomic number of Oxygen atom denoted by O is 8, therefore its ground state electronic configuration is:

${\text{1s}}^{\text{2}}{\text{2s}}^2{\text{ 2p}}^4$

According to Hund’s Rule, when several orbitals having equal energy are available, then electrons are filled singly with parallel spins. No two electrons can have same spin in a given orbital. The most stable arrangement of electrons is the one in which two electrons present in two different orbitals have parallel spins. The total number of orbitals in a given sublevel are given by 2l+1, where l = 0,1,2,3 for s, p, d and f sublevels respectively. In case of Oxygen, since the electrons are present in both s and p-sublevel, l=0 for s and l=1 for p-sublevels

This means 2l+1 = 2(0) +1= 1

Hence one orbital is present for each s-sublevel.

For a p-sublevel, the total number of orbitals are

2(1) +1

2+1

3

This means that three orbitals are present in each p-sublevel of Oxygen atom.

The orbital diagram for its electronic configuration is shown below:

$\begin{array}{l}1s\text{ 2s 2p }\\ \begin{array}{c}\upharpoonleft \downharpoonright \end{array}\begin{array}{c}\upharpoonleft \downharpoonright \end{array}\begin{array}{ccc}\upharpoonleft \downharpoonright & \upharpoonleft & \upharpoonleft \end{array}\end{array}$

Interpretation Introduction

(c)

Interpretation:

The orbital diagram for Co should be written.

Concept introduction:

The simplest method for describing the arrangement of electrons in an atom is by writing its electronic configuration. Since the set of four quantum numbers is used to describe the atomic orbitals in an atom, therefore by writing the electronic configuration one can get details of the number of electrons present in each sublevel. The filling of electrons in the atomic orbitals takes place according to the Aufbau principal which states that when an atom is present in its ground state, electrons are filled in order of increasing energy of the orbitals, which means that firstly lower energy orbitals are filled, and then filling of higher energy orbitals takes place.

To show the distribution of electrons in the various orbitals, orbital diagrams are used.

Answer to Problem 40QAP

The orbital diagram for Co is:

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

Explanation of Solution

When the electronic configuration of an atom is written, it describes the number of electron present in each sublevel by the superscript. Atomic number of an element gives the total number of electrons present in an atom. Since the atomic number of Cobalt atom denoted by Co is 27, therefore its ground state electronic configuration is:

${\text{ 1s}}^{\text{2}}{\text{2s}}^2{\text{ 2p}}^6{\text{ 3s}}^2{\text{3p}}^{\text{6}}{\text{ 3d}}^7{\text{4s}}^{\text{2}}$

According to Hund’s Rule, when several orbitals having equal energy are available, then electrons are filled singly with parallel spins. No two electrons can have same spin in a given orbital. The most stable arrangement of electrons is the one in which two electrons present in two different orbitals have parallel spins. The total number of orbitals in a given sublevel are given by 2l+1, where l = 0,1,2,3 for s, p, d and f sublevels respectively. In case of Cobalt, since the electrons are present in both s and p-sublevel, l=0 for s and l=1 for p-sublevels

This means 2l+1 = 2(0) +1= 1

Hence one orbital is present for each s-sublevel.

For a p-sublevel, the total number of orbitals are

2(1) +1

2+1

3

This means that three orbitals are present in p-sublevel of Cobalt atom.

For d- sublevel since l=2, therefore number of orbitals = 2(2)+1=5

The orbital diagram for its electronic configuration is shown below:

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

Interpretation Introduction

(d)

Interpretation:

The orbital diagram for Cl element should be written.

Concept introduction:

The simplest method for describing the arrangement of electrons in an atom is by writing its electronic configuration. Since the set of four quantum numbers is used to describe the atomic orbitals in an atom, therefore by writing the electronic configuration one can get details of the number of electrons present in each sublevel. The filling of electrons in the atomic orbitals takes place according to the Aufbau principal which states that when an atom is present in its ground state, electrons are filled in order of increasing energy of the orbitals, which means that firstly lower energy orbitals are filled, and then filling of higher energy orbitals takes place.

To show the distribution of electrons in the various orbitals, orbital diagrams are used.

Answer to Problem 40QAP

The orbital diagram for Cl is:

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

Explanation of Solution

When the electronic configuration of an atom is written, it describes the number of electron present in each sublevel by the superscript. Atomic number of an element gives the total number of electrons present in an atom. Since the atomic number of Chlorine atom denoted by Cl is 17, therefore its ground state electronic configuration is:

${\text{1s}}^{\text{2}}{\text{2s}}^2{\text{2p}}^6{\text{3s}}^2{\text{ 3p}}^5$

According to Hund’s Rule, when several orbitals having equal energy are available, then electrons are filled singly with parallel spins. No two electrons can have same spin in a given orbital. The most stable arrangement of electrons is the one in which two electrons present in two different orbitals have parallel spins. The total number of orbitals in a given sublevel are given by 2l+1, where l = 0,1,2,3 for s, p, d and f sublevels respectively. In case of Chlorine, since the electrons are present in s and p-sublevel, l=0 for s and l=1 for p

This means 2l+1 = 2(0) +1= 1

Hence one orbital is present for each s-sublevel.

For a p-sublevel, the total number of orbitals are

2(1) +1

2+1

3

This means that three orbitals are present in each p-sublevel of Cl atom.

The orbital diagram for its electronic configuration is shown below:

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