In this article, we shall study to solve problems on the calculations of the number of electrons, protons, and neutrons in atoms, molecules, and species.

Example 01:

Calculate the charge and mass of 1 mole of electrons.

Solution:

1 mole of electron corresponds to 6.023 x 10²³ electrons

Mass of 1 electron = 9.1 x 10^-31 kg

Mass of one mole of electron = 9.1 x 10^-31 x 6.022 x 10²³ = 5.48 x 10^-7 kg

Charge of 1 electron = 1.602 x 10^-19 C

Charge on one mole of electron = 1.602 x 10^-19 x 6.022 x 10²³ = 9.65 x 10⁴ C

Ans: The mass of one mole of electrons is 5.48 x 10^-7 kg and the charge on one mole of electrons is 9.65 x 10⁴ C

Example 02:

Calculate the total number of electrons in 1 mole of ammonia (NH₃).

Solution:

Nitrogen (N):

Atomic number = Z = 7

Number of electrons = Atomic number = 7

Hydrogen (H):

Atomic number = Z = 1

Number of electrons in 1 atom of hydrogen = Atomic number = 1

Number of electrons in 3 hydrogen atoms = 1 x 3 = 3

Number of electrons in 1 molecule of ammonia = 7 + 3 = 10

1 mole of ammonia contains 6.022 x 10²³ molecules of ammonia

Number of electrons in 1 mole of ammonia = 10 x 6.022 x 10²³ = 6.022 x 10²⁴

Ans: The number of electrons in 1 mole of ammonia = 6.022 x 10²⁴

Example 03:

Calculate the total number of electrons in 1 mole of methane (CH₄).

Solution:

Carbon (C):

Atomic number = Z = 6

Number of electrons = Atomic number = 6

Hydrogen (H):

Atomic number = Z = 1

Number of electrons in 1 atom of hydrogen = Atomic number = 1

Number of electrons in 4 hydrogen atoms = 1 x 4 = 4

Number of electrons in 1 molecule of ammonia = 6 + 4 = 10

1 mole of ammonia contains 6.022 x 10²³ molecules of ammonia

Number of electrons in 1 mole of ammonia = 10 x 6.022 x 10²³ = 6.022 x 10²⁴

Ans: The number of electrons in 1 mole of ammonia = 6.022 x 10²⁴

Example 04:

Find the total number and the total mass of neutrons in 7 m of ¹⁴C. Mass of neutron = 1.675 x 10^-27 kg.

Given mass of carbon = 7 mg = 7 x 10^-3 g

Molecular mass of carbon = 14 g

Number of moles of carbon = (7 x 10^-3)/14 = 5 x 10^-4  mol

Number of carbon atoms in 7 mg of ¹⁴C = 5 x 10^-4 x 6.022 x 10²³ = 3.011 x 10²⁰

Number of neutrons in 1 atom of carbon = A – Z = 14 – 6 = 8

Number of neutrons in 7 mg of ¹⁴C = 8 x 3.011 x 10²⁰ = 2.4088 x 10²¹

Mass of 1 neutron = 1.675 x 10^-27 kg.

Mass of neutron in 7 mg of ¹⁴C = 1.675 x 10^-27 x 2.4088 x 10²¹

Mass of neutron in 7 mg of ¹⁴C = 4.0347 x 10^-6 kg

Example 05:

Find total number and total mass of proton in 34 mg of ammonia at STP. Will the answer change if temperature and pressure are changed?

Given mass of ammonia = 34 mg = 34 x 10^-3 g

Molecular mass of ammonia = 14 + 3 = 17 g

Number of moles of ammonia = (34 x 10^-3)/17 = 2 x 10^-3 mol

Nitrogen (N):

Atomic number = Z = 7

Number of protons = Atomic number = 7

Hydrogen (H):

Atomic number = Z = 1

Number of protons in 1 atom of hydrogen = Atomic number = 1

Number of protons in 3 hydrogen atoms = 1 x 3 = 3

Number of protons in 1 molecule of ammonia = 7 + 3 = 10

Number of molecules in 34 mg of ammonia = 2 x 10^-3 x 6.022 x 10²³ = 1.2044 x 10²¹

Number of protons in 34 mg of ammonia = 10 x 1.2044 x 10²¹= 1.2044 x 10²²

Mass of 1 proton = 1.673 x 10^-27 kg.

Mass of protons in 34 mg of ammonia = 1.673 x 10^-27 x 1.2044 x 10²²

Mass of protons in 34 mg of ammonia = 2.015 x 10^-5 kg

Due to change in temperature and pressure, there is no change in number of moles of the gas. Hence there is no effect o change of temperature and pressure.

Example 06:

Calculate the number of electrons which will together weigh 1 gram.

Mass of 1 electron = 9.1 x 10^-31 kg = 9.1 x 10^-28 g

Number of electrons in 1 gram = 1/(9.1 x 10^-28) = 1.098 x 10²⁷ electrons

Example 07:

2 x 10⁸ atoms of carbon are arranged side by side, calculate the radius of carbon atom, if the length of this arrangement is 2.4 cm

Diameter of carbon atom = 2.4/(2 x 10⁸) = 1.2 x 10^-8  cm

Radius of carbon atom = (1.2 x 10^-8)/2 = 6 x 10^-9 cm = 6 x 10^-11 m = 0.6 x 10^-10  m = 0.6 angstrom

Example 08:

Find the total number and total mass of neutrons in 18 mL of water. The specific gravity of water = 1.

Given volume of water = 18 mL = 18 x 10^-3 L

Mass of water = Volume x density = 18 x 10^-3 L x 1 kg/L = 18 x 10^-3 kg = 18 g

Molecular mass of water= 2 + 16 = 18 g

Number of moles of water = 18/18 = 1 mol

Oxygen (O):

Atomic number = Z = 8

Atomic mass number = A = 16

Number of neutrons = 16 – 8 = 8

Hydrogen (H):

Atomic number = Z = 1

Atomic mass of hydrogen = A = 1

Number of neutrons in 1 atom of hydrogen = 1 – 1 = 0

Number of neutrons in 2 hydrogen atoms = 0 x 2 = 0

Number of neutrons in 1 molecule of water = 8 + 0 = 8

Number of molecules in 18 mL of water = 1  x 6.022 x 10²³ = 6.022 x 10²³

Number of neutrons in 18 mL of water = 8 x 6.022 x 10²³= 4.8176 x 10²⁴

Mass of 1 neutron = 1.675 x 10^-27 kg.

Mass of neutrons in 18 mL of water = 1.675 x 10^-27 x 4.8176 x 10²⁴

Mass of neutrons in 18 mL of water = 8.0695 x 10^-3 kg

Due to change in temperature and pressure, there is no change in number of moles of the gas. Hence there is no effect o change of temperature and pressure.

Science > Chemistry > Atomic Structure > Problems on Calculation of Number of Electrons, Protons, and Neutrons

The post Problems on Calculation of Mass of Electrons, Protons, and Neutrons appeared first on The Fact Factor.

In this article, we shall study to solve problems based on the calculation of atomic number, atomic mass number, and neutron number

Atomic number (Z) :

The number of protons (positive charge) present in the nucleus of an atom of a particular element is called the atomic number of that element. It is denoted by letter ‘Z’.

Neutron number (N): 

The number of neutrons present in the nucleus of an atom is known as neutron number. It is denoted by ‘N’

Mass number (A): 

The total number of protons and neutrons present in the nucleus of an atom of the element is called mass number. The mass number is denoted as ‘A’.

A  =   Z + N

Example 01:

Calculate the number of electrons, protons, and neutrons in the following atoms.

Atomic Number = Z = 15

Atomic Mass Number = A = 31

Number of protons = atomic number = 15

Number of electrons = atomic number = 15

Neutron number = Number of neutrons = A – Z = 31 – 15 = 16

Number of nucleons = Atomic mass number = 31

Atomic Number = Z = 12

Atomic Mass Number = A = 24

Number of protons = atomic number = 12

Number of electrons = atomic number = 12

Neutron number = Number of neutrons = A – Z = 24 – 12 = 12

Number of nucleons = Atomic mass number = 24

Atomic Number = Z = 17

Atomic Mass Number = A = 37

Number of protons = atomic number = 17

Number of electrons = atomic number = 17

Neutron number = Number of neutrons = A – Z = 37 – 17 = 20

Number of nucleons = Atomic mass number = 37

Atomic Number = Z = 18

Atomic Mass Number = A = 40

Number of protons = atomic number = 18

Number of electrons = atomic number = 18

Neutron number = Number of neutrons = A – Z = 40 – 18 = 22

Number of nucleons = Atomic mass number = 40

Atomic Number = Z = 35

Atomic Mass Number = A = 80

Number of protons = atomic number = 35

Number of electrons = atomic number = 35

Neutron number = Number of neutrons = A – Z = 80 – 35 = 45

Number of nucleons = Atomic mass number = 80

Atomic Number = Z = 20

Atomic Mass Number = A = 40

Number of protons = atomic number = 20

Number of electrons = atomic number = 20

Neutron number = Number of neutrons = A – Z = 40 – 20 = 20

Number of nucleons = Atomic mass number = 40

Atomic Number = Z = 6

Atomic Mass Number = A = 13

Number of protons = atomic number = 6

Number of electrons = atomic number = 6

Neutron number = Number of neutrons = A – Z = 13 – 6 = 7

Number of nucleons = Atomic mass number = 13

Atomic Number = Z = 8

Atomic Mass Number = A = 16

Number of protons = atomic number = 8

Number of electrons = atomic number = 8

Neutron number = Number of neutrons = A – Z = 16 – 8 = 8

Number of nucleons = Atomic mass number = 16

Atomic Number = Z = 26

Atomic Mass Number = A = 56

Number of protons = atomic number = 26

Number of electrons = atomic number = 26

Neutron number = Number of neutrons = A – Z = 56 – 26 = 30

Number of nucleons = Atomic mass number = 56

Atomic Number = Z = 38

Atomic Mass Number = A = 88

Number of protons = atomic number = 38

Number of electrons = atomic number = 38

Neutron number = Number of neutrons = A – Z = 88 – 38 = 50

Number of nucleons = Atomic mass number = 88

Example 02:

Calculate the number of electrons, protons, and neutrons in the following species.

Atomic Number = Z = 15

Atomic Mass Number = A = 31

Number of protons = atomic number = 15

P^3- → P + 3 e^–

Number of electrons = 15 + 3 = 18

Number of neutrons = A – Z = 31 – 15 = 16

Number of nucleons = Atomic mass number = 31

Atomic Number = Z = 35

Atomic Mass Number = A = 80

Number of protons = atomic number = 35

Br^– → Br + e^–

Number of electrons = 35 + 1= 36

Number of neutrons = A – Z = 80 – 35 = 45

Number of nucleons = Atomic mass number = 80

Atomic Number = Z = 20

Atomic Mass Number = A = 40

Number of protons = atomic number = 20

Ca²⁺ → Ca – 2e^–

Number of electrons = 20 – 2= 18

Number of neutrons = A – Z = 40 – 20 = 20

Number of nucleons = Atomic mass number = 40

H₃PO₄

Hydrogen (H):

Atomic Number = Z = 1

Atomic Mass Number = A = 1

Number of protons per atom= atomic number = 1

Number of electrons per atom = Z = 1

Number of neutrons per atom = A – Z = 1 – 1 = 0

There are 3 hydrogen atoms in the molecule

Total number of protons in 3 hydrogens = 1 x 3 = 3

Total number of electrons in 3 hydrogens = 1 x 3 = 3

Total number of neutrons in 3 hydrogens = 0 x 3 = 0

Phosphorous (P):

Atomic Number = Z = 15

Atomic Mass Number = A = 31

Number of protons per atom= atomic number = 15

Number of electrons per atom = Z = 15

Number of neutrons per atom = A – Z = 31 – 15 = 16

There is 1 phosphorous atom in the molecule

Total number of protons in 1 phosphorous = 15 x 1 = 15

Total number of electrons in 1 phosphorous = 15 x 1 = 15

Total number of neutrons in 1 phosphorous = 16 x 1 = 16

Oxygen (O):

Atomic Number = Z = 8

Atomic Mass Number = A = 16

Number of protons per atom= atomic number = 8

Number of electrons per atom = Z = 8

Number of neutrons per atom = A – Z = 16 – 8 = 8

There are 4 oxygen atom in the molecule

Total number of protons in 4 oxygen= 8 x 4 = 32

Total number of electrons in 4 oxygen = 8 x 4 = 32

Total number of neutrons in 4 oxygen = 8 x 4 = 32

Ans:

Total number of protons in H₃PO₄ = 3 + 15 + 32 = 50

Total number of electrons in H₃PO₄ = 3 + 15 + 32 = 50

Total number of neutrons in H₃PO₄ = 0 + 16 + 32 = 48

NH₄⁺

Nitrogen (N):

Atomic Number = Z = 7

Atomic Mass Number = A = 14

Number of protons per atom= atomic number = 14

Number of electrons per atom = Z = 14

Number of neutrons per atom = A – Z = 14 – 7 = 7

Hydrogen (H):

Atomic Number = Z = 1

Atomic Mass Number = A = 1

Number of protons per atom= atomic number = 20

Number of electrons per atom = Z = 1

Number of neutrons per atom = A – Z = 1 – 1 = 0

There are 4 hydrogen atoms in the species

Total number of protons in 3 hydrogens = 1 x 4 = 4

Total number of electrons in 3 hydrogens = 1 x 4 = 4

Total number of neutrons in 3 hydrogens = 0 x 4 = 0

Total number of protons in NH₄⁺ = 7 + 4 = 11

NH₄⁺ → NH₄ – e^–

Total number of electrons in NH₄⁺ = 7 + 4 – 1 = 10

Total number of neutrons in NH₄⁺ = 7 + 4 = 11

ClO₃^–

Chlorine (Cl):

Atomic Number = Z = 17

Atomic Mass Number = A = 37

Number of protons per atom= atomic number = 17

Number of electrons per atom = Z = 17

Number of neutrons per atom = A – Z = 37 – 17 = 20

Oxygen (O):

Atomic Number = Z = 8

Atomic Mass Number = A = 16

Number of protons per atom= atomic number = 8

Number of electrons per atom = Z = 8

Number of neutrons per atom = A – Z = 16 – 8 = 8

There are 3 oxygen atom in the species

Total number of protons in 4 oxygen= 8 x 3 = 24

Total number of electrons in 4 oxygen = 8 x 3 = 24

Total number of neutrons in 4 oxygen = 8 x 3 = 24

Total number of protons in ClO₃^– = 17 + 24 = 41

ClO₃^– → ClO₃ + e^–

Total number of electrons in ClO₃^– = 17 + 24 + 1 = 42

Total number of neutrons in ClO₃^– = 20 + 24 = 44

Example 03:

Complete the table

Solution:

Zn

Atomic Number = Z = 30

Atomic Mass Number = A = 64

Number of protons = atomic number = 30

Number of electrons = atomic number = 30

Number of neutrons = A – Z = 64 – 30 = 34

Sr²⁺

Atomic Number = Z = 38

Atomic Mass Number = A = 90

Number of protons = atomic number = 38

Sr²⁺ → Sr – 2e^–

Number of electrons = 38 – 2 = 36

Number of neutrons = A – Z = 90 – 38 = 52

Te

Number of protons = 43

Number of Neutrons = N = 56

Atomic number =Number of protons = Z = 43

Atomic mass number = Z + N = 43 + 56 = 99

Number off electrons = Atomic number = 43

Br^–

Number of neutrons = N = 44

Number of electrons = 36

Br^– → Br + e^–

Number of protons = 36 – 1 = 35

Atomic Number = Number of protons = Z = 35

Atomic mass number = Z + N = 35 + 44 = 79

N

Number of neutrons = N = 7

Number of electrons = 7

Number of protons = 7

Atomic Number = Number of protons = Z = 7

Atomic mass number = Z + N = 7 + 7 = 14

Ca²⁺

Atomic Number = Z = 20

Number of neutrons = N = 20

Atomic Mass Number = Z + N = 20 + 20 = 40

Number of protons = atomic number = 20

Ca²⁺ → Ca – 2e^–

Number of electrons = 20 – 2 = 18

O

Atomic Number = Z = 8

Atomic Mass Number = A = 16

Number of protons = atomic number = 8

Number of electrons = atomic number = 8

Number of neutrons = A – Z = 16 – 8 = 8

The completed table is as follows:

Example 04:

The numbers of electrons, protons, and neutrons in a monoatomic species are equal to 36, 35, and 45 respectively. Assign proper symbol.

Solution:

Number of electrons = 36

Number of protons = Atomic number = Z = 35

Number of neutrons = N = 45

Atomic mass number = A = Z + N = 35 + 45 = 80

Charge on species = Z – Number of electrons = 35 – 36 = -1

The species is

Example 05:

The numbers of electrons, protons, and neutrons in a monoatomic species are equal to 18, 16, and 16 respectively. Assign proper symbol.

Solution:

Number of electrons = 18

Number of protons = Atomic number = Z = 16

Number of neutrons = N = 16

Atomic mass number = A = Z + N = 16 + 16 = 32

Charge on species = Z – Number of electrons = 16 – 18 = -2

The species is

Example 06:

Find the number of electrons in fluorine atom, fluorine molecule, and fluoride ion. The atomic number and mass number of fluorine are 9 and 19 respectively.

Solution:

Atomic number = Z = 9

Atomic mass number A = 19

Fluorine Atom (F):

Number of protons = Atomic number = 9

Number of Protons = Atomic number = 9

Number of electrons = Atomic number = 9

Number of neutrons = a – z = 19 – 9 =10

Fluorine Molecule (F₂):

There are 2 atoms in a molecule of fluorine

Number of protons in fluorine molecule= 9 x 2 = 18

Number of electrons in fluorine molecule= 9 x 2 = 18

Number of neutrons in fluorine molecule= 10 x 2 = 20

Fluoride Ion (F^–):

Number of Protons = Atomic number = 9

F^– → F + e^–

Number of electrons = 9 + 1 = 10

Number of neutrons = a – z = 19 – 9 =10

Example 07:

An isotope of atomic mass 24 had 12 neutrons in its nucleus. What is its atomic number? Represent the isotope in symbolic form.

Solution:

Atomic mass number = A = 24

Number of neutrons = N = 12

Number of protons = A – Z = 24 – 12= 12

Atomic number = Number of protons = Z = 12

The element is

Isotopes, Isotones, and Isobars

• Different atoms of the same element having the same atomic number but having different mass numbers are known as isotopes.
• Atoms of the different elements having a different atomic number but having the same mass numbers are known as isobars.
• Atoms of the different elements having the different atomic number, different mass number but having the same neutron number are known as isotones.

Example 08:

Write the complete symbol for the atom with the given atomic number (Z) and atomic mass (A):

• Z = 17 and A = 35

• Z = 92, A = 233

• Z = 4, A = 9

Example 09:

Give an isobar, an isotone, and an isotope of

Isobar
Isotone
Isotope

Science > Chemistry > Atomic Structure > Problems Based on Atomic Number, Mass Number, and Neutron Number

The post Problems Based on Atomic Number, Mass Number, and Neutron Number appeared first on The Fact Factor.

In this article, we shall study the concept of quantum numbers and the model of an atom based on the quantum numbers.

Concept of Quantum Numbers:

In 1926, Erwin Schrodinger put forward a theory of atom called as a quantum mechanical theory. Quantum mechanical theory is based on two principles. a) Wave-particle duality of subatomic particles and b) Uncertainty principle.

Wave-particle duality principle of subatomic particles was put forward by L. DeBroglie it states that “All the subatomic particles electrons, protons, and neutrons possess both particle and wave properties”.

Uncertainty principle was put forward by W. Heisenberg it states that “It is impossible to determine the position and momentum of a particle simultaneously”. It means that if the position of the electron in an atom is known then there is uncertainty about its momentum and when the momentum of an electron in an atom is known there is uncertainty about the position of the electron.

According to this theory, Each electron shell of an atom is composed of several subshells. To specify the position of the electron in an atom each electron is assigned four quantum numbers and these numbers give location and energy of the electron.

The numbers which assign the position and energy to an electron are called quantum numbers. There is a set of four quantum numbers associated with each electron of an atom. The four quantum numbers are a) Principal quantum number ‘n’  b) Azimuthal quantum number ‘l’ c) Magnetic quantum number ‘m’  d) Spin quantum number ‘s’

Principal Quantum Number (n):

This quantum number was introduced by Neil Bohr. It is denoted by the letter ‘n’. It gives the main energy level (shell) to which the electron belongs. It defines the distance of the electron from the nucleus and energy level. These energy levels are also designated by letters as follows:

Azimuthal Quantum Number (l):

It was introduced by Somerfield.  This quantum number refers to the energy sub-level (subshell).  It is denoted by the letter ‘l’. It can have any integral value from 0 to (n – 1). This quantum number indicates the subshell in which the electron is present and the number of subshells present in a principal shell.

It is also responsible for the particular shape of the particular shell because it defines the spatial distribution of the electron cloud around the nucleus. Thus the subshell (Orbital) with l = 0 is called ‘s’ orbital. The subshell (Orbital) with l = 1 is called ‘p’ orbital. The subshell (Orbital) with l = 2 is called ‘d’ orbital. The subshell (Orbital) with l = 3  is called ‘f’ orbital

It is used to calculate angular momentum of an electron in particular shell using the formula

Each subshell contains maximum 2 (2l +1) electrons. It also decides the number of nodal planes of the orbital. For s orbital l = 0, hence s orbital there is no nodal plane. For p orbital l = 1, hence p orbital there is one nodal plane. For d orbital l = 2, hence d orbital there are two nodal planes.

Notes:

For the first orbit:

n = 1, then  l = 0 to (1 – 1) = 0 to 0,  then  l = 0. Thus in the first main energy level, there is only one subshell i.e. ‘s’ orbital.

For the second orbit:

n = 2 then  l = 0 to (2 – 1) = 0 to 1, then l = 0, 1. Thus in the second main energy level, there are two subshells i.e. ‘s’ orbital and ‘p’ orbital.For the Third orbit n = 3 Then  = 0 to (3 – 1) = 0 to 2 then             = 0, 1, 2. Thus in the third main energy level there

For the Third orbit:

n = 3 then  l = 0 to (3 – 1) = 0 to 2 then l = 0, 1, 2. Thus in the third main energy level, there are three subshells i.e. ‘s’ orbital, ‘p’ orbital and ‘d’ orbital.

For the fourth orbit:

n = 4 Then l = 0 to (4 – 1) = 0 to 3 then l  = 0, 1, 2, 3. Thus in the fourth main energy level, there are four subshells i.e. ‘s’ orbital, ‘p’ orbital, ‘d’ orbital, and ‘f’ orbital.

Magnetic Quantum Number (m):

This quqntum number is denoted by letter ‘m’. This number was introduced to explain the splitting of atomic spectra in a magnetic field (Zeeman effect).

For Azimuthal number ‘l’ possible values of magnetic quantum numbers are – l  to + l   through 0. i.e. m =  , …., -3, -2, -1, 0, 1, 2, 3, ….., . This quantum number gives the possible orientation of the orbital in the space. For a given value of l, there are (2l +1) values of m.

This number is also known as the orientation quantum number because it gives the distribution of electron clouds in space around the nucleus.

The magnetic quantum number tells to which orbital the electron belongs. It also tells the number of orbitals which corresponds to a particular shell. For example: s subshell l = 0 therefore m = 0. This means that s subshell contains only one orbital.

p subshell l = 1 therefore m = -1, 0, +1. This means that p subshell contains three orbitals. They are designated as Px, Py and Pz,  along the x-axis, along the y-axis and along the z-axis respectively. Thus p orbitals have three orientations in space along co-ordinate axes. In absence of magnetic field, these orbitals are equivalent in energy and are said to have three fold degenerate.

d subshell l = 2 therefore m = -2, -1, 0, +1, +2. This means that d subshell contains five orbitals. They are designated as dxy, dyz, dzx, dx2-y2, and dz2.   Thus d orbitals have five orientations in space. In absence of magnetic field, these orbitals are equivalent in energy and are said to have five fold degenerate.

f subshell l = 3 therefore m = -3, -2, -1, 0, +1, +2, +3. This means that f subshell contains seven orbitals and have seven orientations in space. In absence of magnetic field, these orbitals are equivalent in energy and are said to have seven fold degenerate.

Spin Quantum Number (s):

It is denoted by letter ‘s’. The electron in orbital motion can spin clockwise or anticlockwise about its own axis, hence two values of quantum number are possible.  These values are taken as + 1/2 and – 1/2.

Opposite spins are called antiparallel spins. The same spin is called the parallel spin. The spin quantum number indicates a magnetic moment associated with the electron.

Science > Chemistry > Atomic Structure > Quantum Numbers and Quantum Model of an Atom

The post Quantum Numbers and Quantum Model of an Atom appeared first on The Fact Factor.

In this article, we shall study the concept of quantum numbers and different types of orbitals and their shapes.

Difference Between Shell, Subshell, and Orbital:

• All electrons that have the same value for n (the principal quantum number) are in the same shell.Within a shell (same n), all electrons that share the same  (the angular momentum quantum number, or orbital shape) are in the same subshell. The electrons in subshell have same principal quantum number, same azimuthal quantum number and  differ  in magnetic and spin quantum number
• When electrons share the same n, l, and m, we say they are in the same orbital (they have the same energy level, shape, and orientation). The electrons in orbital differ only in spin quantum number.

Shapes of Orbitals:

• Node: There is a region where the probability of finding the electron is zero. It is called a node.
• Nodal Plane: A plane passing through the nucleus on which the probability of finding an electron is zero is called the nodal plane. The number of nodal planes in an orbital is equal to the azimuthal quantum number.
• Spherical or Radial Node: A spherical surface within an orbital on which the probability of finding the electron is zero is called a spherical or radial node.

Types of Orbitals:

s – orbital:

• For s orbital Azimuthal quantum number  = 0 and the magnetic quantum number m = 0 hence s orbitals have unique orientation in space. Thus s orbital corresponds to spherical shape with the atomic nucleus at its centre.
• For every value of ‘n’, there is one ‘s’ orbital i.e. s orbitals are present in all principal energy levels.  Its radius depends on the value of n. As the value of n increases the size of the s orbital also increases.
• It has no nodal plane.

p – orbital:

• For p orbital Azimuthal quantum number l = 1 and the magnetic quantum number  m = -1, 0, +1. Hence p orbitals have three orientations in space.
• Thus p orbital corresponds to dumb-belled shape with the atomic nucleus at its center.
• p orbitals have two lobes directed on opposite sides of the nucleus. P-orbitals are orientated in three different directions along X, Y and Z axis of the usual coordinate system.  These orbitals are designated as P_x, P_y & P_z orbitals.
• p-orbital have one nodal plane.

d – orbital:

• For d orbital Azimuthal quantum number l = 2 and the magnetic quantum number  m = -2, -1, 0, +1, +2. Hence d orbitals have five orientations in space.
• Thus d orbital corresponds to 4 double dumb-belled shapes (d_xy, d_yz, d_zx, d_x²_y²) with the atomic nucleus at its centre and one dumb belled with dough nut shaped (d_z²).
• d orbital has two nodal planes.

f – orbital:

• For f orbital Azimuthal quantum number l = 3 and the magnetic quantum number m = -3. -2, -1, 0, +1, +2, +3. Hence f orbitals have seven orientations in space.
• f orbital has complex shapes with the atomic nucleus at its centre.
• f orbital has three nodal planes.

Quantum Numbers of Electrons in First Three Shells (Orbits)

Science > Chemistry > Atomic Structure > Quantum Numbers and Shapes of Orbitals

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The distribution of electrons in various shells and orbitals of an atom is known as electronic configuration. Generally electronic configuration is given by nl^x. Where n is shell number, l is subshell and x are the number of electrons.

Aufbau Principle:

The arrangement of the electrons in an atom is called the electronic configuration of an atom. The method of filling up or building up a sequence of energy levels for electrons in an atom is based on Aufbau principle. In German, the word aufbau means building up.

According to this principle, orbitals are filled in the order of increasing energy. The lowest energy level is filled first. The higher energy levels are then filled up one after another. This gives the atom the maximum stability and the atom is said to be in ground state or the normal state.

The order of increasing energy levels is 1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4f < 5d < 6p < 7s < 5f < 6d < 7p < 6f Diagrammatically it is represented as follows

Pauli’s Exclusion Principle:

“No two electrons. in an atom can have the same values for all four quantum numbers”. i.e. to exist in the same atom the two electrons can have three quantum numbers identical but fourth must be different.

Example: Let us consider electrons in the first orbit of helium atom

Here we can see that the three values of quantum number are the same but fourth value is different. Thus Pauli’s exclusion principle helps in calculating the maximum number of electrons present in any energy level (Orbit).

Hund’s Rule of Maximum Multiplicity:

In a given subshell, electrons occupy all the available orbitals singly, before they pair up in any orbital.

Example 1:

Consider electronic configuration of phosphorous. The atomic number of phosphorous is 15. Its basic electronic configuration is  2, 8, 5. The detailed electronic configuration is 1s², 2s² 2p⁶, 3s² 3p³. The box diagram for final orbit configuration for P is as below

Here we can see that the electrons in p orbitals occupy the position singly till all orbitals are singly filled.

Example – 2:

Consider electronic configuration of sulphur. The atomic number of sulphur is 16. Its basic electronic configuration is 2, 8, 6. The detailed electronic configuration is 1s², 2s² 2p⁶, 3s² 3p⁴. The box diagram for final orbit configuration for S is as below

Here we can see that the electrons in p orbital occupy the position singly till all orbitals are singly filled and then the pairing starts.

Stability of Completely Filled and Half Filled Subshells:

The ground state electronic configuration of the atom of an element always corresponds to the state of the lowest total electronic energy.

The electronic configurations of most of the atoms follow the basic rules viz. Aufbau principle, Pauli’s exclusion principle, and Hund’s rule. However, in certain elements such as Cu, or Cr, where the two subshells (4s and 3d) differ slightly in their energies, an electron shifts from a subshell of lower energy (4s) to a subshell of higher energy (3d), provided such a shift results in all orbitals of the subshell of higher energy getting either completely filled or exactly half filled.

The valence electronic configurations of Cr and Cu, therefore, are 3d⁵ 4s¹ and 3d¹⁰ 4s¹ respectively and not 3d⁴ 4s² and 3d⁹ 4s². It has been found that there is extra stability associated with these electronic configurations.

Electronic Configuration of copper (Z = 29):

Expected configuration: 1s², 2s² 2p⁶, 3s² 3p⁶, 4s² 3d⁹.

Actual configuration: 1s², 2s² 2p⁶, 3s² 3p⁶, 4s¹ 3d¹⁰.

Electronic Configuration of chromium (Z = 24):

Expected configuration:  1s², 2s² 2p⁶, 3s² 3p⁶, 4s² 3d⁴.

Actual configuration:  1s², 2s² 2p⁶, 3s² 3p⁶, 4s¹ 3d⁵.

Science > Chemistry > Atomic Structure > Electronic Configuration

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