In the previous section, we saw the models proposed by J.J. Thomson and Ernst Rutherford. In this section we will see details about atomic number and mass number
1. We saw that in an atom, there are electrons, protons and neutrons
• Each electron has a negative charge of -1.602176 ×10-19 C
• Each proton has a positive charge of +1.602176 ×10-19 C
2. The magnitudes of both the charges are equal
• But the signs are opposite
3. So the number of protons and electrons in an atom must be equal
• Otherwise, the atom will not be electrically neutral
4. The number of protons in an atom is called the atomic number of that atom
• The symbol of atomic number is Z
Some examples:
• The number of protons in hydrogen atom is 1
♦ So Z of hydrogen atom is 1
• The number of protons in sodium atom is 11
♦ So Z of sodium atom is 11
5. We can write:
Atomic number (Z) of an atom
= Number of protons in the nucleus of that atom
= Number of electrons in that ‘atom in neutral state’
• ‘Atom in neutral state’ has to be specifically written because, if it is not electrically neutral, the number of electrons will not be equal to the number of protons
6. Next we consider mass number
• Protons and neutrons present in the nucleus are collectively known as nucleons
• The total mass of the nucleons is equal to the mass of the atom
• The total number of nucleons in an atom is called the 'mass number' of that atom
• The symbol of mass number is A
7. We can write:
• Mass number of an atom = Number of protons (Z) + Number of neutrons (n)
• We have seen the details in our previous classes.
8. We know that every element is represented by a unique symbol
• Z and A can be attached to that symbol
♦ Z is written as a subscript on the left side of the symbol
♦ A is written as a superscript on the left side of the symbol
9. Thus, if ‘X’ is the symbol of an element and Z and A the atomic and mass numbers, we can write: $\mathbf\small{\rm{^A_ZX}}$
• Isobars are elements with same mass number (A) but different atomic numbers (Z)
Example: $\mathbf\small{\rm{^{14}_{\;6}C\;\;\text{and}\;\;^{14}_{\;7}N}}$
Details about isotopes can be written in 6 steps:
1. Consider the atom of hydrogen
• It has 1 proton, 0 neutron and 1 electron
♦ So Z = Number of protons = 1
♦ Also, A = (Z+n) = (1+0) = 1
• Symbolically we can represent this as $\mathbf\small{\rm{^1_1H}}$
• It is called protium
2. Consider the atom of deuterium
• It has 1 proton, 1 neutron and 1 electron
♦ So Z = Number of protons = 1
♦ Also, A = (Z+n) = (1+1) = 2
• Symbolically we can represent this as $\mathbf\small{\rm{^2_1D}}$
3. Consider the atom of tritium
• It has 1 proton, 2 neutron and 1 electron
♦ So Z = Number of protons = 1
♦ Also, A = (Z+n) = (1+2) = 3
• Symbolically we can represent this as $\mathbf\small{\rm{^3_1T}}$
4. In all the above 3 cases, Z is the same. But A are different
• The difference in A is due to the difference in n
■ Elements with same atomic number (Z) but different mass numbers (A) are called isotopes
• $\mathbf\small{\rm{^1_1H,\;\;\,^2_1D\;\;\text{and}\;\;^3_1T}}$ are isotopes
5. If we take a sample of naturally occurring hydrogen, we get the following quantities:
(i) 99.985% of that sample will be $\mathbf\small{\rm{^1_1H}}$
(ii) The rest 0.015% will be $\mathbf\small{\rm{^2_1D}}$
(iii) $\mathbf\small{\rm{^3_1T}}$ is found only in traces on earth
6. Let us see some more examples of isotopes:
• Isotopes of carbon:
(i) $\mathbf\small{\rm{^{12}_{\;6}C}}$ has 6 protons and 6 neutrons
(ii) $\mathbf\small{\rm{^{13}_{\;6}C}}$ has 6 protons and 7 neutrons
(iii) $\mathbf\small{\rm{^{14}_{\;6}C}}$ has 6 protons and 8 neutrons
• Isotopes of chorine:
(i) $\mathbf\small{\rm{^{35}_{17}Cl}}$ has 17 protons and 18 neutrons
(ii)$\mathbf\small{\rm{^{37}_{17}Cl}}$ has 17 protons and 20 neutrons
1. The chemical properties of an element are controlled by the number of electrons in it’s atoms
2. The number of neutrons have very little effect on the chemical properties
3. The number of electrons is same as the atomic number (Z)
• So with regard to chemical properties, the number Z is what we consider most
4. All isotopes of an element will have the same Z
• So all isotopes of an element will show the same chemical behaviour
Solved example 2.1
Calculate the number of protons, neutrons and electrons in $\mathbf\small{\rm{^{80}_{35}Br}}$
Solution:
1. Given : Z = 35, A = 80
2. We have: Number of protons = Number of electrons = Z = 35
3. Also we have: Number of neutrons + Number of protons = A
⇒ Number of neutrons = A - Number of protons
⇒ Number of neutrons = (80-35) = 45
Solved example 2.2
The number of electrons, protons and neutrons in a species are equal to 18, 16 and 16 respectively. Assign the proper symbol to the species
Solution:
1. Given:
• Number of electrons = 18
• Number of protons = 16
♦ So Z = 16
♦ From the periodic table, we have: Z = 16 for sulphur (S)
• Number of neutrons = 16
2. We have: A = Number of neutrons + Number of protons
⇒ A = (16+16) = 32
3. In a neutral atom, the number of electrons will be equal to that of the protons
• But in this case, number of electrons is greater by 2
• So there will be an excess of two negative charges
• So the symbol is: $\mathbf\small{\rm{^{32}_{16}S^{2-}}}$
4. Note:
• If the number of protons and electrons are not equal, the given species is an ion
• We have to determine whether it is a cation or an anion
• If Number of protons > Number of electrons
♦ It is a +ve ion (cation)
• If Number of protons < Number of electrons
♦ It is a -ve ion (anion)
Solved example 2.3
(i) Calculate the number of electrons which will together weigh one gram
(ii) Calculate the mass and charge of one mole of electrons
Solution:
Part (i)
• Mass of one electron = 9.10939 ×10-31 kg = 9.10939 ×10-28 g
• So number of electrons in 1 g
= $\mathbf\small{\frac{1}{9.10939\times 10^{-28}}=0.1097768\times 10^{28}=1.098\times 10^{27}}$
Part (ii)
• One mole of electrons will contain 6.022 × 1023 electrons
• Mass of one electron = 9.10939 ×10-31 kg
• So mass of one mole of electrons = (6.022 × 1023 × 9.10939 ×10-31) = 5.48 ×10-7 kg
• Charge of one electron = -1.602176 ×10-19 C
• So charge of one mole of electron = (6.022 × 1023 × -1.602176 ×10-19) = 9.65 ×104 C
Solved example 2.4
(i) Calculate the total number of electrons present in one mole of methane
(ii) Find (a) the total number and (b) the total mass of neutrons in 7 mg of $\mathbf\small{\rm{^{14}C}}$
(Assume that mass of a neutron = 1.675 × 10-27 kg).
(iii) Find (a) the total number and (b) the total mass of protons in 34 mg of NH3 at STP.
Will the answer change if the temperature and pressure are changed ?
Solution:
Part (i)
1. The molecular formula of methane is CH4
• So in one molecule of methane, there are:
♦ 1 atom of C
♦ 4 atoms of H
2. In one atom of C there are 6 electrons
• In one atom of H there is 1 electron
• So in 1 molecule of methane, there are (1×6 + 4×1) = 10 electrons
3. One mole of methane contains 6.022 × 1023 molecules of methane
• So in one mole of methane, there will be 6.022 × 1023 × 10 = 6.022 × 1024 electrons
Part (ii)
1. Molar mass of $\mathbf\small{\rm{^{14}C}}$ is 14 g
⇒ In 14 g of $\mathbf\small{\rm{^{14}C}}$, there will be 6.022 × 1023 atoms
⇒ In 1 g, there will be (6.022 × 1023 / 14) atoms
⇒ In 1 mg there will be (6.022 × 1023 / 14000) atoms
⇒ In 7 mg there will be (6.022 × 1023 / 2000) = 3.011 × 1020 atoms
2. Number of neutrons in 1 $\mathbf\small{\rm{^{14}C}}$ atom = (A-Z) = (14-6) = 8
• So total number of neutrons in 7 mg = 8 × 3.011 × 1020 = 24.088 × 1020
3. Mass of this much neutrons = (24.088 × 1020 × 1.675 × 10-27)
= 40.35 × 10-7
= 4.035 × 10-6 kg
Part (iii)
1. Molar mass of NH3 = (14+3) = 17 g
⇒ 17 g of NH3 at STP will contain 6.022 × 1023 molecules
⇒ 1 g of NH3 at STP will contain (6.022 × 1023 / 17) molecules
⇒ 1 mg of NH3 at STP will contain (6.022 × 1023 / 17000) molecules
⇒ 34 mg of NH3 at STP will contain (6.022 × 2 × 1020) molecules
2. one molecule of NH3 will contain (7+3) = 10 protons
• So number of protons in (6.022 × 2 × 1020) molecules
= 6.022 × 2 × 10 × 1020
= 12.044 × 1021
= 1.2044 × 1022
3. Mass of this much protons
= (1.2044 × 1022 × mass of one proton)
= (1.2044 × 1022 × 1.67262 × 10-27)
= 2.015 × 10-5 kg
Solved example 2.5
How many neutrons and protons are there in the following nuclei ?
$\mathbf\small{\rm{^{13}_{\;6}C,\;\;^{16}_{\;8}O,\;\;^{24}_{12}Mg,\;\;^{56}_{26}Fe,\;\;^{88}_{38}Sr}}$
Solution:
Part (i)
1. Given : Z = 6, A = 13
2. We have: Number of protons = Z = 6
3. We have: Number of neutrons + Number of protons = A
⇒ Number of neutrons = A - Number of protons
⇒ Number of neutrons = (13-6) = 7
Part (ii)
1. Given : Z = 8, A = 16
2. We have: Number of protons = Z = 8
3. We have: Number of neutrons = A - Number of protons
⇒ Number of neutrons = (16-8) = 8
Part (iii)
1. Given : Z = 12, A = 24
2. We have: Number of protons = Z = 12
3. We have: Number of neutrons = A - Number of protons
⇒ Number of neutrons = (24-12) = 12
Part (iv)
1. Given : Z = 26, A = 56
2. We have: Number of protons = Z = 26
3. We have: Number of neutrons = A - Number of protons
⇒ Number of neutrons = (56-26) = 30
Part (v)
1. Given : Z = 38, A = 88
2. We have: Number of protons = Z = 38
3. We have: Number of neutrons = A - Number of protons
⇒ Number of neutrons = (88-38) = 50
Solved example 2.6
Write the complete symbol for the atom with the given atomic number (Z) and atomic mass (A)
(i) Z = 17 , A = 35
(ii) Z = 92 , A = 233
(iii) Z = 4 , A = 9
Solution:
Part (i)
• Given: Z = 17, A = 35
♦ From the periodic table, we have: Z = 17 for chlorine (Cl)
• So symbol is $\mathbf\small{\rm{^{35}_{17}Cl}}$
Part (ii)
• Given: Z = 92, A = 233
♦ From the periodic table, we have: Z = 92 for uranium (U)
• So symbol is $\mathbf\small{\rm{^{233}_{92}U}}$
Part (iii)
• Given: Z = 4, A = 9
♦ From the periodic table, we have: Z = 4 for beryllium (Be)
• So symbol is $\mathbf\small{\rm{^{9}_{4}Be}}$
1. We saw that in an atom, there are electrons, protons and neutrons
• Each electron has a negative charge of -1.602176 ×10-19 C
• Each proton has a positive charge of +1.602176 ×10-19 C
2. The magnitudes of both the charges are equal
• But the signs are opposite
3. So the number of protons and electrons in an atom must be equal
• Otherwise, the atom will not be electrically neutral
4. The number of protons in an atom is called the atomic number of that atom
• The symbol of atomic number is Z
Some examples:
• The number of protons in hydrogen atom is 1
♦ So Z of hydrogen atom is 1
• The number of protons in sodium atom is 11
♦ So Z of sodium atom is 11
5. We can write:
Atomic number (Z) of an atom
= Number of protons in the nucleus of that atom
= Number of electrons in that ‘atom in neutral state’
• ‘Atom in neutral state’ has to be specifically written because, if it is not electrically neutral, the number of electrons will not be equal to the number of protons
6. Next we consider mass number
• Protons and neutrons present in the nucleus are collectively known as nucleons
• The total mass of the nucleons is equal to the mass of the atom
• The total number of nucleons in an atom is called the 'mass number' of that atom
• The symbol of mass number is A
7. We can write:
• Mass number of an atom = Number of protons (Z) + Number of neutrons (n)
• We have seen the details in our previous classes.
8. We know that every element is represented by a unique symbol
• Z and A can be attached to that symbol
♦ Z is written as a subscript on the left side of the symbol
♦ A is written as a superscript on the left side of the symbol
9. Thus, if ‘X’ is the symbol of an element and Z and A the atomic and mass numbers, we can write: $\mathbf\small{\rm{^A_ZX}}$
Isobars
Example: $\mathbf\small{\rm{^{14}_{\;6}C\;\;\text{and}\;\;^{14}_{\;7}N}}$
Isotopes
Details about isotopes can be written in 6 steps:
1. Consider the atom of hydrogen
• It has 1 proton, 0 neutron and 1 electron
♦ So Z = Number of protons = 1
♦ Also, A = (Z+n) = (1+0) = 1
• Symbolically we can represent this as $\mathbf\small{\rm{^1_1H}}$
• It is called protium
2. Consider the atom of deuterium
• It has 1 proton, 1 neutron and 1 electron
♦ So Z = Number of protons = 1
♦ Also, A = (Z+n) = (1+1) = 2
• Symbolically we can represent this as $\mathbf\small{\rm{^2_1D}}$
3. Consider the atom of tritium
• It has 1 proton, 2 neutron and 1 electron
♦ So Z = Number of protons = 1
♦ Also, A = (Z+n) = (1+2) = 3
• Symbolically we can represent this as $\mathbf\small{\rm{^3_1T}}$
4. In all the above 3 cases, Z is the same. But A are different
• The difference in A is due to the difference in n
■ Elements with same atomic number (Z) but different mass numbers (A) are called isotopes
• $\mathbf\small{\rm{^1_1H,\;\;\,^2_1D\;\;\text{and}\;\;^3_1T}}$ are isotopes
5. If we take a sample of naturally occurring hydrogen, we get the following quantities:
(i) 99.985% of that sample will be $\mathbf\small{\rm{^1_1H}}$
(ii) The rest 0.015% will be $\mathbf\small{\rm{^2_1D}}$
(iii) $\mathbf\small{\rm{^3_1T}}$ is found only in traces on earth
6. Let us see some more examples of isotopes:
• Isotopes of carbon:
(i) $\mathbf\small{\rm{^{12}_{\;6}C}}$ has 6 protons and 6 neutrons
(ii) $\mathbf\small{\rm{^{13}_{\;6}C}}$ has 6 protons and 7 neutrons
(iii) $\mathbf\small{\rm{^{14}_{\;6}C}}$ has 6 protons and 8 neutrons
• Isotopes of chorine:
(i) $\mathbf\small{\rm{^{35}_{17}Cl}}$ has 17 protons and 18 neutrons
(ii)$\mathbf\small{\rm{^{37}_{17}Cl}}$ has 17 protons and 20 neutrons
Chemical properties of isotopes
1. The chemical properties of an element are controlled by the number of electrons in it’s atoms
2. The number of neutrons have very little effect on the chemical properties
3. The number of electrons is same as the atomic number (Z)
• So with regard to chemical properties, the number Z is what we consider most
4. All isotopes of an element will have the same Z
• So all isotopes of an element will show the same chemical behaviour
Now we will see some solved examples:
Solved example 2.1
Calculate the number of protons, neutrons and electrons in $\mathbf\small{\rm{^{80}_{35}Br}}$
Solution:
1. Given : Z = 35, A = 80
2. We have: Number of protons = Number of electrons = Z = 35
3. Also we have: Number of neutrons + Number of protons = A
⇒ Number of neutrons = A - Number of protons
⇒ Number of neutrons = (80-35) = 45
Solved example 2.2
The number of electrons, protons and neutrons in a species are equal to 18, 16 and 16 respectively. Assign the proper symbol to the species
Solution:
1. Given:
• Number of electrons = 18
• Number of protons = 16
♦ So Z = 16
♦ From the periodic table, we have: Z = 16 for sulphur (S)
• Number of neutrons = 16
2. We have: A = Number of neutrons + Number of protons
⇒ A = (16+16) = 32
3. In a neutral atom, the number of electrons will be equal to that of the protons
• But in this case, number of electrons is greater by 2
• So there will be an excess of two negative charges
• So the symbol is: $\mathbf\small{\rm{^{32}_{16}S^{2-}}}$
4. Note:
• If the number of protons and electrons are not equal, the given species is an ion
• We have to determine whether it is a cation or an anion
• If Number of protons > Number of electrons
♦ It is a +ve ion (cation)
• If Number of protons < Number of electrons
♦ It is a -ve ion (anion)
Solved example 2.3
(i) Calculate the number of electrons which will together weigh one gram
(ii) Calculate the mass and charge of one mole of electrons
Solution:
Part (i)
• Mass of one electron = 9.10939 ×10-31 kg = 9.10939 ×10-28 g
• So number of electrons in 1 g
= $\mathbf\small{\frac{1}{9.10939\times 10^{-28}}=0.1097768\times 10^{28}=1.098\times 10^{27}}$
Part (ii)
• One mole of electrons will contain 6.022 × 1023 electrons
• Mass of one electron = 9.10939 ×10-31 kg
• So mass of one mole of electrons = (6.022 × 1023 × 9.10939 ×10-31) = 5.48 ×10-7 kg
• Charge of one electron = -1.602176 ×10-19 C
• So charge of one mole of electron = (6.022 × 1023 × -1.602176 ×10-19) = 9.65 ×104 C
Solved example 2.4
(i) Calculate the total number of electrons present in one mole of methane
(ii) Find (a) the total number and (b) the total mass of neutrons in 7 mg of $\mathbf\small{\rm{^{14}C}}$
(Assume that mass of a neutron = 1.675 × 10-27 kg).
(iii) Find (a) the total number and (b) the total mass of protons in 34 mg of NH3 at STP.
Will the answer change if the temperature and pressure are changed ?
Solution:
Part (i)
1. The molecular formula of methane is CH4
• So in one molecule of methane, there are:
♦ 1 atom of C
♦ 4 atoms of H
2. In one atom of C there are 6 electrons
• In one atom of H there is 1 electron
• So in 1 molecule of methane, there are (1×6 + 4×1) = 10 electrons
3. One mole of methane contains 6.022 × 1023 molecules of methane
• So in one mole of methane, there will be 6.022 × 1023 × 10 = 6.022 × 1024 electrons
Part (ii)
1. Molar mass of $\mathbf\small{\rm{^{14}C}}$ is 14 g
⇒ In 14 g of $\mathbf\small{\rm{^{14}C}}$, there will be 6.022 × 1023 atoms
⇒ In 1 g, there will be (6.022 × 1023 / 14) atoms
⇒ In 1 mg there will be (6.022 × 1023 / 14000) atoms
⇒ In 7 mg there will be (6.022 × 1023 / 2000) = 3.011 × 1020 atoms
2. Number of neutrons in 1 $\mathbf\small{\rm{^{14}C}}$ atom = (A-Z) = (14-6) = 8
• So total number of neutrons in 7 mg = 8 × 3.011 × 1020 = 24.088 × 1020
3. Mass of this much neutrons = (24.088 × 1020 × 1.675 × 10-27)
= 40.35 × 10-7
= 4.035 × 10-6 kg
Part (iii)
1. Molar mass of NH3 = (14+3) = 17 g
⇒ 17 g of NH3 at STP will contain 6.022 × 1023 molecules
⇒ 1 g of NH3 at STP will contain (6.022 × 1023 / 17) molecules
⇒ 1 mg of NH3 at STP will contain (6.022 × 1023 / 17000) molecules
⇒ 34 mg of NH3 at STP will contain (6.022 × 2 × 1020) molecules
2. one molecule of NH3 will contain (7+3) = 10 protons
• So number of protons in (6.022 × 2 × 1020) molecules
= 6.022 × 2 × 10 × 1020
= 12.044 × 1021
= 1.2044 × 1022
3. Mass of this much protons
= (1.2044 × 1022 × mass of one proton)
= (1.2044 × 1022 × 1.67262 × 10-27)
= 2.015 × 10-5 kg
Solved example 2.5
How many neutrons and protons are there in the following nuclei ?
$\mathbf\small{\rm{^{13}_{\;6}C,\;\;^{16}_{\;8}O,\;\;^{24}_{12}Mg,\;\;^{56}_{26}Fe,\;\;^{88}_{38}Sr}}$
Solution:
Part (i)
1. Given : Z = 6, A = 13
2. We have: Number of protons = Z = 6
3. We have: Number of neutrons + Number of protons = A
⇒ Number of neutrons = A - Number of protons
⇒ Number of neutrons = (13-6) = 7
Part (ii)
1. Given : Z = 8, A = 16
2. We have: Number of protons = Z = 8
3. We have: Number of neutrons = A - Number of protons
⇒ Number of neutrons = (16-8) = 8
Part (iii)
1. Given : Z = 12, A = 24
2. We have: Number of protons = Z = 12
3. We have: Number of neutrons = A - Number of protons
⇒ Number of neutrons = (24-12) = 12
Part (iv)
1. Given : Z = 26, A = 56
2. We have: Number of protons = Z = 26
3. We have: Number of neutrons = A - Number of protons
⇒ Number of neutrons = (56-26) = 30
Part (v)
1. Given : Z = 38, A = 88
2. We have: Number of protons = Z = 38
3. We have: Number of neutrons = A - Number of protons
⇒ Number of neutrons = (88-38) = 50
Solved example 2.6
Write the complete symbol for the atom with the given atomic number (Z) and atomic mass (A)
(i) Z = 17 , A = 35
(ii) Z = 92 , A = 233
(iii) Z = 4 , A = 9
Solution:
Part (i)
• Given: Z = 17, A = 35
♦ From the periodic table, we have: Z = 17 for chlorine (Cl)
• So symbol is $\mathbf\small{\rm{^{35}_{17}Cl}}$
Part (ii)
• Given: Z = 92, A = 233
♦ From the periodic table, we have: Z = 92 for uranium (U)
• So symbol is $\mathbf\small{\rm{^{233}_{92}U}}$
Part (iii)
• Given: Z = 4, A = 9
♦ From the periodic table, we have: Z = 4 for beryllium (Be)
• So symbol is $\mathbf\small{\rm{^{9}_{4}Be}}$
In the next section we will see some more solved examples. We will also see the drawbacks of Rutherford's model
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