Isotopy can be defined as a phenomenon whereby two or more atoms of the same element possess the same atomic number but different mass number(atomic mass) due to difference in neutrons present in the atoms.

It can also be defined as a phenomenon whereby atoms of the same element possess similar chemical properties but different physical properties due to difference in mass number. Atoms that exhibit isotopy are called isotopes of the element.

An isotope is one of two or more species of atoms of a chemical element with the same atomic number and position in the periodic table and nearly identical chemical behaviour but with different atomic masses and physical properties. Every chemical element has one or more isotopes. For example,

Isobars refer to atoms or ions that have the same mass number (total number of protons and neutrons) but different atomic numbers (number of protons). Isobars do not share similar chemical and physical properties. For example;

1. Carbon-14 and Nitrogen-14: These isotopes share the same mass number but have different atomic numbers. Carbon-14 has 6 protons and 8 neutrons, while Nitrogen-14 has 7 protons and 7 neutrons.
2. Sulfur-34 and Chlorine-34: Both of these isotopes have a mass number of 34 but differ in atomic number. Sulfur-34 has 16 protons and 18 neutrons, while Chlorine-34 has 17 protons and 17 neutrons.
3. Phosphorus-32 and Sulfur-32: These isotopes share a mass number of 32 but have different atomic numbers. Phosphorus-32 has 15 protons and 17 neutrons, while Sulfur-32 has 16 protons and 16 neutrons.
4. Potassium-40 and Calcium-40: Potassium-40 and Calcium-40 are isobars with a mass number of 40. However, they have distinct atomic numbers, with Potassium-40 having 19 protons and 21 neutrons, and Calcium-40 having 20 protons and 20 neutrons.
5. Argon-40 and Calcium-40: Both of these isotopes have a mass number of 40, but they differ in atomic number. Argon-40 has 18 protons and 22 neutrons, while Calcium-40 has 20 protons and 20 neutrons.

Isotones are nuclides or atomic nuclei with the same number of neutrons but different numbers of protons. They belong to the same isotonic series, sharing similar nuclear properties due to their matching neutron counts.

1. Carbon-12 (¹²C) and oxygen-16 (¹⁶O) are isotones with 6 neutrons each but different numbers of protons.
2. Nitrogen-14 (¹⁴N) and fluorine-19 (¹⁹F) are isotones with 7 neutrons each but different numbers of protons.
3. Neon-20 (²⁰Ne) and magnesium-24 (²⁴Mg) are isotones with 10 neutrons each but different numbers of protons.
4. Silicon-28 (²⁸Si) and sulfur-32 (³²S) are isotones with 14 neutrons each but different numbers of protons.
5. Calcium-40 (⁴⁰Ca) and argon-44 (⁴⁴Ar) are isotones with 20 neutrons each but different numbers of protons.

Example 1:The natural abundance of four isotopes of lead with mass numbers 204, 206, 207 and 208 are 0.48%, 23.6%, 22.6% and 53.32% respectively. What is the relative atomic mas of naturally occurring lead?

We can solve for relative atomic mass by multiplying the percentages of the isotopes by their corresponding atomic mass
$$\frac{0.48}{100} {×} {204} { +} \hspace{0.5em} \frac{23.6}{100} {×} {206} { +} \hspace{0.5em} \frac{22.6}{100} {×} {207} { +} \hspace{0.5em} \frac{53.32}{100} {×} {208} \hspace{2em}$$

$$ = {0.9792} + {48.616} + {46.782} + { 110.9056} $$ $$ = 207.28 $$

Example 2: An element X has two isotopes present in the ratio 1:3 as shown below. What is the relative atomic mass of X. $$ \hspace{3em}{^{20}_{10}X} \hspace{5em} {^{22}_{10}X} $$

To solve this, sum up the ratios, then divide each ratio by the total ratio multiplied by each masses and add them together

$$\text{Total ratio} = {1 + 3} = {4} $$ $$ \hspace{3em} \frac{1}{4} {×} {20} + \frac{3}{4} {×} {22} $$ $$ \hspace{3em} = 5 + 16.5 $$ $$ \hspace{3em} = 21.5 $$

Example 3: Chlorine has two isotopes with mass number 35 and 37. The relative atomic mass of a sample of prepared chlorine is 35.5. Calculate the percentage abundance of both isotopes.

To solve this, use the formula and then interchange the percentages
$$ \frac{M_1} {M_2} × {100} $$ $$ \text{where,} $$ M₁ = difference between mass number and atomic mass
M₂ = difference between the mass numbers of the isotopes
$$\text{For Chlorine-35} $$ $${M_1} = {35.5 - 35} = {0.5} $$ $${M_2} = {37 - 35} = 2 $$ $$\text{% of Cl- 35} = \frac{0.5}{2} × {100} $$ $$ = \text{25%} $$ $$\text{ For Cl-37} $$ $$ {M_1} = {37 - 35.5} = {1.5} $$ $$ {M_2} = {2} $$ $$ \text{% of Cl-37} = \frac{1.5} {2} × {100} $$ $$ = \text{75%} $$ $$ \text{interchanging percentages we have} $$ $$\text{Cl-35} = \text{75%} \text{ and }\text{Cl-37} = \text{25%} $$

Example 4: An element with relative atomic mass 16.2 contains two isotopes with relative abundance of 90% and 10% respectively as shown below. What is the value of m? $$\hspace{3em} {^{16}_8X} \hspace{3em} {^m_8X} $$

Solve using the percentage method, $$ \frac{90}{100} × {16} + \frac{10}{100} × {m} = {16.2} $$ $$ = {14.4} + {0.1m} = {16.2}$$ $$ {0.1m} = {16.2 - 14.4} $$ $$ {0.1m} = {1.8} $$ $$ \text{dividing both sides by 0.1} $$ $$ \frac{0.1m}{0.1} = \frac{1.8}{0.1} $$ $$ {m} = {18} $$

Example 5: If 100 atoms of element X contains 70 atoms of ⁹X and 30 atoms of ¹¹X. Calculate the relative atomic mass of X. (WAEC)

Solve using percentage method $$ \text{atomic mass} = \frac{70}{100} × {9} + \frac{30}{100} × {11} $$ $$ = {6.3 + 3.3} $$ $$\text{atomic mass} = {9.6} $$