We find different types of energies in, and around us, like mechanical, thermal, sonic, nuclear, gravitational, cinematic, chemical, electromagnetic, potential energy, ionization energy, etc. Ions are made up of when electrons are added or being removed from an atom. In this article, let’s discuss everything about Ionization energy such as how much energy an ion is needed to form an ion, its units with ionizing energy factors, ionizing energy formula, use of ionization, etc. 

What is Ionization Energy?

An atom consists of three fundamental elementary particles, i.e., positively charged protons, negatively charged electrons, and neutral neutrons. Protons and neutrons recede in the nucleus, and electrons are revolving around the nucleus in circular orbitals. If energy is supplied to electrons, electrons get promoted to higher energy levels. If sufficient energy is supplied, electrons may be removed, resulting in the formation of a positively charged ion, i.e., cation.

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The minimum amount of energy required to remove the most loosely bound electron from a neutral isolated gaseous atom from its ground state to form a cation is called its ionization enthalpy (IE) or ionization energy.

It is represented by \(\Delta_{\mathrm{i}} \mathrm{H}\)

\({\rm{Atom}}({\rm{g}}) \to {\rm{Positive\, ion}}({\rm{g}}) + {{\rm{e}}^ – };\quad {\rm{IE}}\left( {{\Delta _{\rm{i}}}{\rm{H}}} \right)\)

\(\mathrm{Na}(\mathrm{g}) \rightarrow \mathrm{Na}^{+}(\mathrm{g})+\mathrm{e}^{-} ; \quad \mathrm{IE}\left(\Delta_{\mathrm{i}} \mathrm{H}\right)\)

Ionization enthalpy is also called ionization potential since it is the minimum potential difference required to remove the most loosely bound electron from an isolated gaseous atom to form a gaseous cation.

Unit of Ionization Energy

The unit of ionization enthalpy is kilo joule per mole \(\left. {({\rm{kJmo}}{{\rm{l}}^{ – 1}}} \right)\) and that of ionization potential is electron volt \(\left( {{\rm{eV}}} \right)\) per atom. The two are related to each other as:

\(1\) electron volt \(\left( {{\rm{eV}}} \right)\) per atom \(=1.602 \times 10^{-19} \mathrm{~J}\) per atom

\(1\) electron volt \(\left( {{\rm{eV}}} \right)\) per mole \( = 1.602\, \times \,{10^{ – 19}}\, \times \,6.022\, \times \,{10^{23}}\;{\rm{J}}\;{\rm{mo}}{{\rm{l}}^{ – 1}}\)

\(1\) electron volt \(\left( {{\rm{eV}}} \right)\) per mole \(=96472 \mathrm{~J} \mathrm{~mol}^{-1}=96.472 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Successive Ionization Enthalpies

If a gaseous atom is to lose more than one electron, they can be removed one after the other, i.e., in succession and not simultaneously. This is known as successive ionization enthalpy. 

When the gaseous atom loses one electron to form a monovalent cation, the number of electrons in the cation decreases by one, and the remaining electrons are held by the nucleus of the cation with greater force. Therefore, with the same energy, the second electron cannot be removed, and therefore, extra energy is needed to form a divalent cation. In the same manner, the energy required to remove the third electron is expected to be even higher.

Example: The successive ionization enthalpies of \({\rm{Al}}({\rm{g}})\) atom is as follows:

\({\rm{Al}}({\rm{g}}) \to {\rm{A}}{{\rm{l}}^ + }({\rm{g}}) + {\rm{e}};{\Delta _{\rm{i}}}{{\rm{H}}_1} = 577.5\;{\rm{kJ}}\;{\rm{mo}}{{\rm{l}}^{ – 1}}\)

\({\rm{A}}{{\rm{l}}^ + }({\rm{g}}) \to {\rm{A}}{{\rm{l}}^{2 + }}({\rm{g}}) + {{\rm{e}}^ – };{\Delta _{\rm{i}}}{{\rm{H}}_2} = 1816.7\;{\rm{kJ}}\;{\rm{mo}}{{\rm{l}}^{ – 1}}\)

\(\mathrm{Al}^{2+}(\mathrm{g}) \rightarrow \mathrm{Al}^{3+}(\mathrm{g})+\mathrm{e}^{-} ; \Delta_{\mathrm{i}} \mathrm{H}_{3}=2744.8 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

If one mole Al atoms in gaseous are to be converted into one mole gaseous \(\mathrm{Al}^{3+}\) ions, the ionization enthalpy required is \((577.5+1816.7+2744.8) 5139 \mathrm{~kJ} \mathrm{~mol}^{-1}\) which is very high.

Theoretically, any number of electrons can be removed from a gaseous atom but practically, it is not possible to remove more than three electrons because the energy needed is so high that it is not easily available.

Factors Governing the Ionization Enthalpy

Ionization enthalpy depends upon the following factors: a) Nuclear charge b) Atomic size c) penetration effect of the electrons d) screening or shielding effect of inner electrons e) electronic configuration. 

a) Nuclear Charge: As the magnitude of the positive charge in the nucleus of an atom increases, its attraction with the electrons also increases. This means that greater energy is needed to remove the electrons particularly present in the valence shell of the gaseous atom. Therefore, the ionization enthalpy increases with the increase in the magnitude of the nuclear charge.
First ionization enthalpies in \({\rm{kJmo}}{{\rm{l}}^{ – 1}}\) of the elements of second period are as follows:

Element	\({\rm{Li}}\)	\({\rm{Be}}\)	\({\rm{B}}\)	\({\rm{C}}\)	\({\rm{N}}\)	\({\rm{O}}\)	\({\rm{F}}\)	\({\rm{Ne}}\)
Nuclear charge (charge of protons)	\(+3\)	\(+4\)	\(+5\)	\(+6\)	\(+7\)	\(+8\)	\(+9\)	\(+10\)
Ionization enthalpy \({\rm{kJmo}}{{\rm{l}}^{ – 1}}\)	\(520.2\)	\(899.5\)	\(800.6\)	\(1086.5\)	\(1402.3\)	\(1314\)	\(1681\)	\(2080.7\)

b) Atomic size: With the increase in the atomic size, the number of electron shells increases. Therefore, the force that binds the electrons with the nucleus decreases. The ionization enthalpy thus decreases with the increase in the atomic size.

First ionization enthalpies (in \({\rm{kJ\, mo}}{{\rm{l}}^{ – 1}}\) of alkali metal are as follow.

Element	\({\rm{Li}}\)	\({\rm{Na}}\)	\({\rm{K}}\)	\({\rm{Rb}}\)	\({\rm{Cs}}\)
Ionization enthalpy \({\rm{kJ\, mo}}{{\rm{l}}^{ – 1}}\)	\(520.2\)	\(495.8\)	\(418.8\)	\(403.0\)	\(375.7\)

c) Penetration effect of the electron: The ionization enthalpy is also influenced by the penetration effect of the electron, i.e, The greater the probability of finding the electron closer to the nucleus, the more will be its attraction towards the nucleus. Therefore, it is difficult to remove the electron from the atom. The order of penetration of different types of electrons is \({\rm{s > p > d > f}}.\) In general, the greater the penetration of the electron towards the nucleus of the atom, the more will be the ionization enthalpy.

d) Screening or shielding effect: The greater the number of electrons in the inner shells, the larger will be the screening effect. As the screening effect increases, the effective nuclear charge decreases. Consequently, the force of attraction by the nucleus for the valence shell electrons decreases, and hence ionization enthalpy decreases. In simple words, as the shielding or the screening effect of the inner electrons increases, the ionization enthalpy decreases.

e) Electronic configuration: If an atom contains exactly half-filled or completely filled orbitals, then their stability is more. Therefore, the removal of an electron from such an atom requires more energy than expected.

Example: \({\rm{N}}\left( {1\;{{\rm{s}}^2},2\;{{\rm{s}}^2},2{\rm{p}}_{\rm{x}}^1,2{\rm{p}}_{\rm{y}}^1,2{\rm{p}}_{\rm{z}}^1} \right)\) has higher ionization enthalpy than oxygen \(\left( {{\rm{1}}{{\rm{s}}^2},{\rm{2}}{{\rm{s}}^{\rm{2}}},{\rm{2p}}_{\rm{x}}^{\rm{2}},{\rm{2p}}_{\rm{y}}^1,{\rm{2p}}_{\rm{z}}^1} \right).\) This is because N contains exactly half-filled p-orbitals.

Noble gases have the highest ionization enthalpy in their respective periods. Since they have completely filled electronic configuration.

Ionization Enthalpies of Few Elements

Element	Atomic Number	\({\Delta _{\rm{i}}}{{\rm{H}}_1}({\rm{KJ}}/{\rm{mol}})\)	\(\Delta {\rm{i}}{{\rm{H}}_2}\left( {{\rm{KJ/mol}}} \right)\)	\(\Delta_{1} \mathrm{H}_{3}(\mathrm{KJ} / \mathrm{mol})\)
Li	\(3\)	\(520.2\)	\(7298.1\)	\(11815.0\)
Be	\(4\)	\(899.5\)	\(1757.1\)	\(14848.7\)
B	\(5\)	\(800.6\)	\(2427.1\)	\(3659.7\)
C	\(6\)	\(1086.5\)	\(2352.6\)	\(4620.5\)
N	\(7\)	\(1402.3\)	\(2856.0\)	\(4578.1\)
O	\(8\)	\(1313.9\)	\(3388.3\)	\(5300.5\)
F	\(9\)	\(1681.0\)	\(3374.2\)	\(6050.4\)

What is the Ionization Energy Formula?

Ionization energy is the energy required to remove the electron completely from the atom so as to convert it to a positive ion. This means that it is the energy absorbed by the electron in the ground state \({\rm{(n = 1)}}\) so as to jump to infinity \({\rm{n = }}\infty .\) Thus, the ionization energy of the H atom and H like atom is calculated as follows:

Ionization energy, \({\rm{IE = }}{{\rm{E}}_\infty }{\rm{ – }}{{\rm{E}}_{\rm{n}}}\)

Energy is given by the formula: 

\({{\rm{E}}_{\rm{n}}}{\rm{ =\, – }}\frac{{{\rm{2}}{{\rm{\pi }}^{\rm{2}}}{{\rm{m}}_e}{{\rm{e}}^{\rm{4}}}{{\rm{z}}^{\rm{2}}}}}{{{{\rm{n}}^{\rm{2}}}{{\rm{h}}^{\rm{2}}}}}\)

Here \({{\rm{m}}_{\rm{e}}} = \) Mass of electron \(\left(9.1 \times 10^{-31} \mathrm{~kg}\right)\)

\({\rm{e = }}\) Magnitude of charge on electron \(\left(1.6 \times 10^{-19} \mathrm{C}\right)\)

\({\rm{Z = }}\) atomic number (for hydrogen, \({\rm{Z = 1}})\)

\({\rm{n = }}\) No. of electron’s orbit

\({\rm{h = }}\) Planck’ constant \(\left(6.626 \times 10^{-34} \mathrm{~kg} \mathrm{~m}^{2} \mathrm{~s}^{-1}\right)\)

On substituting the values, we get 

\({{\rm{E}}_{\rm{n}}}{\rm{ =\, – 2}}{\rm{.18 \times 1}}{{\rm{0}}^{{\rm{ – 18}}}}{\rm{ \times }}\frac{{{{\rm{z}}^{\rm{2}}}}}{{{{\rm{n}}^{\rm{2}}}}}{\rm{J}}\)

Or,

\({{\rm{E}}_{\rm{n}}}{\rm{ =\, – 1312}}{\rm{.8 \times }}\frac{{{{\rm{z}}^{\rm{2}}}}}{{{{\rm{n}}^{\rm{2}}}}}{\rm{kJmo}}{{\rm{l}}^{{\rm{ – 1}}}}\)

For Hydrogen, \({{\rm{E}}_{\rm{1}}}{\rm{ = – 2}}{\rm{.18 \times 1}}{{\rm{0}}^{{\rm{ – 18}}}}{\rm{ \times }}\frac{{{{\rm{1}}^{\rm{2}}}}}{{{{\rm{1}}^{\rm{2}}}}}{\rm{J}}\)

\({{\rm{E}}_{\rm{1}}}{\rm{ =\, – 2}}{\rm{.18 \times 1}}{{\rm{0}}^{{\rm{ – 18}}}}{\rm{J}}\)

Ionization energy of hydrogen, \({\rm{IE = }}{{\rm{E}}_{\rm{m}}}{\rm{ – }}{{\rm{E}}_{\rm{1}}}\)

\({\rm{IE =\, 0 – }}\left( {{\rm{ – 2}}{\rm{.18 \times 1}}{{\rm{0}}^{{\rm{ – 18}}}}{\rm{J}}} \right)\)

\({\rm{IE =\, 2}}{\rm{.18 \times 1}}{{\rm{0}}^{{\rm{ – 18}}}}{\rm{\;J}}\)

Ionization energy of hydrogen like ion:

Example: Ionization energy of \({\rm{L}}{{\rm{i}}^{2 + }}({\rm{Z}} = 3)\)

\({\rm{IE =\, 19}}{\rm{.62 \times 1}}{{\rm{0}}^{{\rm{ – 18}}}}{\rm{J}}\)

\({\rm{IE = }}\,{{\rm{E}}_{\rm{m}}}{\rm{ – }}\left( {{\rm{ – 2}}{\rm{.18 \times 1}}{{\rm{0}}^{{\rm{ – 18}}}}{\rm{ \times }}\frac{{{{\rm{z}}^{\rm{2}}}}}{{{{\rm{n}}^{\rm{2}}}}}{\rm{J}}} \right)\)

\({\rm{IE = }}\,{{\rm{E}}_{\rm{m}}}{\rm{ – }}\left( {{\rm{ – 2}}{\rm{.18 \times 1}}{{\rm{0}}^{{\rm{ – 18}}}}{\rm{ \times }}\frac{{{{\rm{3}}^{\rm{2}}}}}{{{{\rm{l}}^{\rm{2}}}}}{\rm{J}}} \right)\)

\({\rm{IE =\, 0 – }}\left( {{\rm{ – 2}}{\rm{.18 \times 1}}{{\rm{0}}^{{\rm{ – 18}}}}{\rm{ \times }}\frac{{\rm{9}}}{{\rm{1}}}{\rm{J}}} \right)\)

\({\rm{IE = 19}}{\rm{.62 \times 1}}{{\rm{0}}^{{\rm{ – 18}}}}{\rm{J}}\)

What are the Uses of Ionization Energy Formula?

Ionization Energy Formula helps in the calculation of the amount of energy required to generate the respective cation by removing electrons. By the ionization energy values, we can find the feasibility of the reaction, and how many electrons can be removed from an atom can be calculated.

Summary

In the article, Ionization energy formula you have understood the meaning of ionization energy, its units, successive ionization enthalpies, how ionization depends upon different factors like screening effect, atomic radius, electronic configuration, etc. Apart from this, you are able to calculate the ionization enthalpies of hydrogen and hydrogen-like atoms by solving numerical and also, you know the values of IE of different elements.

FAQs on Ionization Energy Formula

Q.1. What is the formula of ionization energy?
Ans: Ionization energy, \({\rm{IE = }}{{\rm{E}}_\infty }{\rm{ – }}{{\rm{E}}_{\rm{n}}}\)
Energy is given by the formula: 
\({{\rm{E}}_{\rm{n}}}{\rm{ =\,  – }}\frac{{{\rm{2}}{{\rm{\pi }}^{\rm{2}}}{{\rm{m}}_{\rm{e}}}{{\rm{e}}^{\rm{4}}}{{\rm{z}}^{\rm{2}}}}}{{{{\rm{n}}^{\rm{2}}}{{\rm{h}}^{\rm{2}}}}}{\rm{ =\,  – 2}}{\rm{.18 \times 1}}{{\rm{0}}^{{\rm{ – 18}}}}{\rm{ \times }}\frac{{{\rm{ – 2}}}}{{{{\rm{n}}^{\rm{2}}}}}{\rm{J}}\)
Here \({{\rm{m}}_{\rm{e}}}\) is mass of the electron \(\left( {9.1 \times {{10}^{ – 31}}\;{\rm{kg}}} \right),{\rm{e}}\) is the magnitude of charge on an electron \(\left(1.6 \times 10^{-19} \mathrm{C}\right), \mathrm{Z}\) is the atomic number (for hydrogen, \({\rm{Z = 1}})\), n is the number of electron’s orbit, h is the Planck’s constant \(\left(6.626 \times 10^{-34} \mathrm{~kg} \mathrm{~m}^{2} \mathrm{~s}^{-1}\right)\)

Q.2. How do you calculate ionization energy in joules?
Ans: Ionization energy can be calculated in joules using the following formula, 
\({\rm{IE = }}{{\rm{E}}_\infty }{\rm{ – }}{{\rm{E}}_{\rm{n}}}\)
Energy is given by the formula 
\({{\rm{E}}_{\rm{n}}}{\rm{ =\, – 2}}{\rm{.18 \times 1}}{{\rm{0}}^{{\rm{ – 18}}}}{\rm{ \times }}\frac{{{{\rm{z}}^{\rm{2}}}}}{{{{\rm{n}}^{\rm{2}}}}}{\rm{J}}\)
Here, Z is the atomic number (for hydrogen,\({\rm{Z = 1}})\) and n is the number of electron’s orbit.

Q.3. What is the use of Ionization Energy Formula?
Ans: Ionization Energy Formula helps in the calculation of the amount of energy required to generate the respective ions by removing electrons. By the ionization energy values, we can find feasibility of the reaction, and how many electrons can be removed from an atom can be calculated.

Q.4. Which element has the smallest ionization energy?
Ans: Caesium has the smallest ionization energy \(\left(\Delta_{\mathrm{i}} \mathrm{H}=375.70 \mathrm{~kJ} / \mathrm{mol}\right).\)

Q.5. How do you read ionization energy in the periodic table?
Ans: In a period, the first ionization energy increases from left to right and decreases from top to bottom in a group generally.

Q.6. What is ionization energy? Give an example.
Ans: The minimum amount of energy required to remove the most loosely bound electron from an isolated gaseous atom or ion in its ground state to form a cation is called its ionization enthalpy (IE) or ionization energy. It is represented by \(\Delta_{\mathrm{i}} \mathrm{H}\)
Example: \(\mathrm{Na}(\mathrm{g}) \rightarrow \mathrm{Na}^{+}(\mathrm{g})+\mathrm{e}^{-} ; \Delta_{\mathrm{i}} \mathrm{H}=495.8 \mathrm{~kJ} / \mathrm{mol}\)

Q.7. Give an example of an ionization energy equation.
Ans: Ionization energy of \(\mathrm{Li}^{2+}(\mathrm{Z}=3)\)
\({\rm{IE = }}\,{{\rm{E}}_{_\infty }}{\rm{ – }}{{\rm{E}}_{\rm{n}}}\)
\({\rm{IE = }}\,{{\rm{E}}_{_\infty }}{\rm{ – }}\left( {{\rm{ – 2}}{\rm{.18 \times 1}}{{\rm{0}}^{{\rm{ – 18}}}}{\rm{ \times }}\frac{{{{\rm{z}}^{\rm{2}}}}}{{{{\rm{n}}^{\rm{2}}}}}{\rm{J}}} \right)\)
\({\rm{IE = }}\,{{\rm{E}}_{_\infty }}{\rm{ – }}\left( {{\rm{ – 2}}{\rm{.18 \times 1}}{{\rm{0}}^{{\rm{ – 18}}}}{\rm{ \times }}\frac{{{{\rm{3}}^{\rm{2}}}}}{{{{\rm{1}}^{\rm{2}}}}}{\rm{J}}} \right)\)
\({\rm{IE =\, 0 – }}\left( {{\rm{ – 2}}{\rm{.18 \times 1}}{{\rm{0}}^{{\rm{ – 18}}}}{\rm{ \times }}\frac{{\rm{9}}}{{\rm{1}}}{\rm{J}}} \right)\)
\({\rm{IE =\,  + 19}}{\rm{.62 \times 1}}{{\rm{0}}^{{\rm{ – 18}}}}{\rm{J}}\)

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