Choose the correct statement

Ionization energy and electron affinity increases across a period

Ionization energy decreases but electron affinity increases

Ionization energy increases but electron affinity decreases along a period

The ionization energy of nitrogen is more than that of oxygen because

Nitrogen has half filled p-orbitals

Nitrogen contains less number of electrons

Nitrogen is left to the oxygen in the same period of the periodic table

Nitrogen is less electronegative

Ionization Energy - Definition, Formula, Examples, Calculation

Ionization energy is simple terms can be described as a measure of the difficulty in removing an electron from an atom or ion or the tendency of an atom or ion to surrender an electron. The loss of electron usually happens in the ground state of the chemical species.

Alternatively, we can also state that ionization or ionisation energy is the measure of strength (attractive forces) by which an electron is held in a place.

Table of Content

What is Ionization Energy?
Factors Governing the Ionization Energy
Ionization Energy and Bohr’s Atomic Model
Ionization
First and Second Ionization Energy
Ionization Energy Trends in the Periodic Table
How to Determine the Ionization Energy of an Element?

What is Ionization Energy?

In more technical terms we can describe ionization energy as the minimum energy that an electron in a gaseous atom or ion has to absorb to come out of the influence of the nucleus. It is also sometimes referred to as ionization potential and is usually an endothermic process.

What we can deduce further is that ionization energy gives us an idea of the reactivity of chemical compounds. It can also be used to determine the strength of chemical bonds. It is measured either in units of electronvolts or kJ/mol.

Depending on the ionization of molecules which often leads to changes in molecular geometry, ionization energy can be either adiabatic ionization energy or vertical ionization energy.

Factors Governing Ionization Energy

Normally, when the ionization energy is high it will be more difficult to remove an electron. There are also several factors that govern the attraction forces.

If the nucleus is positively charged then the electrons are strongly attracted to it.
If an electron lies near or close to the nucleus then the attraction will be greater than the one when the electron is further away.
If there are more electrons between the outer level and the nucleus the attraction forces are less.
When there are two electrons in the same orbital they experience some form of repulsion. Now, this creates disturbances in the attraction of the nucleus. In essence, ionization energy will be less in paired electrons as they can be removed easily.

Ionization Energy and Bohr’s Atomic Model

Atomic ionization energy can be further be predicted using Bohr’s model of an atom. His model predicts the presence of several paths for the electron to go around the nucleus containing protons and neutrons. Each path or orbit is at a fixed distance from the nucleus. Each orbit also represents fixed energy. Electron is a particle and will have the energy of the orbit present. A particle can absorb energy and jump to the next higher orbits of higher energy. If more energy is available and absorbed, the electron will come out of force of attraction of the nucleus, which means out of the atom.

Ionization

Ionization is a process that involves the removal of electron present in an orbit to outside the atom. As the electron in each orbit has characteristic energy, ionization energy is equal to the difference of energy between the energy of the electron in the initial orbit and the energy of the electron outside the atom (in the infinite orbit from the nucleus).

Energy of an electron in ‘n’th orbit is calculated by Bohr model of an atom as –

\(\begin{array}{l}{{E}_{n}}=-\frac{2{{\pi }^{2}}m{{e}^{4}}}{{{(4\pi {{\varepsilon }^{o}})}^{2}}{{h}^{2}}}\times \frac{{{Z}^{2}}}{{{n}^{2}}}=R\times \frac{{{Z}^{2}}}{{{n}^{2}}}J/atom=-13.6\times \frac{{{Z}^{2}}}{{{n}^{2}}}eV/atom=-2.18\times {{10}^{-18}}\times \frac{{{Z}^{2}}}{{{n}^{2}}}J/atom\end{array} \)

Ionization energy for the removal of an electron from a neutral atom can be calculated, by substituting, the orbit number of the electron before transition as ‘n₁‘ and orbit number of the electron after transition as ‘∞'( infinity) as ‘n₂‘ in Bohr’s energy equation.

Also Read: Bohr’s Theory of Hydrogen Atoms

\(\begin{array}{l}E{{n}_{1}}=-R\times \frac{{{Z}^{2}}}{{{n}^{2}}}\,\,\,\,\,\,\,\,\,E{{n}_{2}}=-R\times \frac{{{Z}^{2}}}{{{\infty }^{2}}}\end{array} \)

\(\begin{array}{l}\Delta E=E{{n}_{2}}-E{{n}_{1}}=R\times{{z}^{2}}\left(\frac{1}{{{n}^{2}}}-\frac{1}{{{\infty }^{2}}} \right)=ionization\,Energy\end{array} \)

Since,

\(\begin{array}{l}\frac{1}{{{\infty}^{2}}}\end{array} \)

is almost zero, it can be neglected.

\(\begin{array}{l}Ionization\,Energy\,=\,\Delta E=E{{n}_{2}}-E{{n}_{1}}=R\times {{z}^{2}}\left( \frac{1}{{{n}^{2}}} \right)=2.18\times {{10}^{-18}}\times {{z}^{2}}\left( \frac{1}{{{n}^{2}}} \right)J\end{array} \)

First and Second Ionization Energy

First ionization energy is the energy that is required to remove the first electron from a neutral atom. It is numerically same as the orbital energy of the electron but of opposite sign.

For hydrogen, first orbit energy is –2.18 × 10^{– 18} J/atom (or – 1312.3 KJ/mole), and the ionization energy is + 2.18 × 10^–18 J/atom (or + 1312.3 KJ/mole).

The energy needed for the removal of the second electron away from the unipositive ion is second ionization energy and so on.

For Example:

M + ∆H_1st→ M⁺ + e^– ; ∆H_1st = First Ionization energy

M⁺ + ∆H_2nd →M²⁺ + e^– ; ∆H_2nd = Second Ionization energy, etc.,

Naturally removing the second electron, from an already positive ion will be difficult. Hence second ionization energy will be larger than the first ionization energy. Third ionization energy will be more than second ionization energy etc.

∆H_1st < ∆H_2nd < ∆H_3rd < …..

Because of the enhanced stability of half-filled and fully filled orbitals, removal of electrons from such systems will have relatively higher ionization than other atoms and ions.

For example, helium is more stable due to completely filled s-orbital than hydrogen. So, the first ionization energy of Helium (2372 KJ/mole) is more than that of hydrogen (1312 KJ/mole).

The first ionization energy of nitrogen (1402KJ/mole) is more than that of its near neighbours, carbon (1086 KJ/mole) and oxygen (1313KJ/mole), because of the higher stability due to half-filled orbitals. Chromium has half-filled s- and d-orbitals and so has much more first ionization than titanium.

Ionization Energy Trends in the Periodic Table

The ionization energy of an electron increases with the atomic number of the atom and decreases for higher energy orbitals. If we look at the periodic table and move from left to right across the elements, the ionization energy increases due to decreasing atomic radius.\

Also Read: Atomic Structure

Whereas, if we move from top to bottom, the ionization energy decreases. This is mainly due to the presence of more electron shells in the elements as we move down the group. Additionally, the electrons are placed at a greater distance from the attractive forces of the nucleus.

How to Determine the Ionization Energy of an Element?

1. Electromagnetic radiation of wavelength 242 nm is just sufficient to ionize the sodium atom. Calculate the ionization energy of sodium in kJ mol^–1.

Answer:

Ionization Energy =

\(\begin{array}{l}E=hv=\frac{hv}{\lambda }=\frac{6.6\times {{10}^{-34}}\times 3\times {{10}^{8}}}{242\times {{10}^{-9}}}=\frac{19.8}{242}\times {{10}^{17}}=8.18\times {{10}^{15}}J/atom\end{array} \)

2. How much energy is required to ionize an H atom if the electron occupies n = 5 orbit? Compare your answer with the ionization enthalpy of H atom ( energy required to remove the electron from n =1 orbit).

Answer:

Ionization refers to the electron being out of the atom or to the infinite orbit. The energy of the electron in the hydrogen atom is given by;

\(\begin{array}{l}\Delta E=2.18\times {{10}^{-18}}\times {{1}^{1}}\left. \left\{ \frac{1}{n{{1}^{2}}}-\frac{1}{n{{2}^{2}}} \right\} \right)J/atom\end{array} \)

a) For ionization from n = 1 to n = ∞,

\(\begin{array}{l}\Delta E=2.18\times {{10}^{-18}}\times {{1}^{1}}\left. \left\{ \frac{1}{{{1}^{2}}}-\frac{1}{{{\infty }^{2}}} \right\} \right)=2.18\times {{10}^{-18}}\times {{1}^{1}}\{1-0\}=2.18\times {{10}^{-18}}J/atom\end{array} \)

b) For ionization from n = 5 to n = ∞,

\(\begin{array}{l}\Delta E=2.18\times {{10}^{-18}}\times {{1}^{1}}\left. \left\{ \frac{1}{{{5}^{2}}}-\frac{1}{{{\infty }^{2}}} \right\} \right)=2.18\times {{10}^{-18}}\times {{1}^{1}}\{\frac{1}{25}-0\}=2.18\times {{10}^{-18}}\times \frac{1}{25}J/atom\end{array} \)

The energy required for ionization from the fifth orbit is 25 times lesser than that required for ionization from the first orbit.

3. Calculate the wavenumber for the shortest wavelength transition in the Balmer series of atomic hydrogen.

Answer:

Shortest wavelength corresponds to the highest energy. Highest energy is equal to the ionization energy from that orbit. Balmer series belongs to the second orbit.

Ionization energy from second orbit =

\(\begin{array}{l}\Delta E=2.18\times {{10}^{-18}}\times {{1}^{1}}\left. \left\{ \frac{1}{{{2}^{2}}}-\frac{1}{{{\infty }^{2}}} \right\} \right)=2.18\times {{10}^{-18}}\{0.25-0\}=2.18\times {{10}^{-18}}\times 0.25J/atom\end{array} \)

Ionization Energy

\(\begin{array}{l}=E=hv=\frac{hc}{\lambda }=\frac{6.6\times {{10}^{-34}}\times 3\times {{10}^{8}}}{\lambda }=5.7\times {{10}^{-19}}J/atom\end{array} \)

Shortest wavelength

\(\begin{array}{l}=\lambda =\frac{19.8}{5.7}\times {{10}^{-7}}m=347nm\end{array} \)

4. Arrange the following elements (whose electronic configurations are given below), in increasing order of first ionization energy.

a) [Ne] 3s² 3p¹

b) [Ar] 3d¹⁰ 4s² 4p²

c) [Ne] 3s² 3p²

d) [Ne] 3s² 3p³

Answer:

First ionization in the given elements is related to the removal of s-electron. Higher the attraction of the nucleus on the electron difficult it is to remove and so, higher the first ionization energy.

Electron to be removed in [Ar] 3d¹⁰ 4s² 4p², is in the 4th orbit while in others it is in the third orbit. As the attraction of the nucleus decreases with increasing orbit number, [Ar] 3d¹⁰ 4s² 4p² will have the lowest ionization energy.

Among, the rest, [Ne] 3s² 3p³ is more stable due to half-filled orbitals. Hence, this will have the highest ionization energy.

Among, [Ne] 3s² 3p¹ and [Ne] 3s² 3p², nuclear charge is more in [Ne] 3s² 3p² increasing the attraction of the nucleus on the s-electron. So, this will have higher ionization energy than [Ne] 3s² 3p¹.

The order of increasing order of ionization energy of the atoms is –

[Ar] 3d¹⁰ 4s² 4p² < [Ne] 3s² 3p¹ < [Ne] 3s² 3p² < [Ne] 3s² 3p³

5. The correct order of ionization energy for the following species is:

1) He < Li⁺ < H^–

2) H^– < Li⁺ < He

3) H < Li⁺ < He

4) H^– < He < Li⁺

Answer:

All, He, Li+, H^– have two electrons in the first orbital. But the nuclear charge is different in them. Ionization is directly related to the attraction of the electron by the nucleus. Hence higher the nuclear charge higher the ionization energy.

Nuclear charge in He = 2, Li⁺ = 3, H^– = 1.

So the order of nucleus charge = order of ionization energy = H^– < He < Li⁺

The right answer is 4.