Ionization Energy

Ionization energy is the amount of energy needed to remove an electron from an isolated gaseous neutral atom. 

To understand this concept, imagine electrons as little pieces of a puzzle that fit around the nucleus of an atom. Ionization energy is like the strength needed to pull one of those puzzle pieces away from the atom.

1.0What Is Meant By Ionization Energy?

Atoms hold onto their electrons because of electric forces, and different atoms hold onto their electrons with varying strengths. Elements with lower ionization energies let go of their electrons more easily, while those with higher ionization energies hold onto them more tightly.

Minimum energy required to remove the most loosely bounded outermost shell e^– in ground state from an isolated gaseous atom is known as Ionization energy. (Isolated means without any bonding with other atom)

Ionization energy, technically defined as the minimum energy required for an electron in a gaseous atom or ion to break free from the influence of the nucleus, is also known as ionization potential. This process is typically endothermic in nature.

Understanding ionization energy provides insights into the reactivity of chemical compounds and offers a means to evaluate the strength of chemical bonds. It is commonly quantified in units of electron volts (eV) or kilojoules per mole (kJ/mol).

2.0Definition of Ionization Energy and Bohr’s Atomic Model

Atomic ionization energy can be further predicted using Bohr’s Atomic 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, and it 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 the force of attraction of the nucleus, which means out of the atom.

What is Ionization Energy?

Ionization energy is the amount of energy needed to remove an electron from an atom. It's like the "strength" of the bond between an electron and its home atom. The higher the ionization energy, the stronger this bond, making it more difficult to separate the electron. This energy value is crucial in understanding how elements interact and react with one another in chemical processes.

The energy of an electron in ‘n’th orbit is calculated by the Bohr model of an atom as:

$\begin{array}{rl}& E_n=-\frac{2{\pi }^{2}m{e}^4}{{\left(4\pi {\epsilon }^{\circ }\right)}^2{h}^2}\times \frac{Z^2}{n^2}\\ & =R\times \frac{Z^2}{n^2}\mathrm{J}/\text{ atom }\\ & =-13.6\times \frac{Z^2}{n^2}\mathrm{eV}/\text{ atom }\\ & =-2.18\times 10^{-18}\times \frac{Z^2}{n^2}\mathrm{J}/\text{ atom }\end{array}$

The calculation of ionization energy for removing an electron from a neutral atom involves using Bohr's energy equation. It requires substituting the orbit number of the electron before transition as 'n1' and the orbit number of the electron after transition as '∞' (infinity), represented as 'n2' in the equation. This mathematical approach helps quantify the energy required to move an electron from its initial orbit to an unbound state infinitely far from the nucleus.

$\begin{array}{rl}& E{n}_1=-R\times \frac{Z^2}{n^2}andE{n}_2=-R\times \frac{Z^2}{{\infty }^2}\\ & \Delta E=E{n}_2-E{n}_1=R\times z^2\left(\frac{1}{n^2}-\frac{1}{{\infty }^2}\right)=\text{ Ionization Energy }\end{array}$

Since, $\frac{1}{{\infty }^2}$ is almost zero, it can be neglected.

Ionization Energy $=\Delta E=E{n}_2-E{n}_1=R\times z^2\left(\frac{1}{n^2}\right)=2.18\times 10^{-18}\times z^2\left(\frac{1}{n^2}\right)J$

Bohr Atomic Model for Hydrogen

The energy levels of hydrogen are governed by the equation developed by Niels Bohr to describe the behavior of electrons in atoms:

$E_n=-\frac{13.6\mathrm{eV}}{n^2}$

Here, E_n​ represents the energy of the electron in the nth energy level, and n is the principal quantum number denoting the energy level. For hydrogen, n can take values starting from 1 (the ground state) and increasing to higher energy levels.

• Ground State (n = 1): This is the lowest energy level, where the electron resides closest to the nucleus. Its energy is -13.6 eV, according to the formula.
• Excited States (n > 1): As n increases, the energy levels move farther from the nucleus, and the energy of the electron becomes less negative. The excited states have higher energies than the ground state.

3.0First and Second Ionisation Energy

The first ionization energy refers to the energy required to remove the outermost electron from a neutral atom in its gaseous state. It's the energy needed to convert an atom into a positively charged ion by removing its most loosely held electron.

$\mathrm{M}+\Delta {\mathrm{H}}_{1\text{ st }}\to \mathrm{M}++{\mathrm{e}}^{-};\Delta {\mathrm{H}}_{1\text{ st }}=\text{ First ionisation energy }$

The second ionization energy, on the other hand, pertains to the energy required to remove an electron from an atom that has already been ionized once. This means it involves removing an electron from a positively charged ion to create a higher charged ion.

$M+\Delta H_{2nd}\to M^{+}+{\mathrm{e}}^{-};\Delta H_{2nd}=\text{ Second ionisation energy }$

• Removing a second electron from an already positively charged ion is inherently more challenging. Consequently, the second ionization energy is greater than the first ionization energy. 
• Subsequent ionization energies, such as the third ionization energy and beyond, continue this trend, increasing with each additional removal of an electron. 
• As more electrons are removed, the remaining ones experience an increasingly stronger attraction to the positively charged nucleus, requiring higher energy inputs to detach them from the ion.

$\Delta {\mathrm{H}}_{1\text{ st }}<\Delta {\mathrm{H}}_{2\text{ nd }}<\Delta {\mathrm{H}}_{\text{3rd }}<\dots$

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

4.0Factors Affecting Ionization Energy

Ionization energy, typically higher when removing electrons is more challenging, is influenced by several key factors governing the forces of attraction:

• Nuclear Charge: A positively charged nucleus exerts a strong pull on electrons, increasing the attraction between the nucleus and the electrons orbiting it.
• Electron Proximity: Electrons closer to the nucleus experience stronger attraction compared to those farther away. The closer an electron is to the nucleus, the greater the force of attraction.
• Electron Shielding: The presence of more inner-shell electrons between the outer electron and the nucleus reduces the effective nuclear charge felt by the outer electron. Consequently, the attraction forces diminish.
• Electron-Electron Repulsion: Electrons occupying the same orbital experience repulsion, creating a disturbance in the overall attraction to the nucleus. Paired electrons, due to this repulsion effect, exhibit lower ionization energy as they can be more easily removed compared to unpaired electrons.

5.0 Applications of Ionisation Energy 

(A) Metallic and non metallic character :

Generally for metals Ionisation Energy is low & for Non-metals Ionisation Energy is high.

Metallic character ∝ 1/IE

(B) Reactivity of metals :

Reactivity of metals increases down the group as ionization energy decreases.

Reactivity of metals  ∝ 1/IE

(C)  Stability of oxidation states of an element :

1. If the difference between two successive ionization energies of an element ≥ 16eV, then its lower oxidation state is stable.

Example-

Na (g)    →  Na⁺ (g)                                        

Na⁺ (g)   →  Na⁺² (g)

∆IE = 42.7 eV

Difference between ionization energy > 16 eV. So, Na⁺ is more stable.

2. If the difference between two successive ionization energies of an element < 11 eV, then  its higher oxidation state is stable

Mg (g)      →  Mg⁺   (g)                                               

Mg⁺ (g)    →  Mg⁺²  (g)

∆IE = 7.4 eV

Difference of ionization energy < 11 eV.   So, Mg⁺² is more stable.

3. If the difference in IP is between 11eV to 16eV then also lower oxidation state is stable.

Example-

Al (g)  →  Al⁺ (g)

Al (g)  →  Al⁺² (g)

∆IE = 12.8 eV

So Al⁺ is more stable

Al⁺ (g)  →  Al⁺² (g)

Al⁺² (g)  →  Al⁺³ (g)

∆IE = 6.0 eV

So, Al⁺³ is more stable

Overall order of stability is -    Al⁺³ >Al⁺ > Al⁺² 

6.0Trends in Periodic Table

1. Across a Period (from left to right): Ionization energy generally increases as you move from left to right across a period. This trend is due to the increasing nuclear charge. 

• As you move from left to right, the number of protons in the nucleus increases, resulting in a stronger attraction between the nucleus and the electrons in the same shell. Consequently, more energy is required to remove an electron.

2. Down a Group (from top to bottom): Ionization energy tends to decrease as you move down a group in the periodic table. This trend is because electrons are located in higher energy levels or shells as you move down a group. 

• The outermost electrons are farther away from the nucleus, which weakens the attractive force between the nucleus and these electrons. As a result, it takes less energy to remove an electron.