# Ionization Energy

### What is Ionization Energy?

Ionization energy can be explained as an amount of the trouble or difficulty in eliminating an electron from an atom or ion. It can also be defined as the tendency of an atom or an ion to surrender an electron.

Generally, electron elimination occurs in the ground state of the chemical types.

There are also some technical ways to describe the ionization energy as the least amount of energy required for an electron in a vaporous atom or ion to consume to come out of the effect of the nucleus.

Many definitions also describe it as the ionization potential and commonly, this ionization process is endothermic in nature.

We can figure out the reactivity of chemical compounds by applying the concept of ionization energy.

The strength of chemical bonds can also be governed by using this concept. The measurement unit of ionization energy is either in the units of electron-volts or kJ/mol.

### How to Calculate Ionization Energy?

For ionization energy, the concept is to study a quantity of the tendency of an atom or ion to give up an electron or the effectiveness of the electron binding.

If the ionization energy is at its best and optimum value, it will be very hard to remove an electron.

We may consider the ionization energy as a pointer of the reactivity of an atom or ion. If there is low ionization energy in an element, then it will be a reducing agent and it will combine with the anions instead of the cations to form the salt.

As we know high ionization energy will bring more difficulties to eliminate an electron from the grasp of its nucleus, we are governing several factors which direct the attraction forces.

• If the charge of the nucleus is positive, the attraction of the electrons is so strong towards it.

• If an electron exists adjacent to or near to the nucleus, the attraction will be superior to the one when the electron is a little far.

• If the presence of electrons is more than one in between the outer level and the nucleus, there will be fewer attraction forces.

• At the time of the presence of the two electrons in the identical orbital, they produce some sort of repulsive force.

The repulsion generates some conflict in the nucleus attraction. As per the principle, ionization energy will be in a less amount in paired electrons so that they can be eliminated with ease.

The process of ionization which includes the elimination of electrons present in an orbit external to the atom.

We know that each orbit’s electron has its distinctive energy. So the ionization energy is equal to the energy difference between the energy of the electron in the preliminary orbit and the energy of the electron outside the atom or in the infinite orbit from the nucleus.

The energy of an electron can be calculated for the nth orbit by Bohr model of an atom as –

En=- $\frac{2π^{2}me^{4}}{(4πε^{0})^{2}h^{2}}$ ×  $\frac{z^{2}}{n^{2}}$

=R×$\frac{z^{2}}{n^{2}}$J/atom=-13.6×$\frac{z^{2}}{n^{2}}$eV/atom=-2.18×10-18$\frac{z^{2}}{n^{2}}$  J/atom

As we studied earlier, the procedure to remove an electron from its orbit with a certain amount of energy expenditure is known as the Ionization energy. It can be calculated by replacing the electron’s orbit number before the transition denoted as ‘n1’ and orbit number of the electron after transition as ‘∞’ (infinity) denoted as ‘n2’ in Bohr’s energy equation.

En1= -R×$\frac{z^{2}}{n^{2}}$En2= -R× $\frac{z^{2}}{∞^{2}}$ΔE  = En2-En1= R×z2($\frac{1}{n^{2}}-\frac{1}{∞^{2}}$) = Ionization energy

So the amount of ionization energy will be

Ionization Energy =ΔE = En2-En1= R × z2$\frac{1}{n^{2}}$ = 2.18×10-18×z2$\frac{1}{n^{2}}$ J

### Ionization Energy Example

Numerically, we describe ionization energy as the orbital energy of the electron with the reverse sign.

Let’s consider for hydrogen,

The first orbit energy = – 2.18 × 10– 18 J/atom (or – 1312.3 KJ/mole)

The ionization energy = + 2.18 × 10–18 J/atom (or + 1312.3 KJ/mole).

So the amount of optimum energy required for the elimination of the second electron away from the unipositive ion is termed as the second ionization energy. The procedure is the same for more outputs.

### For Illustration:

First Ionization energy ∆H1st = M + ∆H1st → M+ + e

Second Ionization energy = ∆H2nd = M+ + ∆H2nd →M2+ + e and so on.

Eliminating the second electron from a previously positive ion will be problematic. If you do the math, the ionization energy for the second electron will be greater than the first ionization energy.

Likewise, it will be repeated for the third ionization energy such as it will be more than that of the second ionization energy and so on.

Here is the mathematical expression

∆H1st < ∆H2nd < ∆H3rd <….<Hnth

### Ionization Energy Periodic Table

In the periodic table, the ionization energy of an electron upsurge with the atomic number of the atom and drops for higher energy orbitals. If we spectate on the periodic table and travel from left to right across the elements, the ionization energy will rise due to a declining atomic radius.

[Periodic Table will be Uploaded Soon]

On the other hand, if we go from head to toe, the ionization energy will reduce. This is mostly because of the existence of other electron shells in the elements as we move down the group.

Also, the electrons are kept at a longer distance from the forces of attraction of the nucleus.

Q1. Can You distinguish the element Which has the Highest Ionization Energy?

Ans: The first ionization energy differs expectedly across the periodic table. The ionization energy reduces from top to bottom in groups and rises from left to right across a period.

Therefore, Helium has the largest first ionization energy while Francium has one of the lowest.

Q2. Calculate the net Energy that All Aluminum Atoms desire to turn into Al3+ that are Existing in 0.620g of Aluminum Vapours. As per the Aluminium, the Ionization Energies for the First, the Second, and the Third are 568, 1807, and 2735 KJmol-1.

Ans: We know that the net energy to make the change in Al to Al3+ will be

= IE1 + IE2 +IE3

= 568 + 1807 + 2735 kJmol−1kJmol−1

= 5110 kJmol−1

Now, the number of mole present in 0.620 gm of Al= 0.620/27 = 0.02mole

= 2 × 10−2 mole

Hence, 3 × 10−2 mole of Al will require = 5110 × 2 × 10−2 = 110.2 kJ

Q3. Explain the Working Procedure of the Ionization Energy.

Ans: The working process of ionization energy can be explained as the amount of energy that must be absorbed by an isolated, gaseous atom in the ground electronic state to release an electron.

This process results in cation production. This energy is usually expressed in kJ/mol.

Q4. What is the Possibility of the Ionization Energy; Positive or Negative?

Ans: The answer is the ionization energy is always positive because every solitary electron is bound and therefore, the electron has negative energy.

So the energy required to remove an electron must be positive.