Ionization energy trends in periodic table

Ionization energy trends in the periodic table

Ionization energy defined as the amount of energy required to remove the most loosely bound electron or the outermost electron from an isolated gaseous atom of an element in its lowest energy state to produce a cation.

In the periodic table, ionization energy trends depend on several factors. The factor which influences IE is

  1. Atomic radius
  2. Atomic number
  3. Completely-filled and half-filled orbitals
  4. Shielding electrons of an atom
  5. Overall charge on the ionizing species
Ionization energy trends
Ionization energy trends in the periodic table

Why ionization energy is an endothermic process?

Electrons are raised to higher energy levels by the transfer of energy from external sources. But if energy transfer to electrons sufficient, electrons go completely out of the influence of the nucleus of an atom.

M\left (g \right )+IE\xrightarrow{+\Delta H}M^{+}\left (g \right )+e

Thus the process of ionization is an endothermic process because, during the process, energy is consumed by atoms.

Ionization generally represented I or IE and measured in electron volt or kilocalories per gram atom.

The simple explanation of electron volt

The energy consumption by an electron falling through a potential difference of one volt defined as an electron volt. It is simply represented as eV.

∴ 1 eV = charge of an electron × 1 volt
= (1.6 × 10-19 coulomb) × (1 volt)
= 1.6 × 10-19 Joule

1 eV = 1.6 × 10-12 erg

First, second and third ionization energy

The amount of energy required for removal of the first electron from a gaseous atom called its first ionization.

M (g) + IE1 → M+ (g) + e

But if the energy consumption for removal of the second electron from a cation called second ionization.

M+ (g) + IE2 → M+2 (g) + e

Similarly, we have third, fourth ionization.
M+2 (g) + IE3 → M+3 (g) + e
M+3 (g) + IE4 → M+4 (g) + e

The ionization energy of hydrogen and helium

The energy transfer for completely removing an electron from hydrogen energy levels is called IE.

Simply the energy corresponding to the transition from n = ∞ to n = 1 gives the ionization energy of the hydrogen.

So the IE of the hydrogen

\frac{2\pi ^{2}me^{4}}{h^{2}}\left ( \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right )where n1 = 1 and n2 = ∞.

∴ EH = 2.179 × 10-11 erg
= 2.179 × 10-18 Joule

\frac{2.179\times 10^{-18}}{1.6\times 10^{-19}}\, eV

∴ EH = 13.6 eV

Second ionization of helium

The electron configuration of helium
1S2

Thus the second IE means the removal of the second electron from the 1S orbital against the nuclear charge of +2.

\therefore IE_{He}=\frac{2\pi ^{2}mZ^{2}e^{4}}{h^{2}}\left ( \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right )

= Z2 × IEH

Thus second IE of helium
= 22 × 13.6 eV
= 54.4 eV

Atomic radius in the periodic table

Greater the atomic radius of an element, the weaker will be the attraction. Hence the required energy for the removal of the electron lower.

Thus if an atom raised to an excited state by promoting one electron to a higher energy level, the excited electron more easily detached because the distance between the electron and nucleus increases.

Atomic radius and ionization energy

  1. The atomic radius decreases from left to right along a period in the periodic table. Thus when we move left to right along a period, ionization energy normally increases.
  2. But when we moving from top to bottom in a group the IE of the elements decreases with the increasing size of the atom.

Atomic number and ionization energy

With the increasing atomic number charge on the nucleus increases and more difficult to remove an electron from an atom. Hence grater would be the value of IE.

Normally the value of ionization energy increases in moving from left to right in a period of the periodic table. Because with the increasing atomic number, the change in the nucleus also increases.

Thus the electrostatic attraction between the outermost electrons and the nucleus of an atom IE also increases. So the removal of an electron from the nucleus is more difficult.

Half filled and completely filled orbitals

According to Hund’s rule, half-filled or completely filled orbital comparatively more stable. Hence for such an atom, more energy required to remove an electron.

Thus the ionization of such an atom is relatively more difficult than expected normally from their position in the periodic table.

Exceptions to the ionization energy trends

Few exceptions in the value of the ionization energy trends in the periodic table can be explained on the basis of the half-filled and completely filled orbitals.

Group-15 elements have higher IE than the group-16 elements and group-2 elements have higher than the group-3 elements in the periodic table. Thus the first IE of nitrogen greater than oxygen and phosphorus greater than sulfur.

Nitrogen and phosphorus in group-15 elements with atomic numbers 7 and 15 have the electron configuration

1S2 2S2 2P3
1S2 2S2 2P6 3S2 3P3

Thus the removal of an electron from half-filled 2P and 3P suborbital of nitrogen and phosphorus required more energy.

Removal of an electron from the group-2 element of beryllium and magnesium with completely-filled S-subshell required more energy.

Shielding effect and ionization energy

Electrostatic attraction between the electrons and nucleus shows that an outer electron attracted by the nucleus and repelled by the electrons of the inner shell.

This combines attractive and repulsive force acting on the outer electron experiences less attraction from the nucleus. Thus this effect is known as the shielding effect.

The larger the number of electrons in the inner shell, the lesser the attractive force for holding the outer electron. Hence the radial distribution functions of the S, P, d subshell show that for the same principal quantum number the S-subshell most shielding than the p-subshell and least shielding the d – orbital.

∴ Shielding efficiency
S〉P〉d

But as we move down a group, the number of inner-shells increases and hence the ionization energy tends also decreases.

Thus for group-2 elements
Be〉Mg〉Ca〉Sr〉Ba

Ionization of atoms and the periodic table

The greater the charge on the nucleus of an atom the more energy required for removing an electron from the atom. But with the increasing atomic number electrostatic attraction between the outermost electrons and the nucleus of an atom increases.

Thus the ionization of an atom is more difficult. Again the values of IE generally increase in moving left to right in a period.

Due to the presence of a completely filled and half-filled orbital of beryllium and nitrogen, the ionization of beryllium and nitrogen slightly higher than the neighbor element boron and oxygen. Thus the ionization trends in the periodic table for the second period

LiㄑBㄑBeㄑCㄑOㄑN〈FㄑNe

Overall charge on the ionizing species

An increase in the overall charge on the ionizing species (M+, M+2, M+3, etc) will enormously influence the ionization.

During ionization electron withdrawal from a positively charged species more difficult than from a neutral atom. But the first ionization of the elements varies with their positions in the periodic table.

  1. In each of the tables, the noble gas has the highest value.
  2. The alkali metals the lowest value for the ionization energy.

How ionization energy impacts ionic bonding?

The study of the ionization of the element in a particular group is essential for the properties of the elements.

Hence lithium, sodium, potassium, rubidium, cesium, and alkaline earth metal have a low value of ionization energy. Thus reactivity of alkali alkaline metals for the formation of chemical bonding is greater than the other elements of the periodic table.