Ionization potential and periodic table

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Ionization energy trend in the periodic table

The study of ionization energy is an important article in inorganic chemistry for school and college-level courses.

The electrons are raised to higher energy levels by absorption of energy from external sources. If energy supplied to electrons sufficient, electrons go completely out of the influence of the nucleus of an atom.

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 or ground state to produce a cation is known ionization energy.

M (g) + Ionization energy → M⁺ (g) + e

The process of ionization is an endothermic process since energy is supplied during ionization. Ionization energy generally represented I or IE and measured in electron volt or kilocalories per gram atom.

What is electron volt simple definition?

One electron volt is the energy consumption by an electron falling through a potential difference of one volt. Electron volt simply represented 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

Removal of an electron from the hydrogen atom

The energy required for removing an electron from energy levels of the hydrogen atom called the ionization energy. Simply the energy corresponding to the transition from n = ∞ to n = 1 gives the ionization energy of the hydrogen atom.

The ionization energy of the hydrogen atom
= (2π²me⁴/h²)[(1/n₁²) - (1/n₂²)]
where n₁ = 1 and n₂ = ∞.

∴ EH = 2.179 × 10⁻¹¹ erg
= 2.179 × 10⁻¹⁸ Joule
= (2.179 × 10⁻¹⁸)/(1.6 × 10⁻¹⁹) eV.

∴ EH = 13.6 eV.

Second and third ionization energy

  1. The electrons are removed in stages one by one from an atom. The amount of energy required to remove the first electron from a gaseous atom called its first ionization energy.
  2. M (g) + IE₁ → M⁺ (g) + e
  3. The energy required to remove the second electron from a cation called second Ionization energy.
    M⁺ (g) + IE₂ → M⁺² (g) + e
  4. Similarly, we have third, fourth ionization.
    M⁺² (g) + IE₃ → M⁺³ (g) + e
    M⁺ (g) + IE₄ → M⁺⁴ (g) + e
Question
How to calculate the second ionization energy of helium atom if ionization energy of hydrogen is13.6 eV?

Answer
The ground state electronic configuration of helium 1S². The second ionization potential means the removal of the second electron from the 1S orbital against the nuclear charge of +2.

∴ IEHe = (2π²mZ²e⁴/h²)[(1/n₁²) - (1/n₂²)]
= Z² × IEH.

∴ Second Ionization energy of helium = 2² × 13.6
= 54.4 eV.

First ionization energy trend in the periodic table

First ionization energy trend in the periodic table
Ionization energy and the periodic table
The trend of ionization energy in periodic table influenced by the following factors
  • Atomic radius.
  • Change on the nucleus or atomic number.
  • Completely-filled and half-filled orbitals.
  • Shielding effect of the inner electrons.
  • Overall charge on the ionizing species.

How are the atomic radius and ionization energy related?

Greater the atomic radius or the distance of an electron from the positive charge nucleus of an element, the weaker will be the attraction. Hence the energy required to remove the electron lower.

An atom raised to an excited state by promoting one electron to a higher energy level than the excited electron was more easily detached because of the distance between the electron and nucleus increases.

The Atomic radius decreases from left to right along a period of the periodic table because of the increasing charge on the nucleus of an atom. Thus when we move left to right along with period normally ionization energy increases.

When we moving from top to bottom in a group the ionization energy of the elements decreases with the increasing size of the atom.

Atomic number and ionization energy relationship

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

Normally the value of ionization increases in moving from left to right in a period since with the increasing atomic number the change on the nucleus also increases.

The increase in the magnitude of ionization due to the increase in the electrostatic attraction between the outermost electrons and the nucleus of an atom. Thus it becomes more difficult to remove an electron.

Half filled and completely filled orbitals

According to Hund's rule, an atom having half-filled or completely filled orbital comparatively more stable and hence more energy consumption to remove an electron from such atom.

The ionization of such an atom is therefore relatively difficult than expected normally from their position in the periodic table.
Few exceptions in the value of ionization energy in the periodic table can be explained on the basis of the half-filled and completely filled orbitals.

The ionization energy of group-15 elements is higher than the group-16 elements and group-2 elements are higher than the group-3 elements in the periodic table.
Boron and nitrogen in the second period and magnesium and phosphorus in the third period have a slightly higher value of ionization energy than those normally expected.

Nitrogen and phosphorus in group-15 elements with atomic number 7 and 15 have the electronic configuration

1S² 2S² 2P³
1S² 2S² 2P⁶ 3S² 3P³.

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.
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Shielding effect of the inner electrons

The study of 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.
The combined effect of this attractive and repulsive force acting on the outer electron experiences less attraction from the nucleus. This is known as the shielding effect.
Thus a larger number of electrons in the inner shell, lesser the attractive force for holding outer electron.

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.

As we move down a group, the number of inner-shells increases and hence the ionization tends to decreases.

Group-2 elements
Be〉Mg〉Ca〉Sr〉Ba.

What trend in ionization energy occurs across a period?

What trend in ionization energy occurs across a period?
Ionization energy trend
The greater the charge on the nucleus of an atom the more energy consumed for removing an electron from the atom.

More energy consumed means more energy required for removing an electron from the atom. With the increase atomic number electrostatic attraction between the outermost electrons and the nucleus of an atom increases. Thus removal of an electron from an atom more difficult.

The values of ionization energy generally increase in moving left to right in a period since the nuclear charge of an element also increases in the same direction.

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

LiㄑBㄑBeㄑCㄑOㄑN〈FㄑNe.

The overall charge of an atom

An increase in the overall charge on the ionizing species (M⁺, M⁺², M⁺³, etc) will enormously influence the ionization since electron withdrawal from a positively charged species more difficult than from a neutral atom.

The first ionization of the elements varies with their positions in the periodic table. In each of the tables, the noble gas has the highest value and the alkali metals the lowest value for the ionization energy.

Describe how ionization energy impacts ionic bonding

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

Lithium, sodium, potassium, rubidium, and cesium or alkali metal in the periodic table with a low value of ionization energy, point to the high reactivity of alkali metals for the formation of the ionic bonding.

Study online trend of ionization energy and periodic table for school and college-level courses, atomic number, atomic radius, shielding effect

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