In an atom, electrons go to higher energy levels by absorption of energy from an external source and a stage comes when the electron goes completely out of the influence of the nucleus and produces a positive ion. Based on this phenomenon, the concept of Ionization potential or Ionization energy was introduced.
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What is Ionization Potential?
Ionization potential can be defined as the amount of energy required to remove the most loosely bound electron (i.e. outer most electron) from an isolated gaseous atom of an element in its lowest energy state (ground state) to produce a cation is known as ionization potential or Ionisation energy of that element.
Ionization is an endothermic process and it is generally represented as I or I.P and is measured in electron – volts (eV) or kJ mole – 1 (1 eV per atom = 96.48 kJ mole – 1 = 23.06 kcal mole – 1).
The ionization potential of an atom A can be expressed as follows-
Where, A(g) represents the isolated gaseous atom and A(g)+ represents the cation produced from this process.
Energy required to remove the first electron from a gaseous atom is called the first ionization potential. The energy required to remove the 2nd electron from the cation is called second Ionisation Potential. Similarly, successive removal of electrons gives 3rd, 4th, …Ionization Potential. For example, in the case of Al (Aluminium), the successive Ionization Potentials are shown as follows:
(1st Ionization Energy)
(2nd Ionization Energy)
(3rd Ionization Energy)
The second Ionization (2nd) Energy is always greater than the first Ionization Energy. Similarly, the 3rd Ionization Energy is always greater than the second Ionization Energy. Thus, higher ionization energy values are always greater than the lower ionization energy values as higher energy is always required to take out an electron from more positively charged species. Thus, IP1 < IP2 < IP3<….
Factors On Which Ionization Potential Depends
The magnitude of Ionisation Potential depends on the following factors.
- Charge of the Nucleus
- Atomic Radius or Atomic Size
- Principle Quantum Number and Screening effect of inner shells
- Effect of Full-Filled and Half-filled Orbitals on Ionization Potential Values
Charge of the Nucleus
The greater the charge on the nucleus of an atom, the more difficult it would be to remove an electron from the atom and hence greater would be the value of ionization potential. Thus the value of ionization potential generally increases, when we move from left to right in a period as the nuclear charge of the elements (atomic number) also increases in the same direction.
For example, the table below shows that ionization potential increases from Li to Ne in the first raw of the periodic table as the nuclear charges increase from Li to Ne.
|Atomic Nuclear Charge||Li|
|Ionisation Potential (eV)||5.39||9.32||8.30||11.26||14.53||13.61||17.42||21.56|
With the increase in the atomic radius, the ionization potential decreases. This is because of the fact that the attraction between the nucleus and the outer most electron is less for large size atom, hence it is easier to remove an electron from a larger atom than from a smaller atom. Thus on moving from top to bottom in a group in the periodic table, the ionization potential of the atoms decreases with the increase of their atomic radii.
For example, the table below shows that ionization potential decreases from Lithium (Li) to Caesium (Cs) in Group-IA due to the increase in their atomic radii.
|Elements of Group IA||Li||Na||K||Rb||Cs|
|Atomic Radii (Aº)||1.55||1.90||2.35||2.48||2.67|
|Ionisation Potential (I.P) in (eV)||5.39||5.14||4.34||4.18||3.89|
Principal Quantum Number (n) and Screening Effect of Inner Shells
Principal quantum number (n) determines the location of an electron from the nucleus. Higher the value of n, lower will be the amount of energy required to remove the electron and hence lower would be the value of ionization potential.
Further, when the principal quantum number increases, the inner shells very effectively screen outermost electron, hence the effective nuclear charge decreases. The decrease in effective nuclear charge leads to a decrease in the attraction between the nucleus and the outermost shell. Therefore, with the increase in principal quantum number and screening effect, the ionization potential decreases.
Effect of Full-Filled and Half-filled Orbitals on Ionization Potential Values
According to Hund’s Rule atoms having half-filled or completely filled orbitals are comparatively more stable and hence more energy is needed to remove an electron from such type of atoms. The ionization potential of such atoms is therefore relatively higher than expected normally from their position in the periodic table. For example, the first ionization potential of Nitrogen (N) (14.53 eV) is greater than Oxygen (O) (13.61 eV) due to the half-field orbital of Nitrogen (N) atom.
Variation of Ionization Potential (I.P) in the Periodic Table
In general, I.P gradually increases along a period (from left to right) and reaches a maximum with the noble gas with minor fluctuations. After each noble gas element there occurs a sharp fall of I.P with next alkali metal. The same sequence is then repeated. On moving from left to right in a period, since the nuclear charge of the atoms (atomic number) increases and atomic radii decreases, the ionization potential gradually increases.
The variation of I.P. along a period is shown graphically in the following figure for the 1st, 2nd and 3rd periods of elements.
On moving across a group from top to bottom, the I.P. of the elements decreases with the increase of their atomic radii.
Questions and Answers on Ionization Potential
According to Hund’s Rule atoms having half-filled or completely filled orbitals are comparatively more stable. Therefore, more energy required to remove an electron from such types of atoms. The ionization potentials of such atoms are higher than expected normally from their position in the periodic table.
In the case of B (2s2 2p1), we have to remove an electron from an incompletely filled 2p orbital to get B+ ion while in the case of Be (2s2), we have to remove an electron from a completely filled 2S shell to have Be+ ion. For this reason, the first I.P of the Be is greater than that of B. Similarly, to remove the first electron from stable half-filled 2p orbital of N (2s2 2p3) is required more energy than from incompletely filled 2p orbital of O (2s2 2p4) atom. Therefore, the first ionization potential of the O atom less than of N atom.
On descending a group, the I.P value generally decreases because of the pronounced increase in the size of the atoms. For this reason, the 1st I.P of Ag is less than that of Cu. But due to the effect of lanthanide contraction, there is practically no increase in the size of Au, whereas its nuclear charge is higher. Hence, due to the higher nuclear charge of Au as compared to Ag, the I.P of Au becomes higher than that of Ag i.e. Cu ˃ Ag ˂ Au.