Ionization Energy Formula

Ionization is a process that involves the removal of the electrons that are present in an orbit to the outside of the atom. As the electron present in each orbit has some characteristic energy, that is the ionization energy which is equal to the difference of energies that are present in between the energy of the electron in the initial orbit and the energy of the electron outside the atom.

The ionization energy formula or ionisation energy formula is the representation of the minimum amount of energy that is required to remove the most loosely bound electrons of an isolated neutral gaseous atom or molecule. It is quantitatively expressed as :

X(g) + energy → X⁺(g) + e^-

From the formula of ionisation energy, X represents an atom or molecule, X+represents the resultant ion when the original atom was stripped of a single electron, and e- is the removed electron. This process is generally an endothermic process. As a rule, the higher the atom's ionization energy is experienced when the outermost electrons are closer to the nucleus of the atom.

There are different units for ionization energy in Physics and Chemistry. In physics, the unit of the ionisation energy formula is the amount of energy that is required to remove a single electron from a single atom or molecule it is expressed as electron volts. In chemistry, the unit is the amount of energy that is required for all of the atoms present in a mole of substance to lose one electron each. It is also known as molar ionization energy or approximately enthalpy it is expressed as kilojoules per mole (KJ/mol) or kilocalories per mole (Kcal/mol).

Factors Affecting the Ionization Potential Formula

The term ionization potential formula is an older and obsolete term used for ionization energy because it is the oldest method of measuring the ionization energy and it was based on ionizing a sample and accelerating the electron that is removed using an electrostatic potential.

The Most Notable Factors That are Affecting the Formula for Ionisation Energy Include:

• Electron Configuration: This accounts for most of the element's ionization energy, as all of their physical and chemical characteristics can be ascertained just by determining their respective electron configuration.

• Nuclear Charge: Let us assume the nuclear charge or the atomic number is greater, then the electrons are said to be held more tightly by the nucleus and thus the ionization energy will be greater.

• A Number of Electron Shells: Due to the presence of more shells, if the size of the respective atom is greater, then the electrons that are said to be held by the nucleus are held less tightly and thus the ionization energy will be lesser.

• Effective Nuclear Charge: It is denoted as Z_eff . If the magnitude of shielding of electrons and penetration is greater, then the electrons that are held by the nucleus are held less tightly, then Z_eff of the electron and the ionization energy will be lesser.

• Type of Orbital to be Ionized: An atom that consists of a more stable electronic configuration then it is always prone to less tendency to lose electrons and consequently it has higher ionization energy.

• Electron Occupancy: The highest occupied orbital by the electron is doubly occupied, then it is easier to remove an electron from that particular orbital.

These are the factors that have to be considered when we calculate ionization energy.

Ionization Enthalpy Formula

The enthalpy change that is associated with the removal of the first electron from an isolated gaseous atom in its ground state is called the first ionization enthalpy. The ionization enthalpy of an element can be defined as the amount of energy required to remove an electron from an isolated gaseous atom in its gaseous state. 

Ionization Enthalpy Depends on the Following Factors:

• Penetration Effect: Penetration effect means it is the proximity of an electron present in an orbital to the nucleus. For each shell and the subshell, it can be observed that the penetration effect is the relative density of electrons present near the nucleus of an atom. Now if we look into the radial probability distribution functions, we can see that the electron density of s-orbitals is closer than that of p-orbitals and d-orbitals.

The order of penetration power will be 2s > 2p > 3s > 3p > 4s > 3d.

• Shielding Effect: The shielding effect can be defined as the effect in which the inner electrons can develop a shield for the electrons present in outer shells which do not let the appropriate nuclear charge towards the outermost electrons. Due to this, the electrons present in the outermost shell can experience a low effective nuclear charge and not the actual nuclear charge. 

The Effective Nuclear Charge Can Be Given As: 

Z_effective = Z-S

Z_effective = Effective Nuclear Charge 

Z → Actual Nuclear charge

S( ) → Screening Constant

• Electronic Configuration: Elements that are having half-filled and fully filled orbitals are said to be in a stable state. So if we try to remove an electron from these orbitals, then it will make these electrons experience less stable in nature. Hence, more energy is required to remove an electron from these orbitals. Thus, higher ionization energy.

Ionization Energy Equation Trends in the Periodic Table

Across a Period: In the periodic table while moving from left to right in a period, the atomic radius of an atom decreases. So if the size of an atom decreases, then the attractive forces that are present in between the nucleus and the increase of the outermost electron. Due to this reason across a period in the periodic table, the ionization energy can generally increase.

In the second period, there is a discrepancy in the trend of ionization enthalpy from boron atom to beryllium atom. Ideally, the ionization enthalpy of the boron atom should be more than that of beryllium, but it happens oppositely. The reason being the beryllium atom has fully filled subshells and the second reason is due to the penetration effect. The boron atom has electrons in the 2s and 2p orbitals whereas beryllium have electrons in the 2s orbital only. The penetration power of 2s orbital is more than that of 2p orbitals. Thus, the removal of an electron from the 2p subshell will be easier compared to that of the 2s subshell in beryllium. Hence, due to these two factors, the ionization enthalpy of beryllium will be more compared to that of boron.

In a Group: While moving down in a group the ionization energy of elements decreases as the number of shells increases down the group. The electrons present in the outermost shell will be far away from the nucleus, and the effective nuclear charge will be less. Secondly, the shielding effect can also increase down the group as the number of shells present increases which in turn results in decreasing ionization energy.

First and Second Ionization Energy Equation

The first ionization energy formula can be represented as the energy that is required to remove the first electron from a neutral atom. It is numerically the same as the orbital energy of the electron but of the opposite sign.

For hydrogen, the first orbit energy is -2.1810^-18 J/atom and the ionization energy is +2.1810^-18 J/atom.

The energy needed for the removal of the second electron away from the unipositive ion is second ionization energy and so on.

For Example:

M + ΔH_1st →M⁺+e^-

Here ΔH_1st represents the first ionization energy equation.

M⁺ + ΔH_2nd → M²⁺+e^-

Here ΔH_2nd represents the second ionization energy.

Removing the second electron naturally from an already positive ion will be difficult. Hence the second ionization energy will be larger than the first ionization energy. The third ionization energy will be more than that of the second ionization energy etc.

H_3rd<H_2nd<H_1st<.......

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

For example, the helium atom is more stable due to the fact that it is completely filled with s-orbital than that hydrogen. So, the first ionization energy of the Helium atom is 2372 KJ/mole is more than that of the hydrogen atom is 1312 KJ/mole.

The first ionization energy of the nitrogen atom is 1402KJ/mole is more than that of its near neighbours, the ionization energy of the carbon atom is 1086 KJ/mole and the oxygen atom is 1313KJ/mole, because of the higher stability due to half-filled orbitals. Chromium has half-filled s- and d-orbitals and so has much more first ionization than titanium.

Solved Example

1. Which of the Following Order is Correct With Respect to the First Ionization Energies of the Following Elements?

1. B < Be < N < O 

2. B < Be < O < N

3. Be < B < N < O 

4. B < O < Be < N

Ans:  Ionization energy, in general, increases with a decrease in the atomic radius across the period from left to right. However, there are certain exceptions. The s-orbitals have greater penetration power than p-orbitals. Hence the removal of electrons from s-orbitals requires more energy. The electrons in half-filled orbitals are stable as they experience less repulsion. If there is another electron sharing the same orbital, then they will be destabilized due to repulsion from each other. 

The Above Atoms Belong to the Same Period of the Periodic Table.

B represents the Boron, it is smaller than beryllium, the Be atom. Hence we can expect an increase in ionization energy from Be to B. However, the Be atom has greater ionization energy compared to the B atom. The reason is in the case of Beryllium, the last electron of this atom is in the s-orbital, and in Boron, the last electron is in the p-orbital. We know that the removal of electrons from the s-orbital requires more energy than the electron from the p-orbital.

The 2s² 2p³ configuration is more stable than 2s² 2p⁴ due to the half-filled p-sublevel. The electrons in the p-orbital in the Nitrogen atom experience less repulsion, hence more stable and more ionization energy is found.  Whereas in the case of the Oxygen atom the fourth p-electron atom can be removed more easily as it experiences more repulsion from the electrons in the p-orbitals already present.  Hence nitrogen, an Oxygen atom has less ionization energy than a Nitrogen atom.

The correct ionization energy order is: B < Be < O < N