Chemistry Formulas

Ionization Energy Formula

What is Ionization Energy?

This article deals with the ionization energy formula. Ionization energy refers to the minimum amount of energy required to remove the electron that is most loosely bound, the valence electron of an atom or molecule that is isolated neutral gaseous. Earlier experts called this energy as the ionization potential, but that’s no longer in usage.

Ionization Energy Formula

Furthermore, experts always start with an outer shell when measuring the ionization energy, then they move inward toward the nucleus. Interestingly, as the removal of electrons from an atom take place, it becomes more difficult to remove electrons and requires more ionization energy since the change has taken place in the charge of an atom.

Derivation of Ionization Energy Formula

Use of different measures of ionization energy takes place in the sciences of physics and chemistry. In physics, the unit is the amount of energy that can remove a single electron from a single molecule or single atom. The expression of the unit in physics takes place in electronvolts (eV).

In contrast, the unit in chemistry refers to the amount of energy that all the atoms in a mole of substance require so as to lose one electron each. Moreover, in chemistry, this energy refers to the molar ionization energy or enthalpy and its expression takes place in kilocalories per mole (kcal/mol) or kilojoules per mole (kJ/mol).

Analogs of Ionization Energy

Usage of the term ionization energy is mostly for gas-phase molecular or atomic species. Furthermore, a number of analogous quantities exist that consider the amount of energy that is needed to remove an electron from various physical systems.

Electron binding energy refers to a generic term for the ionization energy. Experts make use of this term for any charge state species. For example, the chloride ion-electron binding energy refers to the minimum amount of energy to bring about an electron removal from the chlorine atom when it carries a -1 charge.

In this particular example, the magnitude of electron binding energy is the same as that of the neutral chlorine atom’s electron affinity. The highest specific binding energy is that of nickel and it is 8.8 MeV.

Solved Questions For You

Question 1- Find out the energy that all Al atoms require to convert to \(Al^{3+}\) that are available in 0.720g of Al vapors. In respect of Al, the ionization energies first, second, and third, are 578, 1817, and 2745 \(kJ mol^{-1}\).

Answer 1- The total energy to bring about the change \(Al\rightarrow Al^{3+}\)

= IE1 + IE2 +IE3

= 578 + 1817 + 2745 \(kJ mol^{-1}\)

= 5140 \(kJ mol^{-1}\)

Now, number of mole present in 0.720 gm of Al=

0.720/27 = 0.03mole

= 3 × \(10^{-2}\) mole

Hence, 3 × \(10^{-2}\) mole of Al will require 5140 × \(10^{-2}\) = 51.40 kJ

Question 2- In both potassium and copper, the removal of an outer electron has to take place from the 4s- orbital. Why the copper first ionization energy 745 \(kJ mol^{-1}\) is higher than the first ionization energy for potassium 418 \(kJ mol^{-1}\)?

Answer 2- In both these elements of potassium and copper, the external electron is in the 4s level.  However, in potassium, the curtailment of the nucleus’ positive charge (Z=19) in potassium is by the argon core: 1s22s22p63s3p6. Furthermore, in copper, the nuclear charge (z=29) in copper is curtained by the argon core and 3d electrons.

In copper, the ten extra electrons are in the 3d level. The diffused, and strongly directional d-electrons, that are diffused, protect the 4s electron from the nuclear charge. However, this protection from the nuclear charge is somewhat poor. Consequently, copper experiences a nuclear charge that is highly effective and the ionization energy of copper is far greater in comparison to potassium.

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