[Explained] Ionization Energy & Successive Ionization Energies [First,Second,Third]

As we all know that an atom is composed of a positively charged nucleus and negatively charged electrons revolving around it. The nucleus exerts an appreciable force of attraction on the electrons that revolve around it. 

If we wish to remove an electron from the atom, we will have to supply energy to overcome the force of attraction that the nucleus exerts on the electrons. This energy is known as Ionization Energy. It is also known as Ionization Enthalpy. We will also cover successive Ionization Energies later in this article.

Let us first look at how Ionization Energy is defined as:

The amount of energy required to remove the most loosely bound electron from an isolated gaseous atom to form a gaseous ion is called the ionization enthalpy or ionization energy of that atom.

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

Here M denotes a metal ion. Metals most easily lose their electrons because, they have a tendency to complete their octet. You can see what trend the Ionization energy follows across a period and also how ionization energy changes for elements if we move down the group.

There are various Factors that affect Ionization Energy.

Ionization Potential

The ionization energy or ionization enthalpy can be measured in a discharge tube containing the gas under study at low pressure. At a certain voltage between the cathode and the anode, the gas gets ionized. The ionization of the gas is indicated by a sudden large increase in the flow of the current through the gas. 

The voltage at which the ionization of the gas occurs is known as the ionization potential and is numerically equal to the ionization energy of the gas under study in the unit of eV (Electron Volts). The ionization potential is very often expressed in terms of electron volts and is used to convey the sense of ionization energy. 

One electron volt (eV) is the energy acquired by an electron under a fall of potential of 1 volt. Thus,

                                         1 eV = 1.0621 x 10^-19 J = 1.6021 x 20^-22 kJ

Successive Ionization Energies

Second Ionization Energy

The removal of the most loosely bound electron from a neutral atom forms a monopositive cation. The ionic radius of this cation is much smaller than the atomic radius of the parent atom. Due to smaller size of the cation, the remaining electrons experience a greater pull of the nucleus. Therefore, a higher energy will be required to remove an electron from the unipositive cation. This energy is called the second ionization energy and is represented as IE₂. 

Third Ionization Energy

The dipositive ion thus formed will be further smaller and still higher energy will be required to remove an electron from it to form a triposititve ion. This energy is called the third ionization energy and is represented as IE₃.

General Formula for Successive Ionization Energy

                        M (g)  + IE₁            → M⁺ (g) + e^-

                        M⁺ (g)  + IE₂          → M²⁺ (g) + e^-

                        M²⁺ (g)   + IE₃        → M³⁺ (g) + e^-

The second, third and fourth etc. ionization energies are known as successive ionization energies.

                                                 IE₃ > IE₂ > IE₁ 

It is comparatively more difficult to remove electrons from completely or half filled shells. This is why helium possesses much higher value of IE₁ (2372.6 kJ mol^-1) as compared to that of hydrogen (1312 kJ mol^-1). 

If the removal of an electron results in the removal of the valence shell, the successive ionization energy will be much higher. For example, in lithium, the removal of an electron removes its valence shell. Therefore, IE₂ of Li (7297 kJ mol^-1) is much higher than IE₁ (520.3 kJ mol^-1) 

A very common question regarding the Second ionization energy is:

Which of the following has the highest second ionization energy, sodium or magnesium?

The answer to this question is Sodium.

Let us understand why this is so. Sodium is an element which is a member of the Alkali Metal Elements. It is denoted by the symbol Na. It has an electronic configuration of [Ne] 3s¹

This means that it only has one electron in its outermost shell. Or if we put it in noble gas terms. It only has a single electron more than the nearest noble gas Neon.

As we know that, when an atom gains the noble gas electronic configuration it is in its most stable form.

So when we want to remove the second electron from the now positively charged sodium ion. We need to supply a large amount of energy to remove the electron a very stable ion.

Magnesium is not the answer because it is a member of Alkaline Earth Metals and it has 2 electrons in the outermost shell.

 So we can expect the third ionization energy to be high on elements of the second group.

Which element has the highest second ionization energy: Li, K, or Be?

The answer for this surely be Lithium, Li, because as we have learnt that the second ionization energy of Alkali metals are very high due to having just an extra electron more than the noble gas configuration.

The ionization energy of an atom decreases as we move down the group. 

As Lithium is on top of the group. It is the element which has the highest second ionization energy.

 

Numerically, IE₂ of Lithium is equal to 7298.1 kJ mol^-1.

Read: [Solved] Arrange These Elements According To First Ionization Energy

           [Solved] Which Reaction Below Represents The First Ionization Of O(Oxygen)?

           [Solved] Write Equations That Show The Process For The First Two Ionization