# Ionization energy (or Ionisation energy) of group 1 (alkali metals) elements

periodic table ionization energy (I.E.) is an important property for the elements. It helps to understand the affinity of an element for electrons as well as the tendency of the elements to form an ionic or covalent bond. So what is the I.E.?

## Ionization energy (I.E.)

The energy needed to remove the most loosely attached 1 mole electron of an isolated gaseous 1 mole atom to form one mole cation. That implies that 1 mole electrons from 1 mole metal must be removed from the energy. The gaseous state of the 1 mole metal.

We can write a general equation to describe the removal of an electron from the metal.

M(g) + energy = M+(g) + e

here, M= the group 1 elements (alkali metals)

The ionization energy of atoms in the periodic table reveals two patterns:

generally-

1. increases as one moves from left to right within a given period.
2. decreases as one moves down a given group.

## What is 1st, 2nd, 3rd Ionization energy (I.E.)

Now, it can be 1st, 2nd, 3rd, etc…

the energy needs to remove a mole of an electron from the gaseous one-mole atoms is first I.E. Same as the energy needs to remove a mole of the electron from the gaseous one mole of the species with a charge of +1.

 1st ionization energy: M → M+ + e− 2nd ionization energy: M+ → M2+ + e− 3rd ionization energy: M2+ → M3+ + e− nth ionization energy: M(n-1)+ → Mn+ + e−

So nth I.E. relates to the quantity of energy needed to remove an electron with a charge of +(n-1) from the species.

## 1st Ionization energy Of group I

 Period Name (Symbol of elements) Atomic Number (Z) Simple Electronic Configuration Atomic Radius in pm 1st Ionization Energy (kJ/mol) decreasing ↓ 2 Lithium (Li) 3 2,1 152 526 3 Sodium (Na) 11 2,8,1 186 504 4 Potassium (K) 19 2,8,8,1 231 425 5 Rubidium (Rb) 37 2,8,18,8,1 244 410 6 Cesium (Cs) 55 2,8,18,18,8,1 262 380 7 Francium (Fr) 87 2,8,18,32,18,8,1 — 370

From the above table, we can see that the 1st I.E. of the group 1 elements are low and gradually decreasing with the increasing atomic radius.

## Reason for decreasing 1st Ionization energy Of group I with radius

As the last electron of the group 1 elements is loosely attached to the nucleus, it’s easy to remove the electron from the atom. That is why the I.E. of the group 1 elements are low. Besides when the atomic radius is increasing with the period, the distance between the last electron which is loosely attached to the nucleus and nucleus increases. So the nucleus is not heavily able to attract the last electron As a result, the ionization becomes easier and it needs less energy.

## Low 1st Ionization Energy Group I Reason

The reason for the low 1st I.E. is hidden in the electronic configuration of the elements.

 Period Name (Symbol of elements) Atomic Number (Z) Simple Electronic Configuration of element First Ionization Energy in kJ/mol 2 Lithium (Li) 3 2,1 526 3 Sodium (Na) 11 2,8,1 504 4 Potassium (K) 19 2,8,8,1 425 5 Rubidium (Rb) 37 2,8,18,8,1 410 6 Cesium (Cs) 55 2,8,18,18,8,1 380 7 Francium (Fr) 87 2,8,18,32,18,8,1 370

From the above table, we can see that the last shell of the atom has only one electron. The single electron is in the last shell and very weakly attached as well as it is long away from the nucleus. So the ionization process needs low energy

## 2nd Ionization energy Of group I

 Period Name (Symbol) Atomic Number of elements (Z) Simple Electronic Configuration Atomic Radius (pm) Second Ionization Energy (kJ/mol) decreasing ↓ 2 Lithium (Li) 3 2,1 152 7296 3 Sodium (Na) 11 2,8,1 186 4563 4 Potassium (K) 19 2,8,8,1 231 3069 5 Rubidium (Rb) 37 2,8,18,8,1 244 2650 6 Cesium (Cs) 55 2,8,18,18,8,1 262 2420 7 Francium (Fr) 87 2,8,18,32,18,8,1 2170

From the above table, we can learn that the 2nd I.E. also decreasing with the increasing atomic radius. But the energy needed is for second ionization is very large.

## Reason for the elevated energy of the 2nd Ionization

Like low 1st ionization, the reason for high 2nd I.E. also depends on the electron configuration of the elements. As we know that the 2nd I.E. relates to the quantity of energy needed to remove an electron from the + 1 charged species. Therefore, the group 1 element received + 1 charge after the first ionization.

 Name of elements electronic configuration After 1st Ionization Electronic configuration of ions Lithium (Li) [He],1 Li+ [He] Sodium (Na) [Ne],1 Na+ [Ne] Potassium(K) [Ar],1 K+ [Ar] Rubidium(Rb) [Kr],1 Rb+ [Kr] Cesium (Cs) [Xe],1 Cs+ [Xe] Francium (Fr) [Rn],1 Fr+ [Rn]

From above we can understand that after the 1st ionization the group 1 elements lose one electron. As a result, the electronic configuration of the group 1 elements having a +1 charge is the same as the inert gas. That means the electronic configuration is very stable and they do not want to loose another electron. As a result, the 2 ionization process needs much more energy than the 1st ionization process.

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Click on the reaction to learn to balance redox reaction easily by ion-electron method

KMnO4 + H2O2 + H2SO4 = O2 + MnSO4 + K2SO4 + H2O

FeSO4 + KMnO4 + H2SO4 = Fe2(SO4)3 + MnSO4 + H2O + K2SO4

NaCl + KMnO4 + H2SO4 = Cl2 + MnSO4 + Na2SO4 + K2SO4 + H2O

K2Cr2O7 + FeSO4 + H2SO4 = Cr2(SO4)3 + Fe2(SO4)3 + K2SO4 + H2O

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