Question

What is the order when Li, Na and K are placed in decreasing order of their ionisation energy?
${\text{A}}{\text{.}}$ Li, K, Na
${\text{B}}{\text{.}}$ K, Na, Li
${\text{C}}{\text{.}}$ Na, Li, K
${\text{D}}{\text{.}}$ Li, Na, K

Answer 1

Hint- Here, we will proceed by defining the term ionization energy. Then, we will mention the relation between ionization energy and atomic radius. Finally, we will see the trend Li, K and Na follows in terms of their atomic radii.

Complete answer:
Ionization energy is the amount of energy that an isolated, gaseous atom in the electronic state of the ground will consume to discharge an electron, leading to a cation.
This energy is usually expressed in kJ / mol, or the amount of energy it takes to lose one electron each of all the atoms in a mole.
In general, the further away an electron is from the nucleus, the easier it is for it to be expelled. In other words, ionization energy is a function of atomic radius; the larger the radius, the smaller the amount of energy required to remove the electron from the outermost orbital.
When considering an atom that is initially neutral, expelling the first electron needs less energy than expelling the second, the second requires less energy than the third, etc. Each successive electron requires more energy to be released. This is because the atom 's overall charge becomes positive after the first electron is lost, and the electron 's negative forces will be attracted to the positive charging of the newly formed ion. The more missing electrons, the more positive this ion becomes, the easier it is to detach the electrons from the atom.
As we move from top to bottom in the group, the ionization energy will be decreased, because the atomic radius will be increased so the removal electron will be easy.
Li, Na, K are IA group elements of the periodic table. Li will be having the smallest atomic radius and K will be having the largest atomic radius when only Li, Na, K. Therefore, the ionization energy will be decreased when moving from Li to K. This means that Li will be having the highest ionization energy and K will be having the lowest ionization energy.
Ionization Energies of Li, Na and K are 520 KJ/mol, 496 KJ/mol and 419 KJ/mol respectively.
Therefore, the correct order of ionization energy is Li>Na>K.

Hence, option D is correct.

Note- Typically, group 2 elements have ionization energy greater than group 13 elements and group 15 elements have greater ionization energy than group 16 elements. Groups 2 and 15 respectively have complete and semi-filled electronic configuration, thereby taking more energy to extract an electron from fully filled orbitals than incomplete orbitals.