Question

With reference to ionization potential which one of the following sets is correct?
(a)- Li > K > B
(b)- B > Li > K
(c)- Cs > Li > K
(d)- Cs < Li < K

Answer 1

Hint: The ionization potential depends on the outer electronic configuration of the element. Size is also the factor on which the ionization potential depends. Ionization potential is inversely proportional to the size of the atom.

Complete answer:
Let us first understand the ionization enthalpy of ionization potential.
Ionization enthalpy is also known as ionization potential because it is the minimum potential difference which is used to remove the most loosely bound electron from an isolated gaseous atom to convert it into a gaseous cation.
The factors on which the ionization potential depends are nuclear charge, size, half or fully filled orbitals etc.

As the nuclear charge increases the ionization potential also increases.
As the size increases the ionization potential decreases.
In the fully filled or half orbitals the ionization potential is exceptionally high due to extra stability.
Hence, along the period the ionization potential increases and down the group the ionization potential decreases.
Let us now see the properties of the elements one by one.

Li– lithium:
It belongs to group 1and it is the first element. Its electronic configuration is\[1{{s}^{2}}2{{s}^{1}}\]. Its atomic radius is 152pm.
Its ionization potential is 520 kJ/mol.

K– potassium:
It also belongs to group 1and is the third element. Its electronic configuration is\[1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}4{{s}^{1}}\]. Its atomic radius is 227pm.
Its ionization potential is 419 kJ/mol.

Cs- cesium:
It also belongs to group 1and is the fifth element Its electronic configuration is\[1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}3{{d}^{10}}4{{s}^{2}}4{{p}^{6}}4{{d}^{10}}5{{s}^{2}}5{{p}^{6}}6{{s}^{1}}\]. Its atomic radius is 265pm.
Its ionization potential is 376 kJ/mol.

B- boron:
It belongs to group 13and it is the first element. Its electronic configuration is\[1{{s}^{2}}2{{s}^{2}}2{{p}^{1}}\]. Its atomic radius is 85pm.
Its ionization potential is 801 kJ/mol.

So, with these observations we can conclude, the ionization potential has the following order:
B > Li > K > Cs

So, the correct option is (b)- B > Li > K

Note: You may get confused that boron would have the least ionization potential because it is the element of group 13 and the rest of them belong to group 1. But while moving along the period the size of the atom decreases. Hence. Boron is the smallest atom among them.