Question

Explain why I.P values of post lanthanides $(Ag,Au)$ have higher values down the group?

Answer 1

Hint: As we know that ionisation potential or ionisation enthalpy is the minimum amount of energy which should be supplied to remove an electron from an isolated, gaseous atom to form a gaseous cation but it is affected by shielding effect and effective nuclear charge of electrons.

Complete answer
As we know that down the group, ionisation energy or ionisation potential tends to decrease with increase in the distance between the outermost valence electron and the nucleus which is commonly called the shielding effect, as well as because of the effective nuclear charge which is basically the charge experience by the outermost valence shell electron.

But this is not the case with post lanthanides like gold and silver. The ionisation energy of gold is higher than that of the silver and the reason is Lanthanoid contraction. As we know the electronic configuration of silver is $[Kr]4{d^{10}}5{s^1}$ and the electronic configuration of gold is $[Xe]4{f^{14}}5{d^{10}}6{s^1}$.

When we move down the group, the electrons are filled one by one in $4f$ orbital of gold due to which the protons are generated in the nucleus which attracts the electrons in $4f$ orbital. But due to the poor shielding effect of electrons in $4f$ orbital, effective nuclear charge is increased which attracts the electron in $6s$ orbital, resulting in smaller atomic radius and consequently the $6s$ electron will be held more tightly by the nucleus.

Hence, the ionisation energy or ionisation potential of gold is higher than that of silver down the group.

Note:So, when we move from $3d$ to $4d$ transition states the atomic radius and ionic radius increases due to increase in the number of shells but when we move from $4d$ to $5d$ transition states the atomic radius and ionic radius become equal because of Lanthanoid contraction.