Read the following passage and answer the next five questions based on it. (37-41)
Transition Series Elements:

Sc Ti V Cr MnFeCoNiCuZn

YZrNbMoTcRuRhPdAgCd
La Hf Ta WReOsIrPtAuHg

In any transition series, as we move from left to right the d-orbitals are progressively filled and their properties vary accordingly.
f-block Elements:

Ce Pr Nd PmSmEuGdTbDyHoErTmYbLu
Th Pa UNpPuAmCmBkCfEsFmMdNoLr

The above are the two series of f-block elements in which the chemical properties won’t change much. The 5f-series elements are radioactive in nature and mostly are artificially synthesized in laboratories and thus much is not known about their chemical properties.

Identify the statement:

The correct option is: Second ionization enthalpy of Ag is greater than second ionization enthalpy of Pd.

Let us analyze the electronic configurations of Palladium (Pd) and Silver (Ag) to understand why the second ionization enthalpy of Ag is greater than that of Pd.

Palladium (Pd) is in the second transition series (4d series) with atomic number 46. Its ground-state electronic configuration is:
[Kr] 4d¹⁰ 5s⁰

Silver (Ag) is adjacent to Palladium with atomic number 47. Its ground-state electronic configuration is:
[Kr] 4d¹⁰ 5s¹

To find the second ionization enthalpy, we first look at the first ionization step, which removes one electron to form the mono-positive cations:
For Pd: Pd → Pd⁺ + e^-
The configuration of Pd⁺ becomes: [Kr] 4d⁹
For Ag: Ag → Ag⁺ + e^-
The configuration of Ag⁺ becomes: [Kr] 4d¹⁰

Now, we consider the second ionization step, which involves removing a second electron from these cations:
For Pd⁺: Pd⁺ ([Kr] 4d⁹) → Pd²⁺ ([Kr] 4d⁸) + e^-
For Ag⁺: Ag⁺ ([Kr] 4d¹⁰) → Ag²⁺ ([Kr] 4d⁹) + e^-

The Ag⁺ ion has a completely filled d-subshell (4d¹⁰). A completely filled d-orbital configuration is exceptionally stable due to high exchange energy and symmetrical distribution of electron density. Removing an electron from this highly stable 4d¹⁰ configuration requires a very large amount of energy.
In contrast, the Pd⁺ ion has a 4d⁹ configuration, which is not completely filled, and removing an electron from it is relatively easier compared to removing one from the stable 4d¹⁰ shell of Ag⁺.

Therefore, the second ionization enthalpy of Ag is significantly greater than the second ionization enthalpy of Pd.