Read the passage carefully and answer the following questions.

There are mainly four series of the transition metals, 3d series (Sc to Zn), 4d series (Y to Cd), 5d series (La and Hf to Hg) and 6d series which has Ac and elements from Rf to Cn. The two series of the inner transition metals; 4f (Ce to Lu) and 5f (Th to Lr) are known as lanthanoids and actinoids respectively. Their chemical properties are transitional between those of s and p-block elements. The transition metals (with the exception of Zn, Cd and Hg) are very hard and have low volatility. Their melting and boiling points are high. Transition metals have incomplete d subshell either in neutral atom or in their ions. With partly filled d orbitals these elements exhibit certain characteristic properties such as display of a variety of oxidation states, formation of coloured ions and entering into complex formation with a variety of ligands. The transition metals and their compounds are known for their catalytic activity.

In general the electronic configuration of outer orbitals of transition metals is $(n-1)d^{1-10} ns^{1-2}$ but Pd shows an exceptional electronic configuration as......

$4d^{10}5s^0$

The correct answer is Option (2) → $4d^{10}5s^0$

Palladium is one of the most famous "exceptions" in the periodic table regarding electronic configuration. While transition metals typically follow the general outer orbital formula of $(n-1)d^{1-10}ns^{1-2}$, Palladium deviates entirely from the $s$-orbital filling.

1. Expected vs. Actual Configuration

• Expected (by Aufbau Principle): Based on its atomic number ($Z=46$), you would expect $[Kr]4d^8 5s^2$.
• Actual (Observed): $[Kr]4d^{10} 5s^0$.

2. Why does this happen?

The primary reason for this exception is the extra stability associated with a completely filled $d$-subshell ($d^{10}$).

• In Palladium, the energy gap between the $4d$ and $5s$ orbitals is very small.
• The energy required to promote two electrons from the $5s$ to the $4d$ orbital is more than compensated for by the exchange energy and the symmetry gained by having a full $4d^{10}$ configuration.
• This results in the $5s$ orbital being completely empty ($5s^0$).