Animated Solution for Chemistry - d and f-Block Elements: The correct order of the first ionisation enthalpies is

First Ionization Enthalpy

When we talk about the first ionization enthalpy ($I{E}_1$), we are essentially measuring how tightly an atom holds onto its outermost electron. Imagine an isolated atom in its gaseous state; the ionization enthalpy is the exact amount of energy you need to supply to overcome the nucleus's attractive pull and rip that most loosely bound electron completely away.

For the elements in the $3d$ transition series—like Titanium (Ti), Manganese (Mn), Nickel (Ni), and Zinc (Zn)—the outermost electrons reside in the $4s$ orbital. Even though the $3d$ subshell is filled after the $4s$ subshell according to the Aufbau principle, the $4s$ electrons are physically further from the nucleus. Therefore, when it's time to remove an electron, the $4s$ electrons are always the first to go.

To determine the order of ionization enthalpies, we must look at what happens as we move from left to right across the periodic table. As the atomic number ($Z$) increases, we are adding more protons to the nucleus, which increases the positive charge pulling on the electrons.

Simultaneously, we are adding more electrons. However, in the $3d$ series, these new electrons are entering the inner $3d$ subshell rather than the outermost $4s$ shell. This is where the shielding effect comes into play. Inner electrons act like a shield, blocking some of the nuclear pull from reaching the outer electrons.

But here is the catch: $d$-orbitals have a highly diffused, spread-out shape. Because of this geometry, $3d$ electrons are notoriously poor at shielding the outer $4s$ electrons from the growing nuclear charge. As a result, the net positive charge felt by the $4s$ electrons—known as the effective nuclear charge ($Z^{\ast }$)—steadily increases as we move from Titanium to Zinc.

Let's write down the raw setup for our specific elements:

Ti ($Z=22$): $[Ar]3d^{2}4s^2$ Mn ($Z=25$): $[Ar]3d^{5}4s^2$ Ni ($Z=28$): $[Ar]3d^{8}4s^2$ Zn ($Z=30$): $[Ar]3d^{10}4s^2$

Because $Z^{\ast }$ increases continuously from left to right ($Ti<Mn<Ni<Zn$), the nucleus grips the $4s^2$ electrons tighter and tighter.

A tighter grip means you have to work harder (supply more energy) to remove that first electron. Therefore, the first ionization enthalpy is directly proportional to the effective nuclear charge ($I{E}_1\propto Z^{\ast }$).

Following this logic, the energy required to remove an electron increases in the exact order of their atomic numbers: $Ti<Mn<Ni<Zn$. Zinc, sitting at the end of the series, not only has the highest effective nuclear charge among these four but also boasts a highly stable, fully filled $3d^{10}4s^2$ configuration, making its ionization enthalpy exceptionally high.