Ionization Enthalpies of d-Block
Decode the energy required to strip electrons from transition metals. Master the thermodynamic stability traps, the Pt vs Ni comparison, and the massive 5d series anomaly.
Module Focus: The Energy of Oxidation
Ionization Enthalpy ($IE$) is the energy required to remove an electron from an isolated gaseous atom. In the d-block, the $ns$ electrons are lost first, followed by the $(n-1)d$ electrons. The trends here are complex because the removal of an electron alters the relative energies of the $4s$ and $3d$ orbitals. Furthermore, the sum of successive ionization enthalpies directly dictates the thermodynamic stability of various oxidation states.
1. Trend Along a Period (3d Series)
As we move from left to right across the 3d series, the nuclear charge increases. This generally increases the Ionization Enthalpy, but the increase is not smooth.
The first ionization enthalpy ($IE_1$) generally increases from Sc to Zn. This is due to the increasing effective nuclear charge ($Z_{eff}$) holding the $4s$ electrons more tightly.
However, the increase is much smaller than in s- and p-blocks because the added d-electrons provide significant shielding.
Elements with stable configurations resist losing electrons.
- Zn ($3d^{10} 4s^2$): Has a very high $IE_1$ because the $4s$ subshell is fully filled.
- Mn ($3d^5 4s^2$): Has a relatively high $IE_3$ because removing the third electron means breaking the highly stable half-filled $3d^5$ core.
2. Trend Down a Group: The 5d Anomaly
Normally, ionization enthalpy decreases down a group as the atomic size increases. However, the d-block presents a massive exception that is a favorite for NEET examiners.
The ionization enthalpies of the elements in the 5d series (third transition series) are significantly HIGHER than those of the 3d and 4d series elements in the same group.
Before the 5d orbitals are filled, the 4f orbitals are filled. The highly diffused 4f electrons offer extremely poor shielding. The massive increase in nuclear charge (+14 protons) pulls the outer electrons very tightly inward. Because they are held so tightly by the nucleus, it requires much more energy to remove them.
General Trend: $IE_1 \text{ of 5d} > \text{3d} > \text{4d}$
3. Thermodynamic Stability of Oxidation States
To determine which oxidation state is more stable in standard conditions, we must look at the sum of the ionization enthalpies required to reach that state. A lower sum indicates a more thermodynamically stable oxidation state.
This is a classic NCERT comparative example testing the sum of ionization energies.
Sum of first two ionization enthalpies ($IE_1 + IE_2$):
Ni requires less energy.
Therefore, $Ni^{2+}$ is more stable than $Pt^{2+}$.
Sum of first four ionization enthalpies ($\sum IE_{1 \to 4}$):
Pt requires less energy.
Therefore, $Pt^{4+}$ is more stable than $Ni^{4+}$.
This thermodynamic reality explains why $K_2[PtCl_6]$ (where Pt is +4) is a highly stable, well-known compound, while the corresponding nickel compound does not exist.
NEET Grand Test: Ionization Enthalpies
15 High-Yield Questions testing the 5d anomaly, sum of IE calculations, and stability configurations.
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