Transition Metal Electron Configurations and Ionization Energies

Identification of Transition Metals with Unpaired d-Electrons

In the first transition series, transition metals fill their $(n-1)d$ subshells sequentially. To determine which transition metal possesses two unpaired electrons in its $(n-1)d$ subshell, Hund's rule of maximum multiplicity must be applied across the orbital arrangement.

According to Hund's rule, electrons occupy degenerate orbitals singly with parallel spins before pairing up. Two main ground-state electronic configurations satisfy the condition of having exactly two unpaired electrons in the $(n-1)d$ subshell:

The first candidate is Titanium ($Z = 22$). The ground-state electronic configuration of Titanium is $1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,4s^2\,3d^2$, which is abbreviated in condensed form as $[\text{Ar}]\,4s^2\,3d^2$. In this element, the two electrons in the $3d$ subshell occupy separate degenerate orbitals, resulting in exactly two unpaired electrons.

The second candidate is Nickel ($Z = 28$). The ground-state electronic configuration of Nickel is $[\text{Ar}]\,4s^2\,3d^8$. In the $3d^8$ configuration, three of the five d-orbitals contain paired electrons, while the remaining two orbitals contain single unpaired electrons, yielding a total of two unpaired electrons.

Detailed Analysis of Titanium Ionization Energies

In problem scenarios examining dramatic spikes in ionization energy, Titanium ($Z = 22$) serves as the primary model. Titanium has a total of four valence electrons distributed across its outer subshells: two electrons in the $4s$ subshell and two electrons in the $3d$ subshell.

The removal of electrons occurs in a step-by-step sequential process. The four valence electrons can be removed successively without disrupting an inner, completely filled energy level:

Removing the first electron yields $\text{Ti}^+$ with the electronic configuration $[\text{Ar}]\,4s^1\,3d^2$. This step corresponds to the first ionization energy.

Removing the second electron yields $\text{Ti}^{2+}$ with the electronic configuration $[\text{Ar}]\,3d^2$. This step corresponds to the second ionization energy.

Removing the third electron yields $\text{Ti}^{3+}$ with the electronic configuration $[\text{Ar}]\,3d^1$. This step corresponds to the third ionization energy.

Removing the fourth electron yields $\text{Ti}^{4+}$ with the electronic configuration $[\text{Ar}]$. This step corresponds to the fourth ionization energy, leaving the titanium ion with a stable noble gas core corresponding to Argon.

Ionization Energy Spikes and Noble Gas Configuration

When all valence electrons are removed from an atom, any subsequent ionization requires removing an electron from a completely filled main energy level or noble gas core. For Titanium, the removal of a fifth electron requires breaking into the filled $3p^6$ subshell of the Argon noble gas core ($[\text{Ar}]$).

Because breaking a filled main energy level requires overcoming exceptionally strong electrostatic attraction from the nucleus on tightly bound inner-shell electrons, the energy required jumps dramatically. This massive increase in required energy produces an abnormally high ionization potential spike.

Therefore, for an element with four valence electrons like Titanium ($[\text{Ar}]\,4s^2\,3d^2$), the highest jump in ionization energy occurs at the fifth ionization energy step ($\text{Ti}^{4+} \rightarrow \text{Ti}^{5+} + e^-$).

Question Solution and Step-by-Step Answer Evaluation

Consider the multiple-choice question: What is the highest/largest ionization energy step for a transition metal containing two unpaired electrons in its $(n-1)d$ subshell?

Option (A) Second Ionization Energy: Incorrect. Removing the second electron involves the $4s$ subshell and requires a normal, moderate amount of energy.

Option (B) Third Ionization Energy: Incorrect. Removing the third electron takes an electron from the $3d$ subshell, which is a standard valence electron removal.

Option (C) Fourth Ionization Energy: Incorrect. Removing the fourth electron empties the $3d$ subshell to achieve a noble gas configuration ($[\text{Ar}]$), which is favorable and achievable under standard strong ionization conditions.

Option (D) Fifth Ionization Energy: Correct. Removing the fifth electron breaks the stable $3p^6$ inner energy level of the noble gas core ($[\text{Ar}]$). This step results in a massive, sharp increase in energy, making the fifth ionization energy extraordinarily large compared to all preceding steps.