The third ionisation enthalpy is minimum for :
(a) Mn
(b) Ni
(c) Co
(d) Fe

Q: Why do we consider the $+2$ oxidation state to find the third ionisation enthalpy?

The third ionisation enthalpy is specifically defined as the energy required to remove the third electron from an atom. By the time you are removing the third electron, the atom has already lost two electrons, meaning it is in the $+2$ oxidation state ($M^{2+}$).

Q: Why is the $3d^5$ configuration considered extra stable?

A $3d^5$ configuration is exactly half-filled, meaning each of the five $d$-orbitals contains exactly one electron. This symmetrical distribution minimizes electron-electron repulsion and maximizes exchange energy, leading to exceptional thermodynamic stability.

Q: Doesn't ionisation enthalpy always increase as we move from left to right across a period?

Generally, yes. As the atomic number increases across a period, the effective nuclear charge increases, holding electrons more tightly. However, this general trend is often broken by the extra stability of half-filled ($d^5$) or fully-filled ($d^{10}$) subshells, which cause significant anomalies.

Q: Why does Nickel have the highest third ionisation enthalpy among these options?

Nickel ($Z=28$) is the furthest to the right among the given options. It has the highest effective nuclear charge and the smallest ionic radius. Since neither $Ni^{2+}$ nor $Ni^{3+}$ achieves a special half-filled stability, the general trend dominates, making it the hardest to ionise.

Q: If $Mn^{2+}$ is so stable, why does Manganese form $+7$ oxidation states?

While $Mn^{2+}$ is highly stable, Manganese can still form higher oxidation states like $+7$ (e.g., in $KMnO_4$) when it bonds with highly electronegative elements like Oxygen. The energy required to remove those electrons is compensated by the massive amount of lattice energy or hydration energy released during compound formation.

Question

The third ionisation enthalpy is minimum for :

Options:

(A) Mn

(B) Ni

(C) Co

(D) Fe

Sigma Percentile · Accepted Answer

### The Core Concept: What is Third Ionisation Enthalpy?

When we talk about the **third ionisation enthalpy** ($IE_3$), we are specifically looking at the energy required to remove the *third* electron from an atom. By the time this happens, the atom has already lost two electrons and exists as a $+2$ cation ($M^{2+}$).

The equation for this process is:
$$M^{2+} \rightarrow M^{3+} + e^-$$

The amount of energy required for this process is heavily dictated by the electronic configuration of the ions involved. If removing an electron disrupts a highly stable configuration, the energy required will be massive. Conversely, if removing an electron *creates* a highly stable configuration, the atom will readily give up that electron, resulting in a very low ionisation enthalpy.

### The Manganese Anomaly: A Fortress of Stability

Let's evaluate Manganese ($Z=25$). Its ground state electronic configuration is $[Ar] 3d^5 4s^2$. When it loses two electrons to form $Mn^{2+}$, the configuration becomes $[Ar] 3d^5$.

This $3d^5$ configuration is exactly **half-filled**. In quantum mechanics, half-filled subshells possess exceptional thermodynamic stability due to symmetrical electron distribution and maximum exchange energy. Because $Mn^{2+}$ is already sitting in a fortress of stability, trying to rip away a third electron to form $Mn^{3+}$ ($[Ar] 3d^4$) requires a tremendous amount of energy. Thus, Manganese has an anomalously high third ionisation enthalpy.

### The Iron Advantage: Eager to Lose

Now, let's look at Iron ($Z=26$). Its ground state configuration is $[Ar] 3d^6 4s^2$. When it forms the $Fe^{2+}$ ion, its configuration is $[Ar] 3d^6$.

In the $3d^6$ state, there are five unpaired electrons and exactly one paired electron. This pairing creates inter-electronic repulsion. If we supply the third ionisation energy to remove this specific paired electron, the ion transforms into $Fe^{3+}$, which has a perfectly half-filled $[Ar] 3d^5$ configuration!

Because the product ($Fe^{3+}$) is exceptionally stable, the $Fe^{2+}$ ion is highly motivated to lose that third electron. Consequently, the energy barrier to remove it is remarkably low. Therefore, Iron has the **minimum** third ionisation enthalpy among the given options.

### The Final Verdict and General Trend

What about Cobalt ($Z=27$) and Nickel ($Z=28$)? As a general rule in the periodic table, ionisation enthalpy increases as we move from left to right across a period. This is because the effective nuclear charge ($Z_{eff}$) increases, pulling the electron cloud tighter and making it harder to extract electrons.

Nickel has the highest effective nuclear charge and the smallest ionic radius among these four elements. Since neither its $+2$ nor $+3$ state involves a special half-filled stability, the general trend dominates, giving Nickel the highest third ionisation enthalpy.

The final order of third ionisation enthalpy is $Ni > Mn > Co > Fe$. The key takeaway is to always watch out for the hidden stability of $d^5$ and $d^{10}$ configurations when dealing with transition metals!

Animated Solution for Chemistry - Periodicity in Properties: The third ionisation enthalpy is minimum for :

Select Answer:

Visualized Solution

$I E_{3} Definition$

$Stability Principle$

$Manganese (M n)$

$Iron (F e^{2 +})$

$Iron Transition$

$Conclusion$

$General Trend$

The Sigma Insight: Periodic Table and Periodic Properties

The Core Concept

The Manganese Anomaly

The Iron Advantage

The Final Verdict and General Trend

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