The atomic numbers of vanadium (V), chromium (Cr), manganese (Mn) and iron (Fe) are, respectively 23

Q: Why is the electron removed from 4s before 3d when 4s is filled first?

Although the 4s orbital is filled before the 3d orbital (following the Aufbau principle), once electrons occupy the 3d orbitals, they shield the 4s electrons from the nucleus. This pushes the 4s orbital to a higher energy level, making its electrons the outermost and easiest to remove.

Q: Why does Chromium have an anomalous configuration of $3d^5 4s^1$?

An exactly half-filled d-subshell ($3d^5$) provides extra stability due to symmetrical electron distribution and maximum exchange energy. Chromium shifts one electron from 4s to 3d to achieve this highly stable state.

Q: Wouldn't Manganese have a higher ionization enthalpy because it has more protons?

While nuclear charge (number of protons) generally increases ionization enthalpy across a period, the exceptional stability of a half-filled configuration outweighs the effect of a single extra proton. $Cr^+$ is $3d^5$, while $Mn^+$ is $3d^5 4s^1$. Removing the 4s electron from $Mn^+$ is much easier than breaking the $3d^5$ core of $Cr^+$.

Q: What is exchange energy?

Exchange energy is the energy released when electrons with the same spin in degenerate orbitals (like the five 3d orbitals) exchange their positions. More possible exchanges mean more energy released and greater stability. A $3d^5$ configuration has the maximum possible exchanges.

Q: How do I quickly solve such questions in the exam?

Always write the configuration of the specific ion from which the electron is being removed. If you spot a $p^3$, $p^6$, $d^5$, or $d^{10}$ configuration, that species will almost certainly have an exceptionally high ionization enthalpy for the next electron removal.

Question

The atomic numbers of vanadium (V), chromium (Cr), manganese (Mn) and iron (Fe) are, respectively 23, 24, 25 and 26. Which one of these may be expected to have the highest second ionisation enthalpy?

Options:

(A) V

(B) Cr

(C) Mn

(D) Fe

Sigma Percentile · Accepted Answer

### The Concept of Second Ionization Enthalpy

When we talk about ionization enthalpy, we are discussing the energy required to rip an electron away from the attractive grip of an atom's nucleus. But here, the question specifically asks for the **second ionization enthalpy** ($IE_2$).

This is a crucial distinction! We are not starting with neutral atoms. The second ionization enthalpy is the energy required to remove an electron from an isolated, gaseous **unipositive ion** ($M^+$).

Mathematically, the process looks like this:

$$M^+_{(g)} \xrightarrow{IE_2} M^{2+}_{(g)} + e^-$$

To figure out which element requires the most energy for this process, we must first determine the electronic configurations of their $+1$ ions.

### Writing the Electronic Configurations

Let's start by writing the ground state electronic configurations for the neutral atoms of the given transition metals. Remember, the $4s$ orbital fills before the $3d$ orbital, but there are exceptions driven by stability.

*   **Vanadium (V, Z=23):** $[Ar] 3d^3 4s^2$
*   **Chromium (Cr, Z=24):** $[Ar] 3d^5 4s^1$ (Notice the anomalous configuration! It shifts an electron from $4s$ to $3d$ to achieve a half-filled state.)
*   **Manganese (Mn, Z=25):** $[Ar] 3d^5 4s^2$
*   **Iron (Fe, Z=26):** $[Ar] 3d^6 4s^2$

Now, to find the configurations of the $M^+$ ions, we must remove one electron. A common pitfall is removing the electron from the $3d$ subshell. **Always remove electrons from the outermost shell first**, which is the $4s$ shell (highest principal quantum number $n=4$).

Let's look at the resulting unipositive ions:

*   **$V^+$:** $[Ar] 3d^3 4s^1$
*   **$Cr^+$:** $[Ar] 3d^5 4s^0$
*   **$Mn^+$:** $[Ar] 3d^5 4s^1$
*   **$Fe^+$:** $[Ar] 3d^6 4s^1$

### The Magic of Half-Filled Stability

Take a close look at the configuration of the $Cr^+$ ion. By losing its single $4s$ electron, it is left with a pure $[Ar] 3d^5$ configuration.

In the realm of quantum mechanics, an exactly half-filled subshell (like $p^3$, $d^5$, or $f^7$) is a state of profound stability. Why? Because of two main reasons:

1.  **Symmetrical Electron Distribution:** The five electrons are evenly distributed across the five degenerate $d$-orbitals, minimizing electron-electron repulsion.
2.  **Maximum Exchange Energy:** Electrons with parallel spins in degenerate orbitals can exchange positions. Each exchange releases energy, stabilizing the system. A $d^5$ configuration allows for the maximum possible number of exchanges (10 exchanges), leading to immense stabilization.

### The Final Verdict

Because the $Cr^+$ ion is sitting in this incredibly stable, low-energy $3d^5$ state, it is highly reluctant to lose another electron.

Disrupting this perfect symmetry to form $Cr^{2+}$ ($3d^4$) requires overcoming a massive energy barrier. Therefore, the energy required to pull that second electron away—the second ionization enthalpy—will be exceptionally high.

Thus, among the given options, **Chromium (Cr)** has the highest second ionization enthalpy.

Animated Solution for Chemistry - Periodicity in Properties: The atomic numbers of vanadium (V), chromium (Cr), manganese (Mn) and iron (Fe) are, respectively 23, 24, 25 and 26. Which one of these may be expected to have the highest second ionisation enthalpy?

Select Answer:

Visualized Solution

$2^{nd} Ionisation Enthalpy$

$Neutral Atom Configurations$

$Unipositive Ion (M^{+}) Configurations$

$Stability of 3 d^{5}$

$Conclusion$

$The Way Forward$

The Sigma Insight: Periodic Table and Periodic Properties

The Concept of Second Ionization Enthalpy

Writing the Electronic Configurations

The Magic of Half-Filled Stability

The Final Verdict

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