Q.Why first ionisation enthalpy of Cr is lower than that of Zn?

The key idea is that transition elements are defined by having partially filled $d$-orbitals in their ground state or common oxidation states. Chromium ($\text{Cr}$, atomic number 24) and zinc ($\text{Zn}$, atomic number 30) are both $d$-block elements, but their electronic configurations explain the difference in ionisation enthalpy.

Reasoning:

1. Electronic configurations:

   $\text{Cr}$ has $[Ar]\,3d^5\,4s^1$ — a half-filled $3d$ subshell (extra stability).

   $\text{Zn}$ has $[Ar]\,3d^{10}\,4s^2$ — a completely filled $3d$ subshell (very stable).

2. Removing the first electron:

   For $\text{Cr}$, the $4s$ electron is relatively easy to remove because the half-filled $d^5$ configuration is already stable. …

The Concept: Stability of Electronic Configurations

First ionisation enthalpy is the energy required to remove the most loosely bound electron from a gaseous atom. The trend across a period generally increases from left to right, but there are exceptions due to special stability associated with half-filled and fully filled subshells.

Chromium (Cr) and Zinc (Zn) are both transition metals in the 3d series. Their atomic numbers are 24 and 30 respectively. The key to understanding their ionisation enthalpies lies in their ground state electronic configurations.

Cr: $1s^2 2s^2 2p^6 3s^2 3p^6 3d^5 4s^1$

Zn: $1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^2$

Notice that Cr has a half-filled $3d^5$ subshell, while Zn has a fully filled $3d^{10}$ subshell. Both are considered stable configurations, but their stability differs in degree and in how they respond to electron removal.

Step-by-Step Reasoning

1. Write the electronic configurations of Cr and Zn atoms.

   Cr (Z=24): $[Ar] 3d^5 4s^1$

   Zn (Z=30): $[Ar] 3d^{10} 4s^2$

   The 4s orbital is higher in energy than 3d for these atoms, so the outermost electron is in the 4s orbital for both.

2. Consider what happens when one electron is removed from each atom.

   For Cr: Removing the 4s electron gives $Cr^+$ with configuration $[Ar] 3d^5$.

   For Zn: Removing one 4s electron gives $Zn^+$ with configuration $[Ar] 3d^{10} 4s^1$.

3. Analyse the stability of the resulting ions.

   $Cr^+$ has a half-filled $3d^5$ subshell — this is a highly stable configuration due to exchange energy and symmetrical distribution of electrons. The atom Cr already has a half-filled $3d^5$ subshell, but it also has one electron in the 4s orbital. Removing that 4s electron leaves behind a perfectly half-filled d-subshell with no s-electrons, which is even more stable than the original configuration.

   $Zn^+$ has a fully filled $3d^{10}$ subshell plus one electron in the 4s orbital. The fully filled $3d^{10}$ is extremely stable, but the $4s^1$ electron is relatively loosely held. However, removing that electron disrupts the symmetry of the full d-subshell? No — the d-subshell remains $3d^{10}$ in $Zn^+$, so that part is still stable. The issue is that the 4s electron in Zn is held more tightly because the effective nuclear charge is higher (Zn has 30 protons vs Cr's 24), and the $4s^2$ pair in Zn has some extra stability from being a filled s-subshell.

4. Compare the energy required.

   The first ionisation enthalpy of Cr is 652 kJ/mol, while that of Zn is 906 kJ/mol. The difference is about 254 kJ/mol.

   Why is Zn's so much higher? Two reasons:

   • Higher nuclear charge: Zn has 30 protons pulling on the 4s electrons, while Cr has only 24. The effective nuclear charge experienced by the 4s electron in Zn is significantly greater. …

Method: Electronic Configuration & Stability Analysis

This method uses electronic configuration and the concept of half-filled / fully-filled orbital stability to explain periodic trends in transition elements.

Steps

1. Write the ground state electronic configurations

   • Cr (Z = 24): $[Ar] 3d^5 4s^1$ (Exception: half-filled $3d$ subshell is extra stable)
   • Zn (Z = 30): $[Ar] 3d^{10} 4s^2$ (Fully-filled $3d$ subshell is very stable)

2. Identify which electron is removed during first ionisation

   • For both Cr and Zn, the first electron removed is from the 4s orbital (not 3d), because 4s is at higher energy than 3d in the neutral atom.

3. Compare the stability of the resulting configuration

   • After losing one electron:
     • Cr⁺ becomes $[Ar] 3d^5$ → half-filled 3d (extra stable)
     • Zn⁺ becomes $[Ar] 3d^{10} 4s^1$ → loses fully-filled 3d stability (less stable)

4. Apply the stability principle

   • Cr loses an electron easily because the resulting Cr⁺ has a stable half-filled 3d⁵ configuration. …

Here are the common mistakes students make when comparing the first ionisation enthalpy of Cr and Zn, along with the correct reasoning and how to avoid each error.

Mistake 1: Assuming IE always increases across a period

The error:

Students see that Cr (atomic number 24) comes before Zn (atomic number 30) in the 3d series and assume that IE must increase from left to right. They then conclude that Cr should have a higher IE than Zn — which is wrong.

Why it’s wrong:

IE does not increase monotonically across a transition series. There are exceptions due to electronic configurations. Cr has a lower IE than Zn because of stability differences.

How to avoid:

• Memorise the exceptions in the 3d series: Cr and Cu have unusually low IE values.
• Always check the electronic configuration before comparing IE.

Mistake 2: Forgetting the electronic configurations

The error:

Students compare IE without writing the actual configurations of Cr and Zn.

Correct configurations:

• Cr (Z=24): $[Ar]\, 3d^5 4s^1$
• Zn (Z=30): $[Ar]\, 3d^{10} 4s^2$

Why this matters:

• Cr has a half-filled d-subshell ($3d^5$), which is extra stable. Removing one electron from the $4s^1$ orbital is relatively easy because it leaves behind the stable $3d^5$ configuration.
• Zn has a fully filled d-subshell ($3d^{10}$) and a filled $4s^2$. Removing an electron from the $4s^2$ requires more energy because it disrupts a stable, filled subshell.

How to avoid:

• Always write the ground state electronic configuration before answering any IE comparison question.
• Remember: half-filled and fully filled subshells confer extra stability.

Mistake 3: Confusing first IE with second or third IE

The error:

Some students think that because Zn has a fully filled d-subshell, its first IE should be lower (since it’s “stable” and doesn’t want to lose an electron). This is a misunderstanding of stability.

The correction:

• Stability means the atom resists change. A stable configuration (like $3d^{10}4s^2$) makes it harder to remove the first electron → higher IE.
• Cr’s $3d^5 4s^1$ is stable but losing the $4s^1$ electron actually increases stability (to $3d^5$), so the first IE is lower.

How to avoid:

• Think: “Stable configuration → high IE (hard to remove).”
• Think: “Removing an electron to achieve a stable configuration → low IE (easy to remove).”

Mistake 4: Ignoring the shielding effect and nuclear charge

The error:

Students only focus on configuration and forget that Zn has a higher nuclear charge ($Z=30$) than Cr ($Z=24$). They then wonder why Zn’s IE is higher.

The correct reasoning: …