In the series Sc (Z = 21) to Zn (Z = 30), the enthalpy of atomisation of zinc is the lowest, i.e., $126\ ext{kJ mol}^{-1}$. Why?

Question

In the series Sc (Z = 21) to Zn (Z = 30), the enthalpy of atomisation of zinc is the lowest, i.e., $126\ 	ext{kJ mol}^{-1}$. Why?

Accepted Answer

> [!TLDR]
> The low enthalpy of atomisation of zinc arises because it has a completely filled 3d¹⁰4s² configuration, which means no unpaired d-electrons are available for metallic bonding. This weakens the metallic bond, making it easier to separate atoms into the gaseous state, hence the lowest value in the series.

The enthalpy of atomisation is the energy required to convert one mole of a solid metal into isolated gaseous atoms. For transition metals, this energy is largely determined by the strength of metallic bonding in the solid state. Stronger metallic bonds mean more energy is needed to break them apart.

**Why does metallic bond strength vary across the 3d series?**  
In transition metals, bonding involves both the 4s electrons and, crucially, the 3d electrons. The more unpaired d-electrons an atom has, the more it can participate in covalent-like bonding with neighbouring atoms. This is because unpaired d-electrons can overlap and form stronger bonds. Conversely, if the d-subshell is completely filled (d¹⁰), all d-electrons are paired and less available for bonding — they are more tightly held by the nucleus and do not contribute effectively to metallic bonding.

Now let’s walk through the reasoning step by step.

1. **Identify the electronic configuration of zinc.**  
   Zinc (Z = 30) has the configuration:  
   $$[	ext{Ar}]\, 3d^{10} 4s^2$$  
   The 3d subshell is completely filled. This is the only element in the series Sc to Zn with a d¹⁰ configuration.

2. **Contrast with other elements in the series.**  
   From Sc (3d¹4s²) to Ni (3d⁸4s²), there are unpaired d-electrons. For example:
   - Sc: 3d¹ (one unpaired)
   - Fe: 3d⁶ (four unpaired in the ground state)
   - Co: 3d⁷ (three unpaired)
   - Ni: 3d⁸ (two unpaired)  
   Even copper (3d¹⁰4s¹) has one unpaired 4s electron available for bonding. Only zinc has both a filled d-subshell and a filled 4s subshell, leaving no unpaired electron at all.

3. **Link electronic structure to metallic bond streng

d-block elements	Transition elements
Groups 3 to 12	Groups 3 to 11 (excluding Zn, Cd, Hg)
All have d electrons	Must have incomplete d sub-shell in atom or common cation

Q.In the series Sc (Z = 21) to Zn (Z = 30), the enthalpy of atomisation of zinc is the lowest, i.e., $126\ \text{kJ mol}^{-1}$ . Why?

Concept understanding — Transition Element Definition

Transition Elements: From Intuition to Definition

The Intuition: Why "Transition"?

The Precise Definition (IUPAC)

1. The "d" sub-shell

2. The "or" part — cations matter

The "d-block" vs "Transition Elements"

Why this formula?

Transition Element Definition: The "Why" Behind the Definition

The Core Definition

Why This Definition? The Reasoning

1. The d-orbital filling pattern

2. The "incomplete d-subshell" condition

3. Why include ions?

The "Formula" — A Decision Tree

Common Exam Exceptions …

Method: Electronic Configuration & Metallic Bonding Analysis

Steps

✗ Mistake 1: Thinking "zinc's full d-subshell is stable, so zinc metal must be strongly bonded"

✗ Mistake 2: Confusing atomisation enthalpy with ionisation enthalpy

✗ Mistake 3: Saying "zinc has low atomisation enthalpy because it has a high melting point"

✗ Mistake 4: Ignoring the trend across the series

Why This Definition Matters

Understanding Metallic Softness

Comparing the Four Metals

Q.In the series Sc (Z = 21) to Zn (Z = 30), the enthalpy of atomisation of zinc is the lowest, i.e., 126 kJ mol−1126\ \text{kJ mol}^{-1}126 kJ mol−1. Why?

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Q.In the series Sc (Z = 21) to Zn (Z = 30), the enthalpy of atomisation of zinc is the lowest, i.e., $126\ \text{kJ mol}^{-1}$ . Why?

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