The transition element having least enthalpy of atomisation is
(a) Zn
(b) V
(c) Fe
(d) Cu

Q: Why does Copper have 1 unpaired electron if its 3d subshell is full?

Copper has an anomalous electronic configuration of $[\text{Ar}] 3d^{10} 4s^1$ to achieve the extra stability of a fully filled d-subshell. The single electron in the 4s orbital is unpaired.

Q: Why is Zinc's enthalpy of atomisation so low?

Zinc has a completely filled $3d^{10} 4s^2$ configuration. With zero unpaired electrons, it forms very weak metallic bonds, making it easy to atomise.

Q: Are Zn, Cd, and Hg considered transition metals?

Strictly speaking, no. According to IUPAC, a transition metal must have a partially filled d-subshell in its ground state or in any of its common oxidation states. Zn, Cd, and Hg have fully filled d-subshells ($d^{10}$) in both, so they are often excluded.

Q: Does atomic size affect the enthalpy of atomisation?

Yes, larger atomic size generally leads to weaker metallic bonding because the valence electrons are further from the nucleus. However, the number of unpaired electrons is the dominant factor in the d-block.

Q: Why does Iron have a high enthalpy of atomisation?

Iron ($[\text{Ar}] 3d^6 4s^2$) has 4 unpaired electrons in its 3d subshell. These unpaired electrons strongly participate in metallic bonding, resulting in a very strong crystal lattice and high enthalpy of atomisation.

Question

The transition element having least enthalpy of atomisation is

Options:

(A) Zn

(B) V

(C) Fe

(D) Cu

Sigma Percentile · Accepted Answer

### Analyzing the Setup
The problem asks us to identify the transition element with the least enthalpy of atomisation among Zinc (Zn), Vanadium (V), Iron (Fe), and Copper (Cu).

To tackle this, we first need to understand what **enthalpy of atomisation** ($\Delta H^\circ_{	ext{atomisation}}$) represents. Imagine a solid block of metal. The enthalpy of atomisation is the amount of energy required to completely break the metallic crystal lattice and separate it into individual, isolated gaseous atoms.

Naturally, the stronger the atoms are bound together in the solid state, the more energy it will take to pull them apart. Therefore, the enthalpy of atomisation is a direct measure of the **strength of the metallic bonding**.

### The Master Equation
What determines the strength of a metallic bond in transition metals? It all comes down to the electrons.

In transition metals, the strength of metallic bonding is directly proportional to the **number of unpaired electrons** ($n$) present in their outermost $ns$ and $(n-1)d$ orbitals.

$$ \Delta H^\circ_{	ext{atomisation}} \propto 	ext{Strength of Metallic Bonding} \propto 	ext{Number of Unpaired Electrons } (n) $$

More unpaired electrons mean that more electrons can be delocalized to form the "sea of electrons" that holds the positively charged metal kernels together. This leads to stronger metallic bonds.

### Final Calculation
Let's write down the electronic configurations for the given elements to find their number of unpaired electrons:

1. **Vanadium (V, $Z=23$)**: 
   Configuration: $[	ext{Ar}] 3d^3 4s^2$
   Unpaired electrons ($n$) = 3

2. **Iron (Fe, $Z=26$)**: 
   Configuration: $[	ext{Ar}] 3d^6 4s^2$
   Unpaired electrons ($n$) = 4

3. **Copper (Cu, $Z=29$)**: 
   Configuration: $[	ext{Ar}] 3d^{10} 4s^1$
   Unpaired electrons ($n$) = 1 (in the $4s$ orbital)

4. **Zinc (Zn, $Z=30$)**: 
   Configuration: $[	ext{Ar}] 3d^{10} 4s^2$
   Unpaired electrons ($n$) = 0

Zinc has a completely filled $3d$ subshell and a completely filled $4s$ subshell. Because it has **zero unpaired electrons**, its atoms cannot participate effectively in metallic bonding. The forces holding Zinc atoms together in the solid state are relatively weak.

Consequently, it requires the least amount of energy to break the crystal lattice of Zinc, giving it the **least enthalpy of atomisation** among the given options.

### The Way Forward
This fundamental lack of unpaired electrons is also the reason why Zinc, along with Cadmium (Cd) and Mercury (Hg), have unusually low melting and boiling points compared to other transition metals. In fact, because their $d$-orbitals are fully filled both in their ground state and in their common oxidation states, they are often not considered true transition elements!

Animated Solution for Chemistry - d and f-Block Elements: The transition element having least enthalpy of atomisation is

Select Answer:

Visualized Solution

$Enthalpy of Atomisation$

$Role of Unpaired Electrons$

$Electronic Configurations$

$Conclusion for Zinc$

$The Way Forward$

The Sigma Insight: d-block Elements

Analyzing the Setup

The Master Equation

Final Calculation

The Way Forward This fundamental lack of unpaired electrons is also the reason why Zinc, along with Cadmium (Cd) and Mercury (Hg), have unusually low melting and boiling points compared to other transition metals

Similar Questions

The correct order of the first ionisation enthalpies is

The element that usually does not show variable oxidation states is

The set that contains atomic numbers of only transition elements, is

Four successive members of the first row transition elements listed below with atomic numbers. Which one of them is expected to have the highest $E_{M^{3 +} / M^{2 +}}^{\circ}$ value?

What is the correct order of the following elements with respect to their density ?

In context with the transition elements, which of the following statements is incorrect?

Which one of the following metals forms interstitial hydride easily ?

The correct order of $E_{M^{2 +} / M}^{\circ}$ values with negative sign for the four successive elements Cr, Mn, Fe and Co is

The pair that has similar atomic radii is

Number of electrons that vanadium ( $Z = 23$ ) has in p-orbitals is equal to ……… .