The transition element that has the lowest enthalpy of atomisation is:

We begin by recalling what “enthalpy of atomisation” means. The enthalpy of atomisation, usually written as $$\Delta H_{\text{atomisation}}$$, is the amount of energy required to break all the bonds in one mole of a substance to obtain separated gaseous atoms. In metals, this energy is directly linked to the strength of metallic bonding. Stronger metallic bonding means more energy is needed, giving a higher $$\Delta H_{\text{atomisation}}$$, while weaker metallic bonding means a lower value.

For transition elements, the strength of metallic bonding depends mainly on the number of unpaired $$d$$-electrons that can participate in the formation of metallic (delocalised) bonds. More unpaired electrons produce stronger metallic bonding because each unpaired electron can overlap with those on neighbouring atoms, increasing the cohesive force in the metallic lattice. Conversely, if an element has all its $$d$$ orbitals completely filled, there are no or very few unpaired electrons available for additional bonding, so the metallic bonding becomes weaker.

Now, let us examine each option by writing its outer electronic configuration:

Option A: V (Vanadium), atomic number $$23$$, has $$[Ar]\,3d^3\,4s^2$$. So, number of unpaired $$d$$-electrons = $$3$$.

Option B: Fe (Iron), atomic number $$26$$, has $$[Ar]\,3d^6\,4s^2$$. Writing the $$3d$$ subshell explicitly, we obtain $$3d^{5\uparrow}3d^{1\downarrow}$$, giving $$4$$ unpaired $$d$$-electrons.

Option C: Zn (Zinc), atomic number $$30$$, has $$[Ar]\,3d^{10}\,4s^2$$. All ten $$d$$-electrons are paired, so number of unpaired $$d$$-electrons = $$0$$.

Option D: Cu (Copper), atomic number $$29$$, has $$[Ar]\,3d^{10}\,4s^1$$. Here, the $$3d$$ subshell is completely filled and paired, leaving only one unpaired $$4s$$-electron to contribute to metallic bonding.

Among these, zinc (Zn) has a completely filled $$3d^{10}$$ configuration with no unpaired $$d$$-electrons at all. Because unpaired $$d$$-electrons are essentially absent, the metallic bonding in Zn is substantially weaker than in V, Fe, or even Cu (which still has one unpaired $$4s$$-electron). Therefore, the energy required to break the lattice into individual atoms—its enthalpy of atomisation—will be the least for Zn.

Hence, the correct answer is Option C.