Chemical Equations

To solve the problem, we need to determine which element in the 3d series has the lowest enthalpy of atomisation and explain why.

1. Understand Enthalpy of Atomisation:
Enthalpy of atomisation is the energy required to convert one mole of a solid metal into gaseous atoms. It depends on the strength of metallic bonds, which is influenced by the number of unpaired electrons available for delocalisation in the metallic lattice.

2. Identify the 3d Series Elements:
The 3d series includes Sc, Ti, V, Cr, Mn, Fe, Co, Ni, Cu, and Zn. We need to compare their electronic configurations and bonding strengths.

3. Analyze Electronic Configurations:
- Sc: , 1 unpaired d-electron.
- Ti: , 2 unpaired.
- V: , 3 unpaired.
- Cr: , 6 unpaired (maximum).
- Mn: , 5 unpaired.
- Fe: , 4 unpaired.
- Co: , 3 unpaired.
- Ni: , 2 unpaired.
- Cu: , 1 unpaired.
- Zn: , 0 unpaired (fully filled).

4. Relate to Metallic Bonding:
Stronger metallic bonding (more unpaired electrons for delocalisation) leads to higher enthalpy of atomisation. Zn has a fully filled configuration, with no unpaired d-electrons for delocalisation, resulting in weaker metallic bonding compared to other 3d elements.

5. Compare Enthalpy Values:
Typical values confirm Zn has the lowest enthalpy of atomisation (~130 kJ/mol), while elements like Cr (~397 kJ/mol) and V (~515 kJ/mol) are much higher due to more unpaired electrons.

Final Answer:
Zinc (Zn) has the lowest enthalpy of atomisation in the 3d series because its fully filled configuration results in no unpaired d-electrons for delocalisation, leading to weaker metallic bonding.