Question

Which of the following element of first transition series has the lowest atomization enthalpy?

• A. Ti
• B. V
• C. Cr
• D. Mn

Hint:

Manganese and Zinc show a "dip" in properties like atomization enthalpy and melting point due to their stable electron configurations ($d^5$ and $d^{10}$ respectively) which inhibit strong metallic bonding.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
Atomization enthalpy is the energy required to break the metallic lattice into individual atoms. It depends on the strength of metallic bonding, which is linked to the number of unpaired d-electrons available for bonding.
Step 2: Key Formula or Approach:
Generally, atomization enthalpy increases with the number of unpaired electrons. However, Manganese (Mn) is an exception.
Step 3: Detailed Explanation:
Manganese has a \( 3d^5 4s^2 \) configuration. Although it has 5 unpaired d-electrons, it has an abnormally low enthalpy of atomization. This is because the half-filled d-subshell is very stable, and the electrons are held tightly by the nucleus, making them less available for metallic bonding. This results in weaker metallic bonds compared to its neighbours Ti, V, and Cr.
Step 4: Final Answer:
Manganese (Mn) has the lowest atomization enthalpy in this series.