Question

The number of valence electrons present in the metal among Cr, Co, Fe and Ni which has the lowest enthalpy of atomisation is

• A. 8
• B. 9
• C. 6
• D. 10

Hint:

In the 3d transition series, the enthalpy of atomization usually peaks around the middle (Cr/Mo/W) due to the maximum number of unpaired electrons and decreases towards the end of the series as electrons pair up.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
Enthalpy of atomization depends on the strength of the metallic bond. In transition metals, the strength of the metallic bond is generally proportional to the number of unpaired electrons available for interatomic metallic bonding.

Step 2: Detailed Explanation:
Let's look at the electronic configurations of the given 3d series elements ($4s$ and $3d$ are valence electrons):
• Cr (\(Z=24\)): \(3d^5 4s^1\) (6 valence electrons, 6 unpaired) - Very high enthalpy.
• Fe (\(Z=26\)): \(3d^6 4s^2\) (8 valence electrons, 4 unpaired).
• Co (\(Z=27\)): \(3d^7 4s^2\) (9 valence electrons, 3 unpaired).
• Ni (\(Z=28\)): \(3d^8 4s^2\) (10 valence electrons, 2 unpaired). As we move from Cr to Ni, the number of unpaired electrons decreases because pairing starts in the $3d$ subshell. Among the given options, Nickel (Ni) has the fewest unpaired electrons, leading to the weakest metallic bonding and thus the lowest enthalpy of atomization.

Step 3: Final Answer
The metal is Nickel (Ni), which has $8 + 2 = 10$ valence electrons.