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 general, for transition metals, a lower number of valence electrons leads to weaker metallic bonding, hence a lower enthalpy of atomisation.

Answer 1

The Correct Option is C

Step 1: Atomic Configuration and Valence Electrons
Chromium (Cr):
Atomic number = 24. Its electron configuration is \([ \text{Ar} ] 3d^5 4s^1\), giving it 6 valence electrons
Cobalt (Co):
Atomic number = 27. Its electron configuration is \([ \text{Ar} ] 3d^7 4s^2\), giving it 9 valence electrons
Iron (Fe):
Atomic number = 26. Its electron configuration is \([ \text{Ar} ] 3d^6 4s^2\), giving it 8 valence electrons
Nickel (Ni):
Atomic number = 28. Its electron configuration is \([ \text{Ar} ] 3d^8 4s^2\), giving it 10 valence electrons
Step 2: Relation to Enthalpy of Atomisation
Enthalpy of atomisation generally decreases as the number of valence electrons increases, because more electrons lead to stronger metallic bonding and thus require more energy to break the bonds. 
Therefore, the element with fewer valence electrons will have the lowest enthalpy of atomisation. Since Cr has 6 valence electrons, it will have the lowest enthalpy of atomisation compared to Co, Fe, and Ni. 
Thus, the correct answer is 6 valence electrons.