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

The question asks about the number of valence electrons in the metal among chromium (Cr), cobalt (Co), iron (Fe), and nickel (Ni) which has the lowest enthalpy of atomization. The enthalpy of atomization is a measure of the strength of the bonds between atoms in a metal's lattice; a lower value suggests weaker bonding.

To answer this question, let's analyze the provided options and elements:

1. Chromium (Cr): Its electronic configuration is \([Ar] 3d^5 4s^1\). Valence electrons: 6 (in 3d and 4s).
2. Cobalt (Co): Its electronic configuration is \([Ar] 3d^7 4s^2\). Valence electrons: 9 (in 3d and 4s).
3. Iron (Fe): Its electronic configuration is \([Ar] 3d^6 4s^2\). Valence electrons: 8 (in 3d and 4s).
4. Nickel (Ni): Its electronic configuration is \([Ar] 3d^8 4s^2\). Valence electrons: 10 (in 3d and 4s).

Enthalpy of atomization is typically influenced by how fully the d subshell is filled and the strength of d-d overlap. Nickel, having a nearly full d subshell, exhibits weaker metallic bonding compared to the other elements listed. Therefore, it generally has the lowest enthalpy of atomization among these elements.

The number of valence electrons in nickel is 10, aligning with the correct answer.

Thus, the metal among Cr, Co, Fe, and Ni with the lowest enthalpy of atomization has 10 valence electrons.