JEE Chemistry 2025 Question with Solution

Given: The metals are Cr, Co, Fe, and Ni.

Find: The number of valence electrons in the metal having the lowest enthalpy of atomisation.

From the solution, the electronic configurations and valence electrons are:

• Chromium (Cr): $[\text{Ar}]\,3d^5 4s^1$, so valence electrons $= 6$
• Cobalt (Co): $[\text{Ar}]\,3d^7 4s^2$, so valence electrons $= 9$
• Iron (Fe): $[\text{Ar}]\,3d^6 4s^2$, so valence electrons $= 8$
• Nickel (Ni): $[\text{Ar}]\,3d^8 4s^2$, so valence electrons $= 10$

Using the stated trend in the solution, the metal with fewer valence electrons is taken to have weaker metallic bonding and hence lower enthalpy of atomisation. Therefore, among the given metals, Cr is identified as the one with the lowest enthalpy of atomisation.

So, the required number of valence electrons is $6$. Therefore, the correct option is C.

Given: Compare Cr, Co, Fe, and Ni on the basis of valence electrons.

Find: Which one has the lowest enthalpy of atomisation, and then report its valence electrons.

List the configurations one by one:

1. Cr: $[\text{Ar}]\,3d^5 4s^1$
2. Fe: $[\text{Ar}]\,3d^6 4s^2$
3. Co: $[\text{Ar}]\,3d^7 4s^2$
4. Ni: $[\text{Ar}]\,3d^8 4s^2$

Now count the valence electrons:

• Cr $\to 6$
• Fe $\to 8$
• Co $\to 9$
• Ni $\to 10$

According to the extracted solution logic, the smallest valence-electron count corresponds to the lowest enthalpy of atomisation in this set. The smallest count is for Cr.

Hence, the number of valence electrons is $6$, so the correct option is C.