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

The enthalpy of atomisation of a transition metal depends on the strength of its metallic bonding.
Stronger metallic bonding arises when there are many unpaired electrons available for formation of the metallic (d-d) bonds.

Across the 3d-series (Sc → Zn) the general trend is: enthalpy of atomisation rises from Sc to V, shows a small fall at Cr, drops sharply at Mn, rises again up to Ni, and then falls towards Zn.

Standard data (in $$\text{kJ mol}^{-1}$$) for the elements asked in the question are: Cr = 397, Fe = 416, Co = 423, Ni = 430.

Thus, among Cr, Fe, Co and Ni, the lowest enthalpy of atomisation belongs to chromium (Cr).

To find the required option we now count the valence electrons of Cr.
In transition elements the valence electrons are all the electrons present in the outermost $$ns$$ subshell and the penultimate $$(n-1)d$$ subshell.

Electronic configuration of chromium: $$\text{Cr}: [\text{Ar}]\,3d^{5}\,4s^{1}$$

Number of valence electrons $$= \,(\text{electrons in }3d) + (\text{electrons in }4s)$$ $$= 5 + 1 = 6$$

Therefore the metal having the lowest enthalpy of atomisation (Cr) possesses $$6$$ valence electrons.

Option C is correct.