The number of valence electrons present in the metal among $\mathrm{Cr}, \mathrm{Co}, \mathrm{Fe}$ and Ni which has the lowest enthalpy of atomisation is :

<p>To determine the number of valence electrons in the metal with the lowest enthalpy of atomization among Chromium (Cr), Cobalt (Co), Iron (Fe), and Nickel (Ni), we need to examine their respective electron configurations.</p> <p>Chromium has the electron configuration:</p> <p>$ \text{Cr} = [\text{Ar}] 3d^5 4s^1 $</p> <p>Based on this configuration, Chromium has a total of six valence electrons (1 from the 4s orbital and 5 from the 3d orbital). </p> <p>Chromium is noted for having the lowest enthalpy of atomization among the given metals. Thus, the number of valence electrons in Chromium, the metal with the lowest enthalpy of atomization, is 6.</p>