JEE Main 2025 · 7 April · morning · Single correct · +4 / −1
The number of valence electrons present in the metal among Cr, Co, Fe and Ni which has the lowest enthalpy of atomisation is :
- (a) 10
- (b) 6
- (c) 9
- (d) 8
Answer and solution
Answer (b) 6
Solution
Chemistry ma’am · 3 steps
Find the metal from the atomisation data first, then count its 3d and 4s electrons together.
- What atomisation measures. Enthalpy of atomisation is the energy needed to break the solid metal into free gaseous atoms. It is low when the metallic bonding is weak.
Metal Configuration Valence electrons (3d + 4s) Δₐ H° (kJ mol⁻¹) Cr [Ar] 3d⁵ 4s¹ 5 + 1 = 6 397, lowest Fe [Ar] 3d⁶ 4s² 6 + 2 = 8 416 Co [Ar] 3d⁷ 4s² 7 + 2 = 9 425 Ni [Ar] 3d⁸ 4s² 8 + 2 = 10 430 - Why chromium dips. Chromium has an extra-stable half-filled 3d⁵ subshell. Its d electrons take less part in metallic bonding, so its atomisation enthalpy falls below that of Fe, Co and Ni. Manganese, the next element with 3d⁵, shows the same dip even more sharply.
- So. The metal with the lowest enthalpy of atomisation is Cr. With the configuration 3d⁵ 4s¹ it has 6 valence electrons, which is option (b).
The idea to keep. For transition metals, valence electrons means (n-1)d plus ns electrons. Look for dips in atomisation enthalpy at the d⁵ elements Cr and Mn, where the half-filled d subshell weakens metallic bonding.
Answer from the NTA final answer key. Solution by PrepWiser.