Enthalpy of atomization of transition elements can be arranged in increasing order as
A. \(V\)
B. \(Cr\)
C. \(Mn\)
D. \(Fe\)
E. \(Co\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(Mn < Cr < Fe < Co < V\)
Transition metals have high enthalpies of atomisation because their unpaired d electrons, along with the s electrons, take part in strong metallic bonding; the more electrons that join in, the stronger the bonding.
C. Mn (first, lowest) — its stable half-filled 3d5 set (Mn is [Ar]3d54s2) holds the d electrons tightly, so fewer of them join in metallic bonding.
B. Cr (second) — lower than its neighbours V and Fe in the series.
D. Fe (third) and E. Co (fourth) — the values rise again after the Mn dip.
A. V (last, highest) — with three unpaired 3d electrons and two 4s electrons all available, it has the strongest metallic bonding of the five.
Shortcut: Mn has the lowest value and V the highest. Options (1) and (3) start with Mn and end with V; Cr is below Fe, which leaves Option (1).
NCERT (2026-27 edition):
Electronic Configurations and some other Properties of the First Series of Transition Elements (row of Table 4.2)
| Sc | Ti | V | Cr | Mn | Fe | Co | Ni | Cu | Zn | |
|---|---|---|---|---|---|---|---|---|---|---|
| Enthalpy of atomisation, ΔaH°/kJ mol–1 | 326 | 473 | 515 | 397 | 281 | 416 | 425 | 430 | 339 | 126 |