Which of the following transition elements of \(3d\) series has the lowest melting point?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → Mn (At. No - 25)
Transition metals have high melting points because, besides the ns electrons, electrons of the (n−1)d subshell take part in metallic bonding. Across a series the melting points rise to a maximum near d5 and then fall, except for the anomalous values of Mn and Tc. Manganese (3d5 4s2) has a stable half-filled 3d subshell; its d electrons are held more tightly and take less part in metallic bonding, so its melting point (about 1519 K) is much lower than that of its neighbours.
Melting points: Ti about 1941 K, V about 2183 K, Cr about 2180 K, Mn about 1519 K. So Mn has the lowest melting point of the four.
Why the other options are incorrect
Option (1) Ti (At. No - 22) — Incorrect. 3d2 4s2; four electrons take part in metallic bonding, melting point about 1941 K.
Option (2) V (At. No - 23) — Incorrect. 3d3 4s2; strong metallic bonding, melting point about 2183 K, among the highest in the series.
Option (3) Cr (At. No - 24) — Incorrect. 3d5 4s1; six unpaired electrons give strong metallic bonding, melting point about 2180 K.
NCERT (2026-27 edition):
Physical Properties – “… The high melting points of these metals are attributed to the involvement of greater number of electrons from (n-1)d in addition to the ns electrons in the interatomic metallic bonding. In any row the melting points of these metals rise to a maximum at d5 except for anomalous values of Mn and Tc and fall regularly as the atomic number increases.”