The correct option with order of melting points of the pairs $(\mathrm{Mn}, \mathrm{Fe}),(\mathrm{Tc}, \mathrm{Ru})$ and $(\mathrm{Re}, \mathrm{Os})$ is :

(C) $\mathrm{Mn}<\mathrm{Fe}, \mathrm{Tc}<\mathrm{Ru}$ and $\mathrm{Os}<\mathrm{Re}$

(A) $\mathrm{Fe}<\mathrm{Mn}, \mathrm{Ru}<\mathrm{Tc}$ and $\mathrm{Re}<\mathrm{Os}$

(B) $\mathrm{Mn}<\mathrm{Fe}, \mathrm{Tc}<\mathrm{Ru}$ and $\mathrm{Re}<\mathrm{Os}$

(D) $\mathrm{Fe}<\mathrm{Mn}, \mathrm{Ru}<\mathrm{Tc}$ and $\mathrm{Os}<\mathrm{Re}$

JEE Main · 2025 · 29 Jan 2025 · Shift 1 · d and f Block Elements

The correct option with order of melting points of the pairs

(Mn, Fe), (Tc, Ru)

and

(Re, Os)

is :

(A) $Fe < Mn, Ru < Tc$ and $Re < Os$
(B) $Mn < Fe, Tc < Ru$ and $Re < Os$
(C) $Mn < Fe, Tc < Ru$ and $Os < Re$ ✓ Correct
(D) $Fe < Mn, Ru < Tc$ and $Os < Re$

Answer: (C)

Solution

Metallic bond strength across a transition series tracks the number of unpaired d electrons available for delocalised metallic bonding, which is why melting points climb through the middle of a series and why the very late-filling elements can buck a simple trend. Following the standard 3d/4d/5d comparison: Fe has more bonding d electrons available than Mn, so

Mn < Fe

; the same pattern holds one row down,

Tc < Ru

; but in the 5d series the exceptionally strong 5d–5d orbital overlap makes Re, with its half-filled

5 d^{5}

configuration, more refractory than Os, reversing the trend:

Os < Re

. The correct order is

Mn < Fe

Tc < Ru

and

Os < Re

JEE Main · 2025 · 29 Jan 2025 · Shift 1 · Single correct option

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JEE Main · 2025 · 29 Jan 2025 · Shift 1 · d and f Block Elements

Solution

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