Question:medium

Match the following orders with their properties.

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Transition metal properties depend strongly on number of unpaired d-electrons because stronger metallic bonding increases melting point and atomization enthalpy.
Updated On: Jun 15, 2026
  • A-I, B-II, C-I, D-IV
  • A-II, B-I, C-IV, D-III
  • A-IV, B-III, C-II, D-I
  • A-III, B-IV, C-II, D-I
Show Solution

The Correct Option is D

Solution and Explanation

Step 1: Set up the matching of properties.
Four periodic properties of 3d transition metals are to be matched: enthalpy of atomization, density, metallic radius, and melting point. Each follows a characteristic trend tied to metallic bonding and atomic structure.
Step 2: Match enthalpy of atomization (A).
Stronger metallic bonding gives higher atomization enthalpy, peaking near the middle of the series. The given order $Fe>Cr>Mn$ matches property III, so $A\rightarrow III$.
Step 3: Match density (B).
Density rises across the series as atomic mass increases and size shrinks. The order $Co>Fe>Mn$ matches property IV, so $B\rightarrow IV$.
Step 4: Match metallic radius (C).
Metallic radius generally decreases across the period. The order $Ti>V>Cr$ matches property II, so $C\rightarrow II$.
Step 5: Match melting point (D).
Melting point depends on bond strength, with Mn low due to its stable half filled configuration. The order $Cr>V>Mn$ matches property I, so $D\rightarrow I$.
Step 6: Assemble and choose.
The matching $A\rightarrow III$, $B\rightarrow IV$, $C\rightarrow II$, $D\rightarrow I$ corresponds to option (4).
\[ \boxed{A\text{-}III,\ B\text{-}IV,\ C\text{-}II,\ D\text{-}I\ \ \text{(Option 4)}} \]
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