Question:

Which of the following does not correctly represent the order of the property indicated against it?

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Chromium possesses exceptionally strong metallic bonding because of its electronic configuration, whereas manganese shows anomalously lower melting point due to stable half-filled configuration.
Updated On: May 20, 2026
  • $\mathrm{Ti^{3+} < V^{3+} < Cr^{3+} < Mn^{3+}}$ [Increasing order of magnetic moment]
  • $\mathrm{Ti < V < Cr < Mn}$ [Increasing order of melting point]
  • $\mathrm{Ti < V < Cr < Mn}$ [Increasing order of highest oxidation state]
  • $\mathrm{Ti < V < Mn < Cr}$ [Increasing order of second ionisation enthalpy]
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The Correct Option is B

Solution and Explanation


Concept: Transition elements show periodic variation in properties such as:
• Magnetic moment
• Melting point
• Oxidation state
• Ionisation enthalpy Among these, melting point depends strongly on metallic bonding which is influenced by the number of unpaired electrons available for bonding.

Step 1:
Checking Option (A) Magnetic moment depends on number of unpaired electrons. Electronic configurations: \[ \mathrm{Ti^{3+} = 3d^1} \] \[ \mathrm{V^{3+} = 3d^2} \] \[ \mathrm{Cr^{3+} = 3d^3} \] \[ \mathrm{Mn^{3+} = 3d^4} \] Number of unpaired electrons increases in the same order. Hence magnetic moment also increases as: \[ \mathrm{Ti^{3+} < V^{3+} < Cr^{3+} < Mn^{3+}} \] Thus, option (A) is correct.

Step 2:
Checking Option (B) Melting point generally increases from Ti to Cr because metallic bonding becomes stronger due to increased participation of unpaired d-electrons. However, manganese has: \[ 3d^5 4s^2 \] configuration. Due to the exceptionally stable half-filled configuration, metallic bonding becomes weaker in manganese. Therefore, manganese has lower melting point than chromium. Actual order is approximately: \[ \mathrm{Ti < V < Mn < Cr} \] or chromium has higher melting point than manganese. Hence: \[ \mathrm{Ti < V < Cr < Mn} \] is incorrect. Thus, option (B) does not correctly represent the property.

Step 3:
Checking Option (C) Highest oxidation states are: \[ \mathrm{Ti = +4} \] \[ \mathrm{V = +5} \] \[ \mathrm{Cr = +6} \] \[ \mathrm{Mn = +7} \] Clearly increasing order is: \[ \mathrm{Ti < V < Cr < Mn} \] Hence option (C) is correct.

Step 4:
Checking Option (D) Second ionisation enthalpy generally follows: \[ \mathrm{Ti < V < Mn < Cr} \] Thus option (D) is also correct. Hence, the incorrect order is: \[ \boxed{(B)} \]
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