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NEET CHEMISTRYHard

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

A

Ti3+<V3+<Cr3+<Mn3+Ti^{3+} < V^{3+} < Cr^{3+} < Mn^{3+} : increasing magnetic moment

B

Ti<V<Cr<MnTi < V < Cr < Mn : increasing melting point

C

Ti<V<Mn<CrTi < V < Mn < Cr : increasing 2nd ionization enthalpy

D

Ti<V<Cr<MnTi < V < Cr < Mn : increasing number of oxidation states

Step-by-Step Solution

We analyze each order based on NCERT data:

  1. Magnetic Moment: The number of unpaired electrons (nn) increases from Ti3+(d1,n=1)Ti^{3+}(d^1, n=1) to Mn3+(d4,n=4)Mn^{3+}(d^4, n=4). Since magnetic moment μ=n(n+2)\mu = \sqrt{n(n+2)}, the order is correct .
  2. Melting Point: The melting points of transition metals generally rise to a maximum near the middle of the series (CrCr). However, Manganese (MnMn) shows an anomalous low melting point (1246 K1246 \text{ K}) compared to Chromium (2180 K2180 \text{ K}) and Vanadium (1910 K1910 \text{ K}), due to its stable d5d^5 configuration leading to weaker metallic bonding. Thus, the order Ti<V<Cr<MnTi < V < Cr < Mn is incorrect because Mn is lower than V and Cr .
  3. 2nd Ionization Enthalpy: The second ionization enthalpy involves removing an electron from the singly charged ion. For Cr+(3d5)Cr^+ (3d^5), removing an electron disrupts the stable half-filled configuration, leading to a very high value (1592 kJ mol11592 \text{ kJ mol}^{-1}). For Mn+(3d54s1)Mn^+ (3d^5 4s^1), the electron is removed from 4s4s, leaving the stable 3d53d^5 intact (1509 kJ mol11509 \text{ kJ mol}^{-1}). Thus, Cr>MnCr > Mn, making the order Ti<V<Mn<CrTi < V < Mn < Cr correct .
  4. Oxidation States: The number of oxidation states increases from Ti (+2 to +4) to Mn (+2 to +7). The order is correct .
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