What NEET Asks
- NEET often features questions on identifying oxidation states of transition metals in compounds/ions.
- Expect questions on reasons for variable oxidation states and trends across the period/group.
- Understanding stability of different oxidation states (e.g., d⁰, d⁵, d¹⁰ configurations) is crucial for 3-4 marks.
Key Points
- Definition: Transition elements exhibit variable oxidation states due to the participation of both (n-1)d and ns electrons in bonding.
- Reason: The energies of (n-1)d and ns orbitals are very close, allowing both sets of electrons to be available for bond formation.
- Common Oxidation States: The most common oxidation state for many transition metals is +2 (loss of two ns electrons).
- Trends: Oxidation states generally increase to a maximum at the middle of the series (e.g., Mn exhibits +7) and then decrease.
- Stability: High oxidation states are more stable with highly electronegative elements like oxygen and fluorine (e.g., Mn₂O₇, K₂Cr₂O₇).
- Examples: Sc (+3 only), Ti (+2, +3, +4), V (+2, +3, +4, +5), Cr (+2, +3, +6), Mn (+2, +3, +4, +6, +7), Fe (+2, +3), Co (+2, +3), Ni (+2, +3), Cu (+1, +2), Zn (+2 only).
Must-Know Formula / Reaction
Calculating Oxidation State: For a neutral compound, sum of oxidation states = 0. For an ion, sum of oxidation states = charge on ion.
Example: In KMnO₄ K (+1) + Mn(x) + 4 × O(-2) = 0 1 + x - 8 = 0 x = +7
Common Mistakes
- Students often forget that Sc and Zn exhibit only one common oxidation state (+3 and +2 respectively).
- Don't confuse the stability of high oxidation states with the commonness of low oxidation states (e.g., +2 for Mn is common, but +7 is highest).
- Incorrectly calculating oxidation states in complex ions by ignoring the charge on ligands (e.g., CN⁻, C₂O₄²⁻).
Rapid Revision
Variable oxidation states arise from small energy difference between ns and (n-1)d orbitals. Highest state generally found with O/F. Sc (+3) and Zn (+2) are exceptions to variability. Mn shows highest (+7) in the 3d series.