Oxidation State Trend In 3d Series
Across the first transition series, the highest oxidation state increases up to manganese and then decreases; what underlies this pattern?
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About This Question
- Subject
- chemistry
- Chapter
- d- and f-block elements
- Topic
- oxidation state trend in 3d series
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
The number of available d and s electrons rises then their stability falls
The trend in highest oxidation state across the first transition series follows the number of electrons available for bonding and how tightly they are held. From scandium to manganese, the number of 3d and 4s electrons that can participate in bonding increases, allowing progressively higher oxidation states, reaching a maximum of +7 at manganese, which can use all its 3d^5 4s^2 electrons. Beyond manganese, the increasing nuclear charge binds the d-electrons more tightly and pairs begin to form, so fewer electrons are available for high oxidation states, and the highest attainable oxidation state falls toward the end of the series. The option that atomic mass decreases is false, since mass increases across the series. The option that ions become larger is wrong, as radii generally decrease. The option that all show only one oxidation state contradicts the variable valency of the series. This pattern is a standard NCERT d-block trend. This concept also bridges to Atomic Structure and Redox Reactions and Electrochemistry, so mastering it strengthens performance on linked questions from those topics as well. Plausibility check: manganese exhibiting +7 at the midpoint, with lower maxima for iron and beyond, matches the rise-then-fall pattern, confirming the reasoning.
This medium difficulty chemistry question is from the chapter d- and f-block elements, covering the topic of oxidation state trend in 3d series. It appeared in the 2025 exam.
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