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MHT-CET Chemistry · Transition and Inner Transition Elements

Oxidation States of Transition Elements

Transition elements show several oxidation states because their (n−1)d and ns electrons are close in energy and can be lost in steps; the range is widest in the middle of the 3d series, at manganese (+2 to +7), narrowest at the ends — scandium +3 (and +2), zinc only +2 — and the highest state in the whole block is +8, reached by osmium in the 5d series.

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Why this matters

6 PYQs, none HARD. Two ask which 3d element shows the most oxidation states (Mn), one which shows the fewest (Zn), one Scandium's states, one the highest state in the third series (+8), and one the Stock notation of MnO₂. One card.

Concept 1 of 1

How Many Oxidation States, and the Highest One

Intuition

The highest oxidation state an early 3d element can reach is the number of its 4s plus 3d electrons: Sc +3, Ti +4, V +5, Cr +6, Mn +7. After manganese the d electrons start pairing and hold on more tightly, so the maximum falls again (Fe +6, Co +4, Ni +4, Cu +2, Zn +2). Manganese therefore has the widest range, +2 to +7. The ends are narrow: scandium mostly +3, zinc only +2. Going down a group higher states become MORE stable, so the 5d series reaches +8 in OsO₄. Stock notation just writes the state in Roman numerals: MnO₂ has Mn at +4, Mn(IV)O₂.

Definition

  • Most states in the 3d series: Mn, +2 to +7 (4s² 3d⁵ — all seven outer electrons can be used).
  • Fewest: Zn, only +2 (fewer than Sc, Ti or Cu).
  • Scandium: +2, +3 (+3 the stable one, giving Sc³⁺ d⁰).
  • Highest state in the d-block: +8, by Os in OsO4\text{OsO}_4 (5d); the maximum in the 3d series is +7 (Mn in MnO4−\text{MnO}_4^-).
  • Stock notation: oxidation state in Roman numerals — MnO2\text{MnO}_2: x+2(−2)=0x + 2(-2) = 0, so Mn(IV)O₂; KMnO4\text{KMnO}_4 is Mn(VII).

Maximum oxidation state, Sc to Mn

max O.S.=n(4s)+n(3d): Sc +3, Ti +4, V +5, Cr +6, Mn +7\text{max O.S.} = n(4s) + n(3d):\ \text{Sc } {+3},\ \text{Ti } {+4},\ \text{V } {+5},\ \text{Cr } {+6},\ \text{Mn } {+7}

Worked example

Give the oxidation state of Cr in K₂Cr₂O₇ and write it in Stock notation.
Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 1Transition and Inner Transition ElementsEASY
Which element from following exhibits highest number of various different possible oxidation states?

[Q76 · 11th May Shift 1 · 2023]

Picking the element with most d electrons

Zinc and copper have the most d electrons but the fewest states — a full or nearly full d subshell does not give up electrons. The widest range belongs to the middle, Mn.
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Formulas (1)

  • How Many Oxidation States, and the Highest One

    Maximum oxidation state, Sc to Mn

    max O.S.=n(4s)+n(3d): Sc +3, Ti +4, V +5, Cr +6, Mn +7\text{max O.S.} = n(4s) + n(3d):\ \text{Sc } {+3},\ \text{Ti } {+4},\ \text{V } {+5},\ \text{Cr } {+6},\ \text{Mn } {+7}

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