Explain the oxidation states of 4d series elements.

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Q. Explain the oxidation states of 4d series elements.

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Concept understanding — Variable Oxidation States of Transition Metals

Scandium, the first 3d element, exhibits only one oxidation state, +3. Every other transition element shows variable oxidation states, because the (n-1)d and ns orbitals lie so close in energy that electrons from either (or both) can be lost. Across the 3d series, the number of accessible oxidation states rises then falls: it rises because more electrons become available through the first half of the series, and falls again as the number of paired d electrons increases toward the end. Scandium (first) has just one state (+3); manganese (middle) has the most, six states from +2 to +7; copper (last) is restricted to +1 and +2.

The relative stability among a metal's different oxidation states correlates with the extra stability of half-filled/fully-filled configurations -- Mn²⁺ (3d⁵) is markedly more stable than Mn⁴⁺ (3d³). For the heavier 4d/5d series, oxidation states range even more widely, from +3 (Y, La) up to +8 (Ru, Os) -- the highest oxidation state anywhere in the d-block -- typically only reached in compounds with strongly electronegative elements (RuO₄, OsO₄, WCl₆). Moving down a group, stability of the higher oxidation state increases while the lower state's stability decreases, the reverse of the p-block pattern.

Among the 3d metals, in the first half of the series (Ti through Mn), the +2 state becomes progressively more accessible/stable as atomic number rises, because more electrons are available to lose from ns and (n-1)d and the +2 ion's configuration approaches the stabilising d⁵ point at Mn²⁺. Copper is unique in showing a stable +1 state, though even Cu⁺ is thermodynamically prone to disproportionate into Cu²⁺ and metallic Cu(0), since (per the Frost diagram) elemental copper sits lower in free energy than either ionic state. Transition metals in high oxidation states characteristically act as oxidising agents -- e.g. Fe³⁺ (E° = +0.77 V) is a moderate oxidant, capable of oxidising Cu to Cu²⁺ (E° = +0.34 V).

Step 1. The 4d series runs from yttrium (Y) to cadmium (Cd), directly below the 3d series.

Step 2. Yttrium and lanthanum-adjacent 4d elements show the lower end of the range, +3, matching their group-3 position and single accessible oxidation state pattern.

Step 3. Moving across the series, accessible oxidation states widen considerably, reaching a maximum of +8 at ruthenium (Ru), roughly at the midpoint of the row -- the highest oxidation state found anywhere among the d-block elements, matched only by osmium (Os) in the corresponding 5d position.

Step 4. These very highest oxidation states are only realised in compounds with strongly electronegative elements -- oxygen, fluorine and chlorine -- for example RuO₄ (ruthenium tetroxide, Ru in +8).

Step 5. As with the 3d series, moving down a group within the d-block generally increases the stability of the higher oxidation state while decreasing the stability of the lower one -- the reverse of the pattern seen in p-block groups.

✓Final answer

4d series oxidation states range from +3 (Y) to +8 (Ru), typically reached in oxides/chlorides such as RuO₄.

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2024Set ANNUAL1 markMCQ
$Q. Which of the following transition elements does not exhibit variable oxidation state?$
(a) Sc
(b) Fe
(c) Mn
(d) Co
›Reveal solutionSolution

Sc has only one stable oxidation state, +3, because its electron configuration leaves no other energetically accessible partial-loss options.

Sc (Z = 21) has configuration [Ar] 3d $^1$ 4s $^2$ . It loses all three of these electrons (3d $^1$ + 4s $^2$ ) to attain the stable, empty 3d $^0$ configuration of Sc $^{3+}$ — isoelectronic with Ar. There is no other oxidation state that gives comparable stability, so Sc shows only +3.

By contrast, Fe, Mn and Co all have several d-electrons that can be lost in varying numbers (e.g. Fe $^{2+}$ /Fe $^{3+}$ , Mn $^{2+}$ up to Mn $^{7+}$ , Co $^{2+}$ /Co $^{3+}$ ), because multiple partially-filled d-configurations are close enough in energy to be accessible — this is the hallmark of transition-metal variable oxidation states.

✓Final answer
$Sc does not exhibit variable oxidation state (shows only +3).$
CBSE 2023Set ANNUAL1 mark
$Q. Which of the 3d series transition elements exhibits the largest number of oxidation states? What is the highest oxidation state of that element? OR Write the general electronic configuration of lanthanides.$
›Reveal solutionSolution

Mn, sitting at the middle of the 3d series with the maximum number of unpaired d and s electrons available for bonding, shows oxidation states from +2 right up to +7.

Across the 3d transition series, the number of accessible oxidation states rises from Sc to Mn and then falls again toward Zn, because Mn ([Ar]3d5 4s2) has the maximum number of electrons (5 d + 2 s = 7) that can be progressively involved in bonding/oxidation. This gives Mn the widest oxidation-state range of any 3d element: +2, +3, +4, +5, +6, and +7.

Its highest oxidation state, +7, is seen in compounds such as KMnO4 (potassium permanganate) and Mn2O7.

✓Final answer
$Manganese; highest oxidation state = +7 (as in KMnO4).$
CBSE 2022Set ANNUAL1 markMCQ
$Q. Variable valency is the main characteristic of$
(a) alkali metals
(b) alkaline earth metals
(c) noble gases
(d) transition elements
›Reveal solutionSolution
Transition (d-block) elements characteristically show variable oxidation states because their (n-1)d and ns electrons are close in energy and can both participate in bonding.

Reasoning: Transition elements have partially filled d-orbitals, and the energy difference between the (n-1)d and ns subshells is small. This allows a variable number of electrons from both subshells to be lost/involved in bonding, giving transition metals multiple stable oxidation states (e.g. Fe: +2, +3; Mn: +2 up to +7; Cu: +1, +2).

Alkali metals (Group 1) show only +1.

Alkaline earth metals (Group 2) show only +2.

Noble gases are largely inert, showing essentially no common oxidation states.

Only transition elements are characteristically known for variable valency.

✓Final answer
$(d) transition elements — variable oxidation states arise because both (n-1)d and ns electrons are of comparable energy and can be involved in bond formation.$
CBSE 2019Set ANNUAL1 markMCQ
$Q. Which of the following transition elements shows +7 oxidation state?$
(a) Cr
(b) Mn
(c) Fe
(d) Ni
›Reveal solutionSolution

Manganese reaches +7 in permanganate.

Manganese shows a maximum oxidation state of +7 (e.g. in $KMnO_4$ , $Mn_2O_7$ ), which is the highest oxidation state shown by any first-row transition element, matching its total number of valence electrons ( $3d^5 4s^2$ ). Cr's maximum is +6, Fe's practical maximum is +6 (rare), and Ni's is typically +2/+4.

✓Final answer
$(b) Mn.$

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Q.Explain the oxidation states of 4d series elements.

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