Why Transition Metals Show Variable States
A transition element is one with an incompletely filled $d$ subshell in its ground state or in at least one of its common oxidation states. The chemical consequence of this incomplete $d$ shell is the appearance of many oxidation states. The cause, stated plainly by NCERT and the NIOS supplement (§21.4.1), is energetic: in addition to the $ns$ electrons, these elements can use their inner $(n-1)d$ electrons for bonding because the difference in energy between the $ns$ and $(n-1)d$ levels is very small.
Contrast this with a typical $s$-block metal. Sodium has one $3s$ electron above a stable noble-gas core; removing it gives $\ce{Na+}$, and there is no economical way to reach a higher state. A transition atom is different. After the $ns$ electrons are removed, the $(n-1)d$ electrons lie close enough in energy that one, two, or more of them can also be removed or shared at modest extra energetic cost. The number of $d$ electrons that participate is variable, so the oxidation state is variable.
Because the $d$ electrons are removed one at a time, the resulting states differ from each other by unity. Vanadium illustrates the pattern cleanly, running through $\ce{V^2+}$, $\ce{V^3+}$, $\ce{V^4+}$ and $\ce{V^5+}$. This unit-step behaviour is a key fingerprint that separates the $d$-block from the $p$-block, where states usually differ by two.
In an $s$-block atom only the lone $ns$ electron is energetically accessible, fixing a single oxidation state. In a transition atom the $4s$ and $3d$ levels are nearly degenerate, so several $d$ electrons join the bonding — giving the ladder of states that differs by unity.
Oxidation States Across the 3d Series
NCERT Table 4.3 lists the common oxidation states of the first transition (3d) series. Reading it from $\ce{Sc}$ to $\ce{Zn}$, the number of available states rises from the start, peaks at manganese in the middle, and then falls away as the $d$ shell fills. The lowest common state for the run from $\ce{Ti}$ to $\ce{Zn}$ is $+2$ (loss of the two $4s$ electrons); the maximum climbs to $+7$ at manganese.
| Element | Config (atom) | Minimum common O.S. | Maximum O.S. | Notable species |
|---|---|---|---|---|
| Sc | 3d¹4s² | +3 | +3 | $\ce{Sc2O3}$ (no variable state) |
| Ti | 3d²4s² | +2 | +4 | $\ce{TiO2}$, $\ce{TiX4}$ |
| V | 3d³4s² | +2 | +5 | $\ce{VO2^+}$, $\ce{V2O5}$, $\ce{VF5}$ |
| Cr | 3d⁵4s¹ | +2 | +6 | $\ce{CrO4^{2-}}$, $\ce{Cr2O7^{2-}}$ |
| Mn | 3d⁵4s² | +2 | +7 | $\ce{MnO4^-}$, $\ce{Mn2O7}$ |
| Fe | 3d⁶4s² | +2 | +3 (usual) | $\ce{Fe^2+}$, $\ce{Fe^3+}$ |
| Co | 3d⁷4s² | +2 | +3 | $\ce{Co^2+}$, $\ce{Co^3+}$ |
| Ni | 3d⁸4s² | +2 | +2 (usual) | $\ce{Ni^2+}$ |
| Cu | 3d¹⁰4s¹ | +1 | +2 | $\ce{Cu+}$, $\ce{Cu^2+}$ |
| Zn | 3d¹⁰4s² | +2 | +2 | $\ce{Zn^2+}$ (no d involvement) |
The shape of this table is not accidental. Elements that show the greatest variety of states sit in or near the middle of the series; the fewer states at the two ends arise either from too few electrons to lose or share (Sc, Ti) or from too many $d$ electrons, which leaves fewer orbitals available for sharing at high valence (Cu, Zn). Scandium does not exhibit variable oxidation states at all — its only common state is $+3$. At the far end, zinc shows only $+2$ because its $3d$ subshell is full and the $d$ electrons take no part in bonding.
The envelope of maximum oxidation states rises steadily from $\ce{Sc}$ ($+3$) to the peak at $\ce{Mn}$ ($+7$), then drops sharply across $\ce{Fe}$–$\ce{Zn}$ as the filling $d$ shell becomes harder to ionise.
The Maximum Oxidation State and Why Mn Reaches +7
Up to manganese, the maximum oxidation state of reasonable stability equals the sum of the $s$ and $d$ electrons. Thus titanium reaches $+4$ in $\ce{Ti^{IV}O2}$, vanadium $+5$ in $\ce{VO2^+}$, chromium $+6$ in $\ce{CrO4^{2-}}$, and manganese — with the $3d^5 4s^2$ configuration, a total of seven valence electrons — reaches $+7$ in $\ce{MnO4^-}$. Manganese therefore exhibits every state from $+2$ to $+7$, the widest range in the 3d series.
Beyond manganese the picture changes abruptly. The third and higher ionisation enthalpies become so large that the full $s+d$ count can no longer be reached, and the typical species become $\ce{Fe^2+}/\ce{Fe^3+}$, $\ce{Co^2+}/\ce{Co^3+}$, $\ce{Ni^2+}$, $\ce{Cu+}/\ce{Cu^2+}$ and $\ce{Zn^2+}$. The collapse of high states past Group 7 is one of the most testable features of the table.
"Maximum state = group number" only holds up to manganese
It is tempting to extend the rule "highest oxidation state = group number" all the way across the row. It works for $\ce{Sc}$ (3) through $\ce{Mn}$ (7), but breaks beyond it: iron does not commonly show $+8$, and the highest iron oxide stops at $\ce{Fe2O3}$ (ferrates $\ce{FeO4^{2-}}$ form only in alkali and decompose readily).
Use the $s+d$ electron sum, valid only Sc → Mn. After Mn, the high states are unattainable in ordinary chemistry.
Why the +2 State Is So Common
With the exception of scandium, the most common oxidation state across the 3d row is $+2$, arising from loss of the two $4s$ electrons. The reason, as NIOS §21.4.1 notes, is that after scandium the $(n-1)d$ orbitals become more stable than the $ns$ orbital. The $4s$ electrons are therefore the outermost and most easily removed, while the $d$ electrons are held more tightly and need much larger ionisation energies to leave.
To form $\ce{M^2+}$ from a gaseous atom, the enthalpy of atomisation plus the first and second ionisation enthalpies must be supplied. The dominant term is the second ionisation enthalpy, which is unusually high for $\ce{Cr}$ and $\ce{Cu}$ — where the $\ce{M+}$ ion already has the stable $d^5$ and $d^{10}$ configurations respectively — and correspondingly low for $\ce{Zn}$, where ionisation removes a $4s$ electron to leave the stable $d^{10}$ core. These same energetics make the $+2$ state appear for nearly every element from titanium to zinc.
The bonding character also tracks the oxidation state. Compounds in the $+2$ and $+3$ states are largely ionic, whereas higher states are essentially covalent — in the permanganate ion $\ce{MnO4^-}$ the $\ce{Mn-O}$ bonds are covalent. The acid–base behaviour of the oxides follows the same logic: as the oxidation state rises, the basic character of the oxide falls. Thus $\ce{MnO}$ ($+2$) is basic, $\ce{Mn2O3}$ is weakly basic/amphoteric, and $\ce{Mn2O7}$ ($+7$) is acidic.
The oxidation-state pattern feeds directly into colour, magnetism and catalysis. See the full overview in General Properties of Transition Elements.
Stability Trends: Mn²⁺, Cr²⁺ and Fe³⁺
Electronic stability — especially the extra stability of half-filled ($d^5$) and fully filled ($d^{10}$) configurations — controls which states are favoured. Manganese(II), $\ce{Mn^2+}$, has the $3d^5$ configuration and is therefore unusually stable. This stability is why the third ionisation enthalpy of manganese is high (removing an electron from the half-filled $d^5$ is difficult) and why $\ce{Mn^3+}$ is a strong oxidising agent: the change $\ce{Mn^3+ -> Mn^2+}$ delivers the very stable half-filled $d^5$.
The same configurational logic resolves a classic comparison. Both $\ce{Cr^2+}$ and $\ce{Mn^3+}$ have the $d^4$ configuration, yet they behave oppositely:
| Ion | Config | Change on reaction | Driving force | Behaviour |
|---|---|---|---|---|
| $\ce{Cr^2+}$ | $d^4$ | $d^4 \rightarrow d^3$ (loses e⁻) | $d^3$ has a half-filled $t_{2g}$ level | Reducing |
| $\ce{Mn^3+}$ | $d^4$ | $d^4 \rightarrow d^5$ (gains e⁻) | $d^5$ is a stable half-filled shell | Oxidising |
$\ce{Cr^2+}$ is reducing because it readily loses an electron to reach the more stable $d^3$ ($t_{2g}^3$). $\ce{Mn^3+}$ is oxidising because gaining an electron takes it to the stable half-filled $d^5$. Iron(III) is stabilised by the same effect: $\ce{Fe^3+}$ has the $d^5$ configuration, which is why the third ionisation enthalpy of iron is comparatively low and $\ce{Fe^3+}$ is the favoured higher state of iron.
Q. Why is $\ce{Cr^2+}$ reducing while $\ce{Mn^3+}$ is oxidising, although both are $d^4$?
$\ce{Cr^2+}$ ($d^4$) loses one electron to become $\ce{Cr^3+}$ ($d^3$), which has a half-filled $t_{2g}$ set and so is favoured — hence $\ce{Cr^2+}$ acts as a reducing agent. $\ce{Mn^3+}$ ($d^4$) gains one electron to become $\ce{Mn^2+}$ ($d^5$), a stable half-filled shell — hence $\ce{Mn^3+}$ acts as an oxidising agent. The driver in both cases is the stability of the product configuration.
Relative Stability and Disproportionation
When one oxidation state is markedly less stable than its neighbours, it can disproportionate — a single species is simultaneously oxidised and reduced. The textbook example is copper(I) in water. Although $\ce{Cu^I}$ halides exist in the solid state, many copper(I) compounds are unstable in aqueous solution:
$$\ce{2Cu+(aq) -> Cu^2+(aq) + Cu(s)}$$
The thermodynamic reason is the hydration enthalpy. The much more negative $\Delta_{hyd}H^\circ$ of $\ce{Cu^2+}(aq)$ compared with $\ce{Cu+}(aq)$ more than compensates for the large second ionisation enthalpy of copper, so the $+2$ state is the stable one in solution. This is exactly why NEET 2023 asked which factor makes $\ce{Cu^2+}$ more stable than $\ce{Cu+}$ — the answer is hydration energy, not the ionisation enthalpy.
A related stability argument explains why $\ce{Cu^2+}$ oxidises iodide. Copper(II) halides are all known except the iodide, because $\ce{Cu^2+}$ oxidises $\ce{I-}$ to iodine:
$$\ce{2Cu^2+ + 4I- -> Cu2I2(s) + I2}$$
In disproportionation the same element splits into a higher and a lower state. For copper the dominant hydration enthalpy of $\ce{Cu^2+}$ tips the balance so that $\ce{Cu+}$ collapses to $\ce{Cu^2+}$ and $\ce{Cu}$ metal.
Highest States in Oxides and Fluorides
A recurring NCERT intext question (Q 4.6) asks why the highest oxidation state of a metal is exhibited in its oxide or fluoride only. The answer rests on the special character of oxygen and fluorine, the two most electronegative elements, which can stabilise high oxidation states.
Fluorides. Fluorine stabilises high states through high lattice energy (as in $\ce{CoF3}$) or high bond enthalpy in higher covalent compounds (as in $\ce{VF5}$ and $\ce{CrF6}$). The highest halides of the 3d metals are $\ce{TiX4}$, $\ce{VF5}$ and $\ce{CrF6}$; vanadium's $+5$ among the halides is represented only by $\ce{VF5}$. Fluorides are also notably unstable in low oxidation states. Note that the $+7$ state of $\ce{Mn}$ is not found in any simple halide — the highest manganese fluoride is $\ce{MnF4}$.
Oxides. Oxygen does everything fluorine does and more: it can form multiple ($\pi$) bonds to the metal. The highest oxidation number in the oxides coincides with the group number, from $\ce{Sc2O3}$ to $\ce{Mn2O7}$. Because oxygen's stabilising ability exceeds that of fluorine, the very highest states appear in oxides — the contrast for manganese is decisive:
| Metal | Highest fluoride | Highest oxide | What it shows |
|---|---|---|---|
| Mn | $\ce{MnF4}$ (+4) | $\ce{Mn2O7}$ (+7) | Oxygen reaches a far higher state than F |
| V | $\ce{VF5}$ (+5) | $\ce{V2O5}$ (+5) | Both reach the group maximum |
| Cr | $\ce{CrF6}$ (+6) | $\ce{CrO3}$ (+6) | Both reach $+6$ |
The structural reason is multiple bonding: in covalent $\ce{Mn2O7}$ each $\ce{Mn}$ is tetrahedrally surrounded by oxygens with a $\ce{Mn-O-Mn}$ bridge, and tetrahedral $\ce{[MO4]^{n-}}$ ions are known for $\ce{V^V}$, $\ce{Cr^{VI}}$, $\ce{Mn^V}$, $\ce{Mn^{VI}}$ and $\ce{Mn^{VII}}$. Two of these tetrahedral oxoanions, $\ce{MnO4^-}$ and $\ce{Cr2O7^{2-}}$, are the workhorse oxidants of this chapter.
Oxide can beat fluoride for the top state
Students often assume fluorine, being the most electronegative element, always pins the highest oxidation state. For $\ce{Mn}$ it is the oxide that wins: $\ce{Mn2O7}$ ($+7$) versus $\ce{MnF4}$ ($+4$). Oxygen's ability to form $\pi$ (multiple) bonds is the deciding factor.
Top states live in oxides and fluorides; where they differ, oxygen usually reaches higher because of multiple bonding.
Contrast With the p-Block
The variability of $d$-block oxidation states differs from $p$-block behaviour in two clean ways, both popular in NEET assertion–reason items.
First, the step size. In the $d$-block, successive states differ by unity ($+2, +3, +4, +5$), because $d$ electrons are removed singly. In the $p$-block they normally differ by two — for example $+3$ and $+5$ for phosphorus or arsenic — reflecting the loss of paired $p$ electrons.
Second, the group trend. In the $p$-block the heavier members favour the lower oxidation state (the inert pair effect). In the $d$-block the opposite is true within a group: heavier members favour higher states. In Group 6, $\ce{Mo(VI)}$ and $\ce{W(VI)}$ are more stable than $\ce{Cr(VI)}$ — so dichromate $\ce{Cr2O7^{2-}}$ in acid is a strong oxidant, whereas $\ce{MoO3}$ and $\ce{WO3}$ are not. Finally, very low and even zero oxidation states appear when ligands with $\pi$-acceptor character are present, as in $\ce{Ni(CO)4}$ and $\ce{Fe(CO)5}$, where the metal is in the $0$ state.
Variable oxidation states in 10 lines
- Variable states arise because $ns$ and $(n-1)d$ orbitals are close in energy, so a variable number of $d$ electrons bonds.
- States differ by unity (e.g. V: +2, +3, +4, +5); contrast the $p$-block, where they differ by two.
- The number of states peaks in the middle: Mn shows +2 to +7; ends (Sc, Ti, Cu, Zn) show fewer.
- Maximum state $=$ sum of $s+d$ electrons up to Mn ($\ce{MnO4^-}$, +7); collapses sharply after Group 7.
- $+2$ is near-universal (loss of two $4s$ electrons); after Sc the $d$ orbitals are more stable than the $4s$.
- $\ce{Mn^2+}$ ($d^5$) and $\ce{Fe^3+}$ ($d^5$) are stabilised by the half-filled shell.
- Both $d^4$: $\ce{Cr^2+}$ is reducing ($\to d^3$); $\ce{Mn^3+}$ is oxidising ($\to d^5$).
- $\ce{2Cu+ -> Cu^2+ + Cu}$: disproportionation driven by the large $-\Delta_{hyd}H^\circ$ of $\ce{Cu^2+}$.
- Highest states live in oxides/fluorides; for Mn the oxide $\ce{Mn2O7}$ (+7) beats $\ce{MnF4}$ (+4) via $\pi$-bonding.
- Higher states are more covalent and more acidic; $+2$/+3 states are ionic and basic.