Variable Oxidation States & Stability

CHEPosition & General Properties

Transition metals show variable oxidation states because $ns$ and $(n-1)d$ electrons have similar energies and can be lost in steps. Stability of a given state is judged by configuration ( $d^0$ , $d^5$ , $d^{10}$ favoured) and by standard electrode potentials.

Stable lanthanoid states from f-subshell symmetry

Eu^{2+}:[Xe]4f^7 \quad Ce^{4+}:[Xe]4f^0 \quad Gd^{3+}:[Xe]4f^7

half-filled (

4f^7

) and empty (

4f^0

) subshells give the extra stability behind these non-

+3

states.

Reducing/oxidising tendency from product stability

Cr^{2+}(d^4)\rightarrow Cr^{3+}(d^3)\ \text{(reducer)} \qquad Mn^{3+}(d^4)\rightarrow Mn^{2+}(d^5)\ \text{(oxidiser)}

each ion moves toward a more stable configuration —

Cr^{3+}

is the stable

d^3

and

Mn^{2+}

the half-filled

d^5

Variable oxidation states arise because the $(n-1)d$ and $ns$ orbitals are close in energy, so a variable number of $d$ electrons (beyond the $s$ electrons) can participate in bonding.
The highest state ( $Mn$ as $+7$ in $MnO_4^-$ ) appears only with oxygen or fluorine; it is covalent (via M=O bonds), not present in simple ionic salts because no metal can spare seven electrons as an ion.
$Cr^{2+}$ ( $d^4$ ) is a strong reducing agent — it is readily oxidised to the stable $d^3$ $Cr^{3+}$ ; $Mn^{3+}$ ( $d^4$ ) is a strong oxidising agent — it is readily reduced to the stable half-filled $d^5$ $Mn^{2+}$ .
Copper has a positive $E^\circ(Cu^{2+}/Cu) = +0.34$ V because the high sum of its first two ionisation enthalpies and its atomisation enthalpy is not balanced by its hydration enthalpy, so it does not displace $H_2$ from acids.
$Eu^{2+}$ is stabilised by the half-filled $4f^7$ configuration and $Ce^{4+}$ by the empty $4f^0$ ( $[Xe]$ core), explaining their stability despite $+3$ being the usual lanthanoid state.
Across a series the maximum (group-equal) oxidation state rises to a peak (e.g. $Mn$ , $+7$ ) then falls, because after $d^5$ the pairing of $d$ electrons makes them harder to remove.
Lower oxidation states (e.g. $+2$ ) are generally ionic and basic, while higher states (e.g. $+6$ , $+7$ ) are covalent and acidic, appearing as oxides or oxoanions.
Adjacent oxidation states usually differ by one unit (e.g. $+2$ and $+3$ for Fe), unlike $p$ -block elements where states often differ by two — a hallmark of $d$ -block redox chemistry.
The relative stability of $M^{2+}$ vs $M^{3+}$ is set by ionisation enthalpy and the configuration of the product: $Fe^{3+}$ ( $d^5$ ) is favoured over $Fe^{2+}$ ( $d^6$ ), but $Mn^{2+}$ ( $d^5$ ) is favoured over $Mn^{3+}$ ( $d^4$ ).
Manganese shows the widest range of oxidation states ( $+2$ to $+7$ ) in the $3d$ series because it has the most $3d+4s$ electrons ( $3d^5 4s^2$ ) available for bonding.

Where the marks go

Claiming $Mn^{7+}$ exists as a simple ion; the $+7$ state only survives as covalent $MnO_4^-$ where oxygen stabilises it through $\pi$ bonding.
Mixing up which $d^4$ ion is the reducer and which is the oxidiser: $Cr^{2+}$ reduces (goes up to stable $Cr^{3+}$ ), $Mn^{3+}$ oxidises (goes down to stable $Mn^{2+}$ ).
Saying $Fe^{2+}$ is more stable than $Fe^{3+}$ — in aqueous/air conditions $Fe^{3+}$ ( $d^5$ ) is the more stable, so $Fe^{2+}$ is easily air-oxidised.
Attributing $Eu^{2+}$ / $Ce^{4+}$ stability to $d$ -orbitals; it is the $4f$ subshell symmetry ( $4f^7$ and $4f^0$ ) that matters for lanthanoids.
Assuming oxidation states must change in steps of two like the $p$ -block; in the $d$ -block consecutive states differing by one are common.

How the board asks it

Give reasonsrelative stability of $M^{2+}$ vs $M^{3+}$ from $d^5$ configuration
Account for the following: $Mn^{2+}$ is more stable than $Mn^{3+}$ , whereas $Fe^{3+}$ is more stable than $Fe^{2+}$ .
Give reasonshighest oxidation state surviving only as a covalent oxoanion
Explain why manganese exhibits its highest oxidation state of $+7$ only in the oxoanion $MnO_4^-$ and not in a simple ionic salt.
Distinguish $Cr^{2+}$ as reducer vs $Mn^{3+}$ as oxidiser, both $d^4$
$Cr^{2+}$ is a strong reducing agent while $Mn^{3+}$ is a strong oxidising agent, although both are $d^4$ ions. Give reasons for this difference in behaviour.
Define / stateclose energies of $(n-1)d$ and $ns$ orbitals
State why transition elements exhibit variable oxidation states, and name the element of the $3d$ series that shows the maximum number of oxidation states.
Predict the productair oxidation of the less stable lower state
A solution containing $Fe^{2+}$ is left exposed to air and slowly changes. Name the species formed and write the balanced ionic equation for the change.
Give reasonslanthanoid stability from $4f^7$ and $4f^0$ symmetry
Give reasons: $Eu^{2+}$ and $Ce^{4+}$ are unusually stable, even though $+3$ is the common oxidation state of the lanthanoids.

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Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.