Variable oxidation states — reasons: Question bank

This deck sharpens the ideas built in the parent note: the closeness of $(n - 1) d$ and $n s$ energies, the energy-repayment condition, and the trends across the $3 d$ series.

Figure — Variable oxidation states — reasons

Definition The energy words we lean on (with signs and units)

All energies below are enthalpies in kJ/mol, and the sign tells you the direction:

$I E$ = ionisation enthalpy — the energy you must spend to pull one electron off, so it is always a positive cost ( $I E > 0$ ). $I E_{1}$ removes the first electron, $I E_{2}$ the second (from the now-positive ion), and so on; higher subscript = harder, because you are prying an electron off an ever more positive ion.
$Δ H_{atom}$ = atomisation (sublimation) enthalpy — positive cost to turn the solid metal into free gaseous atoms.
$Δ H_{eg}$ = electron-gain (anion-formation) enthalpy — energy change when the non-metal grabs the electrons to become the anion; usually a release (negative), sometimes a small cost.
$U_{lattice}$ = lattice energy — the energy released (so negative) when gaseous ions snap together into an ionic solid. In solution the parallel release term is hydration energy; in a covalent molecule it is covalent bond-formation energy. These three are alternative "pay-back" terms depending on the setting — never mix them up as one thing.

The gentle staircase of successive $I E$ values is the whole story: no cliff means many states are affordable. The next figure shows how the realised maximum oxidation state rises to Mn and then collapses.

For the middle metals there is an extra pay-back term worth knowing — ligand field / crystal field stabilisation energy (CFSE) — sketched below and used in the "Why" section.

True or false — justify

True or false: Every transition metal shows variable oxidation states.

False. Scandium is essentially only

+ 3

(it just empties

3 d^{1} 4 s^{2}

) and zinc only

+ 2

(

3 d^{10}

is filled and stable) — the variety peaks in the middle of the series, not at the ends.

True or false: The maximum oxidation state equals the group number for the whole

3 d

series.

False. It holds only up to Mn (

+ 7

= group 7). After Mn rising effective nuclear charge (see Effective Nuclear Charge & Shielding) buries the

d

-electrons, so Fe (group 8) usually maxes at

+ 3

, not

+ 8

True or false: An oxidation state is stable purely because its ionisation enthalpy is low.

False. A state exists when the ionisation cost is repaid by lattice, hydration, or covalent bond energy — it is a balance, not the

I E

alone. A moderately high

I E

is fine if a strong release term pays it back.

True or false: s-block metals have no

d

orbitals, so they cannot vary their oxidation state.

False. They do have empty

d

orbitals, but those sit in a higher shell at much higher energy. The real blocker is the huge gap down to the filled inner shell, making the next

I E

unrecoverable.

True or false: Higher oxidation states of a metal give more ionic compounds.

False. Higher OS concentrates charge on the small metal centre and pulls electron density inward, giving more covalent bonding — that is why

Mn_{2} O_{7}

is covalent and acidic while

MnO

is ionic and basic.

True or false: Because

3 d

and

4 s

are close in energy, all successive ionisation enthalpies of a

d

-metal are equal.

False. They rise gradually, not stay equal — each removed electron leaves a more positive ion, so the next removal is harder. "Close in energy" only means the rise is gentle enough to be repayable, not zero.

True or false: A metal in a high oxidation state can form a stable iodide as easily as a stable fluoride.

False. Iodide

I^{-}

is large and easily oxidised; a high-OS metal would oxidise it to

I_{2}

and drop to a lower state. Only small, electronegative O and F stabilise high OS (contrast

VF_{5}

with the non-existent

VI_{5}

True or false: Copper only ever shows

+ 2

in the

3 d

series.

False. Cu shows both

+ 1

(cuprous, e.g.

Cu_{2} O

, CuCl) and

+ 2

(cupric). The

+ 1

state is stabilised because removing one electron from

3 d^{10} 4 s^{1}

leaves the specially stable fully filled

3 d^{10}

configuration.

Spot the error

Spot the error: "Mn shows the widest range because it has the most protons in the

3 d

series."

The reason is not proton count but that Mn (

3 d^{5} 4 s^{2}

) has 7 valence electrons, all singly occupied/accessible — every one can be engaged, giving

+ 2

through

+ 7

. Zn has more protons yet only

+ 2

Spot the error: "Fe

^{3 +}

is more stable than Fe

^{2 +}

everywhere, so Fe always prefers

+ 3

Stability of

+ 2

versus

+ 3

depends on the environment — the ion, the ligand/counter-anion, and the medium — as detailed in Stability of +2 and +3 states (Fe, Mn). In some conditions

Fe^{2 +}

is favoured; there is no universal winner.

Spot the error: "Sodium could show

+ 2

if we just supplied enough energy for the second ionisation."

You can remove a second electron, but it comes from the filled

2 p

inner shell, so

I E_{2}

is enormous and no lattice/hydration term can repay it. The compound would then have

Δ H_{form} > 0

, so it does not form — energetics, not mere possibility, decides.

Spot the error: "The energetics condition only involves ionisation enthalpy and lattice energy."

It also includes atomisation

Δ H_{atom}

of the metal and the electron-gain term

Δ H_{eg}

— all four terms of

Δ H_{form}

shown in the formula above. Neglecting these breaks the Born–Haber Cycle & Lattice Energy balance and can wrongly predict a state stable or unstable.

Spot the error: "Zn is a transition metal that simply chooses not to use its

d

-electrons."

Its

3 d^{10}

subshell is completely filled and low-lying, so those electrons are not energetically accessible for bonding — only the two

4 s

electrons are. Zn is often excluded from the "true transition metal" definition for exactly this reason (see d-Block Overview).

Spot the error: "Cu

^{+}

is more stable than Cu

^{2 +}

in water because it has the tidy

3 d^{10}

configuration."

In aqueous solution the very large hydration energy of the smaller, higher-charged

Cu^{2 +}

ion overturns the configuration advantage, so

Cu^{+}

actually disproportionates to

Cu

and

Cu^{2 +}

— a reminder that stability is set by the full energy balance, not electron configuration alone.

Spot the error: "Because

MnO_{4}^{-}

exists, Mn

^{7 +}

ions float around freely in the solution."

There is no bare

Mn^{7 +}

cation; the

+ 7

is a formal oxidation state achieved through strong covalent Mn–O bonds. Such a highly charged bare cation would have astronomically high

I E

and could never exist as a free ion.

Why questions

Why do successive ionisation enthalpies of transition metals rise only gently?

Because the electrons come out of

(n - 1) d

and

n s

orbitals that are close in energy, so each removal costs only a little more than the last — no sudden shell jump interrupts the sequence. See Ionisation Enthalpy of Transition Metals.

Why does the highest attainable oxidation state fall after Mn across the

3 d

series?

Effective nuclear charge keeps rising, pulling

3 d

electrons in tightly and forcing pairing, so their ionisation enthalpies climb steeply — the top states can no longer be repaid by bonding energy and become unreachable.

Why are the

+ 2

and

+ 3

states so common for the middle metals (Fe, Co, Ni)?

Losing the two

4 s

electrons gives

+ 2

cheaply, and pulling one further

3 d

electron to reach

+ 3

is still moderate in cost. In coordination compounds an extra pay-back term — crystal/ligand field stabilisation energy (CFSE), shown in the figure above — further favours these

d

-electron counts, while going higher would disrupt the stable

d

configuration at a price hydration/lattice cannot repay.

Why does CFSE make

+ 2

+ 3

ions of the middle metals extra stable?

When ligands surround the ion, the five

d

orbitals split into a lower and a higher set; electrons that drop into the lower set release energy (CFSE), an extra negative term in

Δ H_{form}

. Partly filled

d

configurations (typical of

+ 2

+ 3

Fe, Co, Ni) reap the most CFSE, tipping the energy balance in their favour.

Why does copper show a

+ 1

state at all, unlike its neighbours?

Copper's ground configuration is

3 d^{10} 4 s^{1}

; losing just the lone

4 s

electron leaves the extra-stable, fully filled

3 d^{10}

core, so

+ 1

is energetically accessible — a low-OS quirk absent in Fe, Co, Ni whose

d

shells are not full.

Why do oxides and fluorides stabilise the highest oxidation states?

O and F are small and strongly electronegative, forming many strong bonds that release enough energy to repay the large ionisation cost of a high-OS centre — larger, less electronegative partners cannot.

Why does the s-block stop at one oxidation state while the d-block does not?

The s-block's next electron lies in a deep, filled inner shell (a huge

I E

that is unrecoverable), so it stops dead; the d-block's next electrons lie in near-degenerate

(n - 1) d

n s

orbitals, so removal stays affordable across several states.

Why are high-oxidation-state oxides acidic while low-OS oxides of the same metal are basic?

Higher OS pulls electron density strongly toward the metal, making the M–O bonding covalent and the oxide acidic; low OS leaves the bonding ionic and the oxide basic — the trend is developed in Acidic vs Basic character of Oxides across OS.

Edge cases

Edge case — Sc: why does it show essentially only

+ 3

despite being a transition element?

After losing all three valence electrons (

3 d^{1} 4 s^{2}

) it reaches the stable argon core; a fourth electron would come from that core at unrecoverable cost, so

+ 3

is the only realised state.

Edge case — Cu: why is

+ 1

its low-OS boundary while

+ 2

is its everyday state?

Cu^{+}

(

3 d^{10}

) is stabilised by the filled shell and favoured in solids and low-hydration settings, but in water the huge hydration energy of

Cu^{2 +}

wins, so the boundary between the two states literally shifts with the medium.

Edge case — Zn: why is its single

+ 2

state not the same "fixed OS" reason as sodium's?

Both are fixed, but for opposite structural reasons: Na is limited because its next electron is in a deep inner shell, whereas Zn is limited because its

3 d^{10}

subshell is already filled and stable — only the

4 s^{2}

pair is available.

Edge case — the very top of a group: does the theoretical maximum

+ 8

(group 8) ever appear?

Only in exceptional, small/electronegative environments (e.g. certain Os/Ru oxides down the group), and essentially not for Fe itself, whose rising

Z_{eff}

makes

+ 8

energetically unreachable — group number is a ceiling, not a guarantee.

Edge case — what happens to the energetics inequality when the anion is large and weakly bonding (like I⁻)?

The release term (

U_{lattice}

or bond energy) is too small to repay a large

\sum I E_{i}

, so high oxidation states give

Δ H_{form} > 0

and do not form; the metal settles into a low OS or oxidises the iodide instead.

Edge case — is a "stable" oxidation state the same as an "existing" compound?

Not always the same wording, but the criterion is identical: the state occurs when

Δ H_{form}

is negative. A thermodynamically unfavourable state simply is not isolated, even if formally drawable.

Recall One-line summary of every trap

Variable OS is an energy-balance phenomenon riding on the near-degeneracy of $(n - 1) d$ and $n s$ ; group number caps but doesn't guarantee the maximum; O/F stabilise highs, big/soft anions and water favour lows; high OS means covalent and acidic, not "more metallic." Sc, Cu, Zn and Na sit at the boundaries for four different structural reasons, and CFSE gives the middle $+ 2$ / $+ 3$ ions an extra nudge.

Connections

d-Block Overview
Ionisation Enthalpy of Transition Metals
Born–Haber Cycle & Lattice Energy
Stability of +2 and +3 states (Fe, Mn)
Acidic vs Basic character of Oxides across OS
Effective Nuclear Charge & Shielding

Question bank — Variable oxidation states — reasons