Variable oxidation states — reasons: Visual walkthrough

We only need one idea to start: an oxidation state is just how many electrons an atom has handed away. Lose one → $+ 1$ . Lose two → $+ 2$ . That's the whole vocabulary. Everything else we earn as we go.

Step 1 — Electrons live on "shelves" at different heights

WHAT. Picture every electron in an atom as a book sitting on a shelf. The height of the shelf = the energy of the electron. A book high up is loosely held (easy to take away); a book low down is gripped tightly (hard to take away).

WHY this picture. "Oxidation state" means removing electrons. To know how many you can remove, you must know how hard each one is to lift off its shelf. Height = difficulty. So the shelf diagram is the whole problem, drawn.

PICTURE. Look at the figure. The red book is the easiest one to remove — the highest shelf. The gap above a book tells you how far you'd have to reach for the next book; the gap below tells you how deep the next electron sits.

Step 2 — Sodium: one lonely high book, then a locked basement

WHAT. Sodium's configuration is $[Ne] 3 s^{1}$ : exactly one book on the top shelf ( $3 s$ ). Directly below it there is a big empty stretch, and then a packed, deep shelf — the filled $2 p$ of neon.

WHY this matters. After you take the single $3 s$ book (cheap, that's $I E_{1}$ ), the next book you'd have to grab lives far down in the filled inner shell. Reaching that far costs a huge amount — $I E_{2}$ for sodium is about nine times $I E_{1}$ . Nothing an atom does later (bonding, dissolving) pays back that much. So sodium simply stops after one.

PICTURE. The red arrow shows the tiny, affordable jump to remove the $3 s$ electron. The black arrow shows the cliff-drop to the next electron — the "locked basement." One easy, then a wall.

Step 3 — Manganese: a whole row of books at almost the same height

WHAT. Manganese is $[Ar] 3 d^{5} 4 s^{2}$ . That's 7 books sitting above the argon core: two on the $4 s$ shelf and five on the $3 d$ shelf. The crucial fact — the entire point of the page — is that the $3 d$ shelf and the $4 s$ shelf are at nearly the same height.

WHY this changes everything. With sodium, book #2 was in a basement. With manganese, books #2, #3, #4, … are all on shelves right next to each other. So each successive removal costs only a little more than the last — no cliff. There is no wall to stop the process early.

PICTURE. All 7 removable books (red) cluster in a narrow height band; the deep argon core (black) is far below. Compare the flat cluster here to sodium's lone book + cliff in Step 2.

Step 4 — The successive-IE staircase: gentle for Mn, a cliff for Na

WHAT. Plot the cost of each removal in order: $I E_{1}, I E_{2}, I E_{3}, \dots$ For sodium this is a staircase with one giant step. For manganese it is a gently rising ramp.

WHY plot it this way. We want to see "how far can I keep removing electrons?" The answer is: keep going while the next step is still affordable. A cliff (Na) means stop after step 1. A gentle ramp (Mn) means many stops are all reachable — many oxidation states.

PICTURE. Two curves. The black curve (Na) shoots up after the first electron. The red curve (Mn) climbs slowly — so we can stop at $+ 2$ , $+ 3$ , $+ 4$ … anywhere along it.

Step 5 — But cost alone isn't the whole story: the payback

WHAT. Spending $I E$ to make an ion is only worth it if something pays you back. When the ion joins a crystal (lattice), forms bonds, or dissolves in water, energy is released. That release is the payback.

WHY we need this term. An isolated $Mn^{7 +}$ ion floating in space would be absurdly expensive. It only exists inside $MnO_{4}^{-}$ because oxygen's electrons and strong bonding return most of that energy. So the real question is not "is $I E$ small?" but "is $I E$ smaller than what I get back?"

PICTURE. A see-saw / balance. On the left pan: energy you spend ( $\sum I E_{i}$ , the ionisation staircase, plus atomising the metal). On the right pan: energy you get back (lattice energy $U$ , bond/hydration energy). The oxidation state forms only when the right pan wins.

Formula The balance sheet (Born–Haber style)

$Δ H_{form} = turn metal to gas Δ H_{atom} + remove n electrons i = 1 \sum n I E_{i} + make the anion Δ H_{anion} + build crystal, < 0 U_{lattice}$

$Δ H_{atom}$ — cost to free the metal atom (sublimation). Always positive (you pay).
$\sum_{i = 1}^{n} I E_{i}$ — the staircase from Step 4, added up to the state $+ n$ . Positive.
$Δ H_{anion}$ — energy to build the negative partner (e.g. electron affinity of O, F).
$U_{lattice}$ — energy released when ions snap into a crystal. Negative (this is the payback).

The oxidation state $+ n$ exists when $Δ H_{form} < 0$ , i.e. $\sum_{i = 1}^{n} I E_{i} < ∣ U_{lattice} ∣ + (other release terms)$ Since Mn's $\sum I E_{i}$ rises slowly (Step 4), this inequality stays true for many values of $n$ → many stable states. See Born–Haber Cycle & Lattice Energy.

Step 6 — Why the maximum OS rises then falls across the row

WHAT. Walk Sc → Ti → V → Cr → Mn → Fe → … → Zn. The highest reachable OS climbs to a peak at Mn (+7), then falls to Zn (+2).

WHY the rise, then why the fall.

Rise (Sc→Mn): each new element adds one more accessible $d$ electron of the same near-degenerate height, so the ceiling goes up by one: Sc $+ 3$ , Ti $+ 4$ , V $+ 5$ , Cr $+ 6$ , Mn $+ 7$ . Max OS = total $(3 d + 4 s)$ electrons.
Fall (Mn→Zn): past Mn, the nuclear charge keeps growing but the added electrons go into already-occupied $d$ orbitals (pairing). The rising effective nuclear charge drags the $3 d$ shelf downward — the books get gripped harder, $I E$ shoots up, and the top states become unpayable. By Zn ( $3 d^{10} 4 s^{2}$ ) the $d$ shelf is a full, deep, locked basement again — only the two $4 s$ books come off → $+ 2$ only.

PICTURE. A tent/mountain: max OS on the vertical axis rising to a red peak at Mn, then sliding down to Zn. Annotated with why on each slope.

Step 7 — The degenerate/edge cases: Sc, Zn, and the "iodide can't"

WHAT. Three boundary situations complete the picture.

WHY include them. A rule you can't apply at the edges isn't understood yet. We check the ends of the series and the choice of partner.

PICTURE. Three mini-panels:

Sc $[Ar] 3 d^{1} 4 s^{2}$ : only 3 removable books → $+ 3$ and essentially nothing else (empty $d$ after, a stable "clean" shelf). One state, but it's the maximum one.
Zn $[Ar] 3 d^{10} 4 s^{2}$ : the $d$ shelf is full and deep (locked), so only the two $4 s$ books leave → $+ 2$ only. This is the right-hand mirror of sodium.
Partner matters: the same high-OS metal can only survive next to a partner that pays back enough. Small, greedy O and F (red) stabilise high OS; big, easily-robbed I⁻ (black) gets oxidised to $I_{2}$ instead — so $MnO_{4}^{-}$ and $VF_{5}$ exist but $MnI_{7}$ never does.

The one-picture summary

Everything on one canvas: on the left, sodium — a lone high book over a cliff → one state. On the right, manganese — a flat cluster of 7 books at nearly one height → a ramp of cheap removals → many states, provided the payback see-saw tips negative. The red thread running through is always the same: how big is the jump to the next electron? Small jump ⇒ many oxidation states; big jump ⇒ one.

Recall Feynman retelling — the whole walkthrough in plain words

Think of each atom as a bookshelf. Every electron is a book, and how high its shelf is tells you how easy it is to take away. Sodium keeps one book on a tall shelf, and the next book down is locked in a deep basement — so sodium can only ever give away that one book. It's always $+ 1$ ; nothing can afford to dig into the basement. Manganese is different. Its top seven books all sit on shelves at almost the same height (that's the $3 d$ and $4 s$ being close in energy). So you can hand over two, three, four… up to seven books, each only a little harder than the last — like walking up a gentle ramp instead of hitting a wall. That's variable oxidation states. But there's a catch: giving away books costs energy. You only do it if the atom gets paid back — by snapping into a crystal, forming strong bonds, or dissolving. That's the see-saw. High oxidation states (many books gone) need a generous partner to pay you back, and only small, grabby atoms like oxygen and fluorine are generous enough — which is why the very highest states show up as oxides and fluorides, never as iodides (big iodide just gets robbed and turns into $I_{2}$ ). Finally, as you walk across the row, the pile of same-height books grows up to manganese (peak $+ 7$ ), then the growing nuclear pull yanks the $d$ shelf down into basement territory again, so the ceiling falls — all the way to zinc, whose $d$ shelf is full and locked, leaving only two $4 s$ books and a fixed $+ 2$ . Same shelf → flexible; cliff → stuck. That's the whole story.

Connections

← Back to parent topic
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

Visual walkthrough — Variable oxidation states — reasons