Variable oxidation states — reasons — Chemistry notes

WHAT is happening

WHAT to compare:

Block	Valence config	OS pattern	Why
s (Na)	$3s^1$	only +1	next e⁻ in deep $2p$ → too costly
d (Mn)	$3d^5 4s^2$	+2 … +7	$3d$ ≈ $4s$ energy → easy stepwise loss

WHY transition metals do it (first-principles)

Formula Energetics condition for a stable oxidation state

Derive from a Born–Haber-style balance for $M \to M^{n+}$ in a compound: $\Delta H_{\text{form}} = \underbrace{\Delta H_{\text{atom}}}_{\text{sublimation}} + \underbrace{\sum_{i=1}^{n} IE_i}_{\text{ionisation}} + \underbrace{\Delta H_{\text{anion}}}_{\text{e.g. EA}} + \underbrace{U_{\text{lattice}}}_{\text{(negative)}}$ Why each term? To put the metal in OS $+n$ you must (1) atomise it, (2) remove $n$ electrons (sum of first $n$ ionisation enthalpies), (3) form the anion, (4) assemble the lattice (releases energy, so $U<0$ ). An oxidation state $n$ is stable (occurs) when $\Delta H_{\text{form}}<0$ , i.e. $\sum_{i=1}^{n} IE_i \;<\; |U_{\text{lattice}}| + (\text{other release terms})$ Because for d-metals $\sum IE_i$ grows slowly, the inequality holds for several values of $n$ .

Figure — Variable oxidation states — reasons

HOW to read the trends (the 80/20 core)

Steel-manned mistakes

Recall Feynman: explain to a 12-year-old

Imagine each atom has electrons stacked like books on shelves. For sodium, the top book sits alone on a high shelf and the next book is in a locked basement — so sodium can only ever hand over that one book (always +1). For iron, the top few books all sit on shelves at almost the same height, so iron can give away 2, or 3, or sometimes more — whatever the situation pays it to do. That "same height" is the $3d$ and $4s$ being close in energy, and that's why transition metals are so flexible (variable oxidation states).

Active recall

Fundamental reason transition metals show variable oxidation states

(n-1)d

and

ns

orbitals are close in energy, so successive IEs rise gradually and extra electrons can be removed cheaply.

Why does the s-block NOT show variable OS

After the valence

ns

electrons, the next electron is in a much deeper filled shell → next IE is huge → not recoverable.

Which element has the widest OS range in the 3d series and why

Mn (

3d^54s^2

), because all 7 valence electrons (2×4s + 5×3d) are unpaired/accessible → +2 to +7.

General condition for an oxidation state to be stable

\sum IE_i

must be repaid by lattice/bond/hydration energy, i.e.

\Delta H_{\text{form}}<0

Why does max OS fall after Mn across the series

Rising effective nuclear charge tightens

3d

electrons (pairing, high IE), so highest states become unreachable.

Why are highest oxidation states found mainly in oxides and fluorides

O and F are small, very electronegative → can stabilise many shared/lost electrons via strong bonds.

Why don't high-OS iodides exist (e.g. no MnI₇)

I⁻ is large and easily oxidised; a high-OS metal would oxidise it to I₂, so only low OS iodides form.

Highest OS of Zn and why so limited

+2 only;

3d^{10}

is filled and stable, so only the two

4s

electrons are available.

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

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, transition metals (Fe, Mn, Cr…) ka khaas funda yeh hai ki inke $(n-1)d$ aur $ns$ orbitals ki energy lagbhag same hoti hai. Isliye jab $4s$ ke do electron nikal jaate hain (+2), to ek aur $3d$ electron nikaalna bahut zyada mehnga nahi padta — bas thodi si extra ionisation enthalpy. Aur yeh thodi si energy lattice energy / bond energy / hydration energy se "wapas" mil jaati hai. Isi wajah se ek hi metal +2, +3, +4… kai oxidation states dikhata hai. Sodium jaisa s-block metal yeh nahi kar sakta, kyunki uske top electron ke baad agla electron ekdum andar wale bhare hue shell se nikalna padta hai — uski IE itni high hoti hai ki koi energy use repay nahi kar paati. Isliye Na hamesha +1 hi rehta hai.

Series mein left se right chalo: Sc(+3), Ti(+4), V(+5), Cr(+6), Mn(+7) — yahan tak max OS badhti jaati hai, kyunki $(3d+4s)$ ke saare electrons available hote hain (Mn ke paas 7 hain, isliye highest +7). Mn ke baad effective nuclear charge badhne se $3d$ electrons andar ki taraf tightly bind ho jaate hain (pairing + high IE), to top states reach nahi hote — max OS girne lagti hai, aur Zn pe sirf +2 bachta hai ( $3d^{10}$ full ho gaya).

Ek practical point yaad rakho: highest oxidation states sirf O aur F ke saath milte hain (Mn₂O₇, VF₅), kyunki yeh chote aur strongly electronegative atoms zyada electrons stabilise kar sakte hain. Bade halides jaise I⁻ high OS ko handle nahi kar paate — woh khud oxidise ho jaate hain, isliye iodides mostly low OS mein hote hain. Bas yeh do ideas — "energy gap chhota hai" aur "kaun sa partner high OS sambhaal sakta hai" — pakad lo, pura topic clear ho jayega.

Variable oxidation states — reasons

WHAT is happening

WHY transition metals do it (first-principles)

HOW to read the trends (the 80/20 core)

Steel-manned mistakes

Active recall

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — d-Block (Transition Metals) & f-Block

Connections