Variation of oxidation state across the table: Question bank

Before we start, one reminder of the two words that decide every trap below:

Oxidation state = the pretend charge an atom would have if we handed each shared electron pair entirely to the more electronegative partner (see Electronegativity).
Valence electrons $n_{v}$ = the electrons in the outermost shell, the only ones cheap enough to give away (see Electronic Configuration).

True or false — justify

Each statement is either a rule stretched too far or a truth in disguise. Decide, then give the reason.

Oxygen always has oxidation state

- 2

False. Oxygen only takes

- 2

when it is the most electronegative atom present. In

OF_{2}

fluorine wins, so O is

+ 2

; in peroxides (

H_{2} O_{2}

) two oxygens share, giving

- 1

each; in superoxides (

KO_{2}

) it averages

- \frac{1}{2}

Fluorine's maximum positive oxidation state equals its group valence,

+ 7

False. Nothing is more electronegative than F, so no atom can pull electrons away from it. F is stuck at

- 1

(in compounds) or

0

(as

F_{2}

) — it never reaches its "group max".

The maximum positive oxidation state of a main-group element always equals its group number.

False as a universal law. It holds across period 3 (Na

+ 1

→ Cl

+ 7

), but O and F never reach it (no more-EN partner), and the Inert Pair Effect makes heavy p-block elements (e.g. Pb, Tl, Bi) avoid their group max.

Removing valence electrons is limited because the next electrons live in a full, tightly bound inner shell.

True. This is why a maximum positive state exists at all — once the valence shell is stripped, further removal costs far more energy than any chemical reaction supplies (see Ionisation Energy).

A transition metal shows many oxidation states because its valence electrons are spread over orbitals of very different energies.

False — it's the opposite. The

3 d

and

4 s

orbitals are close in energy, so electrons peel off one at a time in small steps. Main-group

s

and

p

energies are far apart, so states there tend to jump by 2.

The inert-pair effect means heavy elements simply have fewer valence electrons.

False. The electrons are still there; the

n s^{2}

pair is just poorly shielded (by intervening

d

f

electrons) and so held too tightly to bond. The count is unchanged — only the willingness to use them drops.

Down group 14 the most stable oxidation state falls from

+ 4

(C, Si) to

+ 2

(Pb).

True. This is the inert-pair pattern: the stable state drops by 2 descending the group, which is why

PbCl_{2}

is stable while

PbCl_{4}

decomposes.

An oxidation state must always be a whole number.

False. It's an average over identical atoms. In

N_{2} O

, superoxides (

- \frac{1}{2}

), or

Fe_{3} O_{4}

, the per-atom average is fractional even though each real atom carries an integer charge.

Spot the error

Each line contains one flawed step. Name it and fix it.

"In

H_{2} O_{2}

, sum

= 2 (+ 1) + 2 (- 2) = - 2 \neq = 0

, so the formula must be wrong."

The error is assuming O

= - 2

. Here the two oxygens bond to each other, so neither out-competes the other for that shared pair — each O is

- 1

. Then

2 (+ 1) + 2 (- 1) = 0

. ✓

"Sulphur's max is

+ 6

, so in

H_{2} S

sulphur is

+ 6

Wrong extreme. In

H_{2} S

, S is more electronegative than H, so it gains electrons:

x + 2 (+ 1) = 0 \Rightarrow x = - 2

, its most-negative state

- (8 - 6)

, not its max.

"For

Cr_{2} O_{7}^{2 -}

2 x + 7 (- 2) = 0 \Rightarrow x = + 7

The right side is wrong. The sum equals the ion charge

- 2

, not

0

2 x - 14 = - 2 \Rightarrow x = + 6

. Also

+ 7

is impossible — Cr has only 6 removable electrons (

3 d^{5} 4 s^{1}

"Chlorine's minimum is

- (8 - 7) = - 1

, so chlorine can never be positive."

The minus doesn't forbid the plus. With a more-EN partner (O or F) Cl gives electrons up to

+ 7

, as in

HClO_{4}

. An element's range runs from its minimum to its maximum; both are allowed.

"Nitrogen in

HNO_{3}

+ 1 + x + 3 (- 2) = 0 \Rightarrow x = + 6

Arithmetic slip:

+ 1 + x - 6 = 0 \Rightarrow x = + 5

. And

+ 5

is nitrogen's true max (

n_{v} = 5

);

+ 6

would exceed its valence electrons, so the wrong answer is also physically impossible.

"Aluminium can reach

- 5

because its minimum is

- (8 - 3) = - 5

The min formula only applies to non-metals completing an octet. Al is a metal with low ionisation energy — it loses electrons (

+ 3

) and effectively never gains 5. The floor formula is for atoms that want to fill their shell.

"Manganese's max is

+ 7

, which comes from its

3 d^{5}

electrons only."

Incomplete. The

+ 7

counts both the

4 s^{2}

and

3 d^{5}

electrons:

2 + 5 = 7

. Forgetting the

4 s

pair under-counts every transition-metal maximum.

Why questions

The "how" you can memorise; the "why" is the understanding. Answer in one breath.

Why does the maximum positive oxidation state increase by exactly

+ 1

across a period?

Each step right adds one valence electron while staying in the same shell. One more electron to hand off means the ceiling rises by one, giving Na

+ 1

→ Cl

+ 7

Why can oxygen almost never be positive, yet chlorine easily reaches

+ 7

Positive means donating electrons to a more-EN partner. Almost nothing out-pulls O (only F), so it rarely donates; Cl has many more-EN partners (O, F) willing to accept, so its high positive states form readily.

Why do metals on the left show only positive oxidation states?

Their Ionisation Energy is low (electrons leave easily) and their Electron Affinity is weak (they don't want to gain many). So they give but don't grab — positive states only.

Why do p-block elements in the middle show the widest variety of states?

They can either lose their

p

and

s

valence electrons (giving several high positive states) or gain a few to complete the octet (giving negative states), so both directions are open to them.

Why do transition-metal states usually differ by 1 while main-group states jump by 2?

Transition metals draw from near-degenerate

3 d

4 s

orbitals, removable one at a time. Main-group atoms use

s

and

p

levels that are far apart, so a stable configuration is reached by removing a pair, hence steps of 2 (see Transition Metal Chemistry).

Why does the inert-pair effect strengthen as you go down a group?

Each row down adds filled

d

(and later

f

) shells that shield poorly. The

n s^{2}

pair feels more of the nucleus, binds tighter, and grows more reluctant to bond — so the effect intensifies with depth.

Why is knowing an element's oxidation states useful before you even see a reaction?

It predicts formulas (why

NaCl

not

NaCl_{2}

), tells you which species can be oxidised or reduced, and lets you balance Redox Reactions by tracking oxidation-number change.

Edge cases

Where the tidy rules bend. These are the boundaries a good student checks before trusting a formula.

What is oxygen's oxidation state in

OF_{2}

, and why the sign flip?

+ 2

. Fluorine is more electronegative than oxygen, so F takes the negative number (

- 1

each) and O is forced positive:

x + 2 (- 1) = 0 \Rightarrow x = + 2

In the superoxide

KO_{2}

, what is each oxygen's oxidation state?

- \frac{1}{2}

. K is

+ 1

, so the two oxygens together carry

- 1

, averaging

- \frac{1}{2}

per atom — a legitimate fractional average over the

O_{2}^{-}

unit.

What is the average oxidation state of Fe in magnetite

Fe_{3} O_{4}

+ \frac{8}{3}

. From

3 x + 4 (- 2) = 0 \Rightarrow x = + \frac{8}{3}

; physically it's a mix of

Fe^{2 +}

and

Fe^{3 +}

(one

+ 2

and two

+ 3

), whose average is

\frac{8}{3}

What oxidation state does any element take as a free element, and why?

0

. There is no more-electronegative partner — bonds are between identical atoms, so no one wins the shared electrons. Hence

O_{2}

Na

S_{8}

are all

0

Nitrogen ranges from

- 3

+ 5

; what is its state in

N_{2} O

and is it "allowed"?

+ 1

(from

2 x - 2 = 0

). Perfectly allowed — it's an average lying between the extremes, showing that intermediate/low positive states are common for versatile nitrogen.

Fluorine has 7 valence electrons but its only negative state is

- 1

; why not

- (8 - 7)

being larger?

- (8 - 7) = - 1

is its minimum — it needs just one electron to complete the octet. It cannot go more negative because it has no room, and it cannot go positive because nothing out-pulls it.

Can a noble gas like Xe ever have a positive oxidation state?

Yes, at the boundary of the rule. Bonded to the most electronegative atoms (F, O), Xe donates electrons — e.g.

+ 2

XeF_{2}

, up to

+ 6

XeF_{6}

— even though its shell is already full.

Recall One-line summary of every trap

The single question behind all of these ::: "Who is the more electronegative atom here, and how many valence electrons can actually move?" — answer that honestly and no trap can catch you.

Connections

Electronegativity — decides who takes the negative number in every trap above.
Ionisation Energy — why left-side metals stay positive.
Electron Affinity — why right-side non-metals reach negative states.
Electronic Configuration — source of $n_{v}$ and $d$ – $s$ spacing.
Inert Pair Effect — the heavy-element exceptions.
Transition Metal Chemistry — the "steps of 1" edge cases.
Redox Reactions — where correct oxidation states pay off.