Oxoacids of halogens — HClO, HClO₂, HClO₃, HClO₄ — acidity trend: Question bank

Before we spring any traps, let's lock down the ideas the whole page leans on — each with a picture, so nothing below is a symbol you haven't seen drawn.

The anchors (pictures first)

Anchor 0 — What $K_{a}$ and $p K_{a}$ actually mean

Before any "strong" or "weak" talk, we need the one number that measures acid strength.

Anchor 1 — What a Lewis structure of these acids looks like, and how to count $q$

A Lewis structure is just a dot-and-line map of a molecule: a line is a shared pair of electrons (a bond), and dots are unshared (lone) pairs. In these acids Cl sits in the middle, some O's hang off it, and exactly one O carries the acidic H.

Figure — Oxoacids of halogens — HClO, HClO₂, HClO₃, HClO₄ — acidity trend

What to observe: Cl sits centre; the blue O on the right always carries the acidic H (the O–H group); the pink double-bonded O's on the left are the terminal O's. Count the pink ones — that number is $q$ , printed under each acid.

Anchor 2 — How oxidation states ( $+ 1, + 3, + 5, + 7$ ) are assigned

Anchor 3 — The two effects that stabilise the anion (with arrows)

What to observe: the yellow straight arrows show each terminal O pulling electron density off the central Cl (inductive); the pink curved arrow shows the negative charge hopping around the ring so every O carries only $- \frac{1}{4}$ (resonance). No single O owns the double bond — that is the point of drawing four equal $- \frac{1}{4}$ labels.

Anchor 4 — Where Pauling's rule ( $p K_{a} \approx 8 - 5 q$ ) comes from, and the actual numbers

Formula Building the rule from two pieces

Recall from Anchor 0 that a smaller $p K_{a}$ means a stronger acid. Two pieces build the line:

The intercept, $8$ : an oxoacid with no terminal O ( $q = 0$ ) — a bare O–H on the central atom — sits near $p K_{a} \approx 8$ . That's the starting height.
The slope, $- 5$ — where its magnitude comes from: adding one terminal O does two things at once. (i) Inductive: it withdraws electron density, stabilising the anion. (ii) Delocalisation: it gives one more equivalent site to share the charge, so the charge per O falls (from $- 1$ to $- \frac{1}{2}$ to $- \frac{1}{3}$ …), and electrostatic self-energy scales with the square of charge-per-site — so each halving-ish of charge is a large energy drop. Empirically these two combine to shift the anion's free energy by about $2.303 R T \times 5 \approx 28 kJ mol^{- 1}$ per O, which by $Δ (p K_{a}) = Δ G^{\circ} / (2.303 R T)$ is exactly the $\approx 5$ $p K_{a}$ units. In short: the sign is negative because more O ⇒ stronger acid; the size ( $\approx 5$ ) is set by how much a single terminal O lowers the anion's energy.

Put together: $p K_{a} \approx 8 - 5 q .$

What to observe: the yellow line is Pauling's prediction, dropping a steady 5 units per terminal O; the blue dashed points are the measured values. They track the line closely — confirming both the negative slope and its ~5-per-oxygen size — with $H C l O_{4}$ running a bit stronger than the straight-line estimate.

True or false — justify

TF1. "

H C l O_{4}

is the strongest of the four acids because it has the most H atoms to donate."

False. All four have exactly one acidic H.

H C l O_{4}

wins because its anion

C l O_{4}^{-}

spreads the

- 1

charge over 4 equivalent O's, making the conjugate base very stable.

TF2. "Acidity increases

H C l O < H C l O_{2} < H C l O_{3} < H C l O_{4}

True. Each extra terminal O withdraws electron density and delocalises the anion's charge, so the conjugate base gets progressively more stable and the acid progressively stronger.

TF3. "Oxidising power follows the same order as acidity:

H C l O < H C l O_{2} < H C l O_{3} < H C l O_{4}

False. Oxidising power runs the opposite way:

H C l O > H C l O_{2} > H C l O_{3} > H C l O_{4}

. Low Cl oxidation state (

+ 1

in HClO) is most eager to grab electrons.

TF4. "In

H C l O_{2}

the acidic proton sits directly on chlorine."

False. The proton always sits on an O–H group. The structure is

O = C l - O - H

; a direct Cl–H bond would not ionise usefully.

TF5. "A higher Cl oxidation state makes the molecule bigger and therefore a weaker, sluggish acid."

False. Higher oxidation state (

+ 7

H C l O_{4}

) means more O pulling charge away, so it is the strongest acid. Oxidation state and acidity rise together here.

TF6. "

H C l O

is a weak acid, so it is a weak/harmless oxidiser too."

False. Weak acid ≠ weak oxidiser.

H C l O

(Cl at

+ 1

) is actually the strongest oxidiser of the four.

TF7. "

H C l O_{3}

and

H B r O_{3}

have exactly equal acidity."

False (nearly, but not exactly). Both have

q = 2

so Pauling's rule predicts near-equal, but Cl is more electronegative than Br, making

H C l O_{3}

slightly stronger. Oxygen count dominates; electronegativity is the tie-breaker.

TF8. "

H_{2} S O_{4}

is a stronger acid than

H_{2} S O_{3}

True.

H_{2} S O_{4}

has

q = 2

terminal O's,

H_{2} S O_{3}

has

q = 1

; more terminal O ⇒ more stable anion ⇒ stronger acid — identical logic to the halogen oxoacids (see Oxoacids of sulfur and nitrogen).

TF9. "The extra oxygens in

H C l O_{4}

make the O–H bond stronger."

False. They make it weaker/easier to break: the electronegative terminal O's pull electron density off the O–H oxygen, loosening its grip on the H.

TF10. "Because

H_{2} S O_{4}

has two H's, both come off equally easily."

False.

H_{2} S O_{4}

is diprotic but the two steps differ hugely: the first H leaves a

H S O_{4}^{-}

(strong,

p K_{a 1} \approx - 3

), but the second must leave an already-negative ion

S O_{4}^{2 -}

, which resists the extra charge, so

p K_{a 2} \approx 2

. The halogen oxoacids avoid this issue by being monoprotic (one H only).

Spot the error

SE1. "Since

H C l O_{4}

has 4 oxygens and

H C l O

has 1, the ratio of their acid strengths is 4:1."

Error: acid strength is not proportional to oxygen count. Pauling's rule is about

p K_{a}

(a log scale): each terminal O lowers

p K_{a}

by ~5, i.e. multiplies

K_{a}

by ~

1 0^{5}

, so the real gap is many orders of magnitude, not 4×.

SE2. "

q

for

H C l O_{4}

is 4, so

p K_{a} \approx 8 - 5 (4) = - 12

Error:

q

counts terminal O's = total O minus the one in O–H =

4 - 1 = 3

. Correct:

p K_{a} \approx 8 - 5 (3) = - 7

SE3. "The negative charge in

C l O_{2}^{-}

sits fully on one oxygen, so it's as unstable as

C l O^{-}

Error: in

C l O_{2}^{-}

resonance spreads the charge over 2 equivalent O's (

- \frac{1}{2}

each), less charge per atom than

C l O^{-}

's full

- 1

, so

C l O_{2}^{-}

is more stable and

H C l O_{2}

is the stronger acid.

SE4. "Because

H C l O_{4}

is the strongest acid, it must also be the strongest oxidiser and the least thermally stable."

Error: two of three are wrong.

H C l O_{4}

is the strongest acid but the weakest oxidiser and the most thermally stable; only the oxidiser/stability pair go together (both decrease with more O).

SE5. "

C l O^{-}

has no resonance at all, so

H C l O

isn't an acid."

Error:

C l O^{-}

has the charge localised on its single O (no equivalent O to share with), which makes it a poor (unstable) conjugate base — so

H C l O

is a weak acid, not a non-acid.

SE6. "Using

p K_{a} \approx 8 - 5 q

, water has

q = 0

p K_{a} \approx 8

, meaning water is an acid stronger than

H C l O

Error: the rule predicts

H C l O

(

q = 0

) at

p K_{a} \approx 8

too, and both are weak; the rule is a rough baseline for the

q = 0

case, not a claim that water out-acidifies

H C l O

(experimentally water's

p K_{a} \approx 15.7

, weaker).

Why questions

WHY1. Why does spreading the same

- 1

charge over more oxygens lower the anion's energy?

Concentrated charge stores more electrostatic (self-repulsion) energy; splitting it into

- \frac{1}{4}

fractions across 4 atoms (the curved-arrow picture in Anchor 3) lowers that energy, so the delocalised anion is more stable — see Resonance and charge delocalisation.

WHY2. Why is the acidic H on oxygen and never on chlorine?

O is electronegative enough to leave with the bonding electrons as

O^{-}

, releasing

H^{+}

; a Cl–H bond inside the molecule would not ionise to give a stable anion, so it wouldn't be acidic.

WHY3. Why do we judge acid strength by the anion's stability rather than by the acid molecule itself?

An acid's strength is how far

H A ⇌ H^{+} + A^{-}

sits to the right; a more stable

A^{-}

pulls the equilibrium right — see Conjugate acid–base pairs.

WHY4. Why does the inductive effect of terminal O's both weaken O–H and stabilise the anion?

The straight arrows in Anchor 3 pull electron density toward the terminal O's, leaving the O–H oxygen electron-poor (bond easier to break); once the H leaves, that same pull disperses the resulting negative charge — see Inductive effect.

WHY5. Why does oxidising power fall as oxygen count rises, opposite to acidity?

Oxidising power reflects how readily Cl grabs electrons to drop its oxidation state; low states like

+ 1

(HClO) are hungriest for electrons, while

+ 7

(HClO₄) is already "content" — see Oxidising power of oxoacids.

WHY6. Why does Pauling's rule use

q

(terminal O) and not total O?

Only the terminal O's participate in withdrawing charge and delocalising it in the anion; the O in the O–H group is spectator to the stabilisation, so it's excluded — see Pauling rules for oxoacids.

WHY7. Why is electronegativity only the tie-breaker between

H C l O_{3}

and

H B r O_{3}

, not the main driver?

When

q

differs, charge delocalisation dominates and swamps electronegativity; only when

q

is equal does the smaller electronegativity difference (Cl vs Br) become the deciding factor.

WHY8. Why is the slope of Pauling's rule negative (each O lowers

p K_{a}

A stronger acid has a larger

K_{a}

and therefore a smaller

p K_{a} = - lo g_{10} K_{a}

; since each terminal O makes the acid stronger, it must push

p K_{a}

down — hence the

- 5

per oxygen.

Edge cases

EC1. Edge: what does Pauling's rule predict for an oxoacid with

q = 0

, and does it mean "not acidic"?

It gives

p K_{a} \approx 8

, i.e. a weak acid, not zero acidity.

H C l O

(

q = 0

) fits: weak but real.

EC2. Edge:

H C l O_{4}

's predicted

p K_{a} \approx - 7

but measured is

\sim - 10

. Does the rule fail?

No — Pauling's rule is an approximation (

\pm

a couple of units). It correctly signals a very strong acid; the exact value just runs a bit stronger than the linear estimate.

EC3. Edge: compare

H C l O_{3}

and

H C l O_{4}

for both acidity and thermal stability.

H C l O_{4}

is the stronger acid but

H C l O_{3}

is the more powerful oxidiser and less thermally stable — the two properties split in opposite directions across the pair.

EC4. Edge: for two different central atoms with the same

q

(e.g. Cl vs Br oxoacids), what breaks the tie?

The more electronegative central atom gives the slightly stronger acid (it helps pull charge off the O–H), but the effect is small next to the

q

term.

EC5. Edge: does adding an O–H group (not a terminal O) increase acidity the way a terminal O does?

No. An extra O–H is another potential proton donor but doesn't add a charge-sharing terminal O, so it doesn't boost strength per proton the way a terminal O does. It's the terminal O, not any O, that counts.

EC6. Edge: is

H C l O

safe to handle just because it's a weak acid?

No — it is the strongest oxidiser of the four, so "weak acid" says nothing about hazard; oxidising power is a separate, oppositely-trending property.

EC7. Edge: in a diprotic oxoacid like

H_{2} S O_{3}

, does Pauling's

q

change between the first and second deprotonation?

The geometric

q

(terminal O's) is the same molecule, but the second proton leaves an already-negative anion, so

p K_{a 2}

is always much larger (weaker) than

p K_{a 1}

— the rule best predicts the first ionisation.

Connections

◀ Back to parent: acidity trend
Inductive effect — the electron-withdrawal half of the story
Resonance and charge delocalisation — the charge-spreading half
Conjugate acid–base pairs — strength = base stability
Oxidising power of oxoacids — the opposite trend
Pauling rules for oxoacids — the $p K_{a} \approx 8 - 5 q$ shortcut
Oxoacids of sulfur and nitrogen — same logic for S and N
p-Block — group 17 context

Acid	$q$	$p K_{a} \approx 8 - 5 q$	measured $p K_{a}$	Cl state
$H C l O$	0	$8$	$7.5$	$+ 1$
$H C l O_{2}$	1	$3$	$2.0$	$+ 3$
$H C l O_{3}$	2	$- 2$	$\sim - 1$	$+ 5$
$H C l O_{4}$	3	$- 7$	$\sim - 10$	$+ 7$