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

Before we start, one picture fixes the vocabulary we use in every solution.

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

Level 1 — Recognition

Recall Solution L1.1

Rule: H is $+ 1$ , each O is $- 2$ , molecule is neutral, so Cl fills the gap.

$H C l O$ : $(+ 1) + Cl + (- 2) = 0 \Rightarrow Cl = + 1$
$H C l O_{2}$ : $(+ 1) + Cl + 2 (- 2) = 0 \Rightarrow Cl = + 3$
$H C l O_{3}$ : $(+ 1) + Cl + 3 (- 2) = 0 \Rightarrow Cl = + 5$
$H C l O_{4}$ : $(+ 1) + Cl + 4 (- 2) = 0 \Rightarrow Cl = + 7$ Answer: $+ 1, + 3, + 5, + 7$ .

Recall Solution L1.2

Acidity rises with terminal-oxygen count. Full order: $H C l O < H C l O_{2} < H C l O_{3} < H C l O_{4}$ Weakest = $H C l O$ , strongest = $H C l O_{4}$ .

Recall Solution L1.3

To O. The real skeleton is $(O)_{2} C l - O - H$ : the proton always sits on an O–H bond, never straight on chlorine. That O–H is the bond that snaps to release $H^{+}$ .

Acid	total O	$q$	$p K_{a} \approx 8 - 5 q$
$H C l O$	1	0	$8$
$H C l O_{2}$	2	1	$3$
$H C l O_{3}$	3	2	$- 2$
$H C l O_{4}$	4	3	$- 7$
The numbers step down by 5 each time → acid gets stronger each step, matching L1.2.

$p K_{a} = - lo g_{10} K_{a}$ , so a difference in $p K_{a}$ becomes a power of ten in $K_{a}$ : $\frac{K _{a} ( H C l O _{2} )}{K _{a} ( H C l O )} = 1 0^{p K_{a} (H C l O) - p K_{a} (H C l O_{2})} = 1 0^{7.5 - 2} = 1 0^{5.5}$ $1 0^{5.5} \approx 3.2 \times 1 0^{5}$ $H C l O_{2}$ is about $3 \times 1 0^{5}$ (a few hundred thousand) times stronger than $H C l O$ .

Recall Solution L2.3

Skeletons: $H_{2} S O_{4} = (H O)_{2} S (O)_{2} \Rightarrow q = 2$ ; $H_{2} S O_{3} = (H O)_{2} S (O)_{1} \Rightarrow q = 1$ . More terminal O ⇒ smaller $p K_{a}$ ⇒ stronger: $H_{2} S O_{3} < H_{2} S O_{4} (acidity)$ Estimated first $p K_{a}$ : $H_{2} S O_{3} \approx 8 - 5 (1) = 3$ ; $H_{2} S O_{4} \approx 8 - 5 (2) = - 2$ . Same logic as the halogen series.

Level 3 — Analysis

Recall Solution L3.1

Both are $(H O) X (O)_{2}$ ⇒ same $q = 2$ ⇒ Pauling's rule predicts nearly equal. The tie-breaker is electronegativity (Inductive effect): Cl (EN ≈ 3.0) pulls electron density away from the O–H oxygen more than Br (EN ≈ 2.8). Stronger pull ⇒ weaker O–H ⇒ more stable anion. Answer: $H C l O_{3}$ is slightly stronger. Oxygen count is the main driver; electronegativity only decides ties.

Recall Solution L3.2

After ionisation the anion carries a single $- 1$ charge. Resonance (Resonance and charge delocalisation) shares that charge equally over the terminal O atoms:

$C l O^{-}$ : 1 terminal O ⇒ each carries $- \frac{1}{1} = - 1$ (full charge stuck on one atom).
$C l O_{4}^{-}$ : 4 equivalent terminal O ⇒ each carries $- \frac{1}{4}$ . Electrostatic (self-repulsion) energy scales roughly with (charge per atom) $^{2}$ ; spreading it out lowers energy. $C l O_{4}^{-}$ therefore sits much lower in energy → its acid $H C l O_{4}$ releases $H^{+}$ readily.

Recall Solution L3.3

Student B. Oxidising power runs opposite to acidity here (Oxidising power of oxoacids): $oxidising power: H C l O > H C l O_{2} > H C l O_{3} > H C l O_{4}$ A strong oxidiser is one that is eager to grab electrons and drop to a lower oxidation state. Cl in $H C l O$ is only $+ 1$ — a big "hunger" to gain electrons and reach $0$ or $- 1$ . Cl in $H C l O_{4}$ is $+ 7$ and locked in a stable, symmetric, charge-delocalised cage — kinetically sluggish and thermodynamically content, so it is a poor oxidiser though a superb acid.

Level 4 — Synthesis

Recall Solution L4.1

Skeleton $H - O - F$ : the H is on an O–H bond (acidic), and there are no terminal oxygens ⇒ $q = 0$ . $p K_{a} \approx 8 - 5 (0) = 8$ This is the same $q$ as $H C l O$ ( $q = 0$ , $p K_{a} \approx 8$ ). Tie-breaker = electronegativity: F (EN ≈ 4.0) pulls harder than Cl (EN ≈ 3.0), so $H O F$ should be slightly stronger than $H C l O$ — but both are weak. (Experimentally $H O F$ is indeed a weak acid; the model gives the right ballpark.)

Recall Solution L4.2

Tag each with its $q$ (terminal O):

Acid	$q$	note
$H C l O$	0	weakest
$H_{2} S O_{3}$	1
$H B r O_{3}$	2	Br less EN than Cl
$H C l O_{3}$	2	Cl more EN ⇒ edges out $H B r O_{3}$
$H C l O_{4}$	3	strongest
Order by $q$ first, break the $q = 2$ tie by electronegativity:
$H C l O < H_{2} S O_{3} < H B r O_{3} < H C l O_{3} < H C l O_{4}$

Level 5 — Mastery

Recall Solution L5.1

(a) Only the H's on O–H are acidic. Structure has two O–H groups and one P–H (non-acidic, like the Cl–H idea in the parent note). So 2 acidic protons — $H_{3} P O_{3}$ is diprotic, not triprotic. (b) Terminal O count $q = 1$ (the single $P = O$ ). (c) $p K_{a} \approx 8 - 5 (1) = 3$ (experimental first $p K_{a} \approx 1.3$ — right order of magnitude). Wrong assumption: if you thought all three H were on O and read $q$ from "3 oxygens minus 3 OH = 0", you'd predict $p K_{a} \approx 8$ (much too weak) and claim 3 acidic H. Both wrong. Structure beats formula.

Recall Solution L5.2

Acidity $H C l O < H C l O_{2} < H C l O_{3} < H C l O_{4}$ : more terminal O ⇒ the anion $C l O_{x}^{-}$ delocalises its $- 1$ over more atoms ⇒ more stable conjugate base ⇒ stronger acid.
Oxidising power $H C l O > H C l O_{2} > H C l O_{3} > H C l O_{4}$ : fewer O ⇒ lower Cl oxidation state ⇒ Cl is hungrier for electrons ⇒ stronger oxidiser.
Thermal stability $H C l O_{4} > H C l O_{3} > H C l O_{2} > H C l O$ : more O ⇒ higher, symmetric, charge-delocalised cage around Cl ⇒ molecule itself is lower-energy and harder to decompose. One-liner: more oxygens = better anion (acidity ↑) and sturdier cage (stability ↑), but a less electron-hungry Cl (oxidising power ↓).

Recall Solution L5.3

$\frac{K _{a} ( H C l O _{4} )}{K _{a} ( H C l O )} = 1 0^{p K_{a} (H C l O) - p K_{a} (H C l O_{4})} = 1 0^{7.5 - (- 10)} = 1 0^{17.5} \approx 3.2 \times 1 0^{17}$ Physical meaning: at equal concentration, $H C l O_{4}$ ionises essentially completely (a strong acid), while $H C l O$ leaves the vast majority of its molecules intact (a weak acid). The seventeen-orders-of-magnitude gap is what "three extra shared oxygens" buys you.

Wrap-up recall

Recall One-line summary of the whole ladder

Sketch the skeleton → count terminal oxygens $q$ → $p K_{a} \approx 8 - 5 q$ ranks acidity → break ties with electronegativity → remember oxidising power and thermal stability run their own (opposite) ways.

Connections

Parent topic — the theory these drills train.
Pauling rules for oxoacids — the $p K_{a} \approx 8 - 5 q$ engine.
Inductive effect — the electronegativity tie-breaker in L3/L4.
Resonance and charge delocalisation — why spreading the $- 1$ stabilises the anion.
Conjugate acid–base pairs — acidity = base stability.
Oxidising power of oxoacids — the opposite trend in L3/L5.
Oxoacids of sulfur and nitrogen — cross-atom practice in L2/L4.
p-Block — home chapter.

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

Level 1 — Recognition

Level 2 — Application

Level 3 — Analysis

Level 4 — Synthesis

Level 5 — Mastery

Wrap-up recall

Connections

Solve Map