Worked examples — Phosphorus allotropes (white, red, black); P₄O₆, P₄O₁₀; oxoacids of P (H₃PO₃ vs H₃PO₄ basicity)

This page drills the parent note (topic note) through every kind of question the exam can build from three ideas: allotrope structure, oxide building (bridges + caps), and basicity = count the O–H hands.

Before the examples, we lay out a scenario matrix: a checklist of every "case class" this topic hides. Then each worked example is tagged with the cell(s) it clears — so by the end, no scenario is left unseen.

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The scenario matrix

Cell	Case class	The hidden branch	Cleared by
A	Allotrope reactivity/prep	strained $60°$ vs relaxed network	Ex 1
B	Oxide build — LIMITED O₂	only bridges, P stays $+3$	Ex 2
C	Oxide build — EXCESS O₂	bridges + caps, P goes $+5$	Ex 2
D	Hydrolysis product	P–O–P bridge broken by water	Ex 3
E	Basicity, no P–H	all H on O ($H_{3}P{O}_4$)	Ex 4
F	Basicity, some P–H	subtract P–H ($H_{3}P{O}_3$)	Ex 4
G	Basicity, degenerate	max P–H, basicity $=1$ ($H_{3}P{O}_2$)	Ex 5
H	Reducing power ranking	P–H count controls it	Ex 6
I	Oxidation-state arithmetic	assign O, H, solve for P	Ex 7
J	Word / real-world	drying agent, dehydration of $HN{O}_3$	Ex 8
K	Exam twist (disproportionation)	one element, two products	Ex 9
L	Structure counting (bonds)	bridging vs terminal O	Ex 10

Ten examples below hit all twelve cells. Watch the figures — they carry the geometry.

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Example 1 — Allotropes (Cell A)

Forecast: guess the product and one reason before reading on. (Hint: what letter of the rainbow?)

Steps.

1. Identify the starting unit. White P = discrete $P_4$ tetrahedra. Why this step? Every property flows from the unit — you cannot reason about reactivity without naming the building block.
2. Recall the strain. In the tetrahedron each P bonds to the other three, forcing the $\angle \text{P–P–P}=60°$, far below the comfortable $\sim 90$–$109°$. That squeeze stores energy. Why? Stored strain = "loaded spring" = high reactivity, glow, low ignition temperature.

3. Apply heat in inert gas. At $573\text{K}$ the strained tetrahedra crack open and link into long chains → red phosphorus. Why inert (argon)? In air the white P would simply burn ($P_4+5O_2\to P_4{O}_{10}$); argon lets it rearrange instead of combust.
4. Explain the calm. Red P is a polymer with normal bond angles → no stored strain → stable, no glow, ignites only near $260°C$. Why this settles it? Reactivity was never about "being a powder" — it is about angular strain.

Verify: answers — (a) red P; (b) to prevent combustion; (c) strain relieved from $60°$ to normal angle. Sanity check: red is more stable than white → matches "white is wild, red is relaxed."

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Example 2 — Combustion with limited vs excess O₂ (Cells B, C)

Forecast: how many O₂ molecules for each? Which one is $+5$?

Steps.

1. Build $P_4{O}_6$ (limited). Start from the tetrahedron: 6 edges, put one O into each → $P_4{O}_6$. Why? Limited oxygen can only bridge; it cannot spare atoms to cap corners.

2. Balance the O. $P_4{O}_6$ has 6 O; each $O_2$ gives 2 O, so we need $6/2=3$ molecules: $P_4+3O_2\to P_4{O}_6.$
3. Oxidation state (limited). Each of the 4 P shares 6 bridging O equally → each P touches 3 O → $\text{P}=+3$. Why check? The state is the exam's real target.
4. Build $P_4{O}_{10}$ (excess). Now also cap each of the 4 corners with a terminal $=O$: $6+4=10$ oxygens.
5. Balance. $P_4{O}_{10}$ has 10 O → $10/2=5$ molecules: $P_4+5O_2\to P_4{O}_{10}.$
6. Oxidation state (excess). Each P: 3 bridging O + 1 terminal O = 4 O bonds → $\text{P}=+5$.

Verify: atom balance — (a) left P₄ = 4 P, O = $3\times 2=6$; right P₄O₆ = 4 P, 6 O ✓. (b) O = $5\times 2=10$ = P₄O₁₀ ✓. States $+3$ and $+5$ match the table.

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Example 3 — Hydrolysis (Cell D)

Forecast: how many acid molecules from each oxide? Same number or different?

Steps.

1. What water does. Water attacks the P–O–P bridges, inserting H and OH, splitting the cage into monomeric acids. Why? Bridges are the weak, strained P–O–P links.
2. $P_4{O}_6$ hydrolysis. Four P atoms → four acid molecules, each keeping P at $+3$: $P_4{O}_6+6H_{2}O\to 4H_{3}P{O}_3.$
3. $P_4{O}_{10}$ hydrolysis. Same four P, now $+5$: $P_4{O}_{10}+6H_{2}O\to 4H_{3}P{O}_4.$
4. Fiercer grabber. $P_4{O}_{10}$ — it absorbs water so greedily it is a drying agent. Why? Higher oxidation state P is more oxophilic (loves O/water).

Verify: balance $P_4{O}_{10}+6H_{2}O$: left P 4, O $10+6=16$, H 12; right $4H_{3}P{O}_4$ = P 4, O $16$, H 12 ✓. Same for $P_4{O}_6$: O $6+6=12$, right $4H_{3}P{O}_3$ O $=12$, H $12$ ✓.

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Example 4 — Basicity: no P–H vs some P–H (Cells E, F)

Forecast: both have "3 H" — will both neutralise 3 NaOH?

Steps.

1. Golden rule. Only H on O (P–O–H) is releasable as $H^{+}$; H directly on P (P–H) is non-acidic (P–H is non-polar, $E_P\approx 2.19\approx E_H\approx 2.20$). Why? No charge separation → no ionisation.

2. Draw $H_{3}P{O}_4$. P bonded to: one $=O$, three $-OH$. Zero P–H. $\text{basicity}=(\text{total H})-\#(\text{P–H})=3-0=3.$
3. Draw $H_{3}P{O}_3$. P bonded to: one $=O$, one P–H, two $-OH$. $\text{basicity}=3-1=2.$
4. Neutralisation. Basicity = moles NaOH per mole acid: $H_{3}P{O}_4$ needs 3, $H_{3}P{O}_3$ needs 2.

Verify: $H_{3}P{O}_4\to 3-0=3$; $H_{3}P{O}_3\to 3-1=2$. NaOH: 3 and 2 moles respectively.

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Example 5 — Degenerate case: maximum P–H (Cell G)

Forecast: guess the basicity — is this the "limiting" molecule?

Steps.

1. Structure. In $H_{3}P{O}_2$, P is bonded to: one $=O$, two P–H, and one $-OH$. Why two P–H? The lower the oxidation state, the more H sits directly on P.
2. Apply formula. $\text{basicity}=3-2=1(\text{monobasic}).$
3. Why not zero? An acid must donate at least one $H^{+}$; a P with no O–H would not be an acid at all. So basicity $\ge 1$ is the floor for these oxoacids. Why care? This is the degenerate/limiting cell of the basicity scale.
4. Trend check. $H_{3}P{O}_2,H_{3}P{O}_3,H_{3}P{O}_4$ → P–H bonds $2,1,0$ → basicity $1,2,3$. As oxidation state climbs ($+1,+3,+5$), P–H drops and basicity rises.

Verify: $3-2=1$. Oxidation state of P in $H_{3}P{O}_2$: $x+3(+1{)}_H^{\text{O–H is }+1net}$ — compute in VERIFY, expected $+1$.

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Example 6 — Reducing power ranking (Cell H)

Forecast: which end of the list is the strongest reducer?

Steps.

1. Locate the reducing group. The P–H bond is what gets oxidised (P–H → P–OH), so more P–H = stronger reducer. Why? Reducing power = willingness to be oxidised = availability of oxidisable P–H bonds.
2. Count P–H. $H_{3}P{O}_2$: 2; $H_{3}P{O}_3$: 1; $H_{3}P{O}_4$: 0.
3. Rank. $H_{3}P{O}_2>H_{3}P{O}_3>H_{3}P{O}_4.$
4. $A{g}^{+}$ test. $H_{3}P{O}_2$ (strongest) reduces $A{g}^{+}\to Ag$; $H_{3}P{O}_4$ (no P–H) does not reduce at all.

Verify: ordering by P–H count $2>1>0$ ⇒ $H_{3}P{O}_2>H_{3}P{O}_3>H_{3}P{O}_4$.

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Example 7 — Oxidation-state arithmetic (Cell I)

Forecast: will $H_{3}P{O}_3$ come out $+3$?

Steps.

1. Assign known states. H = $+1$, O = $-2$; molecule is neutral (sum = 0). Why? These are fixed reference values that let us solve for the unknown P.
2. $H_{3}P{O}_3$. Let P $=x$: $3(+1)+x+3(-2)=0\Rightarrow x=+3.$
3. $P_4{O}_6$. Let each P $=x$: $4x+6(-2)=0\Rightarrow 4x=12\Rightarrow x=+3.$
4. Cross-check with structure. Both give $+3$ — matches "P touches 3 O" reasoning. Why cross-check? Algebra and geometry must agree; if not, a bond was miscounted.

Verify: $H_{3}P{O}_3$: $3-6+x=0\Rightarrow x=+3$. $P_4{O}_6$: $4x=12\Rightarrow x=+3$.

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Example 8 — Real-world word problem (Cell J)

Forecast: is $P_4{O}_{10}$ acting as an acid, base, or something else here?

Steps.

1. Recall its greed for water. $P_4{O}_{10}$ is a powerful dehydrating agent — it strips water even out of oxoacids. Why it works: high-oxidation-state P is intensely oxophilic.
2. Pull water out of $HN{O}_3$. Two $HN{O}_3$ = $H_{2}O+N_2{O}_5$ "on paper"; $P_4{O}_{10}$ removes that $H_{2}O$: $P_4{O}_{10}+4HN{O}_3\to 4HP{O}_3+2N_2{O}_5.$
3. Role. $P_4{O}_{10}$ is the dehydrating (drying) agent; it grabs the elements of water, leaving anhydrous $N_2{O}_5$. Why not just water? Water would do the opposite — hydrate things, giving $H_{3}P{O}_4$.

Verify: atom balance of $P_4{O}_{10}+4HN{O}_3\to 4HP{O}_3+2N_2{O}_5$: P $4=4$; N $4=2\times 2$; H $4=4$; O left $10+12=22$, right $4\times 3+2\times 5=12+10=22$ ✓.

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Example 9 — Exam twist: disproportionation (Cell K)

Forecast: does one element end up both reduced AND oxidised?

Steps.

1. Starting state. In $P_4$ (element), P = $0$. Why? Uncombined element is always $0$.
2. State in $P{H}_3$. H = $+1$? No — in P–H, H bonded to less-EN P is $+1$ only vs O; here treat H as $+1$ giving $x+3(+1)=0\Rightarrow x=-3$. So P is reduced ($0\to -3$).
3. State in $H_{2}P{O}_2^{-}$. Let P $=x$: $x+2(+1)+2(-2)=-1\Rightarrow x=+1$. So P is oxidised ($0\to +1$).
4. Name it. One element (P at $0$) goes to two different states ($-3$ and $+1$) — simultaneous oxidation and reduction = disproportionation.

Reaction: $P_4+3NaOH+3H_{2}O\to P{H}_3+3Na{H}_{2}P{O}_2.$

Verify: states $0\to -3$ (in $P{H}_3$) and $0\to +1$ (in $H_{2}P{O}_2^{-}$). Electron balance: 1 P gains 3e⁻; 3 P each lose 1e⁻ → $3\times 1=3$ gained = 3 lost ✓. See Oxidation states and disproportionation.

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Example 10 — Structure counting (Cell L)

Forecast: each P touches how many oxygens?

Steps.

1. Split the 10 O. 6 bridging (P–O–P, one per tetrahedron edge) + 4 terminal ($P=O$, one per corner). Why? This split is the whole reason the state is $+5$ not $+3$.
2. Per-P count. Each P sits at a corner touched by 3 bridging O (its three edges) + 1 terminal O = 4 O. Bridging O are shared (count $\frac{1}{2}$ each toward the formula, but each still bonds this P).
3. Formula check. Bridging: $6$ shared among edges; terminal: $4$. Total $=6+4=10$ ✓.
4. State. 4 bonds to O per P → $+5$ (in $P_4{O}_6$ it was 3 bonds → $+3$). Why re-derive? Structure and oxidation state must be one story.

Verify: bridging + terminal $=6+4=10$; per-P O bonds $=3+1=4\Rightarrow$ state $+5$.

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Recall Quick self-test on the matrix

Which cell asks "how many NaOH neutralise $H_{3}P{O}_3$"? ::: Cell F (basicity with one P–H) → 2 NaOH. Which example proves red P is not reactive due to being a powder? ::: Example 1 (Cell A). Which cell is the "limiting/degenerate" basicity? ::: Cell G — $H_{3}P{O}_2$, basicity 1. Which single quantity controls both basicity AND reducing power? ::: The count of P–H bonds.

Connections

• Parent topic note
• Oxidation states and disproportionation
• Reducing agents and P–H bonds
• Hydrolysis reactions
• Oxides — acidic vs basic character
• Allotropy — Carbon, Sulfur, Oxygen comparison