Phosphorus allotropes (white, red, black); P₄O₆, P₄O₁₀; oxoacids of P (H₃PO₃ vs H₃PO₄ basicity): Exercises

This page tests everything from the parent topic. Keep two tools handy: oxidation state counting (from Oxidation states and disproportionation) and the P–OH vs P–H basicity rule.

Level 1 — Recognition

L1.1

State (a) which phosphorus allotrope is most reactive, (b) which is most stable, and (c) how white P is stored.

Recall Solution

(a) White P — its $P_{4}$ tetrahedra have strained $60°$ angles storing energy. (b) Black P — layered graphite-like sheets, lowest energy. (c) Under water, because it ignites in air at about $35° C$ .

L1.2

Give the oxidation state of phosphorus in: $P_{4}$ , $P_{4} O_{6}$ , $P_{4} O_{10}$ , $H_{3} P O_{3}$ , $H_{3} P O_{4}$ .

Recall Solution

$P_{4}$ : element → $0$ .
$P_{4} O_{6}$ : each O is $- 2$ , total from O $= 6 (- 2) = - 12$ ; neutral molecule so $4 x = + 12$ , $x = + 3$ .
$P_{4} O_{10}$ : $10 (- 2) = - 20$ , $4 x = + 20$ , $x = + 5$ .
$H_{3} P O_{3}$ : $3 (+ 1) + x + 3 (- 2) = 0 \Rightarrow x = + 3$ .
$H_{3} P O_{4}$ : $3 (+ 1) + x + 4 (- 2) = 0 \Rightarrow x = + 5$ .

Level 2 — Application

L2.1

Balance the combustion of white P in (a) limited $O_{2}$ and (b) excess $O_{2}$ .

Recall Solution

(a) Limited $O_{2}$ → oxygen only bridges the 6 edges, P stays $+ 3$ : $P_{4} + 3 O_{2} ⟶ P_{4} O_{6}$ Check O: LHS $= 6$ , RHS $= 6$ . ✓ (b) Excess $O_{2}$ → corners also capped, P goes to $+ 5$ : $P_{4} + 5 O_{2} ⟶ P_{4} O_{10}$ Check O: LHS $= 10$ , RHS $= 10$ . ✓

L2.2

Write the hydrolysis of $P_{4} O_{6}$ and $P_{4} O_{10}$ and name each product.

Recall Solution

Water attacks the P–O–P bridges and splits the cage into 4 identical acid units. $P_{4} O_{6} + 6 H_{2} O ⟶ 4 H_{3} P O_{3} (phosphorous acid)$ $P_{4} O_{10} + 6 H_{2} O ⟶ 4 H_{3} P O_{4} (phosphoric acid)$ Mass check (H₃PO₄): LHS H $= 12$ , RHS H $= 4 \times 3 = 12$ ✓; LHS O $= 10 + 6 = 16$ , RHS O $= 4 \times 4 = 16$ ✓. This is a hydrolysis — the oxide is acidic, reacting with water to give an acid.

L2.3

State the basicity of $H_{3} P O_{2}$ , $H_{3} P O_{3}$ , $H_{3} P O_{4}$ using the P–OH rule.

Recall Solution

$basicity = # (P-OH) = (total H) - # (P-H bonds)$

$H_{3} P O_{2}$ : 2 P–H bonds → $3 - 2 = 1$ (monobasic).
$H_{3} P O_{3}$ : 1 P–H bond → $3 - 1 = 2$ (dibasic).
$H_{3} P O_{4}$ : 0 P–H bonds → $3 - 0 = 3$ (tribasic). See the structures in the figure below.

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

Level 3 — Analysis

L3.1

$H_{3} P O_{3}$ and $H_{3} P O_{4}$ both have three hydrogens, yet one is dibasic and the other tribasic. Explain why, using the P–H vs P–O–H distinction and electronegativity.

Recall Solution

Draw both (see figure s01):

$H_{3} P O_{4}$ : P has one $= O$ and three $- O H$ . All three H sit on oxygen → all three ionisable → basicity 3.
$H_{3} P O_{3}$ : P has one $= O$ , one direct P–H, and two $- O H$ . Only the two O–H hydrogens ionise → basicity 2. Why does P–H not ionise? Ionising $H^{+}$ needs a polar bond so the electrons stay behind on the more electronegative atom. For O–H, $E_{O} \approx 3.44 ≫ E_{H} \approx 2.20$ → very polar → H leaves as $H^{+}$ . For P–H, $E_{P} \approx 2.19 \approx E_{H} \approx 2.20$ → the bond is essentially non-polar → no driving force to release $H^{+}$ . So P–H is a "stuck" hydrogen (and, being electron-rich, it acts as a reducing hydrogen instead).

L3.2

Predict the number of terminal ( $P = O$ ) and bridging (P–O–P) oxygens in $P_{4} O_{10}$ , and use them to re-derive the oxidation state $+ 5$ .

Recall Solution

Build from the $P_{4}$ tetrahedron: 6 edges → 6 bridging O, 4 corners → 4 terminal O. Total $6 + 4 = 10$ ✓. Each P touches: 3 bridging O (shared, count as one bond each to that P) + 1 terminal $= O$ (double bond = 2 bonds to O). Bonds to oxygen per P $= 3 + 2 = 5$ ; every P–O bond assigns both electrons to O (P is less electronegative) → each contributes $+ 1$ to P. So P $= + 5$ . Consistent with charge balance ( $10 (- 2) /4 = + 5$ ). ✓

Level 4 — Synthesis

L4.1

$H_{3} P O_{2}$ reduces silver ions to silver metal, but $H_{3} P O_{4}$ does not react with $A g^{+}$ at all. Explain the trend across $H_{3} P O_{2} \to H_{3} P O_{3} \to H_{3} P O_{4}$ and connect it to oxidation state.

Recall Solution

Reducing power comes from the P–H bond: P–H is electron-rich and easily oxidised (P climbs to a higher oxidation state, releasing electrons that reduce $A g^{+} \to A g$ ).

$H_{3} P O_{2}$ : 2 P–H, P is $+ 1$ (far below its max $+ 5$ ) → most room to be oxidised → strongest reducer, reduces $A g^{+}$ : $H_{3} P O_{2} + 4 A g N O_{3} + 2 H_{2} O \to 4 A g + H_{3} P O_{4} + 4 H N O_{3}$ (P goes $+ 1 \to + 5$ , a 4-electron oxidation; each electron reduces one $A g^{+}$ .)
$H_{3} P O_{3}$ : 1 P–H, P is $+ 3$ → moderate reducer.
$H_{3} P O_{4}$ : 0 P–H, P already at maximum $+ 5$ → cannot be oxidised → not a reducing agent. So reducing power ∝ number of P–H bonds ∝ how far below $+ 5$ the P sits. (See Reducing agents and P–H bonds.)

L4.2

White P disproportionates in hot NaOH: $P_{4} + 3 N a O H + 3 H_{2} O \to P H_{3} + 3 N a H_{2} P O_{2}$ Identify the oxidation-state changes and confirm this is a disproportionation.

Recall Solution

Assign P oxidation states:

Reactant $P_{4}$ : $0$ .
$P H_{3}$ : H is $+ 1$ (bonded to less electronegative? — no: here P is more electronegative than H, so H is $+ 1$ , P is $- 3$ ). $x + 3 (+ 1) = 0 \Rightarrow x = - 3$ .
$N a H_{2} P O_{2}$ (the salt of hypophosphorous acid): Na $+ 1$ , O $- 2$ , H $+ 1$ : $1 + 2 (+ 1) + x + 2 (- 2) = 0 \Rightarrow x = + 1$ . One portion of P goes down $0 \to - 3$ (reduced) and another goes up $0 \to + 1$ (oxidised). Same element split into higher and lower states → disproportionation ✓ (see Oxidation states and disproportionation). Electron check: reduction $0 \to - 3$ gains 3 e⁻ per P; oxidation $0 \to + 1$ loses 1 e⁻ per P. To balance electrons, 1 reduced P needs 3 oxidised P → ratio $1 : 3$ , matching $1 P H_{3} : 3 N a H_{2} P O_{2}$ . ✓

Level 5 — Mastery

L5.1

A sample of $P_{4} O_{10}$ weighing $2.84 g$ is completely hydrolysed. (a) How many moles of $H_{3} P O_{4}$ form? (b) What total volume of $0.100 M$ NaOH is needed to completely neutralise the acid produced? (Molar masses: P $= 31$ , O $= 16$ , H $= 1$ .)

Recall Solution

(a) $M (P_{4} O_{10}) = 4 (31) + 10 (16) = 124 + 160 = 284 g mol^{- 1}$ . Moles $P_{4} O_{10} = 2.84/284 = 0.0100 mol$ . Hydrolysis: $P_{4} O_{10} + 6 H_{2} O \to 4 H_{3} P O_{4}$ , so moles $H_{3} P O_{4} = 4 \times 0.0100 = 0.0400 mol$ . (b) $H_{3} P O_{4}$ is tribasic → each mole needs 3 mol NaOH: $H_{3} P O_{4} + 3 N a O H \to N a_{3} P O_{4} + 3 H_{2} O$ Moles NaOH $= 3 \times 0.0400 = 0.120 mol$ . Volume $= n / C = 0.120/0.100 = 1.20 L = 1200 mL$ .

L5.2

Repeat L5.1(b) but for the acid from hydrolysing the same number of moles ( $0.0100$ mol) of $P_{4} O_{6}$ . Why is the NaOH requirement smaller?

Recall Solution

$P_{4} O_{6} + 6 H_{2} O \to 4 H_{3} P O_{3}$ , so moles $H_{3} P O_{3} = 4 \times 0.0100 = 0.0400 mol$ . $H_{3} P O_{3}$ is dibasic (one P–H does not ionise), so: $H_{3} P O_{3} + 2 N a O H \to N a_{2} H P O_{3} + 2 H_{2} O$ Moles NaOH $= 2 \times 0.0400 = 0.0800 mol$ . Volume $= 0.0800/0.100 = 0.800 L = 800 mL$ . Why smaller? Same moles of acid, but $H_{3} P O_{3}$ donates only 2 $H^{+}$ per molecule vs 3 for $H_{3} P O_{4}$ — the extra hydrogen in $H_{3} P O_{3}$ is a stuck P–H. Ratio of NaOH volumes $= 2 : 3 = 800 : 1200$ ✓.

Recall One-line mastery summary

Oxidation state fixes which oxide/acid you have; P–OH count fixes basicity and titration; P–H count fixes reducing power. Three independent counts, all read off the structure, never off the formula.

Connections

p-Block Group 15 (Nitrogen family) trends
Oxidation states and disproportionation
Oxides — acidic vs basic character
Hydrolysis reactions
Reducing agents and P–H bonds
Allotropy — Carbon, Sulfur, Oxygen comparison