AS Chemistry 9701: Formation and Reactions of P4O10 and SO3

Phosphorus, $\text{P}_4$ , and sulfur, $\text{S}_8$ , are both burned separately in an excess of dry oxygen, forming phosphorus(V) oxide and sulfur trioxide respectively. Fresh samples of each solid oxide are then shaken separately with an excess of distilled water, and finally with an excess of aqueous sodium hydroxide.

(a) Construct balanced symbol equations, including state symbols, for the combustion of $\text{P}_4$ and of $\text{S}_8$ in an excess of dry oxygen. [2]

(b) Phosphorus(V) oxide reacts vigorously with an excess of water to form phosphoric(V) acid, $\text{H}_3\text{PO}_4$ . Construct the balanced symbol equation, including state symbols, for this reaction. [2]

(c) State the type of structure and bonding present in solid phosphorus(V) oxide and solid sulfur trioxide, and explain why this structure allows both oxides to react rapidly with water despite consisting of covalently bonded molecules. [1]

(d) Sulfur trioxide also reacts with an excess of aqueous sodium hydroxide. Construct the balanced symbol equation, including state symbols, for this reaction. [2]

(e) Magnesium oxide reacts with dilute acid but not with aqueous sodium hydroxide. Using this contrast, explain, in terms of structure and bonding, why phosphorus(V) oxide and sulfur trioxide are classified as acidic oxides whereas magnesium oxide is classified as a basic oxide. [2]

Part (a): Combustion of phosphorus and sulfur in excess oxygen

Phosphorus burns in excess oxygen to form phosphorus(V) oxide, in which every P atom is fully oxidised: $\text{P}_4\text{(s)} + 5\text{O}_2\text{(g)} \rightarrow \text{P}_4\text{O}_{10}\text{(s)}$ Checking the balance: $4$ P atoms on each side, $10$ O atoms on each side ( $5\times2$ on the left; $10$ in $\text{P}_4\text{O}_{10}$ on the right).

Sulfur burns in excess oxygen to form sulfur trioxide: $\text{S}_8\text{(s)} + 12\text{O}_2\text{(g)} \rightarrow 8\text{SO}_3\text{(g)}$ Checking the balance: $8$ S atoms on each side, $24$ O atoms on each side ( $12\times2$ on the left; $8\times3$ on the right). Because combustion is strongly exothermic, the $\text{SO}_3$ formed leaves the flame as a gas.

Part (b): Phosphorus(V) oxide with water

$\text{P}_4\text{O}_{10}$ reacts vigorously and completely with an excess of water to form phosphoric(V) acid: $\text{P}_4\text{O}_{10}\text{(s)} + 6\text{H}_2\text{O(l)} \rightarrow 4\text{H}_3\text{PO}_4\text{(aq)}$ Checking the balance: $4$ P atoms on each side; O atoms: $10 + 6 = 16$ on the left, $4\times4 = 16$ on the right; H atoms: $6\times2 = 12$ on the left, $4\times3 = 12$ on the right.

Part (c): Structure of the solid oxides and why hydrolysis is fast

Both $\text{P}_4\text{O}_{10}$ and $\text{SO}_3$ are simple molecular solids: discrete covalent molecules held to one another only by weak van der Waals forces. This weak lattice does not slow down the reaction with water. Melting or vaporising the solid is easy, but that isn’t what hydrolysis requires. Instead, hydrolysis depends on the P-O and S-O bonds within each molecule, which are strongly polar because P and S are both bonded to several highly electronegative O atoms, leaving the central P or S atom significantly electron-deficient. This electron-deficient centre is readily attacked by the lone pair on a water molecule’s oxygen atom, so both oxides hydrolyse rapidly and completely, even though the intermolecular forces holding the solid together are weak.

Part (d): Sulfur trioxide with aqueous sodium hydroxide

$\text{SO}_3$ neutralises the base, forming sodium sulfate solution: $\text{SO}_3\text{(g)} + 2\text{NaOH(aq)} \rightarrow \text{Na}_2\text{SO}_4\text{(aq)} + \text{H}_2\text{O(l)}$ Checking the balance: $1$ S atom, $2$ Na atoms on each side; O atoms: $3 + 2 = 5$ on the left, $4 + 1 = 5$ on the right; H atoms: $2$ on each side.

Part (e): Why P4O10 and SO3 are acidic while MgO is basic

$\text{P}_4\text{O}_{10}$ and $\text{SO}_3$ are covalent, simple molecular oxides of P and S, both highly electronegative non-metals. As shown in parts (b) and (d), they react with water to release $\text{H}^+$ ions (forming $\text{H}_3\text{PO}_4$ or $\text{H}_2\text{SO}_4$ ) and they neutralise a base (NaOH) directly. Both are defining behaviours of an acidic oxide.

MgO, by contrast, is an ionic lattice built from $\text{Mg}^{2+}$ and discrete $\text{O}^{2-}$ ions. The $\text{O}^{2-}$ ion reacts readily with $\text{H}^+$ ions supplied by an acid to form water, so MgO neutralises acids, the defining behaviour of a basic oxide. MgO has no electron-deficient covalent centre and no acidic hydrogen to offer to a base, so it does not react with NaOH.

This contrast reflects the broader Period 3 trend: oxides of the strongly electropositive metals on the left of the period are ionic and basic, while oxides of the strongly electronegative non-metals on the right are covalent and acidic.

Final answers

(a) $\text{P}_4\text{(s)} + 5\text{O}_2\text{(g)} \rightarrow \text{P}_4\text{O}_{10}\text{(s)}$ ; $\text{S}_8\text{(s)} + 12\text{O}_2\text{(g)} \rightarrow 8\text{SO}_3\text{(g)}$
(b) $\text{P}_4\text{O}_{10}\text{(s)} + 6\text{H}_2\text{O(l)} \rightarrow 4\text{H}_3\text{PO}_4\text{(aq)}$
(c) Both are simple molecular solids (weak van der Waals forces between molecules); hydrolysis is fast because the polar, electron-deficient P-O/S-O bonds within each molecule are readily attacked by water’s lone pairs, independent of the weak lattice forces.
(d) $\text{SO}_3\text{(g)} + 2\text{NaOH(aq)} \rightarrow \text{Na}_2\text{SO}_4\text{(aq)} + \text{H}_2\text{O(l)}$
(e) $\text{P}_4\text{O}_{10}$ / $\text{SO}_3$ are covalent oxides that release $\text{H}^+$ in water and react with a base -> acidic; MgO is an ionic oxide containing $\text{O}^{2-}$ that reacts with $\text{H}^+$ from an acid -> basic.