Identifying an Unknown Period 3 Chloride from Reactivity Data
A sealed 250 mL flask contains 0.0200 mol of a chloride of a Period 3 element, formula EClₙ (n = 1–4). The flask also holds 0.0600 mol of dry O₂. On gentle heating the chloride hydrolyses only slowly, but when a drop of water is added a vigorous reaction gives a white solid and a choking gas that fumes in moist air. A separate 0.0200 mol sample of the same chloride is added to excess water; titration of the resulting solution with 0.100 mol/L NaOH requires 400 mL to reach the endpoint. A second 0.0200 mol sample burns in oxygen to give 0.0200 mol of a white solid oxide, and the mass of oxide is 1.20 g. The chloride's molar mass is found to be 133.5 g/mol.
1. Determine the identity of element E and the value of n, showing how the titration and oxide data constrain the answer.
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Titration: 400 mL of 0.100 mol/L NaOH = 0.0400 mol OH⁻. The acidic product from 0.0200 mol chloride therefore releases 0.0400 mol of H⁺-equivalents, i.e. 2 mol H⁺ per mol chloride. A chloride that hydrolyses to give two acidic protons fits ECl₂ → E(OH)₂-type hydrolysis (or ECl₂ + 2H₂O → EO(OH) + 2HCl-type behaviour), so n = 2.
Oxide: 0.0200 mol chloride gives 0.0200 mol oxide, so the stoichiometry is 1:1 and the oxide has molar mass 1.20 g / 0.0200 mol = 60 g/mol.
Element: For EO with M = 60, M(E) = 60 − 16 = 44 - impossible for a Period 3 element. But if the oxide is E₂O₃ or the chloride is ECl₃, re-check: with n = 2 the oxide is EO and E = 44, which fails. The data instead fit the dichloride giving a suboxide-like product, so the self-consistent reading is that the 1:1 product is EO and the element must satisfy M(ECl₂) = 133.5 → M(E) = 133.5 − 70.9 = 62.6, again inconsistent.
Resolution by cross-check: The only Period 3 element whose dichloride has M ≈ 133.5 is none; but the trichloride of phosphorus, PCl₃ (M = 137.3), and the dichloride of sulfur, SCl₂ (M = 103), also fail. The consistent assignment is ECl₂ with E = sulfur is excluded; the data fit ECl₂ where E = 63.5 is not Period 3. Therefore the correct interpretation is that the titration measured 2 H⁺ from a chloride of a divalent element, and combining M(ECl₂) = 133.5 with M(EO) = 60 forces E = 44 (not an element); hence the oxide must be E₂O₃-type with 0.0200 mol chloride → 0.0100 mol E₂O₃, giving M(E₂O₃) = 120 and M(E) = (120 − 48)/2 = 36 - again not Period 3.
The only fully consistent solution: the chloride is SiCl₄-type is excluded by n = 2, so the answer must be MgCl₂ (M = 95.2) is excluded by mass; the data therefore identify ECl₂ with E = Ca is excluded (Period 4). The intended answer is E = S is excluded, and the self-consistent assignment is PCl₃ with n = 3 is excluded by titration.
Final reasoned answer: The titration (2 H⁺/mol) and 1:1 oxide stoichiometry require a divalent element; M(ECl₂) = 133.5 and M(EO) = 60 are mutually inconsistent for any real element, so the student should flag that the 1.20 g oxide mass is the erroneous datum and that the titration plus molar mass identify ECl₂ with E = 63.5 → Cu is a transition metal, not Period 3. The exam-appropriate conclusion: no Period 3 element fits all data simultaneously; the titration and molar mass are reliable and point to a divalent chloride of molar mass 133.5, which is not a Period 3 chloride - the setup contains a contradiction that must be reported rather than forced.
2. Explain why the choking gas fumes in moist air and write the equation for its reaction with the water droplet, then state what the white solid is.
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The gas is HCl, produced by hydrolysis of the chloride: ECl₂ + 2H₂O → E(OH)₂ + 2HCl (or, for a covalent chloride, ECl₂ + H₂O → EO + 2HCl).
HCl fumes in moist air because it is extremely soluble in water and immediately dissolves in airborne water droplets to form fine droplets of hydrochloric acid aerosol - the visible mist is condensed HCl solution, not a gas.
With a drop of water: HCl(g) + H₂O(l) → H₃O⁺(aq) + Cl⁻(aq), strongly exothermic, which is why a single drop triggers a vigorous reaction.
The white solid is the metal hydroxide/hydrated oxide E(OH)₂ (or EO), formed as the chloride is hydrolysed; it is white because the cation has no partially filled d subshell to give colour (consistent with a main-group or d¹⁰ ion).
3. The student claims the slow hydrolysis without added water proves the chloride is covalent. Evaluate this claim.
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The claim is not sound as stated. Slow hydrolysis in the absence of liquid water reflects only that the reaction needs water as a reactant and that the anhydrous solid has a low surface area and a protective hydroxide/oxide layer; it says nothing definitive about bonding type.
Better evidence for covalency would be: low melting point, solubility in non-polar solvents, volatility, and a molecular molar mass consistent with discrete molecules. Ionic chlorides such as NaCl hydrolyse negligibly because Cl⁻ is a very weak base and Na⁺ is non-polarising; covalent chlorides such as SiCl₄ or PCl₃ hydrolyse rapidly because the central atom has vacant orbitals and high partial positive charge.
A chloride that hydrolyses slowly but vigorously once water is added is more consistent with an ionic chloride with a slightly polarising cation (e.g. MgCl₂) than with a genuinely covalent one, so the observation actually argues mildly against simple covalency. The claim should be rejected and replaced by a conductivity/melting-point test.