Two Trichlorides on One Shelf: Reading Shape from Dipole Moment and Boiling Point
Two unlabelled, anhydrous trichlorides sit on a stock shelf. Sample J is a colourless gas at room temperature; sample K is a colourless liquid of boiling point 76 °C. Both fume in moist air and give an acidic solution on hydrolysis. Vapour-density measurements relative to hydrogen give 58.6 for J and 68.7 for K. A dipole-moment apparatus records 0.00 D for J and 0.56 D for K. Electron-diffraction data give Cl–X–Cl angles of 120° for J and 100.3° for K. A student proposes that both molecules are pyramidal, arguing that 'three chlorines plus a lone pair means trigonal pyramidal', and suggests that the zero dipole moment of J is a faulty reading. The teacher asks for a structure-based identification of both samples, an explanation of the two bond angles, and a prediction of which sample will react with trimethylamine to give a white solid adduct.
1. Assign J and K using vapour density, dipole moment, shape and physical state, with reasoning.
Read the worked answer
Molar mass = 2 × vapour density. For J, M = 117.2 g mol⁻¹; for K, M = 137.4 g mol⁻¹. The candidates are BCl3 (10.81 + 3 × 35.45 = 117.16) and PCl3 (30.97 + 3 × 35.45 = 137.32). J: μ = 0. Three polar B–Cl bonds can only give zero resultant if they are arranged symmetrically at 120° in a plane. Boron has three valence electrons, forms three σ bonds, has no lone pair and is sp² hybridised; the observed 120° angle confirms trigonal planar geometry. A gas at room temperature fits a small, non-polar molecule (BCl3 boils at about 13 °C). K: μ = 0.56 D, so the bond moments do not cancel and the molecule cannot be planar. Phosphorus has five valence electrons, three used in σ bonds and one lone pair left, giving three bond pairs plus one lone pair, sp³ hybridisation and a trigonal pyramidal shape. The 100.3° angle confirms a non-planar, lone-pair-bearing centre, and the liquid state fits the larger, polar molecule (PCl3, b.p. 76 °C). Conclusion: J is BCl3 and K is PCl3. The student's rule fails because it ignores the electron count on the central atom.
2. Explain both bond angles using VSEPR, including why 100.3° is well below 109.5°.
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In BCl3 there are three electron domains around boron and no lone pair, so the domains adopt the arrangement of maximum separation in a plane: 120°, trigonal planar. All three B–Cl bond dipoles are equal and at 120°, so they cancel and μ = 0. In PCl3 there are four domains (three bonding pairs and one lone pair), so the electron-pair geometry is tetrahedral. Lone-pair–bond-pair repulsion is stronger than bond-pair–bond-pair repulsion because the lone pair is held by only one nucleus and spreads out more, so the three P–Cl bonds are squeezed together and the Cl–P–Cl angle drops below 109.5°. The further fall to 100.3°, larger than the fall seen in NH3 (107°), is explained by the size and lower electronegativity of phosphorus: the P–Cl bonds are long and the bonding pairs are relatively far from P and from each other, so bond-pair–bond-pair repulsion is weak and offers little resistance to compression by the lone pair.
3. Predict which sample forms the white adduct with trimethylamine, and which will bind to a metal centre. Justify in terms of electron pairs and orbitals, and link this to the hydrolysis behaviour.
Read the worked answer
J (BCl3) forms the adduct. Boron in BCl3 has only six valence electrons and an empty 2p orbital perpendicular to the molecular plane. It is electron-deficient, a Lewis acid, and accepts the lone pair on the nitrogen of N(CH3)3 to give Cl3B←N(CH3)3, a white solid. K (PCl3) cannot accept an electron pair: phosphorus already has a full octet and no low-energy vacant orbital. Instead the lone pair makes PCl3 a Lewis base, so it coordinates to transition-metal centres as a ligand. Hydrolysis follows the same logic. With BCl3, a water molecule donates a lone pair to boron, the intermediate eliminates HCl, and boric acid plus HCl are formed - hence the fumes in moist air. With PCl3, water attacks the phosphorus lone-pair region and the product is phosphorous acid, H3PO3, with HCl. So the fuming is a Lewis acid–base reaction in both cases, but the roles are opposite.