IB Chemistry S2.2: The covalent model worked solutions

\ce{NH3} has three bonding pairs and one lone pair on nitrogen. \ce{CO2}, \ce{CH4} and \ce{BF3} have none on the central atom.

The four C– Cl dipoles of \ce{CCl4} point to the corners of a tetrahedron and cancel. The other three molecules are asymmetric, so their dipoles do not cancel.

Hydrogen bonds need a hydrogen bonded to N, O or F in the molecule. Only \ce{CH3NH2} has an N– H bond. \ce{CH3OCH3} and \ce{CH3CHO} have O but no O– H, and \ce{CH3F} has C– H and C– F only.

Diamond: all four valence electrons of every carbon are held in bonds to four neighbours, so no electrons are free to move (1). Graphite: each carbon bonds to three neighbours and the fourth electron is delocalized across the layer, free to move along it (1).

5-0-\tfrac82=+1: nitrogen has five valence electrons, no lone pair, and four bonds (eight bonding electrons).

R_\mathrm{F}=2.4/6.0=0.40. It has no unit, and both distances are measured from the origin line.

\ce{H2O} and \ce{H2S} each have two bonds and two lone pairs, so both are bent. \ce{CO2} is linear and \ce{SO2} is bent; \ce{NH3} is pyramidal and \ce{BF3} planar; \ce{CH4} is tetrahedral and \ce{NH3} pyramidal.

The electron budget is 4+6=10. Put six electrons in the triple bond and one lone pair on each atom, giving both atoms octets. The formal charges are -1 on C and +1 on O; show these on the drawing (1). One shared pair has both electrons supplied by oxygen, so the donation is O to C (1).

B. The shared electron pair has negative charge and attracts both positive bromine nuclei. Intermolecular forces act between different molecules.

D. For the same pair of elements, more shared pairs concentrate electron density between the nuclei. Bond length decreases from single to double to triple.

(a) 6+2(7)=20 electrons: two O– F bonds, two lone pairs on O and three lone pairs on each F (1). (b) Four domains on O, two of them lone pairs, so the shape is bent (1). Lone pairs repel more than bonding pairs, so the angle is squeezed below 109.5^\circ (1). (c) \Delta\chi=0.6, so each bond is polar; the bent shape does not let the dipoles cancel, so the molecule is polar (1).

(a) London forces (1). (b) The branched molecule is compact and touches its neighbours over a smaller area, so its London forces are weaker (1); less energy is needed to separate the molecules (1). (c) Butan-1-ol forms hydrogen bonds through O– H (1), which are stronger than the London forces in pentane at similar molar mass, so more energy is needed (1).

(a) 5+2(6)+1=18 electrons. Each contributor has N=O, N-O^- and one lone pair on N, with the double bond on opposite oxygens in the two drawings (1 for each correct contributor). (b) (1+2)/2=1.5 (1). (c) A localized triene would release 3\times120=360\,\mathrm{kJ\,mol^{-1}} (1); benzene releases 208, so it is 360-208=152\,\mathrm{kJ\,mol^{-1}} more stable (1).

(a) |2.6-2.2|=0.4 with C \delta^- (1); |3.2-2.6|=0.6 with Cl \delta^- (1). (b) Four bonds give tetrahedral geometry (1). The two C–Cl dipoles have a resultant that is not cancelled by the C–H dipoles; the net dipole is non-zero (1).

(a) The O–H group can hydrogen-bond with water (1); the non-polar propyl group has London attractions with hexane (1). This predicts interactions, not unlimited miscibility. (b) \ce{HCl + H2O -> H3O+ + Cl-} (1). The dissolved ions move and carry charge through the solution (1).

(a) Spot travel =2.7 cm and front travel =7.5 cm (1), so R_\mathrm{F}=2.7/7.5=0.36, without units (1). (b) The slower component is more strongly retained by silica relative to hexane (1). Polar groups can interact more strongly with polar silica, so greater polarity is consistent with this result (1); migration alone does not establish identity.

(a) There are five C–H bonds and one O–H bond, plus four heavy-atom connections, so 6+4=10\sigma (1). The triple bond contains two \pi bonds (1). (b) Each triple-bonded C has two domains and is sp (1). Two unhybridized p orbitals on each carbon overlap sideways in perpendicular planes to give the two \pi interactions (1).

(a) Choose \ce{SiO2}: a giant network of strong Si–O covalent bonds (1), requiring much energy to disrupt, hence high melting point (1). (b) Graphite has mobile delocalized electrons, so is not an electrical insulator (1). Iodine consists of discrete \ce{I2} molecules held by London attractions, so separates and sublimes much more readily than a covalent network (1).

(a) 7+3(7)=28 electrons. Three Cl– F bonds use 6; three lone pairs on each F use 18; the remaining 4 are two lone pairs on Cl. Domains =3+2=5 (2). (b) In a trigonal bipyramid an equatorial position has two neighbours at 90^\circ, an axial position has three (1). Lone pairs repel more strongly, so they go where there are fewer 90^\circ repulsions (1). (c) Polar: the T-shape is not symmetric, so the dipoles do not cancel (1). (d) Chlorine is in period 3 and can expand its valence shell beyond eight (1).

(a) 1+4+2(6)+1=18 valence electrons (1). Each contributor has one C=O and one C–O bond; the single-bonded O has three lone pairs and -1 charge, the double-bonded O has two lone pairs and zero charge, and carbon has zero charge (1). Place the double bond on the other O for the second contributor; bracket each ion (1). (b) (1+2)/2=1.5 (1). The actual ion has one delocalized distribution across both C–O bonds, giving equal lengths between localized single and double reference lengths (1). (c) Carbon has three domains, so sp^2 (1); its remaining p orbital overlaps with parallel p orbitals on both oxygens in a planar O–C–O unit (1).

(a) The localized model releases 3(119)=357 kJ mol^{-1} (1). Z releases 205, giving stabilization 357-205=152 kJ mol^{-1} (1), assuming comparable hydrogenated products. (b) Fixed alternating single/double bonds predict two C–C lengths, unlike the observations (1). A planar localized model can also have equal angles, so equal angles alone do not discriminate (1). (c) Substitution replaces a ring H while restoring the delocalized ring (1); addition converts ring carbons to saturated centres and removes part of the conjugated p-orbital system (1).

(a) The fluoride lone pair is donated to phosphorus (1); a neutral \ce{PF5} plus \ce{F-} gives charge -1 (1). (b) 5+6(7)+1=48 electrons; six bonds use 12 and fluorine lone pairs use 36, leaving no central lone pairs (1). Six bonding domains give an octahedral ion (1). (c) Trigonal-bipyramidal \ce{PF5} has ideal angles 90^\circ, 120^\circ, 180^\circ, whereas octahedral \ce{PF6-} has 90^\circ and 180^\circ (1). Nitrogen is a period-2 element and cannot accommodate the 12 electrons required by six ordinary shared pairs (1).

(a) n=9.20/46.0=0.200 mol (1). q=mc\Delta T=9.20(2.44)(40.0)=897.92\simeq898 J (1). (b) Ethanol has an O–H donor and O lone pairs and forms intermolecular hydrogen bonds (1). Dimethyl ether lacks an O–H donor, so cannot hydrogen-bond to itself; more energy is needed to separate ethanol molecules (1). (c) At the boiling point the supplied energy overcomes intermolecular attractions and raises potential energy (1); the average kinetic energy, and therefore temperature, stays constant while both phases coexist at fixed pressure (1).

(a) Draw an arrow from the nitrogen lone pair in \ce{NH3} to \ce{H+} (1). The resulting \ce{NH4+} has four bonding domains and no lone pairs, giving a tetrahedral shape (1). (b) The new N–H covalent bond attracts both nuclei to a shared pair (1). In solid ammonium chloride, electrostatic attractions act between oppositely charged \ce{NH4+} and \ce{Cl-} ions (1). (c) With solid activity one, Q=1/([\mathrm{NH_3}][\mathrm{HCl}]); adding ammonia lowers Q (1). Temperature is unchanged, so K is unchanged and Q<K favours the forward reaction until equilibrium returns (1).

(a) Cyclohexene: 6.4/8.0=0.80 (1); cyclohexanol: 2.4/8.0=0.30 (1). (b) Cyclohexanol has an OH group that can hydrogen-bond to surface groups on silica (1). It spends more time retained relative to the hydrocarbon mobile phase than cyclohexene and travels less far (1). (c) Cyclohexene decolourizes bromine water by addition across its C=C bond; cyclohexanol does not under these test conditions (1). Different compounds may co-migrate, so one matching spot supports identity but does not prove purity (1).