Question 2 of 7: Chemical Bonding

Question 2: Chemical Bonding (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

2.1 — Polarity of Covalent Bonds

A covalent bond is polar when the bonding electron pair is shared unequally, i.e. when the two atoms differ in electronegativity; the more electronegative atom carries a partial negative charge $\delta^-$ and the other a partial positive charge $\delta^+$, giving the bond a dipole moment. A bond is non-polar only when the electronegativity difference is exactly zero, which requires identical atoms.

Of the four, F₂ is the non-polar one: it is a homonuclear diatomic molecule, so both atoms have identical electronegativity and the shared pair is divided exactly evenly ($\Delta\chi=0$). HCl, HF and ClF are all heteronuclear and therefore polar, with HF the most polar of the three (largest electronegativity difference, $\chi_F-\chi_H\approx1.78$).

2.2 — Hydrogen Bonding

A hydrogen bond is a relatively strong (typically 10–40 kJ/mol) secondary, electrostatic attraction between a hydrogen atom covalently bonded to a small, highly electronegative atom (N, O or F) and a lone pair on a nearby electronegative atom. Unlike a covalent bond — a primary, intramolecular bond formed by sharing electron pairs between two atoms, with bond energies of hundreds of kJ/mol — hydrogen bonding is an intermolecular force that does not involve orbital overlap/electron sharing between the bonded molecules themselves; it arises purely from the large dipole created by the H–(N,O,F) bond.

Water has the highest boiling point (100°C, versus 19.5°C for HF and −33°C for NH₃) even though HF's H–F bond has the largest dipole. The reason is network connectivity, not individual bond strength: each H₂O molecule has two H atoms to donate and two lone pairs to accept, so it can form up to four hydrogen bonds and build an extensive three-dimensional network. HF has only one donor H per molecule (forms zig-zag chains, not a 3-D network) and NH₃ has only one acceptor lone pair per molecule; both are more limited in the total hydrogen-bond network they can sustain, so less thermal energy is needed to separate the molecules.

2.3 — RDX Heat of Detonation (Bond-Energy Estimate)

Given. RDX, C₃H₆N₆O₆, is a six-membered ring of three –CH₂– groups alternating with three ring N atoms, each ring N carrying a pendant –NO₂ (nitramine) substituent. The reaction is C₃H₆N₆O₆ + &frac32;O₂ → 3N₂ + 3CO₂ + 3H₂O, with the bond energies tabulated above.

Find. $\Delta H_{\text{rxn}}$ per mole of RDX, and whether the reaction is exothermic or endothermic.

Approach. $\Delta H_{\text{rxn}}=\sum(\text{bonds broken, reactants})-\sum(\text{bonds formed, products})$. Bonds broken: count every bond in one RDX molecule plus 1.5 mol O=O. Bonds formed: count every bond in 3N₂ + 3CO₂ + 3H₂O.

1. Bonds broken in one RDX molecule + 1.5 mol O₂. The ring has 3 C atoms and 3 N atoms alternating, giving 6 ring C–N bonds; each CH₂ contributes 2 C–H bonds (6 total); each ring N bonds to its pendant nitro-N via one N–N bond (3 total); each –NO₂ group is modelled as one N–O single bond plus one N=O double bond (3 of each). Adding 1.5 mol O=O for the oxidant: $$\textstyle\sum E_{\text{broken}} = 6(305)+6(411)+3(167)+3(201)+3(607)+1.5(494)=\boxed{7962\ \text{kJ}}$$
2. Bonds formed in the products. $3\,\text{N}_2$ (3 triple bonds), $3\,\text{CO}_2$ (2 C=O each, 6 total), $3\,\text{H}_2\text{O}$ (2 O–H each, 6 total): $$\textstyle\sum E_{\text{formed}} = 3(945)+6(799)+6(459)=\boxed{10\,383\ \text{kJ}}$$
3. Heat of reaction. $$\Delta H_{\text{rxn}}=\sum E_{\text{broken}}-\sum E_{\text{formed}}=7962-10\,383=\boxed{-2421\ \text{kJ/mol RDX}}$$ Negative $\Delta H$ — the reaction is strongly exothermic, consistent with RDX's real-world role as a high explosive (order of magnitude matches published detonation-energy data, ∼2000–6000 kJ/mol depending on the reference state assumed).