Chemical Bonding and Molecular Structure — Practice Questions with Answers

The octet rule states that atoms tend to gain, lose, or share electrons in order to achieve a stable electronic configuration with eight electrons in their outermost (valence) shell, resembling the nearest noble gas.

Exception example: $BF_3$ (boron trifluoride) — boron has only 6 electrons around it in this molecule (an "electron-deficient" species), not following the octet rule, yet forms a stable compound.

Sigma bond: Formed by direct, head-on (axial) overlap of orbitals along the internuclear axis — generally stronger, due to greater degree of orbital overlap.

Pi bond: Formed by sideways (lateral) overlap of parallel, unhybridised p-orbitals above and below the internuclear axis — generally weaker than a sigma bond, due to comparatively less effective overlap.

Hybridisation is the mixing of atomic orbitals of similar energy (within the same atom) to form new, equivalent hybrid orbitals, better suited for bonding and explaining observed molecular geometries.

In methane, carbon's one 2s and three 2p orbitals mix to form four equivalent $sp^3$ hybrid orbitals, each oriented towards the corners of a tetrahedron (109.5° apart) — each hybrid orbital then overlaps with a hydrogen 1s orbital, forming four equivalent C-H sigma bonds, giving methane its characteristic tetrahedral shape.

• The shape of a molecule is determined by minimising repulsion between electron pairs (both bonding and lone pairs) around the central atom.
• Electron pairs arrange themselves around the central atom to be as far apart as possible, minimising overall repulsion.
• Lone pair-lone pair repulsion is greater than lone pair-bond pair repulsion, which is in turn greater than bond pair-bond pair repulsion.

A coordinate (dative) covalent bond forms when both shared electrons in the bond are contributed by a single atom (the "donor"), rather than one electron from each bonding atom, as in a typical covalent bond — the other atom (the "acceptor") simply accepts this shared electron pair.

Example: Formation of the ammonium ion ($NH_4^+$) — the nitrogen atom in ammonia ($NH_3$) donates its lone pair of electrons to a hydrogen ion ($H^+$, which has no electrons to contribute), forming a coordinate bond and completing the tetrahedral ammonium ion.

Resonance describes situations where a single Lewis structure cannot adequately represent a molecule/ion's actual bonding — instead, the true structure is best described as a hybrid (average/blend) of multiple valid contributing Lewis structures ("resonance structures"), differing only in electron placement, not atom positions.

For carbonate ($CO_3^{2-}$), three equivalent resonance structures can be drawn, each showing one C=O double bond and two C-O single bonds, but with the double bond positioned differently in each structure — the actual carbonate ion's bonding is best described as a resonance hybrid, with all three C-O bonds being equivalent, of intermediate bond length/order between a single and double bond.

Bond order = $\dfrac{1}{2}$(number of electrons in bonding molecular orbitals − number of electrons in antibonding molecular orbitals).

A higher bond order generally indicates a stronger, more stable bond with shorter bond length, since more electrons occupy stabilising bonding orbitals relative to destabilising antibonding orbitals — a bond order of zero indicates the molecule/species is not expected to exist (no net bonding).

Oxygen in water is $sp^3$ hybridised, with two hybrid orbitals forming O-H bonds and the remaining two hybrid orbitals holding lone pairs of electrons.

Since lone pair-lone pair repulsion is stronger than lone pair-bond pair repulsion, which is in turn stronger than bond pair-bond pair repulsion (per VSEPR theory), the two lone pairs on oxygen push the two O-H bonds closer together than the ideal tetrahedral angle would predict — compressing the H-O-H bond angle down to approximately 104.5°, rather than the unperturbed 109.5° tetrahedral angle.

Ionic compounds consist of a rigid, extended lattice structure held together by strong electrostatic (Coulombic) forces of attraction between oppositely charged ions, extending throughout the entire crystal structure.

Melting or boiling an ionic compound requires overcoming these strong, pervasive electrostatic attractive forces throughout the lattice — since these forces are considerably stronger than typical intermolecular forces (like van der Waals forces) found in molecular compounds, significantly more thermal energy (higher temperature) is required to disrupt the ionic lattice, resulting in the characteristically high melting and boiling points of ionic compounds.

Electronegativity is a measure of an atom's tendency to attract shared electrons towards itself within a chemical bond.

A large electronegativity difference between two bonded atoms typically results in an ionic bond (essentially complete electron transfer). A moderate difference results in a polar covalent bond (unequal sharing, creating partial charges). A negligible or zero difference (as between identical atoms, or atoms of very similar electronegativity) results in a non-polar covalent bond (essentially equal sharing of electrons).

Nitrogen has 5 valence electrons; each hydrogen contributes 1. Total valence electrons = $5+3(1) = 8$.

Lewis structure: Nitrogen forms 3 single bonds with 3 hydrogen atoms, and retains 1 lone pair of electrons.

Total electron pairs around nitrogen = 4 (3 bond pairs + 1 lone pair) — arranged tetrahedrally, but since one position is a lone pair (not counted in molecular shape description), the actual molecular shape is trigonal pyramidal, with a bond angle slightly less than 109.5° (approximately 107°) due to lone pair repulsion.

MO configuration for $O_2$ (16 electrons): $\sigma1s^2\,\sigma^*1s^2\,\sigma2s^2\,\sigma^*2s^2\,\sigma2p_z^2\,\pi2p_x^2=\pi2p_y^2\,\pi^*2p_x^1=\pi^*2p_y^1$

Bonding electrons = $2+2+2+2+2 = 10$; Antibonding electrons = $2+2+1+1 = 6$

Bond order $= \dfrac{1}{2}(10-6) = \dfrac{4}{2} = \mathbf{2}$

This bond order of 2 confirms the O=O double bond, and the presence of two unpaired electrons (in the $\pi^*$ orbitals) correctly predicts oxygen's observed paramagnetic behaviour — a notable early success of MO theory over simple Lewis structure predictions.

Sulfur has 6 bond pairs and 0 lone pairs, requiring 6 hybrid orbitals for bonding.

Hybridisation: $sp^3d^2$ (mixing 1 s, 3 p, and 2 d orbitals to give 6 equivalent hybrid orbitals)

Shape: With 6 bond pairs and no lone pairs, the resulting geometry that minimises repulsion is octahedral, with all F-S-F bond angles equal to 90° (adjacent) or 180° (opposite).

Fajans' rules state that covalent character in a predominantly ionic bond increases with: (1) smaller cation size, (2) higher cation charge, and (3) larger anion size — all factors that increase the cation's polarising power and/or the anion's polarisability.

Aluminium ion ($Al^{3+}$) has a notably small ionic radius combined with a high +3 charge — this combination gives $Al^{3+}$ very high polarising power, meaning it strongly distorts/polarises the electron cloud of the larger, more polarisable chloride ions nearby. This significant distortion causes substantial electron density to be pulled back towards aluminium (partial electron sharing, rather than complete transfer), resulting in $AlCl_3$ showing significant covalent character, consistent with Fajans' rules, despite the underlying electronegativity difference between Al and Cl that might otherwise suggest a purely ionic bond.

Formal charge $= (\text{valence electrons}) - (\text{non-bonding electrons}) - \dfrac{1}{2}(\text{bonding electrons})$

Nitrogen (4 bonds total: 1 double + 2 single, no lone pair): FC $= 5 - 0 - \dfrac{1}{2}(8) = 5-4 = \mathbf{+1}$

Double-bonded oxygen (2 lone pairs, 1 double bond): FC $= 6-4-\dfrac{1}{2}(4) = 6-4-2 = \mathbf{0}$

Each single-bonded oxygen (3 lone pairs, 1 single bond): FC $= 6-6-\dfrac{1}{2}(2) = 6-6-1 = \mathbf{-1}$ (for each of the two single-bonded oxygens)

Total charge check: $+1 + 0 + (-1) + (-1) = -1$ ✓, matching the overall $-1$ charge of the nitrate ion.

Each carbon atom in ethene forms 3 sigma bonds (to 2 hydrogen atoms and 1 carbon atom) and participates in 1 pi bond with the other carbon.

Hybridisation: Each carbon is $sp^2$ hybridised — mixing 1 s and 2 p orbitals to form 3 equivalent $sp^2$ hybrid orbitals (used for the 3 sigma bonds), while the remaining unhybridised p-orbital (perpendicular to the plane of the $sp^2$ orbitals) forms the pi bond via sideways overlap with the corresponding p-orbital on the adjacent carbon.

Geometry: Each carbon is trigonal planar, with bond angles of approximately 120°, and the entire ethene molecule is planar overall, due to the specific geometric requirement for effective pi bond (p-orbital) overlap.

$H_2$ (2 electrons): MO configuration $\sigma1s^2$ — both electrons occupy the bonding molecular orbital. Bond order $= \dfrac{1}{2}(2-0) = 1$, indicating a stable single bond, consistent with the well-known existence of $H_2$.

$He_2$ (4 electrons): MO configuration $\sigma1s^2\,\sigma^*1s^2$ — 2 electrons occupy the bonding orbital, but the remaining 2 must occupy the antibonding orbital. Bond order $= \dfrac{1}{2}(2-2) = 0$.

Since the bond order for $He_2$ is exactly zero, there is no net stabilising bonding interaction — this correctly predicts, based on molecular orbital theory, that a stable $He_2$ molecule should not exist, consistent with experimental observation.

Total electron pairs around chlorine = 5 (3 bond pairs + 2 lone pairs) — the underlying electron pair geometry (for 5 total pairs) is trigonal bipyramidal.

To minimise repulsion, the 2 lone pairs preferentially occupy the equatorial positions (which have less severe 90° repulsive interactions compared to axial positions).

With the 2 lone pairs occupying 2 of the 3 equatorial positions, the resulting molecular shape (considering only bonded atoms) is T-shaped, with slight distortion from ideal angles due to the strong repulsive influence of the two lone pairs.

All three molecules have a central atom that is $sp^3$ hybridised, with 4 total electron pairs arranged in an underlying tetrahedral geometry — but they differ in the number of lone pairs versus bond pairs.

Methane (0 lone pairs, 4 bond pairs): No lone pair repulsion effect, giving the ideal, undistorted tetrahedral angle of 109.5°.

Ammonia (1 lone pair, 3 bond pairs): One lone pair's stronger repulsion pushes the three N-H bonds slightly closer together, compressing the angle to approximately 107°.

Water (2 lone pairs, 2 bond pairs): Two lone pairs exert even greater combined repulsive effect, compressing the H-O-H angle further still, to approximately 104.5°.

This progressive decrease in bond angle (109.5° → 107° → 104.5°) directly reflects the increasing number of lone pairs (0 → 1 → 2) and their progressively greater cumulative repulsive effect on the remaining bond pairs.

Bond order and bond length are inversely related — a higher bond order (more shared electron pairs holding the atoms together, one sigma plus additional pi bonds) results in a stronger, shorter bond, since the increased electron density between the nuclei pulls them closer together and holds them more tightly.

$N_2$ has bond order 3 (one sigma + two pi bonds) — the highest of the three, giving the shortest, strongest N≡N bond.

$O_2$ has bond order 2 (one sigma + one pi bond) — intermediate bond order, giving an intermediate O=O bond length, longer than $N_2$'s but shorter than $F_2$'s.

$F_2$ has bond order 1 (single sigma bond only, no pi bonds) — the lowest bond order of the three, giving the longest, weakest F-F bond. This clear inverse relationship between bond order and bond length explains the observed trend: $N_2 < O_2 < F_2$ in bond length.