Chemical Bonding and Molecular Structure

The chapter builds bonding theory in the same order the Class 11 NCERT text does, and this article follows that map so you can revise section by section. Start with the Kossel-Lewis picture: Lewis symbols, the octet rule, and how ionic bonds (electron transfer) and covalent bonds (electron sharing) arise from atoms chasing a noble-gas electron count. Move to writing Lewis structures, using formal charge to choose between candidates, and where the octet rule breaks down. Then see what stabilises an ionic solid (lattice enthalpy), the four bond parameters, and resonance and bond polarity. Next is VSEPR theory, the simplest tool for molecular shape. Then the chapter goes quantum mechanical: valence bond theory (orbital overlap, sigma and pi bonds, hybridisation sp through sp3d2) and molecular orbital theory (bond order, and uniquely, oxygen's paramagnetism). It closes with hydrogen bonding, followed by worked examples, derivations, common traps and a cheat sheet.

Kossel-Lewis approach: Lewis symbols and the octet rule

In 1916 Kossel and Lewis independently gave the first electronic explanation of valence, built on one observation: noble gases are chemically inert, and every other element reacts in a way that pushes it toward a noble gas's electron count. Lewis pictured the atom as a positively charged kernel (nucleus plus inner electrons) surrounded by an outer shell holding at most eight electrons, an octet, and represented an atom's valence electrons with dots around its symbol, a Lewis symbol; the number of dots gives the atom's common valence (or eight minus that number). Kossel showed that sodium and chlorine, flanking the noble gases neon and argon, can each reach a noble-gas configuration by a single electron transfer, and the resulting oppositely charged ions are held together by electrostatic attraction, an electrovalent or ionic bond.

where: the bracketed symbol is the electron configuration left after the change; an ion's electrovalence equals the number of unit charges it carries, so calcium in CaF2 has electrovalence +2 and each fluorine has -1. When two identical or near-identical atoms meet, as in Cl2, H2 or N2, neither can strip an electron from the other, so instead they share a pair, contributing one each, and both attain an octet (or, for hydrogen, a duplet matching helium). Langmuir in 1919 named this a covalent bond. Sharing two electron pairs gives a double bond (as in O=C=O), and sharing three gives a triple bond (as in N2 or ethyne). The rule that atoms combine, by transfer or sharing, until every atom's valence shell holds eight electrons (helium excepted) is the octet rule, the organising principle behind every Lewis structure. Writing a correct Lewis structure follows a fixed procedure: count total valence electrons (sum each atom's group valence, add one electron per unit negative charge, subtract one per unit positive charge); pick a skeletal arrangement with the least electronegative atom (never hydrogen) central, so nitrogen is central in NF3 and carbon in CO32-; place one shared pair between every bonded pair of atoms and complete octets on the outer atoms first with lone pairs; then, if the central atom still lacks an octet, convert outer lone pairs into additional shared pairs (double or triple bonds) until every period-2 atom reaches eight.

Formal charge and the real limits of the octet rule

A Lewis structure does not, by itself, tell you which structure is best when several are chemically reasonable. Formal charge is the accounting tool that decides: the charge an atom would carry if every bonding pair were split exactly equally between the two bonded atoms, regardless of real electronegativity differences.

where: $V$ is the free atom's valence electrons; $L$ is the number of non-bonding (lone-pair) electrons the atom owns in that structure; $B$ is the total number of electrons it shares in bonds (so $\tfrac{1}{2}B$ is its number of bonds). In one resonance structure of ozone, O3, with a central oxygen double-bonded to one terminal oxygen and single-bonded to the other, the central atom works out to $FC = 6 - 2 - \tfrac{1}{2}(6) = +1$, the doubly bonded terminal oxygen to $FC = 6-4-2=0$, and the singly bonded terminal oxygen to $FC = 6-6-1=-1$. Formal charges are not real, separated charges; they are a device for comparing candidate structures, and the structure with the smallest formal charges, with any negative charge on the more electronegative atom, is closest to the true structure. The octet rule is not universal, and three genuine classes of exception matter for exams. An incomplete octet occurs when the central atom simply lacks enough valence electrons to reach eight: lithium, beryllium and boron form LiCl, BeH2 (four electrons around Be) and BCl3 or BF3 (six electrons around B). Odd-electron molecules such as NO and NO2 have an odd total electron count, so one atom is stuck with an unpaired electron, which is exactly why they are reactive free radicals. An expanded octet appears from period 3 onward, where empty 3d orbitals become available, letting the central atom in PF5, SF6 or H2SO4 hold ten or twelve electrons. The octet rule is also silent about molecular shape and energy, and cannot explain why xenon and krypton still form real compounds such as XeF2 and KrF2.

The ionic bond and lattice enthalpy

Whether an ionic bond forms, and how strong it is, depends on two energy questions: how easily the metal loses an electron and the non-metal gains one (ionisation enthalpy and electron gain enthalpy), and how much energy is released when the resulting gaseous ions pack into an ordered crystal lattice. That second quantity is lattice enthalpy, the energy required to completely separate one mole of a solid ionic compound into its gaseous constituent ions.

where: ionisation enthalpy $\Delta_i H$ is always endothermic, while electron gain enthalpy $\Delta_{eg} H$ is usually exothermic, though it can be endothermic. The crucial point, worked in full arithmetic in the first derivation below using sodium chloride's real numbers, is that the sum of ionisation and electron gain enthalpy for forming isolated gaseous ions can be positive overall, and yet the compound still forms readily, because the enormous negative lattice enthalpy released when the gaseous ions condense into a crystal more than compensates. The true measure of an ionic compound's stability is its lattice enthalpy, not merely whether the isolated ions reach an octet.

Bond parameters: length, angle, enthalpy and order

Four measurable numbers characterise every covalent bond. Bond length is the equilibrium distance between two bonded nuclei, and to a good approximation it is additive: each atom contributes a fixed covalent radius, and the bond length is simply the sum.

where: $R$ is the bond length, and $r_A$, $r_B$ are the covalent radii of the two bonded atoms. A related quantity, the van der Waals radius, half the distance between two non-bonded atoms of the same element in a solid, is always larger than the covalent radius. Bond angle, the angle between two bonds sharing a central atom, is the most direct experimental clue to geometry; water's H-O-H angle of 104.5 degrees and ammonia's H-N-H angle of 107 degrees are two of the most frequently tested numbers in this chapter. Bond enthalpy is the energy needed to break one mole of a bond in the gas phase (435.8 kJ/mol for H-H, 946.0 kJ/mol for the N2 triple bond); where successive bonds of the same nominal type break with different energies (the first O-H bond in water costs 502 kJ/mol, the second only 427 kJ/mol), chemists quote a mean bond enthalpy, here $(502+427)/2 = 464.5$ kJ/mol. Bond order is simply the number of shared electron pairs: 1 for H-H, 2 for O=O, 3 for N2 or CO. The single most useful correlation here: as bond order increases, bond enthalpy increases and bond length decreases.

Resonance: when one Lewis structure is not enough

Some molecules cannot be captured accurately by any single Lewis structure. Ozone, O3, is the textbook case: you can draw the terminal double bond on either oxygen, giving two equally valid canonical forms, yet experiment finds both oxygen-oxygen bonds in real O3 have the identical length, 128 pm, sitting between a normal O-O single bond (148 pm) and a normal O=O double bond (121 pm). Neither canonical form alone matches this. The resolution is resonance: when several Lewis structures of similar energy and the same skeletal arrangement can all be drawn for one species, the true structure is a resonance hybrid of all of them, shown with a double-headed arrow. The carbonate ion CO32- is a hybrid of three canonical forms, which is why all three carbon-oxygen bonds in the real ion are identical and intermediate in length; carbon dioxide's measured C-O bond length of 115 pm, between a C=O double bond (121 pm) and a C triple-bonded to O (110 pm), is explained the same way. Three points about resonance cost the most exam marks when misunderstood. The canonical forms have no independent physical existence; the molecule is never one structure for part of the time and another for the rest (that is tautomerism, a genuinely different equilibrium between distinct species); the molecule has, at every instant, exactly one real structure, the hybrid, which cannot be drawn with a single set of dots and lines. Resonance always stabilises a molecule, since the hybrid's energy is lower than any single canonical form's, and it averages bond lengths and bond orders across equivalent positions.

Bond polarity and dipole moment

No real bond is purely ionic or purely covalent. When two identical atoms share an electron pair, as in H2 or N2, the pair sits exactly midway and the bond is non-polar covalent. When the atoms differ in electronegativity, as in HF, the pair is pulled toward the more electronegative atom, producing a polar covalent bond with partial charges on each end and a dipole moment.

where: $\mu$ is the dipole moment, usually reported in debye ($1\ \text{D} = 3.33564 \times 10^{-30}\ \text{C m}$); $Q$ is the separated partial charge; $r$ is the distance between the charge centres. By chemistry convention the dipole is drawn as a crossed arrow, cross on the positive end, arrowhead on the negative end. For a polyatomic molecule the net dipole is the vector sum of every bond dipole, so geometry matters as much as electronegativity: BeF2's two opposed Be-F dipoles cancel exactly (zero net moment); BF3's three B-F dipoles at 120 degrees also sum to zero; water's two O-H dipoles, bent at 104.5 degrees, reinforce, giving a net moment of 1.85 D. A subtler case is NH3 versus NF3: although fluorine is more electronegative, NH3's dipole (1.47 D) exceeds NF3's (0.24 D), because in NH3 the lone pair's own contribution reinforces the bond dipoles, while in NF3 it opposes and partly cancels them.

VSEPR theory: reading shape off electron pairs

Neither the Lewis structure nor formal charge tells you a molecule's three-dimensional shape. The Valence Shell Electron Pair Repulsion (VSEPR) theory, proposed by Sidgwick and Powell in 1940 and refined by Nyholm and Gillespie in 1957, fills that gap: a molecule's shape depends only on the total number of electron pairs, bonding or lone, around the central atom; these pairs, being negatively charged clouds, arrange themselves to maximise the distance between them, as if constrained to a sphere centred on the central atom; a multiple bond counts as one "super pair"; and VSEPR applies to any resonance structure equally. The key refinement is that lone pairs and bonding pairs are not equivalent: a lone pair, held by only one nucleus, occupies more angular space than a bonding pair pulled taut between two nuclei, giving a strict repulsion order, lp-lp greater than lp-bp greater than bp-bp. Every time a lone pair replaces a bonded atom in a symmetric geometry, the remaining bond angles compress below the ideal value. Four electron pairs with none lone give perfect tetrahedral CH4 at 109.5 degrees; one lone pair (NH3) squeezes the angle to 107 degrees; two lone pairs (H2O) squeeze it further to 104.5 degrees. In trigonal bipyramidal geometries a lone pair always prefers an equatorial position (only two 90-degree lp-bp repulsions there, versus three if axial); this single rule correctly predicts the seesaw shape of SF4, the T-shape of ClF3, and the bent shape of XeF2.

Valence bond theory: orbital overlap, and why Lewis alone is not enough

Lewis theory cannot explain why the H-H bond (435.8 kJ/mol, 74 pm) is so much stronger and shorter than the F-F bond (155 kJ/mol, 144 pm), even though both are drawn as an identical single shared pair, nor can it explain molecular shape. Valence bond (VB) theory, introduced by Heitler and London in 1927 and developed by Pauling, supplies the missing physics: a covalent bond forms when the atomic orbitals of two approaching atoms partially overlap, pairing electrons of opposite spin. As two hydrogen atoms approach, new attractive forces (each nucleus for the other's electron) outweigh the new repulsive forces, so potential energy falls, reaches a minimum at the equilibrium bond length (74 pm), then rises sharply if pushed closer; the depth of that minimum is the bond enthalpy. Greater orbital overlap gives a stronger bond. Overlap comes in two geometrically distinct flavours. A sigma bond forms by head-on overlap along the internuclear axis (s-s, s-p, or p-p), giving maximal overlap and the strongest, freely rotatable single bond. A pi bond forms by sideways overlap of two parallel p orbitals perpendicular to the axis, giving two lobes above and below the bond and, with less overlap, a weaker bond that blocks free rotation. Every single bond is one sigma bond; every double bond is one sigma plus one pi; every triple bond is one sigma plus two pi bonds. Sigma bonds are always stronger than pi bonds, which is why the pi bond of a double bond is the reactive site in addition reactions. Simple, unhybridised orbital overlap cannot reproduce observed geometry: carbon's three singly occupied 2p orbitals sit at 90 degrees to each other, so naive overlap predicts 90-degree HCH angles in methane against the measured 109.5 degrees, and similarly wrong angles for ammonia and water. Resolving this is exactly what hybridisation does.

Hybridisation: sp, sp2, sp3, sp3d, sp3d2

Pauling proposed that atomic orbitals of similar energy on one atom can hybridise into a new set of equivalent orbitals pointing in the directions needed to match observed bond angles, and it is these hybrid orbitals that overlap with neighbouring atoms. The hybrid orbitals overlap more effectively than pure atomic orbitals, form stronger bonds, and orient to minimise repulsion, so hybridisation type directly predicts molecular geometry. In sp hybridisation, one s and one p orbital mix into two equivalent orbitals 180 degrees apart (diagonal hybridisation): BeCl2 is linear. In sp2 hybridisation, one s and two p orbitals give three orbitals at 120 degrees: BCl3 and BF3 are trigonal planar, and the unhybridised p orbital on each carbon in ethene forms the C=C pi bond. In sp3 hybridisation, one s and three p orbitals give four orbitals at 109.5 degrees: methane is the parent example, and the same scheme, with one or two positions occupied by lone pairs instead of bonds, gives the pyramidal shape of ammonia (107 degrees) and the bent shape of water (104.5 degrees). From period 3 onward, d orbitals join the mix. In sp3d hybridisation, five orbitals point to a trigonal bipyramid, as in PCl5; the five positions are not equivalent, three equatorial bonds at 120 degrees and two axial bonds at 90 degrees, and because axial bonds suffer more neighbouring repulsion they are slightly longer and weaker, making them the reactive site. In sp3d2 hybridisation, six equivalent orbitals point to an octahedron, as in SF6. A further scheme, dsp2 hybridisation, gives the square planar geometry of complex ions such as [Ni(CN)4]2- and [PtCl4]2-.

Molecular orbital theory: bonding, antibonding and bond order

Valence bond theory keeps electrons localised between specific atom pairs and explains bond formation well, but it cannot predict magnetic behaviour. Molecular orbital (MO) theory, developed by Hund and Mulliken in 1932, instead treats every electron as occupying a molecular orbital spread across two or more nuclei. These are built by the linear combination of atomic orbitals (LCAO): wavefunctions $\psi_A$ and $\psi_B$ combine by addition into a bonding orbital, and by subtraction into an antibonding orbital.

where: $\sigma$ is the bonding combination, where the waves interfere constructively, piling electron density between the nuclei and lowering energy below either parent orbital; $\sigma^*$ is the antibonding combination, where the waves interfere destructively, creating a node between the nuclei and raising energy. Combination requires comparable energy, matching symmetry about the internuclear axis (2pz combines with 2pz, not with 2px or 2py), and appreciable overlap. Orbitals symmetric about the bond axis give sigma molecular orbitals; sideways p-orbital combinations give pi molecular orbitals. For second-period homonuclear diatomics, eight atomic orbitals combine into eight molecular orbitals, filled by the aufbau principle, Pauli's exclusion principle and Hund's rule, in one of two experimentally determined orders. For O2, F2, Ne2: $\sigma1s < \sigma^*1s < \sigma2s < \sigma^*2s < \sigma2p_z < (\pi2p_x = \pi2p_y) < (\pi^*2p_x = \pi^*2p_y) < \sigma^*2p_z$. For Li2 through N2, sigma2pz sits above the pi2p pair instead. Bond order then measures net bonding strength:

where: $N_b$, $N_a$ are the electrons in bonding and antibonding orbitals. Positive bond order predicts a stable molecule; zero or negative predicts it does not exist, as for He2 and Be2. Integral bond orders of 1, 2, 3 match Lewis theory's single, double, triple bonds, and higher bond order still tracks greater bond enthalpy and shorter length. Magnetic behaviour follows directly: fully paired orbitals give diamagnetism, any singly occupied orbital gives paramagnetism. This is where MO theory succeeds where Lewis theory fails: O=O shows every electron paired, predicting diamagnetism, but correct MO filling forces two electrons by Hund's rule into separate, singly occupied pi* orbitals, correctly predicting O2's observed paramagnetism.

Hydrogen bonding

When hydrogen is covalently bonded to fluorine, oxygen or nitrogen, the shared pair is pulled strongly toward that atom, leaving the hydrogen nucleus, essentially a bare proton, carrying a partial positive charge ($\delta^+$) while the electronegative atom carries $\delta^-$. Because the exposed hydrogen nucleus is so small, it can form a genuine additional attraction with a lone pair on an F, O or N atom of a neighbouring molecule (or a distant part of the same molecule); that attraction is the hydrogen bond, drawn as a dotted line, weaker than a covalent bond but distinctly stronger than an ordinary van der Waals attraction. It is strongest in the solid state and weakest in the gas phase, and its cumulative effect on boiling point, density and solubility is large: water's high boiling point relative to H2S and ice's lower density than liquid water are both direct consequences. Hydrogen bonds come in two types. An intermolecular hydrogen bond links two separate molecules, as between successive HF or water molecules. An intramolecular hydrogen bond forms within a single molecule, as in ortho-nitrophenol, where the phenolic O-H bonds to the nearby nitro group's oxygen, which is also why ortho-nitrophenol boils lower than its para isomer, since the intramolecular bond uses up hydrogen-bonding capacity that would otherwise link separate molecules.

Board level - Lewis dot structure, bond count and formal charge of CO (NCERT Problem 4.1 extended)

Write the Lewis dot structure of the carbon monoxide molecule, CO, and confirm it using formal charge.

Step 1. Count total valence electrons. Carbon's outer configuration is $2s^2 2p^2$ (4 valence electrons) and oxygen's is $2s^2 2p^4$ (6 valence electrons), giving $4 + 6 = 10$ electrons available.

Step 2. Write the skeletal structure, C-O, place a single shared pair between them, and complete the octet on oxygen with three lone pairs (6 electrons), using $2+6=8$ electrons and leaving $10-8=2$ electrons as a lone pair on carbon. Carbon now has only one bond pair and one lone pair, well short of an octet.

Step 3. Since carbon's octet is incomplete, convert two of oxygen's lone pairs into two additional shared pairs, forming a triple bond between C and O. Carbon now shares 6 bonding electrons plus 1 lone pair for a full octet, and oxygen shares 6 bonding electrons plus 1 lone pair for a full octet; total electrons used: 6+2+2=10, matching Step 1.

Step 4. Verify with formal charge. Carbon: $FC = 4-2-3 = -1$. Oxygen: $FC = 6-2-3=+1$. These sum to zero, matching the neutral molecule, and this is indeed the accepted structure of CO despite the unusual negative formal charge on carbon. Answer: C is triple-bonded to O, each atom carries one lone pair, and the formal charges are -1 on carbon and +1 on oxygen.

Competitive level (JEE / NEET) - bond order and magnetic character of O2, O2+ and O2- by molecular orbital theory

Using molecular orbital theory, find the bond order of O2, O2+ (formed by removing one electron from O2) and O2- (formed by adding one electron to O2), and state which are paramagnetic.

Step 1. Start from neutral O2's configuration (O2/F2/Ne2 ordering): $(\sigma2s)^2(\sigma^*2s)^2(\sigma2p_z)^2(\pi2p_x^2=\pi2p_y^2)(\pi^*2p_x^1=\pi^*2p_y^1)$, giving $N_b=8$, $N_a=4$, bond order $=2$, with two unpaired pi* electrons, hence paramagnetic.

Step 2. For O2+, remove one electron from a singly occupied pi* orbital (removing from an antibonding orbital, when available, is always favoured). This leaves $N_b=8$, $N_a=3$, bond order $=2.5$. One unpaired electron remains, so O2+ is paramagnetic, and its higher bond order correctly predicts a shorter, stronger bond than O2.

Step 3. For O2-, add one electron, pairing up one pi* orbital: $N_b=8$, $N_a=5$, bond order $=1.5$. One unpaired electron remains, so O2- (superoxide) is also paramagnetic, with a lower bond order predicting a longer, weaker O-O bond than in neutral O2.

Step 4. Rank all three by bond order: O2+ (2.5) strongest, O2 (2.0) next, O2- (1.5) weakest. Answer: bond orders are O2+ = 2.5, O2 = 2, O2- = 1.5; all three species are paramagnetic. Timing note: about 90 seconds once the O2 configuration is memorised. Start from O2's known bond order of 2, then apply "remove from antibonding first raises bond order by 0.5, add to antibonding first lowers it by 0.5" instead of rebuilding the configuration each time. Do not assume O2- is diamagnetic just because it gained an electron; check which orbital it lands in.

Board level - VSEPR shape, hybridisation and bond angle of ClF3

Predict the shape, the hybridisation of the central chlorine atom, and the approximate bond angles in chlorine trifluoride, ClF3.

Step 1. Count central chlorine's valence electron pairs. Chlorine has 7 valence electrons; each of the three fluorine atoms contributes 1 electron to a shared pair, adding 3; total $7+3=10$ electrons, or 5 electron pairs.

Step 2. Split the pairs into bonding and lone: 3 bonding pairs (one to each F) and $5-3=2$ lone pairs on chlorine.

Step 3. Five electron pairs give a trigonal bipyramidal parent geometry, corresponding to sp3d hybridisation of chlorine. Both lone pairs occupy equatorial positions (lower-repulsion sites, per the VSEPR derivation above), leaving the three fluorines in a T-shaped molecular arrangement.

Step 4. Predict bond angle distortion. The two equatorial lone pairs push the bonded fluorines together via strong lp-bp repulsion, bending the ideal 90-degree axial-equatorial angle down to about 87.5 degrees. Answer: ClF3 is T-shaped, the chlorine is sp3d hybridised, and the F-Cl-F bond angles are slightly less than 90 degrees (about 87.5 degrees).

Competitive level (JEE / NEET) - resolving water's net dipole moment from its bond dipoles and bond angle

Each O-H bond in water has an individual bond dipole moment of 1.60 D directed from H (positive end) to O (negative end), and the H-O-H bond angle is 104.5 degrees. Using vector addition, estimate the molecule's net dipole moment, and compare it to the experimentally measured value of 1.85 D.

Step 1. Place oxygen at the origin with the two O-H bonds symmetric about the vertical bisector axis, each bond making a half-angle of $104.5^\circ/2 = 52.25^\circ$ with that bisector, each dipole vector of magnitude 1.60 D.

Step 2. Resolve each bond dipole into components along the bisector and perpendicular to it. The perpendicular components are equal and opposite by symmetry, so they cancel; only the along-axis components add.

Step 3. Each along-bisector component is $1.60 \times \cos(52.25^\circ) = 1.60 \times 0.6129 = 0.980\ \text{D}$.

Step 4. Add the two along-axis components, which reinforce: $\mu_{\text{net}} = 2 \times 0.980 = 1.96\ \text{D}$. Answer: the vector-addition estimate gives $\mu_{\text{net}} \approx 1.96$ D, reasonably close to the measured 1.85 D, the gap explained by the simplified fixed bond-dipole value not fully capturing the lone pairs' own contribution. Timing note: under two minutes if you remember that only components along the bisector survive by symmetry, and each component is $\mu_{\text{bond}}\cos(\theta/2)$, not $\mu_{\text{bond}}\cos\theta$. The most common JEE slip is using the full bond angle instead of the half-angle, silently halving the wrong quantity; always sketch the bisector first.

Confusing formal charge with real, physically separated charge is the single most frequent scoring error: formal charge is a bookkeeping convention assuming perfectly equal electron sharing, and it never represents the true, unequal charge distribution electronegativity actually produces (dipole moment measures that instead). Students regularly forget that VSEPR's "no lone pair" shape names describe the arrangement of bonded atoms only, while the underlying electron-pair geometry, including lone pairs, is what determines bond angles; always count total electron pairs first, then name the shape from bonded-atom positions alone. Another common slip is applying the lp-lp greater than lp-bp greater than bp-bp order backwards, predicting angle expansion instead of compression; every lone pair always shrinks nearby bond angles, never expands them. In hybridisation questions, students forget that sp3 hybridisation gives three different observed shapes (tetrahedral, pyramidal, bent) depending on how many hybrid positions hold lone pairs rather than bonds. In molecular orbital questions, applying the wrong energy-level ordering, sigma2pz-below-pi2p (correct for Li2 through N2) instead of the reversed order for O2, F2 or Ne2, silently produces wrong bond orders and magnetic predictions for exactly the elements most tested. Finally, students often treat resonance structures as if the molecule genuinely oscillates between them, rather than understanding the hybrid is the one true, unchanging structure.

• Write a correct Lewis dot structure for any small neutral molecule or simple polyatomic ion, correctly accounting for total valence electrons including charge adjustments.
• Calculate the formal charge on any atom in a Lewis structure and use it to choose the best structure among several candidates.
• State the three genuine categories of octet rule exception (incomplete octet, odd-electron molecule, expanded octet) with one correct example of each.
• Explain, using ionisation enthalpy, electron gain enthalpy and lattice enthalpy, why an ionic solid can be highly stable even when forming its isolated gaseous ions is energetically unfavourable overall.
• Define and correctly use all four bond parameters, bond length, bond angle, bond enthalpy and bond order, and state how bond order correlates with the other three.
• Identify when a species requires a resonance description, draw its canonical forms, and explain clearly why the hybrid, not any single canonical form, is the real structure.
• Apply the VSEPR algorithm (count total electron pairs, find the parent geometry, place lone pairs to minimise repulsion, then name the resulting molecular shape) to predict the shape and approximate bond angles of any AXnEm species.
• Identify the hybridisation (sp, sp2, sp3, sp3d, sp3d2) of a central atom from its number of sigma-bonded neighbours plus lone pairs, and connect that hybridisation to the resulting molecular geometry.
• Distinguish sigma and pi bonds by their overlap geometry and relative strength, and correctly count the sigma and pi bonds in any molecule containing single, double or triple bonds.
• Write the molecular orbital electronic configuration of any second-period homonuclear diatomic (using the correct energy ordering for that element), then calculate its bond order and predict whether it is paramagnetic or diamagnetic.

Ask the tutor: Want to work through drawing Lewis structures and predicting VSEPR shapes for five more molecules of your choice, with formal-charge checks and hybridisation stated for each?