Chemical Bonding and Molecular Structure

JEE Main and NEET Chemistry notes. RelaxPrep is not affiliated with the NTA or the IITs.

Revision sheet

Ionic bond: complete electron transfer, held by electrostatic attraction. Lattice energy grows with ion charges and falls with ion size, $U \propto \dfrac{q_+ q_-}{r_+ + r_-}$.
Fajans' rules: covalent character rises with a small, highly charged cation, a large, highly charged anion, and a cation with an 18-electron (pseudo noble gas) shell such as $\ce{Ag+}$ or $\ce{Cu+}$.
Formal charge $= V - L - \tfrac{1}{2}B$. The best Lewis structure has charges closest to zero, with any negative charge on the more electronegative atom.
Resonance: the real structure is one hybrid, not a flip between forms. Bond lengths are equal and intermediate, and the hybrid is more stable than any single form.
Steric number (SN) $= \sigma$ bonds $+$ lone pairs on the central atom. SN 2, 3, 4, 5, 6, 7 gives $sp$, $sp^2$, $sp^3$, $sp^3d$, $sp^3d^2$, $sp^3d^3$.
Repulsion order: lone pair-lone pair
gt;$ lone pair-bond pair
gt;$ bond pair-bond pair. Lone pairs sit equatorial in a trigonal bipyramid.
Shapes by (SN, lone pairs): (5,1) see-saw, (5,2) T-shaped, (5,3) linear, (6,1) square pyramidal, (6,2) square planar.
A $\sigma$ bond forms by end-on (axial) overlap. A $\pi$ bond forms by sideways overlap and needs a $\sigma$ bond already in place. A double bond is $1\sigma + 1\pi$, a triple bond $1\sigma + 2\pi$.
Bond order $= \tfrac{1}{2}(N_b - N_a)$. Higher bond order means a shorter, stronger bond.
$\sigma 2p$ lies below $\pi 2p$ only for $\ce{O2}$, $\ce{F2}$ and heavier. For species with 14 or fewer total electrons ($\ce{B2}$, $\ce{C2}$, $\ce{N2}$ and their ions) $\pi 2p$ lies below $\sigma 2p$.
$\ce{O2}$ and $\ce{B2}$ are paramagnetic with 2 unpaired electrons. $\ce{N2}$, $\ce{C2}$, $\ce{F2}$ and $\ce{CO}$ are diamagnetic.
Dipole moment $\mu = q \times d$, a vector, in debye. $\ce{CO2}$, $\ce{BF3}$, $\ce{CH4}$, $\ce{CCl4}$, $\ce{XeF4}$ and $\ce{SF6}$ have zero net moment despite polar bonds.
Hydrogen bond: $\ce{H}$ bonded to $\ce{N}$, $\ce{O}$ or $\ce{F}$ attracted to a lone pair on another $\ce{N}$, $\ce{O}$ or $\ce{F}$. Strength about 5 to 40 kJ/mol, much weaker than a covalent bond.
Bond enthalpy trend: $\ce{C-C}$ about 348, $\ce{C=C}$ about 612, $\ce{C#C}$ about 837 kJ/mol, with lengths 154, 134, 120 pm.

Formal charge: $\text{FC} = V - L - \tfrac{1}{2}B$ ($V$ valence electrons of the free atom, $L$ electrons in lone pairs, $B$ electrons in bonds.)
Steric number of the central atom: $\text{SN} = \tfrac{1}{2}\,(V + M + a - c)$ ($M$ monovalent atoms (H, halogens), $a$ anionic charge, $c$ cationic charge. Doubly bonded O or S count as 0.)
Lattice energy trend: $U \propto \dfrac{q_+\,q_-}{r_+ + r_-}$
Bond order (MOT): $\text{BO} = \dfrac{N_b - N_a}{2}$
Dipole moment: $\mu = q \times d$ (1 D $= 3.336\times10^{-30}$ C m.)
Resultant of two bond moments: $\mu = \sqrt{\mu_1^2 + \mu_2^2 + 2\mu_1\mu_2\cos\theta}$
Percent ionic character: $\% \text{ ionic} = \dfrac{\mu_{\text{observed}}}{q\,d} \times 100$ ($q$ is the full electronic charge $e$ and $d$ the bond length.)
Spin-only magnetic moment: $\mu_s = \sqrt{n(n+2)}\ \text{BM}$ ($n$ is the number of unpaired electrons.)
Reaction enthalpy from bond enthalpies: $\Delta H \approx \sum \text{(bonds broken)} - \sum \text{(bonds formed)}$

Topics

Ionic bonding and lattice energy

An ionic bond forms when one atom gives up electrons completely and another takes them, leaving oppositely charged ions that attract each other. Metals with low ionisation enthalpy (groups 1 and 2) pair with non-metals of high electron affinity (groups 16 and 17). The attraction acts equally in all directions, so a solid ionic compound is a three-dimensional lattice and not a set of separate molecules. The formula such as $\ce{NaCl}$ gives only the simplest ratio of ions.

Forming gaseous ions from atoms usually costs energy, so the driving force is the energy released when the gaseous ions pack into a lattice. Lattice enthalpy is the energy released when one mole of the solid forms from its gaseous ions (the magnitude is often quoted as a positive number for the reverse process, breaking the lattice). It cannot be measured directly. It is found indirectly with a Born-Haber cycle, which applies Hess's law to sublimation, ionisation, bond dissociation, electron gain and formation.

Lattice energy depends mainly on the ionic charges and sizes: it rises as $q_+ q_-$ increases and as $r_+ + r_-$ decreases. That is why $\ce{MgO}$ (2+ and 2-) is far more strongly bound than $\ce{NaCl}$ (1+ and 1-), and why $\ce{LiF}$ has a higher lattice energy than $\ce{CsI}$.

Properties follow from the lattice. Ionic solids have high melting and boiling points, are hard but brittle (a shifted layer brings like charges face to face), do not conduct as solids because the ions are fixed, and conduct when molten or dissolved. Solubility in water depends on whether hydration enthalpy of the ions can pay for the lattice enthalpy.

Key points

Favourable conditions: low ionisation enthalpy of the metal, high electron affinity of the non-metal, high lattice enthalpy.
Lattice enthalpy rises with ion charge and falls with ion size.
A Born-Haber cycle is Hess's law applied to a closed loop of steps, with lattice enthalpy as the unknown.
Ionic solids are brittle and conduct only when molten or in solution.
A compound dissolves readily when the hydration enthalpy of its ions is comparable to or larger than its lattice enthalpy.

Lattice energy dependence: $U \propto \dfrac{q_+\,q_-}{r_+ + r_-}$
Born-Haber cycle for NaCl(s): $\Delta_f H = \Delta_{sub}H + \tfrac12 \Delta_{bond}H + \Delta_{ie}H + \Delta_{eg}H + \Delta_{lattice}H$ (All terms in kJ/mol. The lattice term is negative when defined as formation of the solid from gaseous ions.)

Common traps

Lattice enthalpy has two sign conventions. As formation of the solid from gaseous ions it is negative (exothermic); as breaking the solid into gaseous ions it is positive. Check which one the question uses.
Electron gain enthalpy of Cl is $-349$ kJ/mol (exothermic), but for $\ce{O^{2-}}$ the second step $\ce{O^- -> O^{2-}}$ is endothermic. The lattice energy of the oxide pays for it.
Compounds with a large charge product are not always the most soluble or least soluble. Solubility compares lattice enthalpy with hydration enthalpy, not lattice enthalpy alone.
Ionic compounds are not hard in the sense of being tough. They are hard but brittle.

Shortcuts

Compare charges first, then radii. To rank lattice energies, compare $q_+ q_-$ first, since it changes by factors of 2, 3 or 4 (for example $\ce{NaCl}$ has $1\times1$ and $\ce{MgO}$ has $2\times2$). Only when the charge product is equal compare $r_+ + r_-$, which changes by well under 30 percent down a group.

Fajans' rules and covalent character in ionic bonds

No bond between two different atoms is perfectly ionic. A small, highly charged cation can pull the electron cloud of a neighbouring anion towards itself, distorting it. This distortion is called polarisation, and the cation's ability to cause it is its polarising power. The more the anion cloud is drawn into the region between the nuclei, the more the bond resembles a shared-electron bond.

Fajans' rules list what increases polarisation, and hence covalent character: a small cation, a large cation charge, a large anion, a large anion charge, and a cation whose outer shell is not a noble gas shell. A cation with an $18$-electron outer shell ($\ce{Ag+}$, $\ce{Cu+}$, $\ce{Zn^{2+}}$, $\ce{Hg^{2+}}$) shields its nuclear charge poorly, so it polarises more than a noble-gas-type cation of nearly the same size, such as $\ce{Na+}$ or $\ce{K+}$.

Observable consequences are lower melting points, lower solubility in water, higher solubility in non-polar solvents, and deeper colour. $\ce{AgCl}$ is much less soluble in water than $\ce{NaCl}$, $\ce{CuCl}$ melts at a much lower temperature than $\ce{NaCl}$, and the series $\ce{LiF}$, $\ce{LiCl}$, $\ce{LiBr}$, $\ce{LiI}$ becomes more covalent as the anion enlarges. The sequence $\ce{NaCl}$, $\ce{MgCl2}$, $\ce{AlCl3}$ moves from an ionic crystal to a compound that sublimes near 180 degrees Celsius as charge on the cation grows and its size shrinks.

Key points

Small cation plus large anion gives the most covalent character.
Higher charge on either ion raises covalent character.
For equal charge and similar size, a pseudo-noble-gas (18-electron) cation is more polarising than a noble-gas-type (8-electron) cation.
More covalent character means lower melting point, lower aqueous solubility, and often deeper colour.
Covalent character in the lithium halides rises $\ce{LiF} < \ce{LiCl} < \ce{LiBr} < \ce{LiI}$.

Ionic potential (qualitative index of polarising power): $\phi = \dfrac{z}{r_{\text{cation}}}$ (Valid for comparing cations with the same type of electron shell.)

Common traps

$\ce{AgCl}$ and $\ce{NaCl}$ have cations of similar size, yet $\ce{AgCl}$ is far more covalent. Size and charge alone do not explain it: the shell type matters.
A larger anion is more polarisable, so the iodide of a metal is more covalent than its fluoride, even though the iodide ion carries the same charge.
Fajans' rules compare covalent character between related compounds. They do not give an absolute percentage.
$\ce{LiCl}$ dissolves in ethanol while $\ce{NaCl}$ hardly does, because its covalent character is higher.

Shortcuts

Rank polarising power with z over r. For cations with the same type of shell, rank polarising power by $z/r$ in pm$^{-1}$: $\ce{K+}$ $1/138 = 0.0072$, $\ce{Na+}$ $1/102 = 0.0098$, $\ce{Li+}$ $1/76 = 0.0132$, $\ce{Mg^{2+}}$ $2/72 = 0.0278$, $\ce{Al^{3+}}$ $3/54 = 0.0556$. The larger the number, the more covalent its compounds. Do not use it to compare a noble-gas-type cation with $\ce{Ag+}$ or $\ce{Cu+}$.

Covalent bonding and Lewis structures

A covalent bond is a shared pair of electrons. Atoms of similar electronegativity, usually non-metals, share electrons so that each reaches a stable outer shell, commonly eight electrons (octet) and two for hydrogen. A single, double or triple bond shares 2, 4 or 6 electrons. A coordinate (dative) bond is a shared pair in which both electrons come from one atom, as in $\ce{NH4+}$ (from $\ce{NH3}$ and $\ce{H+}$), $\ce{H3O+}$, $\ce{CO}$ and $\ce{BF3.NH3}$. Once formed it is identical to any other bond of that order.

A Lewis structure shows valence electrons as dots or lines. Count total valence electrons, adjusting for charge; place the least electronegative atom (except hydrogen) in the centre; join the atoms with single bonds; complete the octets of the outer atoms; put any remaining electrons on the central atom; and convert lone pairs into multiple bonds if the central atom lacks an octet.

Three families do not obey the octet rule. Incomplete octets: $\ce{BeCl2}$ (4 electrons around Be), $\ce{BF3}$ (6), $\ce{AlCl3}$ (6, which dimerises to $\ce{Al2Cl6}$ to gain electrons). Odd-electron species: $\ce{NO}$, $\ce{NO2}$, $\ce{ClO2}$. Expanded octets, possible from period 3 onward: $\ce{PCl5}$ (10), $\ce{SF6}$ (12), $\ce{IF7}$ (14). Period 2 elements cannot expand, so $\ce{NCl5}$ and $\ce{OF6}$ do not exist.

Ionic and covalent behaviour are two ends of one range. A bond between two different atoms is polar covalent, with the shared pair displaced toward the more electronegative atom.

Key points

Single, double and triple bonds share 2, 4 and 6 electrons respectively.
A coordinate bond differs from a normal covalent bond only in its origin.
The central atom is usually the least electronegative atom, never hydrogen.
Octet exceptions are of three kinds: electron-deficient, odd-electron and expanded.
Expanded octets need period 3 or lower. Period 2 atoms stop at 8.

Common traps

In $\ce{NH4+}$ all four $\ce{N-H}$ bonds are identical even though one was formed by donation. Do not describe one bond as weaker or longer.
In $\ce{CO}$ one bond is a coordinate bond donated by oxygen, so the formal charges are $\ce{C^-}$ and $\ce{O^+}$ even though oxygen is the more electronegative atom.
$\ce{BF3}$ is stable with 6 electrons on boron, but it readily accepts a pair, as a Lewis acid, to complete the octet in $\ce{BF3.NH3}$ or $\ce{BF4-}$.
Hydrogen forms only one covalent bond, so it is never the central atom of a Lewis structure.

Formal charge and resonance

Formal charge is a bookkeeping device that assigns each bonded atom half of the bonding electrons and all of its own lone-pair electrons, then compares with its free-atom valence: $\text{FC} = V - L - \tfrac12 B$. The charges in a Lewis structure sum to the charge on the species. When several structures are possible, prefer the one with the smallest formal charges, with negative charge on the more electronegative atom, and with no like charges on adjacent atoms.

Some species cannot be drawn with one satisfactory Lewis structure. Ozone, the carbonate ion and benzene are each described by two or more structures that differ only in where the double bonds and charges sit, never in the positions of the atoms. These are resonance (canonical) structures. The real molecule is a single resonance hybrid, with delocalised electrons, intermediate bond lengths and lower energy than any contributing structure. The energy difference is the resonance energy.

The hybrid does not oscillate between its forms. The double-headed arrow means that the true structure is a weighted blend. Equivalent structures contribute equally, so all the $\ce{C-O}$ bonds in $\ce{CO3^{2-}}$ are identical (one third double-bond character). Unequal structures contribute unequally, with the lower-energy one dominating.

The resonance idea explains why the two $\ce{O-O}$ bonds in $\ce{O3}$ are equal (128 pm, between the single-bond 148 pm and the double-bond 121 pm values), why all six $\ce{C-C}$ bonds in benzene are 139 pm, and why carboxylate ions have two equal $\ce{C-O}$ bonds.

Key points

Formal charges sum to the charge on the ion or zero for a neutral molecule.
Resonance structures differ only in electron placement. Atoms do not move, and the number of unpaired electrons does not change.
The resonance hybrid is more stable than any single contributing structure.
Bond order in the hybrid is the average over the equivalent structures: carbonate $4/3$, ozone $1.5$, benzene $1.5$.
More equivalent structures generally means greater stabilisation.

Formal charge: $\text{FC} = V - L - \tfrac{1}{2}B$ ($V$ valence electrons of the neutral atom, $L$ non-bonding electrons, $B$ bonding electrons.)
Average bond order over equivalent bonds: $\overline{\text{BO}} = \dfrac{\text{total number of bonds from one structure to the equivalent atoms}}{\text{number of equivalent bonding positions}}$

Common traps

Formal charge is not the actual charge on an atom. It is a count under the assumption of equal sharing.
Resonance structures are not isomers. Isomers differ in atom positions and exist as separate species. Resonance forms are not separate species.
In ozone the formal charges are $+1$ on the central oxygen and $-1$ on one terminal oxygen, yet the molecule is overall neutral and has a small dipole moment.

Shortcuts

Average bond order of a delocalised ion. Count the total bonds from the centre in one Lewis structure and divide by the number of equivalent outer atoms: $\ce{CO3^{2-}}$ has 4 bonds over 3 oxygens, giving $4/3$; $\ce{NO3-}$ also gives $4/3$; $\ce{O3}$ has 3 bonds over 2 positions, giving $1.5$.

VSEPR theory and molecular shapes

Valence shell electron pair repulsion theory predicts shape from one idea: electron domains around the central atom (bonding pairs and lone pairs) arrange themselves as far apart as possible. A multiple bond counts as one domain because its electrons occupy the same region. The number of domains (steric number) fixes the electron geometry, and the number of lone pairs among them fixes the molecular shape, which describes only the positions of atoms.

The electron geometries are linear (2 domains, $180^\circ$), trigonal planar (3, $120^\circ$), tetrahedral (4, $109.5^\circ$), trigonal bipyramidal (5, $90^\circ$ axial-equatorial, $120^\circ$ equatorial-equatorial, $180^\circ$ axial-axial), octahedral (6, $90^\circ$) and pentagonal bipyramidal (7, $72^\circ$ and $90^\circ$). Replacing bond pairs by lone pairs gives derived shapes: bent, trigonal pyramidal, see-saw, T-shaped, linear, square pyramidal and square planar.

A lone pair is held by only one nucleus, so its cloud is wider and repels more than a bonding pair. The order is lone pair-lone pair

gt;$ lone pair-bond pair

gt;$ bond pair-bond pair. Hence bond angles shrink as lone pairs are added: $\ce{CH4}$ $109.5^\circ$, $\ce{NH3}$ $107^\circ$, $\ce{H2O}$ $104.5^\circ$. In a trigonal bipyramid, a lone pair takes an equatorial position, where it has two neighbours at $90^\circ$ instead of three, so $\ce{SF4}$ is see-saw, $\ce{ClF3}$ is T-shaped and $\ce{XeF2}$ is linear. In an octahedron, two lone pairs sit opposite each other, so $\ce{XeF4}$ is square planar.

Other factors modify angles. If the surrounding atoms are more electronegative, bonding pairs are pulled away from the central atom and repel less, so $\ce{NF3}$ ($\approx 102^\circ$) has a smaller angle than $\ce{NH3}$ ($107^\circ$). In the hydrides of group 15 and 16 the angle falls steeply for heavier central atoms ($\ce{H2O}$ $104.5^\circ$, $\ce{H2S}$ about $92^\circ$) because the bond pairs are farther from the nucleus and the bonding uses nearly pure $p$ orbitals. A multiple bond occupies more space than a single bond, so angles next to a double bond are larger.

Key points

Steric number: number of $\sigma$ bonds plus lone pairs on the central atom. Multiple bonds count once.
Lone pairs repel more than bond pairs and shrink bond angles.
In a trigonal bipyramid lone pairs go equatorial. In an octahedron two lone pairs sit trans to each other.
$\ce{XeF6}$ (7 domains, 1 lone pair) is a distorted octahedron and $\ce{IF7}$ is a pentagonal bipyramid.
Isoelectronic species have the same shape: $\ce{NH4+}$, $\ce{CH4}$ and $\ce{BH4-}$ are tetrahedral.

Steric number: $\text{SN} = \tfrac{1}{2}\,(V + M + a - c)$ ($V$ valence electrons of the central atom, $M$ monovalent atoms attached, $a$ anionic charge, $c$ cationic charge. Terminal O and S joined by double bonds add 0.)

Common traps

The shape of the molecule names only the atoms. $\ce{NH3}$ has tetrahedral electron geometry but pyramidal shape, and $\ce{H2O}$ has tetrahedral electron geometry but bent shape.
$\ce{SO2}$, $\ce{O3}$, $\ce{NO2-}$ are bent with trigonal planar electron geometry (angle near $120^\circ$ or a little less), whereas $\ce{H2O}$ is bent with tetrahedral electron geometry (angle near $104.5^\circ$).
$\ce{NO2+}$ is linear (2 domains, no lone pair) while $\ce{NO2-}$ is bent (3 domains, 1 lone pair), and $\ce{NO2}$ with an odd electron is bent.
$\ce{I3-}$ is linear because its three lone pairs on the central iodine occupy the equatorial plane. $\ce{ICl2-}$ is linear for the same reason, and $\ce{ICl4-}$ is square planar.
In $\ce{PCl5}$ the two axial bonds are longer than the three equatorial bonds. Solid $\ce{PCl5}$ is ionic, made of tetrahedral $\ce{PCl4+}$ and octahedral $\ce{PCl6-}$.

Shortcuts

Steric number without drawing the Lewis structure. Use $\text{SN} = \tfrac12(V + M + a - c)$: $\ce{ClF3}$ gives $(7+3)/2 = 5$ with $5-3 = 2$ lone pairs, so T-shaped; $\ce{XeF4}$ gives $(8+4)/2 = 6$ with 2 lone pairs, so square planar; $\ce{SO4^{2-}}$ gives $(6+0+2)/2 = 4$, so tetrahedral. Lone pairs on the central atom equal SN minus the number of attached atoms.

Shape from (SN, lone pairs). SN 5 with 1, 2, 3 lone pairs gives see-saw, T-shaped, linear. SN 6 with 1, 2 lone pairs gives square pyramidal, square planar. SN 4 with 1, 2 lone pairs gives pyramidal, bent. SN 3 with 1 lone pair gives bent.

Hybridisation of orbitals

Hybridisation is the mixing of atomic orbitals of one atom, with similar energy, into the same number of new equivalent orbitals that point in the directions needed for bonding. Orbitals mix, not electrons, and the total number of hybrid orbitals equals the number of orbitals combined. The model explains why the four $\ce{C-H}$ bonds of methane are identical although carbon has one $2s$ and three $2p$ valence orbitals.

The hybrid type matches the steric number: 2 gives $sp$ (linear, $180^\circ$, 50 percent $s$ character), 3 gives $sp^2$ (trigonal planar, $120^\circ$, 33.3 percent), 4 gives $sp^3$ (tetrahedral, $109.5^\circ$, 25 percent), 5 gives $sp^3d$ (trigonal bipyramidal), 6 gives $sp^3d^2$ (octahedral) and 7 gives $sp^3d^3$ (pentagonal bipyramidal). Hybrid orbitals that hold lone pairs are counted too, so $\ce{NH3}$ and $\ce{H2O}$ are $sp^3$ even though their shapes are pyramidal and bent.

Unhybridised $p$ orbitals form $\pi$ bonds. In ethene each carbon is $sp^2$ with one leftover $p$ orbital that forms the $\pi$ bond; in ethyne each carbon is $sp$ with two leftover $p$ orbitals that form two $\pi$ bonds. More $s$ character means a shorter, stronger bond and a more electronegative carbon: $\ce{C-H}$ lengths are about 106.0 pm ($sp$, ethyne), 108.7 pm ($sp^2$, ethene) and 109.4 pm ($sp^3$, ethane), and the acidity of the $\ce{C-H}$ increases in the order ethane

lt;$ ethene

lt;$ ethyne.

For equivalent hybrid orbitals of fractional $s$ content $s$, the interorbital angle satisfies $\cos\theta = -\dfrac{s}{1-s}$. This gives $180^\circ$, $120^\circ$ and $109.47^\circ$ for 50, 33.3 and 25 percent $s$ character. When lone pairs take more $s$ character than the bonding hybrids, the bonding hybrids have less $s$ character, and the angle falls below $109.5^\circ$.

Key points

Hybridisation type follows from the steric number, counting $\sigma$ bonds plus lone pairs.
Hybrid orbitals form $\sigma$ bonds and hold lone pairs; leftover unhybridised $p$ orbitals form $\pi$ bonds.
$s$ character: $sp$ 50 percent, $sp^2$ 33.3, $sp^3$ 25. Bond strength and carbon electronegativity rise with $s$ character.
Examples: $\ce{BeCl2}$ $sp$; $\ce{BF3}$ $sp^2$; $\ce{CH4}$, $\ce{NH3}$, $\ce{H2O}$ $sp^3$; $\ce{PCl5}$ $sp^3d$; $\ce{SF6}$ $sp^3d^2$; $\ce{IF7}$ $sp^3d^3$.
Allene $\ce{CH2=C=CH2}$ has a central $sp$ carbon and two terminal $sp^2$ carbons. Its two terminal $\ce{CH2}$ planes are perpendicular.

Angle between equivalent hybrids: $\cos\theta = -\dfrac{s}{1-s}$ ($s$ is the fractional $s$ character of each equivalent hybrid orbital (idealised model).)

Common traps

Hybridisation does not equal shape. $\ce{H2O}$ is $sp^3$ but bent, not tetrahedral.
The exam convention gives $sp^3d$ and $sp^3d^2$ for $\ce{PCl5}$ and $\ce{SF6}$ using $d$ orbitals, though modern bonding descriptions assign little role to $d$ orbitals in main-group compounds. Use the syllabus convention in answers.
The hybridisation of an atom in an ion uses the same count: $\ce{NH4+}$ is $sp^3$, $\ce{NO2+}$ is $sp$, $\ce{NO3-}$ is $sp^2$ and $\ce{I3-}$ is $sp^3d$.
Different atoms in one molecule have different hybridisation. In $\ce{CH3-CH=CH-C#N}$ the four carbons are $sp^3$, $sp^2$, $sp^2$, $sp$ and the nitrogen is $sp$.
$\ce{NH3}$ and $\ce{PH3}$ have the same steric number 4 and the same label $sp^3$, but the $\ce{H-N-H}$ angle is $107^\circ$ while $\ce{H-P-H}$ is near $93.5^\circ$, because heavier central atoms bond with nearly pure $p$ orbitals. The syllabus explanation is that the central atom is less electronegative, so bond pairs lie farther from it and repel less.

Shortcuts

Hybridisation of an organic carbon from its pi bonds. Carbon is $sp^3$ with no $\pi$ bond, $sp^2$ with one $\pi$ bond and $sp$ with two $\pi$ bonds, whether as a triple bond or two double bonds. So $\ce{C}$ in $\ce{CO2}$ is $sp$, in $\ce{HCHO}$ is $sp^2$, and in $\ce{CH3OH}$ is $sp^3$.

Hybridisation as steric number to type. Take SN $= \sigma$ bonds $+$ lone pairs, then read: 2 is $sp$, 3 is $sp^2$, 4 is $sp^3$, 5 is $sp^3d$, 6 is $sp^3d^2$, 7 is $sp^3d^3$. Check: $\ce{ClO3-}$ has 3 $\sigma$ bonds and 1 lone pair, so SN 4 and $sp^3$ (pyramidal).

Valence bond theory: sigma and pi bonds

Valence bond theory pictures a bond as the overlap of two half-filled atomic orbitals with paired spins. The shared electron pair is found mainly in the overlap region. The greater the overlap, the stronger the bond, and bonds form along the direction in which the participating orbitals overlap best. This is the reason for directed bonds and for using hybrid orbitals, which are concentrated in one direction and overlap better than pure $s$ or $p$ orbitals.

A $\sigma$ bond forms by head-on (axial) overlap along the internuclear axis: $s-s$, $s-p$, $p-p$ axial, or any hybrid with another orbital. The overlap is cylindrically symmetrical about the axis, so groups can rotate freely about a single bond. A $\pi$ bond forms by sideways overlap of parallel $p$ orbitals, with electron density above and below the axis and a node along it. It is weaker than a $\sigma$ bond because lateral overlap is poorer, and it blocks rotation. A $\pi$ bond can form only after a $\sigma$ bond has formed between the same atoms.

A double bond is one $\sigma$ plus one $\pi$ and a triple bond is one $\sigma$ plus two $\pi$. Counting follows from the Lewis structure: every single bond is a $\sigma$ bond, and every extra bond of a double or triple bond is a $\pi$ bond. Ethene has 5 $\sigma$ and 1 $\pi$, ethyne 3 $\sigma$ and 2 $\pi$, benzene 12 $\sigma$ and 3 $\pi$ in the localised picture.

The theory has limits. It cannot explain why $\ce{O2}$ is paramagnetic, and it describes delocalised bonding only through the extra idea of resonance. Molecular orbital theory fixes both.

Key points

A bond forms by overlap of half-filled orbitals of opposite spin.
$\sigma$ overlap is end-on and rotation-friendly; $\pi$ overlap is side-on and restricts rotation.
A $\sigma$ bond is stronger than a $\pi$ bond between the same atoms.
Double bond $= 1\sigma + 1\pi$; triple bond $= 1\sigma + 2\pi$.
Hybrid orbitals overlap better than pure orbitals. Relative strengths are about $s$ 1, $p$ 1.73, $sp$ 1.93, $sp^2$ 1.99, $sp^3$ 2.00.

Common traps

A $\pi$ bond uses unhybridised $p$ orbitals. Hybrid orbitals never form $\pi$ bonds.
Overlap of two $p$ orbitals along the internuclear axis gives a $\sigma$ bond, while parallel overlap gives a $\pi$ bond. The label depends on the geometry of overlap, not on which orbitals are involved.
A single bond is always $\sigma$, but the $\sigma$ bond in a molecule is not always between hybrid orbitals: in $\ce{H2}$ it is $s-s$, and in $\ce{HF}$ it is $s-p$.
$\pi$ bonds restrict rotation, which is why $\ce{cis}$ and $\ce{trans}$ alkenes do not interconvert at room temperature.

Molecular orbital theory, bond order and magnetism

Molecular orbital theory treats the molecule as one unit. Atomic orbitals combine linearly (LCAO) into molecular orbitals that extend over all nuclei. Combining two atomic orbitals gives two molecular orbitals: a lower-energy bonding orbital (waves add, density between the nuclei) and a higher-energy antibonding orbital (waves cancel, a node between the nuclei). Electrons fill them by the Aufbau principle, Pauli exclusion and Hund's rule.

For second-period homonuclear diatomics the valence order for $\ce{O2}$, $\ce{F2}$ and $\ce{Ne2}$ is $\sigma 2s < \sigma^* 2s < \sigma 2p_z < (\pi 2p_x = \pi 2p_y) < (\pi^* 2p_x = \pi^* 2p_y) < \sigma^* 2p_z$. For $\ce{Li2}$ through $\ce{N2}$ the $2s$-$2p$ energy gap is small, so mixing pushes $\sigma 2p_z$ above the $\pi 2p$ pair: $\sigma 2s < \sigma^* 2s < (\pi 2p_x = \pi 2p_y) < \sigma 2p_z < (\pi^*) < \sigma^* 2p_z$.

Bond order is $\tfrac12 (N_b - N_a)$, with $N_b$ bonding and $N_a$ antibonding electrons. A positive value means a stable species. Higher bond order means a shorter and stronger bond. Fractional values are legitimate, as in $\ce{O2+}$ (2.5) and $\ce{O2-}$ (1.5). Magnetic behaviour follows from the unpaired electrons: unpaired electrons give paramagnetism, none gives diamagnetism, and the spin-only moment is $\sqrt{n(n+2)}$ Bohr magnetons.

Results worth knowing: $\ce{H2}$ 1, $\ce{He2}$ 0 (does not exist), $\ce{Be2}$ 0, $\ce{B2}$ 1 (paramagnetic, two unpaired electrons in the degenerate $\pi$ pair), $\ce{C2}$ 2 (diamagnetic, both bonds are $\pi$ bonds), $\ce{N2}$ 3 (diamagnetic), $\ce{O2}$ 2 (paramagnetic, two unpaired electrons in $\pi^*$), $\ce{F2}$ 1 (diamagnetic). Heteronuclear diatomics with 10 valence electrons ($\ce{CO}$, $\ce{CN-}$, $\ce{NO+}$) behave like $\ce{N2}$ with bond order 3, and $\ce{NO}$ (11 valence electrons) has bond order 2.5 and one unpaired electron.

Changing the electron count shows which type of orbital is affected. Removing an electron from an antibonding orbital raises bond order ($\ce{O2 -> O2+}$ 2 to 2.5), and removing from a bonding orbital lowers it ($\ce{N2 -> N2+}$ 3 to 2.5). Adding an electron to an antibonding orbital lowers bond order ($\ce{O2 -> O2-}$ 2 to 1.5, then $\ce{O2^{2-}}$ 1).

Key points

$n$ atomic orbitals give $n$ molecular orbitals. Bonding lowers energy and antibonding raises it.
Bond order $= \tfrac12 (N_b - N_a)$. Stability and bond strength rise with bond order; bond length falls.
Order of $\sigma 2p$ and $\pi 2p$ depends on the species: $\pi$ below $\sigma$ up to $\ce{N2}$ (14 or fewer total electrons), $\sigma$ below $\pi$ from $\ce{O2}$ onward.
$\ce{O2}$, $\ce{B2}$, $\ce{NO}$, $\ce{O2+}$, $\ce{O2-}$, $\ce{N2+}$ are paramagnetic. $\ce{N2}$, $\ce{C2}$, $\ce{F2}$, $\ce{CO}$, $\ce{O2^{2-}}$ are diamagnetic.
Bond length order for the oxygen family: $\ce{O2+} < \ce{O2} < \ce{O2-} < \ce{O2^{2-}}$.

Bond order: $\text{BO} = \dfrac{N_b - N_a}{2}$
Spin-only magnetic moment: $\mu_s = \sqrt{n(n+2)}\ \text{BM}$ ($n = 1, 2, 3, 4, 5$ gives 1.73, 2.83, 3.87, 4.90, 5.92 BM.)

Common traps

$\ce{N2}$ highest occupied orbital is $\sigma 2p_z$ (bonding), but for $\ce{O2}$ it is $\pi^*2p$ (antibonding). Ionisation of $\ce{N2}$ lowers bond order, while ionisation of $\ce{O2}$ raises it.
$\ce{O2}$ has two unpaired electrons in separate $\pi^*$ orbitals (Hund's rule), not one paired electron pair, and is attracted into a magnetic field.
$\ce{C2}$ has no $\sigma$ bond from the $2p$ set: all four bonding $2p$ electrons are in the $\pi$ pair. It is diamagnetic, and a question may offer paramagnetic as a wrong choice.
$\ce{N2+}$ and $\ce{N2-}$ both have bond order 2.5, but they differ: $\ce{N2+}$ has lost an electron from the bonding $\sigma 2p$ orbital, while $\ce{N2-}$ has gained one in the antibonding $\pi^*$ pair.
$\ce{He2}$ and $\ce{Be2}$ have bond order 0, but $\ce{He2+}$ and $\ce{H2-}$ have bond order 0.5 and can exist as unstable species.

Shortcuts

Bond order from valence electron count v. For a second-period diatomic with $v$ valence electrons, use $\text{BO} = (v-4)/2$ when $4 \le v \le 10$ and for $v \ge 10$ use $\text{BO} = (16 - v)/2$. So $\ce{B2}$ ($v = 6$) gives 1, $\ce{N2}$ ($v=10$) gives 3, $\ce{O2}$ ($v=12$) gives 2, $\ce{O2-}$ ($v=13$) gives 1.5, $\ce{NO}$ ($v=11$) gives 2.5, $\ce{F2}$ ($v=14$) gives 1.

Unpaired electrons and magnetic moment. In the $\pi$ and $\pi^*$ pair, count electrons: a half-filled pair gives 2 unpaired ($\ce{B2}$, $\ce{O2}$), one extra or missing electron gives 1 ($\ce{O2+}$, $\ce{O2-}$, $\ce{NO}$), and a filled pair gives 0. Then $\mu = \sqrt{n(n+2)}$: $n=1$ is 1.73 BM and $n=2$ is 2.83 BM.

Dipole moment and molecular polarity

When two bonded atoms differ in electronegativity, the shared pair is displaced toward the more electronegative atom, creating partial charges $+\delta$ and $-\delta$ separated by the bond length. The dipole moment is $\mu = q \times d$, where $q$ is the magnitude of the partial charge. It is a vector, drawn by convention here from the positive to the negative end. The SI unit is coulomb metre, but the debye is standard: $1\ \text{D} = 3.336\times10^{-30}\ \text{C m}$.

The molecular dipole moment is the vector sum of bond moments (and lone-pair contributions). A molecule with polar bonds can be non-polar if the bond moments cancel by symmetry, as in $\ce{CO2}$ (linear), $\ce{BF3}$ (trigonal planar), $\ce{CH4}$ and $\ce{CCl4}$ (tetrahedral), $\ce{PCl5}$, $\ce{SF6}$ and $\ce{XeF4}$. Bent, pyramidal, see-saw and T-shaped molecules such as $\ce{H2O}$, $\ce{NH3}$, $\ce{SF4}$ and $\ce{ClF3}$ are polar.

For two equal bond moments $m$ with angle $\theta$ between the bonds, the resultant is $2m\cos(\theta/2)$. The experimental moment of $\ce{H2O}$ is 1.85 D with an $\ce{H-O-H}$ angle of $104.5^\circ$, which gives about 1.51 D. This is an effective $\ce{O-H}$ moment that lumps in the lone-pair contribution. A lone pair on the central atom contributes its own moment. In $\ce{NH3}$ (1.47 D) the $\ce{N-H}$ bond moments point toward nitrogen, the same way as the lone-pair moment, so they add. In $\ce{NF3}$ (about 0.2 D) the $\ce{N-F}$ bond moments point toward fluorine, against the lone-pair moment, so they largely cancel.

Dipole moment relates to percent ionic character: $\mu_{\text{observed}} / (e d) \times 100$, where $e d$ would be the moment if one full electron charge were transferred over the bond length. Polarity affects boiling point, solubility (like dissolves like), and the reactivity of molecules. The dipole moments of hydrogen halides fall from $\ce{HF}$ (about 1.8 D) to $\ce{HI}$ (about 0.4 D) as electronegativity difference falls.

Key points

$\mu = q \times d$; it is a vector, so net polarity depends on geometry as well as bond polarity.
Symmetric molecules with identical outer atoms and no lone pair on the centre are non-polar.
$\ce{H2O}$ $\mu = 1.85$ D; $\ce{NH3}$ $1.47$ D; $\ce{CO2}$, $\ce{BF3}$, $\ce{CH4}$ zero.
For disubstituted benzenes, ortho
gt;$ meta
gt;$ para ($\mu = 0$ for identical substituents).
Trans-1,2-dichloroethene has zero moment; the cis isomer is polar.

Dipole moment: $\mu = q \times d$ (1 D $= 3.336\times10^{-30}$ C m.)
Resultant of two bond moments: $\mu_R = \sqrt{\mu_1^2 + \mu_2^2 + 2\mu_1\mu_2\cos\theta}$ ($\theta$ is the angle between the two bond-moment vectors.)
Percent ionic character: $\% \text{ ionic} = \dfrac{\mu_{\text{obs}}}{e\,d}\times 100$

Common traps

Ozone is polar, with $\mu \approx 0.5$ D, even though it consists of one element. The centre atom differs from the end atoms in formal charge and the molecule is bent.
$\ce{CH3Cl}$ (1.87 D) has a larger moment than $\ce{CH3F}$ (1.85 D) because the longer $\ce{C-Cl}$ bond length outweighs the smaller charge. The orders of dipole moment and electronegativity do not always agree.
$\ce{NF3}$ is less polar than $\ce{NH3}$ even though F is far more electronegative than H, because of the direction of the bond moments relative to the lone pair.
$\ce{CO}$ has a very small dipole moment (about 0.1 D). Oxygen pulls the shared electrons toward itself, while the coordinate bond donates a pair from oxygen to carbon, and these two effects oppose.
A zero dipole moment does not mean non-polar bonds. $\ce{CCl4}$ has four polar bonds and zero net moment.

Shortcuts

Zero-moment test. A molecule $\ce{AX_n}$ has $\mu = 0$ when all $n$ outer atoms are identical and the central atom has no lone pair ($\ce{BF3}$, $\ce{CCl4}$, $\ce{PCl5}$, $\ce{SF6}$), or when the lone pairs sit symmetrically as in $\ce{XeF2}$ (3 lone pairs) and $\ce{XeF4}$ (2 lone pairs, opposite each other).

Ortho, meta and para ratio. For a disubstituted benzene with two identical substituents of bond moment $m$, the vector sum gives $\mu_{ortho} = 2m\cos 30^\circ = 1.73m$, $\mu_{meta} = 2m\cos 60^\circ = m$ and $\mu_{para} = 0$. The ratio ortho : meta is $\sqrt3 : 1$.

Hydrogen bonding

A hydrogen bond is an attraction between a hydrogen atom covalently bound to a highly electronegative atom (N, O or F) and a lone pair on another N, O or F atom. The hydrogen is left with a strong partial positive charge and has no inner electrons to shield it, so it can approach the lone pair closely. The usual notation is $\ce{X-H...Y}$. Strengths range from about 5 to 40 kJ/mol, around one tenth of a covalent bond but much larger than ordinary dipole-dipole or dispersion forces.

Intermolecular hydrogen bonds link separate molecules. They raise melting and boiling points and viscosity, and make a molecule soluble in water if it can bond with the solvent. Water boils at 100 degrees Celsius while $\ce{H2S}$ boils near $-60$ degrees Celsius. Similarly $\ce{NH3}$ ($-33$ degrees Celsius) is higher than $\ce{PH3}$ ($-88$ degrees Celsius), and $\ce{HF}$ (about 20 degrees Celsius) is far higher than $\ce{HCl}$ ($-85$ degrees Celsius). Water forms more hydrogen bonds per molecule than $\ce{HF}$ or $\ce{NH3}$ because it has two hydrogens and two lone pairs, so its boiling point is the highest of the three.

Intramolecular hydrogen bonds form within one molecule when the donor and acceptor are close, as between $\ce{OH}$ and $\ce{NO2}$ in ortho-nitrophenol or between $\ce{OH}$ and $\ce{C=O}$ in salicylaldehyde. They lock the molecule into a ring, reduce interaction with other molecules and lower the melting point, boiling point and water solubility. Ortho-nitrophenol (intramolecular) is steam-volatile with a melting point near 45 degrees Celsius, whereas para-nitrophenol (intermolecular) melts near 114 degrees Celsius and is not steam-volatile.

Ice has each oxygen bonded to four others in an open tetrahedral network, with two covalent and two hydrogen bonds per oxygen. The structure has empty channels, so ice is less dense (about 0.92 g/cm$^3$) than liquid water and floats; water has maximum density near 4 degrees Celsius. In biology, base pairing holds the strands of DNA together: adenine-thymine with 2 and guanine-cytosine with 3 hydrogen bonds. Carboxylic acids such as ethanoic acid form hydrogen-bonded dimers in non-polar solvents, giving an apparent molar mass near 120 instead of 60.

Key points

Donor: $\ce{H}$ bonded to $\ce{N}$, $\ce{O}$ or $\ce{F}$. Acceptor: a lone pair on $\ce{N}$, $\ce{O}$ or $\ce{F}$.
Typical strength 5 to 40 kJ/mol; $\ce{F-H...F}$ is the strongest.
Intermolecular bonding raises boiling point and viscosity. Intramolecular bonding lowers them and reduces water solubility.
Ice is less dense than water because of the open hydrogen-bonded lattice.
Anomalous boiling points of $\ce{NH3}$, $\ce{H2O}$, $\ce{HF}$ compared with the hydrides of the heavier group members.

Common traps

$\ce{H2O}$ has a higher boiling point than $\ce{HF}$ although the $\ce{F-H...F}$ bond is the stronger one. A water molecule can take part in up to four hydrogen bonds (two as donor, two as acceptor), while $\ce{HF}$ forms chains with about two. The number of bonds per molecule matters as well as the strength of each.
A hydrogen bond is not a covalent bond. It is an electrostatic interaction between molecules, even if it is drawn as a dotted line connecting two bonds.
Within group 14 the hydrides $\ce{CH4}$ and $\ce{SiH4}$ show no hydrogen bonding, and their boiling points rise regularly with size.
More hydrogen bonding does not always mean a higher boiling point. Check whether the bonding is intermolecular (raises it) or intramolecular (the ortho isomer is more volatile than the para isomer).
Chlorine is as electronegative as nitrogen but forms much weaker hydrogen bonds because of its larger size and lower charge density.

Bond length, bond angle, bond enthalpy and bond order

Bond length is the equilibrium distance between two bonded nuclei. In covalent bonds it is roughly the sum of covalent radii: $\ce{C-Cl}$ is about $77 + 99 = 176$ pm (observed 177 pm). In an ionic solid it is the sum of ionic radii, for example $102 + 181 = 283$ pm for $\ce{NaCl}$. Bond length falls as bond order rises, so $\ce{C-C}$, $\ce{C=C}$, $\ce{C#C}$ are about 154, 134 and 120 pm. For a given bond order, length increases down a group because atoms get larger.

Bond enthalpy is the energy needed to break one mole of a particular bond in the gas phase. For a diatomic molecule it is the bond dissociation enthalpy. For a polyatomic molecule the tabulated number is an average over all bonds of that type, because successive bonds in $\ce{CH4}$ or $\ce{H2O}$ break at different energies. Enthalpy rises with bond order ($\ce{C-C}$ about 348, $\ce{C=C}$ 612, $\ce{C#C}$ 837 kJ/mol) and falls with bond length. Within the hydrogen halides it falls from $\ce{HF}$ (568) to $\ce{HCl}$ (431), $\ce{HBr}$ (366) and $\ce{HI}$ (299 kJ/mol).

Bond angle is the angle between two bonds at a central atom. It depends on hybridisation, lone pairs, the electronegativity of the attached atoms and the size of the central atom, as described under VSEPR. Resonance and molecular orbital theory give fractional bond orders, and bond length for such bonds lies between those of the integer orders, as in carbonate and benzene.

Bond enthalpies give an estimate of reaction enthalpy for gas-phase reactions: $\Delta H \approx \sum$ (enthalpies of bonds broken) $-$ $\sum$ (enthalpies of bonds formed). The estimate is approximate because tabulated values are averages. For $\ce{H2 + Cl2 -> 2HCl}$ the estimate is $436 + 242 - 2(431) = -184$ kJ, close to the measured $-184.6$ kJ.

Key points

Higher bond order gives a shorter bond and a higher bond enthalpy.
Bond length of a covalent bond $\approx$ sum of covalent radii of the two atoms.
Bond enthalpy of a polyatomic bond is an average; only a diatomic has a unique value.
$\ce{HF}$ has the strongest bond among the hydrogen halides, and $\ce{HI}$ the weakest.
$\Delta H$ (gas phase) $=$ energy to break bonds $-$ energy released on forming bonds.

Reaction enthalpy estimate: $\Delta H \approx \sum \Delta H_{\text{bonds broken}} - \sum \Delta H_{\text{bonds formed}}$ (Reactants and products must all be gaseous.)
Covalent bond length: $d_{A-B} \approx r_A + r_B$

Common traps

The $\ce{F-F}$ bond (158 kJ/mol) is weaker than $\ce{Cl-Cl}$ (242 kJ/mol). Repulsion between lone pairs on the small fluorine atoms weakens it, and the same effect lowers $\ce{O-O}$ and $\ce{N-N}$ below $\ce{S-S}$ and $\ce{P-P}$.
Bond enthalpy values are averages. Do not treat the $\ce{O-H}$ enthalpy of 463 kJ/mol as the energy of breaking one specific $\ce{O-H}$ in water.
Bond enthalpy applies to gaseous species. If the reactants or products are liquids or solids, add the enthalpies of vaporisation or sublimation.
Bond length does not decide strength alone. $\ce{HF}$ is the shortest hydrogen halide bond and the strongest, but $\ce{F-F}$ is short and weak.
Bond angle and bond length are separate. Bond angle depends on the number of electron domains, while bond length depends on atomic size and bond order.

Worked examples

Lattice enthalpy of NaCl from a Born-Haber cycle

Using $\Delta_f H(\ce{NaCl}) = -411$, $\Delta_{sub}H(\ce{Na}) = +107$, bond dissociation enthalpy of $\ce{Cl2} = +242$, ionisation enthalpy of $\ce{Na} = +496$ and electron gain enthalpy of $\ce{Cl} = -349$ (all in kJ/mol), find the lattice enthalpy of $\ce{NaCl}$ as formation of the solid from gaseous ions.

Write the cycle. $\ce{Na(s) + 1/2 Cl2(g) -> NaCl(s)}$ can be reached directly or by the route $\ce{Na(s)} \to \ce{Na(g)} \to \ce{Na+(g)}$, $\tfrac12\ce{Cl2(g)} \to \ce{Cl(g)} \to \ce{Cl-(g)}$, then $\ce{Na+(g) + Cl-(g) -> NaCl(s)}$. Hess's law makes both routes equal.
Add the steps before the lattice step. Atomisation of sodium $+107$; atomisation of chlorine $\tfrac12 \times 242 = +121$; ionisation of sodium $+496$; electron gain of chlorine $-349$. Sum $= 107 + 121 + 496 - 349 = +375$ kJ/mol.
Solve for the lattice step. $\Delta_f H = 375 + \Delta_{lattice}H$, so $\Delta_{lattice}H = -411 - 375 = -786$ kJ/mol.
Interpret. The 786 kJ/mol released on building the lattice is more than enough to pay for the $+375$ kJ/mol needed to make the gaseous ions, so $\ce{NaCl}$ is stable by $411$ kJ/mol. As breaking the lattice, the value is $+786$ kJ/mol.

Answer: $-786$ kJ/mol (about $-787$ with a slightly different $\ce{Cl2}$ value), or $+786$ kJ/mol when quoted as the energy to separate the lattice into gaseous ions.

Shape, hybridisation and polarity of ClF3, XeF4 and SF4

For $\ce{ClF3}$, $\ce{XeF4}$ and $\ce{SF4}$ find the steric number, the number of lone pairs, the hybridisation, the molecular shape and whether the molecule is polar.

Apply the steric number formula. $\text{SN} = \tfrac12 (V + M)$. For $\ce{ClF3}$: $(7+3)/2 = 5$. For $\ce{XeF4}$: $(8+4)/2 = 6$. For $\ce{SF4}$: $(6+4)/2 = 5$.
Find lone pairs and hybridisation. Lone pairs equal SN minus attached atoms: $\ce{ClF3}$ has $5-3 = 2$, $\ce{XeF4}$ has $6-4 = 2$, $\ce{SF4}$ has $5-4 = 1$. Hybridisation: $\ce{ClF3}$ is $sp^3d$, $\ce{XeF4}$ is $sp^3d^2$, $\ce{SF4}$ is $sp^3d$.
Place lone pairs. In a trigonal bipyramid lone pairs take equatorial positions, giving T-shaped $\ce{ClF3}$ (2 equatorial lone pairs) and see-saw $\ce{SF4}$ (1 equatorial lone pair). In an octahedron two lone pairs sit opposite each other, giving square planar $\ce{XeF4}$.
Decide polarity. In $\ce{XeF4}$ the four $\ce{Xe-F}$ moments and the two opposite lone pairs cancel, so $\mu = 0$. In $\ce{ClF3}$ and $\ce{SF4}$ the arrangement is lopsided and the moments do not cancel, so both are polar.

Answer: $\ce{ClF3}$: SN 5, 2 lone pairs, $sp^3d$, T-shaped, polar. $\ce{XeF4}$: SN 6, 2 lone pairs, $sp^3d^2$, square planar, non-polar. $\ce{SF4}$: SN 5, 1 lone pair, $sp^3d$, see-saw, polar.

Counting sigma and pi bonds and hybridisation in CH3CH=CHCN

For $\ce{CH3-CH=CH-C#N}$ find the total number of $\sigma$ and $\pi$ bonds, the hybridisation of each carbon and of nitrogen, and the number of atoms that lie on a straight line with the $\ce{CN}$ group.

Count sigma bonds. Every single bond is $\sigma$, and each double or triple bond has one $\sigma$. The skeleton gives $\ce{C-C}$ (1), $\ce{C=C}$ $\sigma$ (1), $\ce{C-C}$ (1) and $\ce{C#N}$ $\sigma$ (1), which is 4. The $\ce{C-H}$ bonds are $3 + 1 + 1 = 5$. Total $\sigma = 9$.
Count pi bonds. The double bond adds 1 $\pi$ and the triple bond adds 2 $\pi$, so total $\pi = 3$.
Assign hybridisation using pi bonds. The methyl carbon has no $\pi$ bond, so $sp^3$. Each carbon of $\ce{C=C}$ has one $\pi$ bond, so $sp^2$. The nitrile carbon has two $\pi$ bonds, so $sp$. Nitrogen has 1 $\sigma$ bond and 1 lone pair (SN 2), so $sp$.
Check the straight line. An $sp$ carbon has a $180^\circ$ bond angle, so the carbon bonded to the nitrile carbon, the nitrile carbon and the nitrogen are collinear, which is 3 atoms.

Answer: 9 $\sigma$ and 3 $\pi$ bonds. Hybridisation: $sp^3$, $sp^2$, $sp^2$, $sp$ for the four carbons in order and $sp$ for nitrogen. Three atoms (the $sp^2$ carbon bonded to $\ce{CN}$, the nitrile carbon, and nitrogen) are collinear.

Bond order, bond length and magnetism across O2+, O2, O2- and O2 2-

Arrange $\ce{O2+}$, $\ce{O2}$, $\ce{O2-}$ and $\ce{O2^{2-}}$ in order of bond length, and state the bond order and magnetic behaviour of each.

Count valence electrons. $\ce{O2}$ has 12 valence electrons. $\ce{O2+}$ has 11, $\ce{O2-}$ has 13 and $\ce{O2^{2-}}$ has 14.
Fill the molecular orbitals. Order for oxygen: $\sigma 2s^2\,\sigma^*2s^2\,\sigma 2p^2\,\pi 2p^4\,\pi^*2p^n$. The $\pi^*$ pair holds 1 electron ($\ce{O2+}$), 2 ($\ce{O2}$), 3 ($\ce{O2-}$) and 4 ($\ce{O2^{2-}}$). Bonding electrons are 8 in every case, and antibonding electrons are $2 + n$.
Compute bond orders. $\text{BO} = (8 - (2+n))/2$: $\ce{O2+}$ $= 2.5$, $\ce{O2}$ $= 2$, $\ce{O2-}$ $= 1.5$, $\ce{O2^{2-}}$ $= 1$.
Unpaired electrons. $\pi^*$ with 1 electron has 1 unpaired; with 2 electrons in the degenerate pair (Hund's rule) has 2 unpaired; with 3 has 1 unpaired; with 4 has 0. So $\ce{O2+}$ (1.73 BM), $\ce{O2}$ (2.83 BM) and $\ce{O2-}$ (1.73 BM) are paramagnetic, and $\ce{O2^{2-}}$ is diamagnetic.
Rank the bond lengths. Bond length falls as bond order rises, so the order is $\ce{O2+}$ (shortest, BO 2.5)
lt;$ $\ce{O2}$ (2)
lt;$ $\ce{O2-}$ (1.5)
lt;$ $\ce{O2^{2-}}$ (longest, BO 1).

Answer: Bond length: $\ce{O2+} < \ce{O2} < \ce{O2-} < \ce{O2^{2-}}$. Bond orders 2.5, 2, 1.5, 1. Paramagnetic: $\ce{O2+}$, $\ce{O2}$, $\ce{O2-}$; diamagnetic: $\ce{O2^{2-}}$.

Why the H-O-H angle is 104.5 degrees: s character in the hybrids

Using $\cos\theta = -s/(1-s)$ for the bonding hybrids of oxygen in water ($\theta = 104.5^\circ$), find the $s$ character of the bonding hybrids and of the two lone-pair orbitals.

Rearrange the angle formula. From $\cos\theta = -\dfrac{s}{1-s}$ we get $s = \dfrac{-\cos\theta}{1 - \cos\theta}$.
Substitute the angle. $\cos 104.5^\circ = -0.2504$, so $s = \dfrac{0.2504}{1.2504} \approx 0.20$. Each $\ce{O-H}$ hybrid is about 20 percent $s$ and 80 percent $p$, which is $sp^4$.
Share out the rest of the s orbital. The oxygen $2s$ orbital contributes one unit of $s$ character in total. The two bonding hybrids take $2 \times 0.20 = 0.40$, leaving $0.60$ for the two lone pairs. Each lone-pair orbital is 30 percent $s$, about $sp^{2.3}$.
Check the p character. Bonding hybrids use $2\times0.80 = 1.6$ units of $p$ and the lone pairs use $2 \times 0.70 = 1.4$, so the total is 3.0, the three $2p$ orbitals of oxygen.

Answer: Bonding hybrids about 20 percent $s$ ($sp^4$). Lone pairs about 30 percent $s$. Lone pairs hold more $s$ character, which closes the bond angle below $109.5^\circ$. This is an idealised model, not a measured composition.

Percent ionic character of HCl from its dipole moment

The measured dipole moment of $\ce{HCl}$ is about 1.03 D (a common textbook value; measured values lie between 1.03 and 1.1 D) and the bond length is 127 pm. Calculate the percent ionic character.

Find the moment of a fully ionic bond. If one full electron charge $e = 1.602\times10^{-19}$ C were transferred over 127 pm, $\mu = e d = 1.602\times10^{-19} \times 127\times10^{-12} = 2.035\times10^{-29}$ C m. In debye this is $2.035\times10^{-29}/3.336\times10^{-30} = 6.10$ D.
Convert the observed moment. $1.03\ \text{D} = 1.03 \times 3.336\times10^{-30} = 3.44\times10^{-30}$ C m.
Take the ratio. $\% \text{ ionic} = \dfrac{1.03}{6.10} \times 100 \approx 16.9\%$ (about 17.7 percent if 1.08 D is used).
Interpret. The partial charge on each atom is about $0.17 e$. The bond is polar covalent, not ionic.

Answer: About 17 percent ionic character (17 to 18 percent across the quoted range of the moment), so the effective charge on $\ce{H}$ is about $+0.17e$ and on $\ce{Cl}$ about $-0.17e$.

Formal charges, resonance and bond order in the carbonate ion

Draw the Lewis structure of $\ce{CO3^{2-}}$, assign formal charges, state the number of resonance structures, and find the average bond order and the charge on each oxygen in the hybrid.

Count electrons and draw a structure. Valence electrons: $4 + 3(6) + 2 = 24$. Carbon is bonded to three oxygens, with one $\ce{C=O}$ and two $\ce{C-O^-}$. All atoms have an octet.
Assign formal charges. Carbon: $4 - 0 - \tfrac12(8) = 0$. Double-bonded oxygen: $6 - 4 - \tfrac12(4) = 0$. Each single-bonded oxygen: $6 - 6 - \tfrac12(2) = -1$. The sum is $0 + 0 - 1 - 1 = -2$, matching the ion.
Count resonance structures. The double bond can be placed on any of the three oxygens, giving 3 equivalent structures.
Average the bonds. Four bonds are shared over three equivalent $\ce{C-O}$ positions, so bond order $= 4/3 = 1.33$. The two negative charges are shared by three oxygens, so each carries $-2/3$.

Answer: Three resonance structures, formal charges $0$, $0$, $-1$, $-1$ in each. Average $\ce{C-O}$ bond order $4/3$, with all three bonds equal in length and a charge of $-2/3$ on each oxygen.

Estimating the enthalpy of combustion of methane from bond enthalpies

Estimate $\Delta H$ for $\ce{CH4(g) + 2O2(g) -> CO2(g) + 2H2O(g)}$ using $\ce{C-H}$ 414, $\ce{O=O}$ 498, $\ce{C=O}$ (in $\ce{CO2}$) 799 and $\ce{O-H}$ 463 kJ/mol.

List the bonds broken. One mole of $\ce{CH4}$ has 4 $\ce{C-H}$ bonds: $4 \times 414 = 1656$. Two moles of $\ce{O2}$ have 2 $\ce{O=O}$ bonds: $2 \times 498 = 996$. Total broken $= 2652$ kJ.
List the bonds formed. $\ce{CO2}$ has 2 $\ce{C=O}$ bonds: $2 \times 799 = 1598$. Two moles of $\ce{H2O}$ have 4 $\ce{O-H}$ bonds: $4 \times 463 = 1852$. Total formed $= 3450$ kJ.
Apply the rule. $\Delta H = 2652 - 3450 = -798$ kJ/mol.
State the limits. The measured value for gaseous water is about $-802$ kJ/mol. The two agree within 1 percent, but the method uses average bond enthalpies and gives only an estimate.

Answer: $\Delta H \approx -798$ kJ/mol (exothermic).

Figure: Five electron-domain geometries drawn as ball-and-line models: linear, trigonal planar, tetrahedral, trigonal bipyramidal and octahedral, each with its bond angles.

Figure: Eight molecular shapes with lone pairs shown as shaded lobes: bent SO2, bent H2O, pyramidal NH3, see-saw SF4, T-shaped ClF3, linear XeF2, square planar XeF4 and square pyramidal BrF5.

Figure: Four orbital overlap diagrams: s-s, s-p and p-p axial overlap forming sigma bonds, and p-p lateral overlap forming a pi bond above and below the bond axis.

Figure: Molecular orbital energy diagram of O2 with sigma 2p below pi 2p, electrons shown as arrows, with two unpaired electrons in the pi star 2p orbitals.

Figure: Molecular orbital energy diagram of N2 with pi 2p below sigma 2p, all 10 valence electrons paired, and empty pi star and sigma star orbitals.

Figure: Bond dipole arrows for CO2, H2O, BF3 and NH3 showing the cancellation in CO2 and BF3 and a net dipole toward oxygen in water and toward nitrogen in ammonia.

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