Summary and Quick Revision

How to Use This Section

This section is for the night before the paper, and again in the queue outside the hall.

Nothing new is taught here. Every card compresses something Sections 1 to 16 worked through properly, in the same notation and with the same numbers. If a line surprises you, go back and reread that section instead of memorising the line.

Eight cards, one mistake checklist, one 60-second panic list. Screenshot the three figures.

Card	Topic	Sections it compresses	Who needs it most
1	Kössel-Lewis approach — Lewis symbols, octet rule, Lewis-structure recipe, formal charge, octet exceptions	1, 2	Everyone
2	Ionic bond — conditions, NaCl energy balance, lattice enthalpy, Fajans' rules	3	Board, JEE
3	Bond parameters — length, angle, enthalpy, order, resonance	4	Everyone
4	Dipole moment and percentage ionic character	5	Everyone
5	VSEPR — the shape table with angles and examples	6	Everyone
6	Valence bond theory and hybridisation — overlap, sigma/pi, sp to sp$^3$d$^2$	7, 8, 9	Everyone
7	Molecular orbital theory — energy orders, bond orders, magnetism	10	JEE, NEET
8	Hydrogen bonding — types, consequences, boiling-point orders	11	Board, NEET

Key Point: The whole chapter is one question asked five ways: why do atoms stick together, and in what shape? Kössel-Lewis answers with electron counting (octets, dots, formal charge). The ionic model answers with energy (lattice enthalpy beats the cost of making ions). VSEPR answers with repulsion (electron pairs stay as far apart as they can). Valence bond theory answers with orbital overlap, plus hybridisation to get the angles right. Molecular orbital theory answers with delocalised orbitals, and predicts bond order and magnetism that nothing else can. Hydrogen bonding is the after-story — what molecules do to each other once they are made. Hold that map and the cards below are just the numbers.

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Card 1 — The Kössel-Lewis Approach

Lewis symbols and the octet rule

Key Point (Definition): A Lewis symbol is the element symbol with its valence electrons shown as dots. Group 1 to 18 main-group elements carry 1, 2, 3, 4, 5, 6, 7 and 8 dots. Period 2 in dots: Li 1, Be 2, B 3, C 4, N 5, O 6, F 7, Ne 8. Valence of a main-group element = number of dots, or 8 minus the number of dots (O: 6 dots, valence 2; Cl: 7 dots, valence 1).

Key Point (Definition): Octet rule (Kössel and Lewis, 1916): atoms combine by transfer or sharing of valence electrons so that each atom acquires eight electrons in its outer shell (the nearest noble-gas configuration). A covalent bond is a shared pair; one shared pair is a single bond, two a double bond, three a triple bond. Electrons not shared are lone pairs.

Molecule	Shared pairs	Lone pairs	Written as
$\mathrm{Cl_2}$	1	3 on each Cl	$\mathrm{Cl{-}Cl}$
$\mathrm{H_2O}$	2	2 on O	$\mathrm{H{-}O{-}H}$
$\mathrm{CO_2}$	4 (two double bonds)	2 on each O	$\mathrm{O{=}C{=}O}$
$\mathrm{C_2H_4}$	6 (one double C=C)	0	$\mathrm{H_2C{=}CH_2}$
$\mathrm{N_2}$	3 (one triple)	1 on each N	$\mathrm{N{\equiv}N}$
$\mathrm{C_2H_2}$	5 (one triple C$\equiv$C)	0	$\mathrm{H{-}C{\equiv}C{-}H}$

The Lewis-structure recipe (five steps)

Step	Do this	Example: $\mathrm{NF_3}$
1	Add up the valence electrons of all atoms; add one per negative charge, subtract one per positive charge	$5 + 3 \times 7 = 26$
2	Write the skeleton: the least electronegative atom is usually central (H and F are never central)	N in the middle, three F around
3	Put one shared pair (2 electrons) in each bond	3 bonds, 6 electrons used
4	Distribute the rest as lone pairs, terminal atoms first, until every atom has an octet	20 electrons left: 3 lone pairs on each F (18), 1 on N (2)
5	If the central atom still lacks an octet, convert lone pairs of neighbours into multiple bonds	not needed here

Counts worth memorising: $\mathrm{NH_4^+}$ has $5 + 4 - 1 = 8$ electrons, four bonds, no lone pair. $\mathrm{CO_3^{2-}}$ has $4 + 18 + 2 = 24$: one $\mathrm{C{=}O}$, two $\mathrm{C{-}O^-}$. $\mathrm{NO_2^-}$ has $5 + 12 + 1 = 18$: one $\mathrm{N{=}O}$, one $\mathrm{N{-}O^-}$, one lone pair on N. $\mathrm{CO}$ has 10: a triple bond, one lone pair on each atom. $\mathrm{O_3}$ has 18: one $\mathrm{O{=}O}$, one $\mathrm{O{-}O}$, central O carries one lone pair.

Formal charge

Key Point (Definition): $$\text{Formal charge} = V - L - \tfrac{1}{2}S$$ where $V$ = valence electrons of the free atom, $L$ = lone-pair (non-bonding) electrons on that atom in the structure, $S$ = shared (bonding) electrons around it. It is a bookkeeping number, not a real charge. Formal charges on all atoms must add up to the charge of the species. The structure with the lowest formal charges (and negative ones on the more electronegative atoms) is the most stable.

The ozone card, $\mathrm{O{=}O{-}O}$ with 18 electrons:

Atom	$V$	$L$	$S$	Formal charge
Central O (one lone pair, one double + one single bond)	6	2	6	$6 - 2 - 3 = +1$
Terminal O of the double bond (two lone pairs)	6	4	4	$6 - 4 - 2 = 0$
Terminal O of the single bond (three lone pairs)	6	6	2	$6 - 6 - 1 = -1$

Sum: $+1 + 0 - 1 = 0$, the charge on $\mathrm{O_3}$. Correct.

The four exceptions to the octet rule

Exception	What happens	Examples
Incomplete octet of the central atom (fewer than 8)	Elements with fewer than four valence electrons cannot reach eight	$\mathrm{LiCl}$, $\mathrm{BeH_2}$ (4 electrons on Be), $\mathrm{BCl_3}$ (6 on B), $\mathrm{AlCl_3}$
Odd-electron molecules	An odd total means one atom cannot have an octet	$\mathrm{NO}$ (11 electrons), $\mathrm{NO_2}$ (17 electrons)
Expanded octet (more than 8)	Period 3 and heavier central atoms use their $d$ orbitals	$\mathrm{PF_5}$ (10 on P), $\mathrm{SF_6}$ (12 on S), $\mathrm{H_2SO_4}$ (12 on S), $\mathrm{IF_7}$ (14 on I)
Noble-gas compounds	Xe and Kr, with complete octets, still form compounds	$\mathrm{XeF_2}$, $\mathrm{XeF_4}$, $\mathrm{XeF_6}$, $\mathrm{XeOF_2}$, $\mathrm{KrF_2}$

Three further limitations: the rule says nothing about the shape of a molecule; it says nothing about the relative stability or energy of molecules; and it rests on the "inertness" of noble gases, which Xe and Kr disobey.

[NEET] $\mathrm{BF_3}$, $\mathrm{BeCl_2}$ and $\mathrm{AlCl_3}$ are electron-deficient (incomplete octet, Lewis acids). $\mathrm{PCl_5}$ and $\mathrm{SF_6}$ are hypervalent (expanded octet). $\mathrm{NO}$ and $\mathrm{NO_2}$ are paramagnetic because of the odd electron.

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Card 2 — The Ionic (Electrovalent) Bond

When an ionic bond forms

Key Point: An ionic bond is the electrostatic attraction between oppositely charged ions formed by complete transfer of one or more electrons. It is favoured when the metal has a low ionization enthalpy (easy to make the cation), the non-metal has a highly negative electron gain enthalpy (easy to make the anion), and the crystal has a high lattice enthalpy (the ions pack tightly and release a lot of energy). Electrovalence = number of unit charges on the ion: Ca is +2, Cl is $-1$ in $\mathrm{CaCl_2}$.

The NaCl energy balance — the numbers

Step	Equation	Enthalpy / $\mathrm{kJ\ mol^{-1}}$
Ionization of sodium	$\mathrm{Na(g) \rightarrow Na^+(g) + e^-}$	$+495.8$
Electron gain by chlorine	$\mathrm{Cl(g) + e^- \rightarrow Cl^-(g)}$	$-348.7$
Net cost of making the gaseous ions		$+147.1$
Lattice formation	$\mathrm{Na^+(g) + Cl^-(g) \rightarrow NaCl(s)}$	$-788$
Net		$147.1 - 788 = -640.9$ (strongly exothermic)

The electron transfer alone is uphill by 147.1 kJ/mol; the crystal forms because the lattice enthalpy (788 kJ/mol) pays for it many times over. The lattice, not the electron transfer, is why NaCl exists.

Lattice enthalpy

Key Point (Definition): The lattice enthalpy of an ionic solid is the energy required to completely separate one mole of the solid ionic compound into gaseous constituent ions. For NaCl it is 788 kJ/mol: $\mathrm{NaCl(s) \rightarrow Na^+(g) + Cl^-(g)}$, $\Delta_{lattice}H = +788$ kJ/mol. (Written the other way, as formation of the lattice from gaseous ions, it is $-788$.)

Factor	Effect on lattice enthalpy	Example
Charge on the ions — the stronger effect	higher charges, larger lattice enthalpy (attraction $\propto q_+ q_-$)	MgO (2+, 2$-$) $\gg$ NaF (1+, 1$-$)
Size of the ions	smaller ions, shorter distance, larger lattice enthalpy (attraction $\propto 1/r$)	LiF > NaF > KF > RbF > CsF; NaF > NaCl > NaBr > NaI
Lattice type (packing)	closer packing releases more energy	a JEE-level refinement; the Born-Haber cycle sums all the steps

Lattice enthalpy is not measured directly; it comes from a Born-Haber cycle (Hess's law) or is estimated from the Coulomb interaction of the ions.

Fajans' rules — covalent character in an ionic bond (one-liners)

Rule	Covalent character increases when
Cation size	the cation is small (high charge density, polarises the anion): $\mathrm{LiCl}$ more covalent than $\mathrm{CsCl}$
Anion size	the anion is large (easily polarised): $\mathrm{LiI}$ more covalent than $\mathrm{LiF}$
Charge	the charge on either ion is high: $\mathrm{AlCl_3}$ more covalent than $\mathrm{NaCl}$
Cation configuration	the cation has a pseudo-noble-gas $d^{10}$ shell ($\mathrm{Cu^+}$, $\mathrm{Ag^+}$, $\mathrm{Zn^{2+}}$) rather than a noble-gas shell ($\mathrm{Na^+}$, $\mathrm{K^+}$)

These are trends, not laws.

[JEE Main] Rank covalent character with the "small cation, large anion, high charge" mantra: $\mathrm{NaF} < \mathrm{NaCl} < \mathrm{NaBr} < \mathrm{NaI}$ and $\mathrm{BeCl_2} > \mathrm{MgCl_2} > \mathrm{CaCl_2}$. Melting point and solubility in water run the other way (more ionic, higher melting point, more soluble in water).

Properties of ionic compounds in one line: crystalline solids, high melting and boiling points, conduct electricity only when molten or dissolved (ions must be free to move), soluble in polar solvents, insoluble in non-polar ones, hard but brittle.

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Card 3 — Bond Parameters and Resonance

The definitions

Parameter	Definition	Units
Bond length	equilibrium distance between the nuclei of two bonded atoms	pm
Covalent radius	half the distance between the nuclei of two like atoms joined by a single covalent bond ($\mathrm{Cl_2}$: 198 pm, so 99 pm)	pm
van der Waals radius	half the distance between the nuclei of two non-bonded atoms of adjacent molecules in the solid ($\mathrm{Cl}$: 180 pm)	pm
Bond angle	angle between the orbitals containing the bonding pairs around the central atom (H-O-H in water 104.5°)	degrees
Bond enthalpy	energy needed to break one mole of a particular bond in gaseous molecules; for polyatomic molecules use the average	kJ/mol
Bond order	number of bonds (shared pairs) between two atoms in a molecule	none

Bond-length data

Bond	Length / pm	Bond	Length / pm
O-H	96	C=O	121
C-H	107	N=O	122
N-O	136	C=C	133
C-O	143	C=N	138
C-N	143	C$\equiv$N	116
C-C	154	C$\equiv$C	120

Molecule	Length / pm	Molecule	Length / pm
$\mathrm{H_2}$	74	$\mathrm{HF}$	92
$\mathrm{F_2}$	144	$\mathrm{HCl}$	127
$\mathrm{Cl_2}$	199	$\mathrm{HBr}$	141
$\mathrm{Br_2}$	228	$\mathrm{HI}$	160
$\mathrm{I_2}$	267	$\mathrm{N_2}$	109
$\mathrm{O_2}$	121	$\mathrm{O_3}$ (both O-O)	128

Rules: single > double > triple in length (C-C 154, C=C 133, C$\equiv$C 120); bond length rises with atom size (HF 92 to HI 160; $\mathrm{F_2}$ 144 to $\mathrm{I_2}$ 267). Bond length of a covalent bond = sum of the two covalent radii.

Bond-enthalpy data

Bond broken	$\Delta_a H$ / $\mathrm{kJ\ mol^{-1}}$
$\mathrm{H_2(g) \rightarrow 2\,H(g)}$	435.8
$\mathrm{O_2(g) \rightarrow 2\,O(g)}$	498
$\mathrm{N_2(g) \rightarrow 2\,N(g)}$	946.0
$\mathrm{H_2O(g) \rightarrow H(g) + OH(g)}$	502
$\mathrm{OH(g) \rightarrow H(g) + O(g)}$	427
Average O-H in water	$(502 + 427)/2 = 464.5$

Rules: enthalpy rises with bond order (H-H single 435.8, O=O double 498, N$\equiv$N triple 946); larger bond enthalpy means stronger, shorter bond. The two O-H bonds of water do not cost the same to break — hence the average.

Bond order — the counting rules

Species	Bond order	Isoelectronic partners with the same bond order
$\mathrm{H_2}$, $\mathrm{Cl_2}$, $\mathrm{F_2}$	1	—
$\mathrm{O_2}$	2	—
$\mathrm{N_2}$	3	$\mathrm{CO}$, $\mathrm{NO^+}$ (14 electrons each)
$\mathrm{CO}$	3	$\mathrm{N_2}$, $\mathrm{NO^+}$

Key Point: Isoelectronic molecules and ions have identical bond orders. As bond order rises, bond enthalpy rises and bond length falls.

Resonance — the rules and the hybrids

Key Point (Definition): When a single Lewis structure cannot describe a molecule, several canonical (resonance) structures are written, and the real molecule is the resonance hybrid. The hybrid is more stable than any single canonical structure by the resonance stabilisation energy. Canonical structures exist only on paper; the molecule does not flip between them.

Rule	Meaning
Same positions of nuclei	only electrons move, never atoms
Same number of unpaired electrons	all canonical forms have the same spin state
Bond lengths in the hybrid are all equal	$\mathrm{O_3}$: both O-O bonds 128 pm, between O-O 148 and O=O 121; $\mathrm{CO_3^{2-}}$: all three C-O bonds equal
Bond order of the hybrid	total bonds between the atom pair, divided by the number of positions

Species	Canonical structures	Bond order of each bond
$\mathrm{O_3}$	2	$(2 + 1)/2 = 1.5$
$\mathrm{CO_3^{2-}}$	3	$(2 + 1 + 1)/3 = 1.33$
$\mathrm{NO_3^-}$	3	$1.33$
$\mathrm{NO_2^-}$	2	$1.5$
$\mathrm{C_6H_6}$ (benzene)	2 (Kekulé)	$1.5$
$\mathrm{CO_2}$	3	2 (the $\mathrm{O{=}C{=}O}$ form dominates)
$\mathrm{SO_2}$	2	$1.5$

[JEE/NEET] In $\mathrm{CO_2}$ the measured C-O length is 115 pm — shorter than a normal C=O (121 pm) and longer than a C$\equiv$O (110 pm); that is the fingerprint of resonance among $\mathrm{O{=}C{=}O}$, $\mathrm{O{\equiv}C{-}O}$ and $\mathrm{O{-}C{\equiv}O}$.

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Card 4 — Polarity, Dipole Moment and Ionic Character

The formula

Key Point (Definition): Dipole moment $\mu = Q \times r$, where $Q$ is the magnitude of the separated charge and $r$ the distance between the charge centres. It is a vector; the arrow points from the positive end to the negative end (crossed at the positive end). Units: debye, $1\ \mathrm{D} = 3.33564 \times 10^{-30}$ C m. For a molecule with several polar bonds, $\mu$ is the vector sum of the bond dipoles and the lone-pair contributions.

Check on water: $1.85\ \mathrm{D} = 1.85 \times 3.33564 \times 10^{-30} = 6.17 \times 10^{-30}$ C m.

Zero or non-zero — the table

Molecule	Shape	$\mu$ / D	Why
$\mathrm{H_2}$, $\mathrm{Cl_2}$	linear diatomic	0	identical atoms, no bond polarity
$\mathrm{HF}$	diatomic	1.78	polar bond; the most polar hydrogen halide
$\mathrm{HCl}$	diatomic	1.07	electronegativity difference falls down the group
$\mathrm{HBr}$	diatomic	0.79
$\mathrm{HI}$	diatomic	0.38
$\mathrm{H_2O}$	bent	1.85	two O-H dipoles add, lone pairs help
$\mathrm{H_2S}$	bent	0.95	smaller electronegativity difference
$\mathrm{NH_3}$	trigonal pyramidal	1.47	lone-pair dipole adds to the N-H resultant
$\mathrm{NF_3}$	trigonal pyramidal	0.23	lone-pair dipole opposes the N-F resultant
$\mathrm{CHCl_3}$	tetrahedral, unsymmetrical	1.04	C-Cl dipoles no longer cancel
$\mathrm{CO_2}$	linear	0	two equal C=O dipoles cancel
$\mathrm{BF_3}$	trigonal planar	0	three B-F dipoles at 120° cancel
$\mathrm{CH_4}$, $\mathrm{CCl_4}$	tetrahedral	0	four identical dipoles cancel
$\mathrm{BeCl_2}$, $\mathrm{PCl_5}$, $\mathrm{SF_6}$, $\mathrm{XeF_2}$, $\mathrm{XeF_4}$	symmetric	0	symmetric arrangement, all lone pairs opposite each other

NH$_3$ versus NF$_3$ — the classic

Both are pyramidal with one lone pair on N. In $\mathrm{NH_3}$ nitrogen is the more electronegative atom, so the three N-H bond dipoles point towards N, in the same direction as the lone-pair dipole; they add, $\mu = 1.47$ D. In $\mathrm{NF_3}$ fluorine is more electronegative, so the N-F dipoles point away from N, against the lone pair; they partly cancel, $\mu = 0.23$ D. Same shape, opposite arithmetic.

Percentage ionic character

Key Point: $$\%\ \text{ionic character} = \frac{\mu_{\text{observed}}}{\mu_{\text{calculated (100\% ionic)}}} \times 100, \qquad \mu_{\text{calc}} = e \times d = 1.6 \times 10^{-19}\ \mathrm{C} \times d$$ Example: HCl with $d = 127$ pm gives $\mu_{\text{calc}} = 1.6 \times 10^{-19} \times 1.27 \times 10^{-10} = 2.03 \times 10^{-29}$ C m $= 6.09$ D; observed 1.07 D, so ionic character $\approx 17.6\%$.

Rule-of-thumb from electronegativity difference: a difference of about 1.7 gives roughly 50% ionic character; larger than 1.7 is mostly ionic, smaller is mostly covalent. A purely covalent bond ($\mathrm{H_2}$) has zero; even the "most ionic" bonds (CsF) fall short of 100%.

[JEE Main] For a symmetric molecule the answer is "zero dipole moment" no matter how polar the bonds are; the question is testing geometry. The order among the hydrogen halides ($\mathrm{HF} > \mathrm{HCl} > \mathrm{HBr} > \mathrm{HI}$) and the pair $\mathrm{NH_3} > \mathrm{NF_3}$ are the two rankings asked most.

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Card 5 — VSEPR Theory

The postulates (Sidgwick and Powell 1940, Nyholm and Gillespie 1957)

#	Postulate
1	The shape depends on the number of valence-shell electron pairs (bonding and non-bonding) around the central atom
2	Electron pairs repel one another because their clouds are negatively charged
3	The pairs occupy positions that minimise repulsion and hence maximise distance
4	The valence shell is taken as a sphere with the pairs at maximum distance on its surface
5	A multiple bond is treated as a single electron pair, and its two or three pairs as one super pair
6	Where a molecule has two or more resonance structures, VSEPR applies to any one of them

The repulsion order

Key Point: $$\text{lone pair-lone pair} > \text{lone pair-bond pair} > \text{bond pair-bond pair}$$ A lone pair belongs to one nucleus only, so it is fatter and closer to the central atom; it pushes the bond pairs together and shrinks the bond angle. One lone pair costs about 2.5°, two cost about 5° in the tetrahedral family: $\mathrm{CH_4}$ 109.5°, $\mathrm{NH_3}$ 107°, $\mathrm{H_2O}$ 104.5°.

VSEPR predicts the molecular shape from the electron-pair geometry, with lone pairs distorting the angles.

The full shape table

Type	Bond pairs	Lone pairs	Electron-pair geometry	Molecular shape	Angle(s)	Examples
$\mathrm{AB_2}$	2	0	linear	linear	180°	$\mathrm{BeCl_2}$, $\mathrm{HgCl_2}$, $\mathrm{CO_2}$
$\mathrm{AB_3}$	3	0	trigonal planar	trigonal planar	120°	$\mathrm{BF_3}$, $\mathrm{BCl_3}$, $\mathrm{SO_3}$, $\mathrm{NO_3^-}$
$\mathrm{AB_2E}$	2	1	trigonal planar	bent (V-shape)	< 120°	$\mathrm{SO_2}$, $\mathrm{O_3}$, $\mathrm{SnCl_2}$
$\mathrm{AB_4}$	4	0	tetrahedral	tetrahedral	109.5°	$\mathrm{CH_4}$, $\mathrm{NH_4^+}$, $\mathrm{SiCl_4}$, $\mathrm{SO_4^{2-}}$
$\mathrm{AB_3E}$	3	1	tetrahedral	trigonal pyramidal	107°	$\mathrm{NH_3}$, $\mathrm{PCl_3}$, $\mathrm{H_3O^+}$
$\mathrm{AB_2E_2}$	2	2	tetrahedral	bent	104.5°	$\mathrm{H_2O}$, $\mathrm{H_2S}$, $\mathrm{OF_2}$
$\mathrm{AB_5}$	5	0	trigonal bipyramidal	trigonal bipyramidal	120° (equatorial), 90° (axial)	$\mathrm{PCl_5}$, $\mathrm{PF_5}$
$\mathrm{AB_4E}$	4	1	trigonal bipyramidal	see-saw	< 120°, < 90°	$\mathrm{SF_4}$
$\mathrm{AB_3E_2}$	3	2	trigonal bipyramidal	T-shape	< 90°	$\mathrm{ClF_3}$, $\mathrm{BrF_3}$
$\mathrm{AB_2E_3}$	2	3	trigonal bipyramidal	linear	180°	$\mathrm{XeF_2}$, $\mathrm{I_3^-}$
$\mathrm{AB_6}$	6	0	octahedral	octahedral	90°	$\mathrm{SF_6}$, $\mathrm{PF_6^-}$
$\mathrm{AB_5E}$	5	1	octahedral	square pyramidal	< 90°	$\mathrm{BrF_5}$, $\mathrm{IF_5}$
$\mathrm{AB_4E_2}$	4	2	octahedral	square planar	90°	$\mathrm{XeF_4}$, $\mathrm{ICl_4^-}$

Where the lone pairs sit

Geometry	Lone-pair position	Why
Trigonal bipyramidal	equatorial	an equatorial pair has only two 90° neighbours; an axial pair has three
Octahedral	any one position; a second one goes opposite (trans)	keeps the two lone pairs 180° apart
Tetrahedral, trigonal planar	any position (all equivalent)	—

In $\mathrm{PCl_5}$ the two axial bonds are longer (and weaker) than the three equatorial ones, because each axial pair suffers three 90° repulsions against only two for an equatorial pair — the reason $\mathrm{PCl_5}$ is so reactive.

[NEET] Angle-ordering questions: $\mathrm{CH_4}$ (109.5°) > $\mathrm{NH_3}$ (107°) > $\mathrm{H_2O}$ (104.5°); $\mathrm{NH_3}$ > $\mathrm{PH_3}$ > $\mathrm{AsH_3}$ (bigger central atom, less electronegative, bond pairs sit farther out, less repulsion); $\mathrm{H_2O}$ > $\mathrm{H_2S}$ > $\mathrm{H_2Se}$ for the same reason; $\mathrm{NH_3}$ > $\mathrm{NF_3}$ (F pulls the bond pairs away from N).

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Card 6 — Valence Bond Theory and Hybridisation

Orbital overlap — the rules

Key Point: (Heitler and London 1927, Pauling and Slater.) A covalent bond forms when two half-filled atomic orbitals with electrons of opposite spin overlap. Greater overlap, stronger bond. As two H atoms approach, attractions (nucleus-electron of the other atom) and repulsions (nucleus-nucleus, electron-electron) compete; the potential energy falls to a minimum at 74 pm, the bond length, and the depth of that well is the bond enthalpy, 435.8 kJ/mol. Energy is released when the bond forms and absorbed when it breaks.

Type of overlap	Along the axis or sideways	Bond formed	Examples
$s$-$s$	end-on (axial)	$\sigma$	$\mathrm{H_2}$
$s$-$p$	end-on	$\sigma$	$\mathrm{HF}$, $\mathrm{HCl}$
$p$-$p$ end-on	along the axis	$\sigma$	$\mathrm{F_2}$, $\mathrm{Cl_2}$
$p$-$p$ sideways	lateral (parallel axes)	$\pi$	the second bond of $\mathrm{O_2}$, the two extra bonds of $\mathrm{N_2}$

Sigma versus pi — the facts

Feature	$\sigma$ bond	$\pi$ bond
Overlap	end-to-end along the internuclear axis	sideways, above and below the axis
Extent of overlap	larger	smaller
Strength	stronger	weaker
Rotation about the bond	free	restricted
Electron cloud	symmetric about the axis	two lobes, a nodal plane containing the axis
Order of formation	first bond of any pair is always $\sigma$	second and third bonds
Can exist alone?	yes	no — only with a $\sigma$

Counting rule: every single bond = 1 $\sigma$; double bond = 1 $\sigma$ + 1 $\pi$; triple bond = 1 $\sigma$ + 2 $\pi$. $\mathrm{C_2H_2}$: 3 $\sigma$, 2 $\pi$. $\mathrm{C_2H_4}$: 5 $\sigma$, 1 $\pi$. $\mathrm{CO_2}$: 2 $\sigma$, 2 $\pi$. Benzene: 12 $\sigma$, 3 $\pi$. $\mathrm{N_2}$: 1 $\sigma$, 2 $\pi$.

Hybridisation

Key Point (Definition): Hybridisation is the intermixing of atomic orbitals of slightly different energies to form the same number of new equivalent orbitals of equal energy and identical shape. Only orbitals of comparable energy mix; half-filled, fully filled and empty orbitals can all take part; hybrid orbitals are more directional and give stronger bonds; the hybridisation is of the central atom and it is a model applied once the shape is known, not an observed event or a cause. Promotion of an electron is not required before hybridisation.

Hybridisation	Orbitals mixed	Hybrids	s-character	Geometry	Angle	Examples
$sp$	one $s$ + one $p$	2	50%	linear	180°	$\mathrm{BeCl_2}$, $\mathrm{BeH_2}$, $\mathrm{C_2H_2}$ (each C), $\mathrm{CO_2}$
$sp^2$	one $s$ + two $p$	3	33.3%	trigonal planar	120°	$\mathrm{BCl_3}$, $\mathrm{BF_3}$, $\mathrm{C_2H_4}$ (each C), $\mathrm{SO_2}$, $\mathrm{NO_3^-}$, benzene C
$sp^3$	one $s$ + three $p$	4	25%	tetrahedral	109.5°	$\mathrm{CH_4}$, $\mathrm{NH_3}$ (107°), $\mathrm{H_2O}$ (104.5°), $\mathrm{NH_4^+}$, $\mathrm{C_2H_6}$
$sp^3d$	one $s$ + three $p$ + $d_{z^2}$	5	20%	trigonal bipyramidal	120°, 90°	$\mathrm{PCl_5}$, $\mathrm{PF_5}$, $\mathrm{SF_4}$, $\mathrm{ClF_3}$, $\mathrm{XeF_2}$
$sp^3d^2$	one $s$ + three $p$ + $d_{x^2-y^2}$ + $d_{z^2}$	6	16.7%	octahedral	90°	$\mathrm{SF_6}$, $\mathrm{XeF_4}$, $\mathrm{BrF_5}$, $\mathrm{[CoF_6]^{3-}}$ (outer-orbital complex)
$d^2sp^3$	two inner $d$ + $s$ + three $p$	6	16.7%	octahedral	90°	$\mathrm{[Co(NH_3)_6]^{3+}}$ (inner-orbital complex)

The steric-number rule: Steric number = number of $\sigma$ bonds around the central atom + number of lone pairs on it. 2 gives $sp$, 3 gives $sp^2$, 4 gives $sp^3$, 5 gives $sp^3d$, 6 gives $sp^3d^2$. $\pi$ bonds are not counted. More $s$-character means a shorter, stronger bond and a larger angle ($sp$ C-H is the shortest and most acidic).

C$_2$H$_6$, C$_2$H$_4$ and C$_2$H$_2$ — the data

Molecule	C hybridisation	C-C bond	C-C length / pm	C-C enthalpy / kJ mol$^{-1}$	C-H length / pm	Angles	$\sigma$ / $\pi$
Ethane $\mathrm{C_2H_6}$	$sp^3$	one $\sigma$ ($sp^3$-$sp^3$)	154	348	109	109.5°	7 / 0
Ethene $\mathrm{C_2H_4}$	$sp^2$	one $\sigma$ ($sp^2$-$sp^2$) + one $\pi$ ($2p$-$2p$)	134	614	108	H-C-H about 117.6°, H-C-C about 121°; planar	5 / 1
Ethyne $\mathrm{C_2H_2}$	$sp$	one $\sigma$ ($sp$-$sp$) + two $\pi$ ($2p_y$, $2p_z$)	120	839	106	180°; linear	3 / 2

The $\pi$ bond in ethene is worth about $614 - 348 = 266$ kJ/mol, less than the $\sigma$ (348) — sideways overlap is the weaker kind.

[JEE Main] $sp^3d$ uses the $d_{z^2}$ orbital; $sp^3d^2$ uses $d_{x^2-y^2}$ and $d_{z^2}$. In $\mathrm{PCl_5}$ the promotion is $3s^2 3p^3 \rightarrow 3s^1 3p^3 3d^1$; in $\mathrm{SF_6}$, $3s^2 3p^4 \rightarrow 3s^1 3p^3 3d^2$. For $\mathrm{XeF_2}$ (5 pairs, 3 lone) the hybridisation is $sp^3d$ and the shape linear; for $\mathrm{XeF_4}$ (6 pairs, 2 lone) $sp^3d^2$ and square planar.

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Card 7 — Molecular Orbital Theory

LCAO and the conditions

Key Point: (Hund and Mulliken, 1932.) Atomic orbitals combine by Linear Combination of Atomic Orbitals: addition $\psi_A + \psi_B$ gives a bonding MO ($\sigma$, lower energy, electron density between the nuclei), subtraction $\psi_A - \psi_B$ gives an antibonding MO ($\sigma^*$, higher energy, a node between the nuclei). Two AOs always give two MOs. The bonding MO is stabilised by about as much as the antibonding is destabilised. Molecular orbitals are polycentric (belong to the whole molecule); atomic orbitals are monocentric. Filling follows Aufbau, Pauli and Hund's rule exactly as for atoms.

Condition for combination	Consequence
Same or nearly the same energy	$1s$ combines with $1s$, not with $2s$
Same symmetry about the molecular axis	$2p_z$ combines with $2p_z$, not with $2p_x$ (with $z$ as the bond axis)
Maximum overlap	greater overlap, greater electron density between the nuclei, stronger bond

Naming with $z$ as the internuclear axis: $2p_z$ orbitals give $\sigma 2p_z$ and $\sigma^* 2p_z$; $2p_x$ and $2p_y$ give the degenerate pairs $\pi 2p_x$, $\pi 2p_y$ and $\pi^* 2p_x$, $\pi^* 2p_y$. A $\sigma$ MO is symmetric about the axis; a $\pi$ MO has a nodal plane containing the axis.

The two energy orderings

Molecules	Order	Why
$\mathrm{O_2}$, $\mathrm{F_2}$ (and $\mathrm{Ne_2}$)	$\sigma 1s < \sigma^* 1s < \sigma 2s < \sigma^* 2s < \sigma 2p_z < (\pi 2p_x = \pi 2p_y) < (\pi^* 2p_x = \pi^* 2p_y) < \sigma^* 2p_z$	large $2s$-$2p$ energy gap, no mixing
$\mathrm{Li_2}$ to $\mathrm{N_2}$ ($\mathrm{B_2}$, $\mathrm{C_2}$, $\mathrm{N_2}$ especially)	$\sigma 1s < \sigma^* 1s < \sigma 2s < \sigma^* 2s < (\pi 2p_x = \pi 2p_y) < \sigma 2p_z < (\pi^* 2p_x = \pi^* 2p_y) < \sigma^* 2p_z$	$2s$-$2p$ mixing pushes $\sigma 2p_z$ above the $\pi 2p$ pair

The only difference: for $\mathrm{B_2}$, $\mathrm{C_2}$, $\mathrm{N_2}$ the $\pi 2p$ pair fills before $\sigma 2p_z$. Get this wrong and $\mathrm{B_2}$ comes out diamagnetic (it is paramagnetic) and $\mathrm{C_2}$ comes out paramagnetic (it is diamagnetic).

Bond order

Key Point (Definition): $$\text{B.O.} = \tfrac{1}{2}\,(N_b - N_a)$$ $N_b$ = electrons in bonding MOs, $N_a$ = electrons in antibonding MOs. Positive bond order means the molecule is stable; zero or negative means it does not exist. Bond order 1, 2, 3 correspond to single, double, triple. Higher bond order: shorter bond, higher bond enthalpy. Any unpaired electron in an MO makes the species paramagnetic; all paired, diamagnetic.

The master table

Species	Electrons	Configuration (valence MOs)	$N_b$	$N_a$	Bond order	Magnetism
$\mathrm{H_2}$	2	$\sigma 1s^2$	2	0	1	diamagnetic
$\mathrm{H_2^+}$	1	$\sigma 1s^1$	1	0	0.5	paramagnetic
$\mathrm{He_2^+}$	3	$\sigma 1s^2\, \sigma^* 1s^1$	2	1	0.5	paramagnetic
$\mathrm{He_2}$	4	$\sigma 1s^2\, \sigma^* 1s^2$	2	2	0	does not exist
$\mathrm{Li_2}$	6	$\mathrm{KK}\, \sigma 2s^2$	4	2	1	diamagnetic
$\mathrm{Be_2}$	8	$\mathrm{KK}\, \sigma 2s^2\, \sigma^* 2s^2$	4	4	0	does not exist
$\mathrm{B_2}$	10	$\mathrm{KK}\, \sigma 2s^2\, \sigma^* 2s^2\, \pi 2p_x^1\, \pi 2p_y^1$	6	4	1	paramagnetic
$\mathrm{C_2}$	12	$\mathrm{KK}\, \sigma 2s^2\, \sigma^* 2s^2\, \pi 2p_x^2\, \pi 2p_y^2$	8	4	2	diamagnetic (both bonds $\pi$)
$\mathrm{N_2}$	14	$\mathrm{KK}\, \sigma 2s^2\, \sigma^* 2s^2\, \pi 2p_x^2\, \pi 2p_y^2\, \sigma 2p_z^2$	10	4	3	diamagnetic
$\mathrm{N_2^+}$	13	remove one from $\sigma 2p_z$	9	4	2.5	paramagnetic
$\mathrm{O_2}$	16	$\mathrm{KK}\, \sigma 2s^2\, \sigma^* 2s^2\, \sigma 2p_z^2\, \pi 2p_x^2\, \pi 2p_y^2\, \pi^* 2p_x^1\, \pi^* 2p_y^1$	10	6	2	paramagnetic (2 unpaired)
$\mathrm{O_2^+}$	15	one fewer $\pi^*$ electron	10	5	2.5	paramagnetic (1)
$\mathrm{O_2^-}$ (superoxide)	17	one more $\pi^*$ electron	10	7	1.5	paramagnetic (1)
$\mathrm{O_2^{2-}}$ (peroxide)	18	$\pi^*$ full	10	8	1	diamagnetic
$\mathrm{F_2}$	18	$\ldots \pi^* 2p_x^2\, \pi^* 2p_y^2$	10	8	1	diamagnetic
$\mathrm{Ne_2}$	20	$\ldots \sigma^* 2p_z^2$	10	10	0	does not exist
$\mathrm{CO}$	14	like $\mathrm{N_2}$	10	4	3	diamagnetic
$\mathrm{NO}$	15	like $\mathrm{O_2^+}$	10	5	2.5	paramagnetic (1)
$\mathrm{NO^+}$	14	like $\mathrm{N_2}$	10	4	3	diamagnetic
$\mathrm{CN^-}$	14	like $\mathrm{N_2}$	10	4	3	diamagnetic

$\mathrm{KK}$ stands for $\sigma 1s^2\, \sigma^* 1s^2$, the non-bonding inner shells.

The rankings that follow

Question	Answer
Bond length order	$\mathrm{O_2^+} < \mathrm{O_2} < \mathrm{O_2^-} < \mathrm{O_2^{2-}}$ (bond order 2.5, 2, 1.5, 1)
Bond strength (enthalpy) order	$\mathrm{O_2^+} > \mathrm{O_2} > \mathrm{O_2^-} > \mathrm{O_2^{2-}}$
Removing an electron from $\mathrm{N_2}$ versus $\mathrm{O_2}$	$\mathrm{N_2 \rightarrow N_2^+}$: bond order 3 to 2.5, bond weakens; $\mathrm{O_2 \rightarrow O_2^+}$: 2 to 2.5, bond strengthens (the electron leaves an antibonding orbital)
Same bond order as $\mathrm{N_2}$	$\mathrm{CO}$, $\mathrm{NO^+}$, $\mathrm{CN^-}$ (all 14-electron, bond order 3)
Highest bond order in the second period	$\mathrm{N_2}$ (3); $\mathrm{N_2}$ has the highest bond enthalpy, 946 kJ/mol
Paramagnetic among $\mathrm{B_2}$, $\mathrm{C_2}$, $\mathrm{N_2}$, $\mathrm{O_2}$, $\mathrm{F_2}$	$\mathrm{B_2}$ and $\mathrm{O_2}$

[NEET] The paramagnetism of $\mathrm{O_2}$ (liquid oxygen sticks to a magnet) is the single fact valence bond theory cannot explain and MO theory explains at once — the two unpaired electrons sit one each in the degenerate $\pi^* 2p_x$ and $\pi^* 2p_y$ orbitals by Hund's rule.

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Card 8 — Hydrogen Bonding

The definition

Key Point (Definition): When hydrogen is bonded to a highly electronegative, small atom (F, O or N), the shared pair is pulled away and the H becomes a bare, positive centre. This H is attracted to the lone pair of an electronegative atom of another molecule (or of the same molecule). That attraction is a hydrogen bond, written with a dotted line: $\mathrm{H^{\delta+}{-}F^{\delta-}} \cdots \mathrm{H^{\delta+}{-}F^{\delta-}}$. It is weaker than a covalent bond (about 10-40 kJ/mol against 400 or more) but much stronger than ordinary van der Waals forces, and it is largely electrostatic.

Strength order: $\mathrm{F{-}H \cdots F} > \mathrm{O{-}H \cdots O} > \mathrm{N{-}H \cdots N}$, following electronegativity (F 4.0 > O 3.5 > N 3.0). Chlorine is electronegative enough (3.0) but too large, so HCl does not hydrogen-bond appreciably.

The two types

Type	Where	Examples	Consequence
Intermolecular	between different molecules	HF (zig-zag chains), $\mathrm{H_2O}$ (each molecule bonds to four neighbours in ice), $\mathrm{NH_3}$, alcohols, carboxylic acid dimers, p-nitrophenol	molecules cling together: higher boiling point, higher solubility in water, association
Intramolecular	within the same molecule, forming a ring (usually 5- or 6-membered)	o-nitrophenol, salicylaldehyde, o-hydroxybenzoic acid	no clinging between molecules: lower boiling point, lower water solubility, more volatile, chelation

The consequences

Observation	Explanation
Water is a liquid, $\mathrm{H_2S}$ a gas, although S is heavier	O-H$\cdots$O hydrogen bonds hold water molecules together; S is too large and not electronegative enough
Ice floats on water	in ice every water molecule is hydrogen-bonded tetrahedrally to four others in an open cage with empty space; on melting some bonds break and molecules pack closer, so liquid water is denser (maximum density at 4 °C)
HF has the highest boiling point among the hydrogen halides	F-H$\cdots$F bonds; HCl, HBr, HI have none, so their boiling points follow molar mass
$\mathrm{NH_3}$ boils higher than $\mathrm{PH_3}$	N-H$\cdots$N bonds; P cannot hydrogen-bond
Ethanol boils at 78 °C, dimethyl ether at $-24$ °C, same formula $\mathrm{C_2H_6O}$	ethanol has O-H and hydrogen-bonds; the ether has no H on O
Ethanol and ammonia dissolve freely in water; the DNA double helix holds together; proteins keep their shape	hydrogen bonds to water; A$\cdots$T (two) and G$\cdots$C (three) hydrogen bonds between bases
o-nitrophenol is steam-volatile, p-nitrophenol is not	o-isomer has an intramolecular bond (no association), p-isomer forms intermolecular chains

Boiling-point orders to write from memory

Series	Order	Reason
Group 16 hydrides	$\mathrm{H_2O} > \mathrm{H_2Te} > \mathrm{H_2Se} > \mathrm{H_2S}$	water anomalous (H-bonding); the rest rise with molar mass (van der Waals)
Group 17 hydrides	$\mathrm{HF} > \mathrm{HI} > \mathrm{HBr} > \mathrm{HCl}$	HF anomalous; the rest rise with molar mass
Group 15 hydrides	$\mathrm{SbH_3} > \mathrm{NH_3} > \mathrm{AsH_3} > \mathrm{PH_3}$	$\mathrm{NH_3}$ anomalous (H-bonding lifts it above $\mathrm{PH_3}$ and $\mathrm{AsH_3}$), but the N-H$\cdots$N bond is the weakest of the three, so the very heavy $\mathrm{SbH_3}$ still boils higher
Group 14 hydrides	$\mathrm{SnH_4} > \mathrm{GeH_4} > \mathrm{SiH_4} > \mathrm{CH_4}$	no hydrogen bonding in any (C not electronegative enough); pure molar-mass order
Water versus HF versus $\mathrm{NH_3}$	$\mathrm{H_2O}$ (100 °C) $> \mathrm{HF}$ (19.5 °C) $> \mathrm{NH_3}$ ($-33$ °C)	water averages two hydrogen bonds per molecule when each bond is counted once — it takes part in four, two through its own H atoms and two through its lone pairs; HF has one H but three lone pairs, $\mathrm{NH_3}$ three H but one lone pair, so each is limited to one

[Board] Two-mark answer in two lines: "Hydrogen bond is the attractive force between the hydrogen atom attached to a highly electronegative atom (F, O, N) of one molecule and the electronegative atom of another molecule. Example: HF, in which $\mathrm{H{-}F \cdots H{-}F}$ chains raise its boiling point above HCl."

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The Mistakes That Cost the Most Marks

Each of these was flagged somewhere in Sections 1 to 16, ordered roughly by how often it shows up in answer scripts.

1. Counting $\pi$ bonds and lone pairs when deciding the hybridisation. Steric number = $\sigma$ bonds + lone pairs on the central atom. A double bond is one $\sigma$. So $\mathrm{CO_2}$ is $sp$ (2 $\sigma$, 0 lone pairs), $\mathrm{SO_2}$ is $sp^2$ (2 $\sigma$ + 1 lone pair), $\mathrm{NH_3}$ is $sp^3$ (3 + 1), $\mathrm{XeF_2}$ is $sp^3d$ (2 + 3), $\mathrm{XeF_4}$ is $sp^3d^2$ (4 + 2).

2. Writing the O$_2$ energy order for N$_2$ (or the N$_2$ order for O$_2$). Up to $\mathrm{N_2}$: $\pi 2p_x = \pi 2p_y$ below $\sigma 2p_z$. From $\mathrm{O_2}$ on: $\sigma 2p_z$ below the $\pi 2p$ pair. Bond orders do not change ($\mathrm{N_2}$ 3, $\mathrm{O_2}$ 2 either way), but the magnetism of $\mathrm{B_2}$ (paramagnetic) and $\mathrm{C_2}$ (diamagnetic) does, and so do the configurations you are asked to write.

3. Calling O$_2$ diamagnetic because "all its electrons are paired in the Lewis structure". The MO configuration ends $\pi^* 2p_x^1\, \pi^* 2p_y^1$: two unpaired electrons, paramagnetic, bond order 2. $\mathrm{B_2}$ is also paramagnetic; $\mathrm{C_2}$, $\mathrm{N_2}$, $\mathrm{F_2}$, $\mathrm{CO}$, $\mathrm{NO^+}$, $\mathrm{O_2^{2-}}$ are diamagnetic; $\mathrm{NO}$, $\mathrm{O_2^+}$, $\mathrm{O_2^-}$ have one unpaired electron.

4. Ranking bond length by charge instead of bond order. Bond order decides: $\mathrm{O_2^+}$ (2.5) $< \mathrm{O_2}$ (2) $< \mathrm{O_2^-}$ (1.5) $< \mathrm{O_2^{2-}}$ (1) in length; the reverse in strength. Removing an electron from $\mathrm{O_2}$ strengthens the bond (antibonding electron leaves); removing one from $\mathrm{N_2}$ weakens it (bonding electron leaves).

5. Giving a polar molecule a zero dipole moment, or a symmetric one a non-zero value. $\mathrm{CO_2}$, $\mathrm{BF_3}$, $\mathrm{CH_4}$, $\mathrm{CCl_4}$, $\mathrm{BeCl_2}$, $\mathrm{PCl_5}$, $\mathrm{SF_6}$, $\mathrm{XeF_2}$, $\mathrm{XeF_4}$: zero (symmetric, dipoles cancel). $\mathrm{H_2O}$ 1.85, $\mathrm{NH_3}$ 1.47, $\mathrm{CHCl_3}$ 1.04, $\mathrm{SO_2}$, $\mathrm{O_3}$, $\mathrm{NF_3}$ 0.23: non-zero. And $\mathrm{NH_3} > \mathrm{NF_3}$ because the lone pair adds in one and opposes in the other.

6. Forgetting the lone-pair angle squeeze. $\mathrm{CH_4}$ 109.5°, $\mathrm{NH_3}$ 107°, $\mathrm{H_2O}$ 104.5° — all $sp^3$, angles shrink because lone pair-bond pair repulsion beats bond pair-bond pair. Do not write "109.5°" for water. And in a trigonal bipyramid the lone pairs go equatorial, so $\mathrm{ClF_3}$ is T-shaped and $\mathrm{XeF_2}$ linear, not bent.

7. Formal charge that does not add up. $\text{FC} = V - L - \tfrac{1}{2}S$. Ozone: central O $+1$, single-bonded O $-1$, double-bonded O $0$; sum 0. In $\mathrm{NH_4^+}$ nitrogen is $5 - 0 - 4 = +1$; in $\mathrm{CO_3^{2-}}$ each single-bonded O is $-1$ and the total is $-2$. If your sum is not the charge on the ion, a count is wrong.

8. Saying the electron transfer is what makes NaCl form. Making the ions costs $495.8 - 348.7 = +147.1$ kJ/mol — endothermic. It is the lattice enthalpy (788 kJ/mol) that makes the overall process exothermic. And lattice enthalpy is the energy to separate one mole of solid into gaseous ions, a positive number.

9. Mixing up bond order in resonance hybrids. Divide total bonds by positions: $\mathrm{O_3}$ 1.5, $\mathrm{NO_2^-}$ 1.5, $\mathrm{SO_2}$ 1.5, benzene 1.5, $\mathrm{CO_3^{2-}}$ 1.33, $\mathrm{NO_3^-}$ 1.33. All bonds in the hybrid are equal (both O-O in ozone 128 pm). Canonical structures do not exist and the molecule does not oscillate between them.

10. Using "octet" for BF$_3$, PCl$_5$ or NO. Boron has six electrons in $\mathrm{BF_3}$ (incomplete), phosphorus ten in $\mathrm{PCl_5}$ and sulphur twelve in $\mathrm{SF_6}$ (expanded), NO has an odd number (11). Period 2 elements can never expand ($\mathrm{NCl_5}$ does not exist; no $d$ orbitals).

11. Sigma/pi miscounts. Single = 1 $\sigma$; double = 1 $\sigma$ + 1 $\pi$; triple = 1 $\sigma$ + 2 $\pi$. Benzene 12 $\sigma$ and 3 $\pi$; $\mathrm{C_2H_2}$ 3 $\sigma$ and 2 $\pi$; $\mathrm{CO_2}$ 2 $\sigma$ and 2 $\pi$; a $\pi$ bond never exists without a $\sigma$; $\sigma$ is stronger than $\pi$.

12. Hydrogen bonding with the wrong atom, or the wrong isomer. Only H bonded to F, O, N counts — not Cl (too large), not C. HF boils above HCl, water above $\mathrm{H_2S}$, $\mathrm{NH_3}$ above $\mathrm{PH_3}$; but $\mathrm{CH_4}$ is the lowest in its group. o-nitrophenol (intramolecular) is more volatile and less soluble than p-nitrophenol (intermolecular). Ice floats because its hydrogen-bonded cage is open.

Key Point: Two more that cost single marks: writing the debye conversion as $3.336 \times 10^{-30}$ C m (not J or C/m), and giving the covalent radius of Cl as 198 pm instead of 99 pm (half the Cl-Cl distance; the van der Waals radius is 180 pm).

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