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Chemistry, Class 11 and 12 formula book

Every chapter on one page, to read straight through or print. 10 chapters, 171 formulas with every symbol defined, 31 memory hooks, and the mistakes that cost the most marks.

1Mole Concept and Stoichiometry

The mole is the chapter everything else in chemistry leans on. Physical chemistry numericals almost always begin by converting what you were given into moles, so the few conversions here are worth more marks than their size suggests.

Counting particles

WhatFormulaSymbols and notes
Avogadro's numberNA = 6.022 × 10²³ mol−1particles in one mole
Moles from massn = given mass / molar massthe first line of most numericals
Moles from particlesn = N / NAN is the number of atoms or molecules
Moles of a gas at STPn = volume / 22.4 Lat 273 K and 1 bar, for an ideal gas
Number of particlesN = n × NAatoms, molecules or ions, so say which

Concentration

WhatFormulaSymbols and notes
MolarityM = moles of solute / volume of solution in litreschanges with temperature, because volume does
Molalitym = moles of solute / mass of solvent in kgindependent of temperature
Mole fractionx = ncomponent / ntotalthe fractions of all components add to 1
NormalityN = gram equivalents / volume in litresN = M × n-factor
DilutionM₁V₁ = M₂V₂moles of solute do not change on adding water
Parts per millionppm = (mass of solute / mass of solution) × 10⁶used for very dilute solutions

Formulas and yield

WhatFormulaSymbols and notes
Empirical formulathe simplest whole number ratio of atomsfound from percentage composition
Molecular formulamolecular formula = n × empirical formulan = molar mass / empirical formula mass
Percentage composition% of element = (mass of element / molar mass) × 100per mole of compound
Percentage yield(actual yield / theoretical yield) × 100theoretical yield comes from the limiting reagent
Limiting reagentthe reactant giving the fewest moles of productdivide moles by the coefficient and take the smallest

How to remember this chapter

“Everything through moles”

How to start any numerical

Grams, litres and particle counts cannot be compared directly. Convert each to moles first, use the balanced equation, then convert back to whatever the question wants.

“Molarity has volume, molality has mass”

Telling the two apart

Molarity is per litre of solution, molality per kilogram of solvent. Only molality is safe when the temperature changes.

“Smallest share wins”

Finding the limiting reagent

Divide each reactant's moles by its coefficient in the balanced equation. The smallest answer is the limiting reagent, and it sets the yield.

The mistakes that cost the most

  • Dividing moles by the wrong volume. Molarity uses the volume of the whole solution, not of the solvent.
  • Using 22.4 litres per mole for a gas that is not at STP, or for a liquid or solid.
  • Forgetting to balance the equation before comparing mole ratios.
  • Assuming the reactant present in the smallest mass is the limiting reagent. It is the smallest mole-to-coefficient ratio that decides.
  • Reporting the empirical formula when the molecular formula was asked for.

Open Mole Concept and Stoichiometry on its own page

2Atomic Structure

Atomic structure is where chemistry borrows from physics. The Bohr results and the quantum numbers carry the marks, and the rules for filling orbitals are asked in every board paper in one form or another.

Waves and photons

WhatFormulaSymbols and notes
Wave relationc = νλc = 3 × 10⁸ m s−1
Photon energyE = hν = hc/λh = 6.626 × 10−34 J s
Photoelectric effecthν = hν₀ + ½mv²ν₀ is the threshold frequency
de Broglie wavelengthλ = h/mvevery moving particle has one
Heisenberg uncertaintyΔx · Δp ≥ h/4πposition and momentum cannot both be exact

The Bohr atom

WhatFormulaSymbols and notes
Energy of the nth levelEn = −13.6Z²/n² eVfor hydrogen-like species only
Radius of the nth orbitrn = 0.529n²/Z Ågrows as n²
Angular momentummvr = nh/2πthis quantisation is Bohr's postulate
Rydberg equation1/λ = R(1/n₁² − 1/n₂²)R = 1.097 × 10⁷ m−1
Number of spectral linesn(n − 1)/2from the nth level down to the ground state

Quantum numbers and filling

WhatFormulaSymbols and notes
Principal, nshell number, 1, 2, 3, …decides size and most of the energy
Azimuthal, l0 to n − 1, shapes s, p, d, fdecides the shape of the orbital
Magnetic, m−l to +lgives 2l + 1 orbitals in a subshell
Spin, s+½ or −½two electrons per orbital, opposite spins
Electrons in a shell2n²2, 8, 18, 32 for the first four shells
Electrons in a subshell2(2l + 1)s 2, p 6, d 10, f 14
Aufbau order by (n + l)lower n + l fills first; on a tie, lower nthis is why 4s fills before 3d
Hund's rulefill singly before pairingall with parallel spins
Pauli exclusionno two electrons share all four quantum numbershence two per orbital

How to remember this chapter

“n plus l, then n”

The order in which orbitals fill

Lower n + l fills first. If two are equal, the one with the smaller n goes first. That single rule puts 4s before 3d without memorising the diagonal chart.

“Bus seats”

Hund's rule

People take empty seats on a bus before sitting next to someone. Electrons fill each orbital of a subshell singly before any pairs up.

“Half and full are happiest”

Chromium and copper

Cr is 3d⁵4s¹ and Cu is 3d¹⁰4s¹, not the expected 3d⁴4s² and 3d⁹4s², because half-filled and fully filled subshells are extra stable.

The mistakes that cost the most

  • Using the Bohr formulas for a multi-electron atom. They apply to hydrogen and hydrogen-like ions only.
  • Writing the ground state of chromium and copper as the Aufbau order predicts.
  • Forgetting that 4s fills before 3d but empties first when the ion forms.
  • Giving l a value equal to n. It runs from 0 to n − 1.
  • Treating the uncertainty principle as a limit of the instrument. It is a property of nature.

Open Atomic Structure on its own page

3States of Matter and Gas Laws

The gas laws are the most formula-dense part of Class 11 chemistry, and the most predictable. Almost every question is one of the named laws or the ideal gas equation, and the deviations at the end supply the theory mark.

The named laws

WhatFormulaSymbols and notes
Boyle's lawp₁V₁ = p₂V₂at constant temperature and amount
Charles's lawV₁/T₁ = V₂/T₂at constant pressure; T in kelvin
Gay-Lussac's lawp₁/T₁ = p₂/T₂at constant volume
Combined gas lawp₁V₁/T₁ = p₂V₂/T₂the three above in one line
Avogadro's lawV ∝ nequal volumes hold equal numbers of molecules

The ideal gas

WhatFormulaSymbols and notes
Ideal gas equationpV = nRTR = 8.314 J mol−1 K−1 = 0.0821 L atm mol−1 K−1
In terms of densitypM = ρRTa quick way to find molar mass from density
Dalton's law of partial pressuresptotal = p₁ + p₂ + …for non-reacting gases
Partial pressure from mole fractionpi = xi × ptotalthe most used form of Dalton's law
Graham's law of diffusionr₁/r₂ = √(M₂/M₁)lighter gases diffuse faster

Real gases and speeds

WhatFormulaSymbols and notes
van der Waals equation(p + an²/V²)(V − nb) = nRTa corrects for attraction, b for molecular volume
Compressibility factorZ = pV/nRTZ = 1 for an ideal gas
Root mean square speedurms = √(3RT/M)M in kg per mole
Average speeduav = √(8RT/πM)slightly less than urms
Most probable speedump = √(2RT/M)the smallest of the three
Critical temperatureabove Tc a gas cannot be liquefiedhowever great the pressure

How to remember this chapter

“Kelvin or nothing”

Every gas law

Every gas calculation needs absolute temperature. Using °C in Charles's law is the single most common error in the chapter.

“Light gases are quick”

Graham's law

Rate goes as one over the square root of molar mass. Hydrogen diffuses four times faster than oxygen, because 32 over 2 is 16 and its square root is 4.

“a for attraction, b for bulk”

The van der Waals constants

a corrects the pressure for intermolecular attraction; b corrects the volume for the space the molecules themselves take up.

The mistakes that cost the most

  • Using °C anywhere in a gas law. Kelvin only.
  • Mixing units of R. Use 0.0821 with litres and atmospheres, 8.314 with cubic metres and pascals.
  • Using grams per mole for M in the speed formulas, which need kilograms per mole.
  • Applying the ideal gas equation at high pressure and low temperature, where real gases deviate most.
  • Forgetting that Dalton's law needs the gases not to react with each other.

Open States of Matter and Gas Laws on its own page

4Chemical Thermodynamics

Thermodynamics in chemistry answers one question: will this reaction go by itself? Enthalpy, entropy and Gibbs energy together give the answer, and the sign conventions are the whole difficulty.

First law and enthalpy

WhatFormulaSymbols and notes
First lawΔU = q + wq heat added to the system, w work done on the system
Work of expansionw = −pextΔVnegative when the gas expands
EnthalpyH = U + pV, ΔH = ΔU + ΔngRTΔng is the change in moles of gas
At constant volumeqv = ΔUno expansion work is done
At constant pressureqp = ΔHthe usual laboratory condition
Heat capacity relationCp − Cv = Rper mole, for an ideal gas

Thermochemistry

WhatFormulaSymbols and notes
Hess's lawΔH is the same by any routeso thermochemical equations can be added
From formation enthalpiesΔH = ΣΔHf(products) − ΣΔHf(reactants)elements in their standard state have ΔHf = 0
From bond enthalpiesΔH = Σ(bonds broken) − Σ(bonds formed)breaking costs energy, forming releases it
ExothermicΔH negativeheat is released, the beaker warms
EndothermicΔH positiveheat is absorbed, the beaker cools

Entropy and Gibbs energy

WhatFormulaSymbols and notes
Entropy changeΔS = qrev/Tunit J K−1 mol−1
Gibbs energyΔG = ΔH − TΔSthe test for spontaneity
SpontaneousΔG < 0at the given temperature
At equilibriumΔG = 0the forward and reverse tendencies balance
Gibbs energy and KΔG° = −2.303 RT log Klinks thermodynamics to equilibrium
Second lawthe entropy of the universe always increasesfor any spontaneous change

How to remember this chapter

“Negative means it goes”

Reading ΔG, ΔH and ΔS

Negative ΔG means spontaneous. Negative ΔH means heat released. Positive ΔS means more disorder. Only ΔG decides whether a reaction happens.

“Both negative, cold helps; both positive, heat helps”

When temperature decides

If ΔH and ΔS are both negative, low temperature makes ΔG negative. If both are positive, high temperature does. If signs differ, temperature cannot change the answer.

“Break costs, make pays”

Bond enthalpy calculations

Breaking bonds always absorbs energy and forming them always releases it. Bonds broken minus bonds formed gives ΔH, in that order.

The mistakes that cost the most

  • Mixing sign conventions for work. In chemistry w is the work done on the system, which is the opposite of the physics convention.
  • Using ΔH = ΔU without the ΔngRT term when the number of moles of gas changes.
  • Deciding spontaneity from ΔH alone. Endothermic reactions can be spontaneous if ΔS is large enough.
  • Using °C in ΔG = ΔH − TΔS. It needs kelvin.
  • Forgetting that ΔH is usually in kilojoules and ΔS in joules per kelvin, so one must be converted.

Open Chemical Thermodynamics on its own page

5Chemical and Ionic Equilibrium

Equilibrium is two chapters treated as one: the equilibrium constant, and what happens in water. Between them they supply pH, buffers and solubility, which are among the most reliably asked numericals in the whole subject.

The equilibrium constant

WhatFormulaSymbols and notes
In concentrationsKc = [products]/[reactants], each raised to its coefficientpure solids and liquids are left out
In partial pressuresKp = Kc(RT)ΔnΔn is moles of gaseous products minus reactants
Reaction quotientQ compared with KQ < K goes forward, Q > K goes backward, Q = K is equilibrium
Reversed reactionK' = 1/Kand for a reaction multiplied by n, K becomes Kn
Le Chatelier's principlea system at equilibrium opposes any change imposed on itpressure, concentration or temperature

Acids, bases and pH

WhatFormulaSymbols and notes
Ionic product of waterKw = [H⁺][OH⁻] = 10−14at 25 °C
pHpH = −log[H⁺]7 is neutral at 25 °C
pOH and the relationpOH = −log[OH⁻], pH + pOH = 14at 25 °C
Weak acid[H⁺] = √(KaC)C is the initial concentration
Degree of dissociationα = √(Ka/C)Ostwald's dilution law: dilution increases α
pK valuespKa = −log Kaa smaller pKa means a stronger acid

Buffers and solubility

WhatFormulaSymbols and notes
Acidic bufferpH = pKa + log([salt]/[acid])the Henderson-Hasselbalch equation
Basic bufferpOH = pKb + log([salt]/[base])then pH = 14 − pOH
Solubility product, ABKsp = s²for example AgCl
Solubility product, AB₂Ksp = 4s³for example CaF₂
Solubility product, A₂B₃Ksp = 108s⁵for example Ca₃(PO₄)₂ type salts
Precipitationprecipitates when the ionic product exceeds Kspotherwise the salt stays dissolved
Common ion effectadding a common ion suppresses dissociationwhich is how buffers resist pH change

How to remember this chapter

“Solids and liquids sit out”

Writing an equilibrium constant

Pure solids and pure liquids never appear in K. Their concentration does not change, so it is absorbed into the constant.

“Push it, it pushes back”

Le Chatelier's principle

Add a reactant and the reaction moves forward; raise the pressure and it shifts to the side with fewer gas molecules; heat an exothermic reaction and it goes backwards.

“Buffer is pKa plus the log of the ratio”

Henderson-Hasselbalch

When salt and acid are equal the log is zero, so pH equals pKa. That is the best buffering point, and it is worth remembering on its own.

“s squared, 4s cubed, 108s to the five”

Solubility product expressions

AB gives s², AB₂ gives 4s³, A₂B₃ gives 108s⁵. The coefficients come from the stoichiometry, not from memory alone.

The mistakes that cost the most

  • Including pure solids or liquids in the equilibrium constant expression.
  • Forgetting that Kp equals Kc only when Δn is zero.
  • Using pH + pOH = 14 at a temperature other than 25 °C, where Kw is different.
  • Treating a strong acid with the weak acid formula. A strong acid dissociates completely, so [H⁺] equals its concentration.
  • Writing Ksp = s² for every salt. The exponents follow the formula of the salt.

Open Chemical and Ionic Equilibrium on its own page

6Redox Reactions

Redox looks like bookkeeping and it is. Oxidation numbers, then balancing, then the language of oxidising and reducing agents. Every question in this chapter can be answered by assigning oxidation numbers correctly first.

Oxidation numbers

WhatFormulaSymbols and notes
Free element0for example H₂, O₂, Na, S₈
Hydrogen+1, except −1 in metal hydridesNaH has hydrogen at −1
Oxygen−2, except −1 in peroxides and +2 in OF₂H₂O₂ has oxygen at −1
Group 1 and 2 metals+1 and +2always, in compounds
Fluorine−1 alwaysthe most electronegative element
Sum ruleneutral compound sums to 0, an ion sums to its chargethis is how unknown states are found

The language

WhatFormulaSymbols and notes
Oxidationloss of electrons, oxidation number risesthe substance oxidised is the reducing agent
Reductiongain of electrons, oxidation number fallsthe substance reduced is the oxidising agent
Oxidising agentitself gets reducedfor example KMnO₄, K₂Cr₂O₇
Reducing agentitself gets oxidisedfor example H₂, CO, metals
Disproportionationone element both oxidised and reducedfor example 2H₂O₂ → 2H₂O + O₂

Balancing and equivalents

WhatFormulaSymbols and notes
Ion electron methodbalance atoms, then O with H₂O, then H with H⁺, then charge with electronsin basic medium add OH⁻ at the end
Oxidation number methodequalise the total rise and fall in oxidation numberthen balance the rest by inspection
n-factor of an acidnumber of replaceable H⁺H₂SO₄ has n-factor 2
n-factor in redoxelectrons gained or lost per formula unitKMnO₄ in acid has n-factor 5
Equivalent weightmolar mass / n-factorused in titration calculations
TitrationN₁V₁ = N₂V₂normality times volume, on both sides

How to remember this chapter

“OIL RIG”

Oxidation and reduction

Oxidation Is Loss of electrons, Reduction Is Gain. Say it once at the start of every redox question and the agents follow.

“The agent suffers the opposite”

Naming the agents

The oxidising agent is itself reduced; the reducing agent is itself oxidised. Whatever the agent does to the other, it has the reverse done to itself.

“KMnO₄ is 5 in acid, 3 in neutral”

n-factor of permanganate

In acidic medium manganese goes from +7 to +2, so five electrons. In neutral or faintly alkaline medium it goes to +4, so three.

The mistakes that cost the most

  • Giving oxygen −2 in peroxides. In H₂O₂ and other peroxides it is −1.
  • Naming the agents the wrong way round. The oxidising agent is the one reduced.
  • Balancing charge before balancing atoms. Atoms first, then oxygen, then hydrogen, then charge.
  • Using n-factor 1 for KMnO₄ without checking the medium.
  • Forgetting that the oxidation number of an element in its free state is zero, whatever the molecule.

Open Redox Reactions on its own page

7Solutions and Colligative Properties

Colligative properties depend on how many particles are dissolved, not on what they are. That one sentence explains every formula here, and the van't Hoff factor is what turns it from theory into a numerical.

Expressing concentration

WhatFormulaSymbols and notes
MolarityM = moles of solute / litres of solutiontemperature dependent
Molalitym = moles of solute / kg of solventtemperature independent, so used in these formulas
Mole fractionx = ni/ntotaladds to 1 over all components
Mass percentage(mass of solute / mass of solution) × 100the everyday way of stating strength

Vapour pressure

WhatFormulaSymbols and notes
Raoult's lawp = xsolvent × p°p° is the vapour pressure of the pure solvent
Relative lowering(p° − p)/p° = xsoluteequal to the mole fraction of the solute
Ideal solutionΔHmix = 0, ΔVmix = 0obeys Raoult's law at all compositions
Positive deviationweaker new interactions, higher vapour pressurefor example ethanol and acetone
Negative deviationstronger new interactions, lower vapour pressurefor example chloroform and acetone
Henry's lawp = KHxsolubility of a gas rises with pressure

The four colligative properties

WhatFormulaSymbols and notes
Boiling point elevationΔTb = iKbmKb for water is 0.52 K kg mol−1
Freezing point depressionΔTf = iKfmKf for water is 1.86 K kg mol−1
Osmotic pressureπ = iCRTC is molarity; the most sensitive of the four
Relative lowering of vapour pressure(p° − p)/p° = ixsolutethe fourth colligative property
van't Hoff factori = observed value / calculated value2 for NaCl, 3 for CaCl₂, less than 1 for association
Molar mass from a colligative propertyM = iwK/(ΔT × W)w solute mass, W solvent mass in kg

How to remember this chapter

“Count particles, not kind”

What colligative means

One mole of sugar and one mole of NaCl do not behave alike, because NaCl gives two particles. It is the number of particles in solution that matters, never their identity.

“i is how many it breaks into”

The van't Hoff factor

NaCl gives 2, CaCl₂ gives 3, glucose gives 1. If the solute associates instead, as benzoic acid does in benzene, i falls below 1.

“Salt on ice, sugar in the pot”

Which way each effect goes

Adding solute lowers the freezing point and raises the boiling point. Salt melts ice on a road; sugar syrup boils above 100 °C.

The mistakes that cost the most

  • Using molarity in the boiling point and freezing point formulas. Both need molality.
  • Leaving out the van't Hoff factor for an ionic solute, which halves or thirds the right answer.
  • Getting the direction wrong: boiling point goes up, freezing point goes down.
  • Applying Raoult's law to a non-volatile solute's own vapour pressure. It has none.
  • Using the molar mass formula with the solvent mass in grams. It must be in kilograms.

Open Solutions and Colligative Properties on its own page

8Electrochemistry

Electrochemistry joins redox to energy. The Nernst equation and the link between cell potential and Gibbs energy carry the numericals, and conductance supplies the rest. Sign conventions decide most of the marks.

Cells and potentials

WhatFormulaSymbols and notes
Cell potentialE°cell = E°cathode − E°anodeboth as reduction potentials
Galvanic cellspontaneous, E°cell positiveanode is negative, cathode is positive
Electrolytic cellnon-spontaneous, needs an external supplyanode is positive, cathode is negative
Nernst equationE = E° − (0.0591/n)log([products]/[reactants])at 298 K
At equilibriumE = 0 and E° = (0.0591/n)log Kwhich links potential to the equilibrium constant
Gibbs energyΔG° = −nFE°cellF = 96500 C mol−1

Conductance

WhatFormulaSymbols and notes
ConductanceG = 1/Runit siemens
Conductivityκ = G × (l/A)l/A is the cell constant, unit m−1
Molar conductivityΛm = κ × 1000/CC in mol per litre
On dilutionΛm rises, κ fallsmore ions are free but they are further apart
Kohlrausch's lawΛ°m = λ°+ + λ°−limiting molar conductivity adds from the ions
Degree of dissociationα = Λm/Λ°mfor a weak electrolyte

Electrolysis

WhatFormulaSymbols and notes
Faraday's first lawm = ZItZ is the electrochemical equivalent
In terms of equivalentsm = (E × I × t)/96500E is the equivalent weight
Faraday's second lawm₁/m₂ = E₁/E₂for the same charge passed
One faraday96500 C deposits one gram equivalentone mole of electrons

How to remember this chapter

“Red Cat, An Ox”

What happens at each electrode

REDuction at the CAThode, ANode is OXidation. True in both galvanic and electrolytic cells, even though the signs of the electrodes swap.

“Cathode minus anode”

Calculating cell potential

E°cell = E°cathode − E°anode, with both written as reduction potentials. Never reverse a sign and subtract as well; do one or the other.

“0.0591 over n”

The Nernst equation at room temperature

At 298 K the factor 2.303RT/F is 0.0591 volts. Memorise the number and the equation becomes one line.

The mistakes that cost the most

  • Reversing the sign of the anode potential and then also subtracting it. Do one, not both.
  • Assuming the electrode signs are the same in galvanic and electrolytic cells. The chemistry is the same; the signs are opposite.
  • Using concentration instead of the correct ratio in the Nernst equation; products over reactants, each to its power.
  • Thinking conductivity rises on dilution. Molar conductivity rises, conductivity falls.
  • Forgetting that n in ΔG° = −nFE° is the number of electrons in the balanced cell reaction.

Open Electrochemistry on its own page

9Chemical Kinetics

Kinetics asks how fast, not how far. Order, rate constant and half-life are the whole chapter, and the first order results are asked so often that they are worth memorising as a block.

Rate and order

WhatFormulaSymbols and notes
Rate of reactionrate = −Δ[R]/Δt = +Δ[P]/Δtdivided by the stoichiometric coefficient
Rate lawrate = k[A]x[B]yx and y come from experiment, not from the equation
Orderorder = x + ycan be zero, fractional or negative
Molecularitythe number of species colliding in one stepalways a whole number, never zero
Units of k, zero ordermol L−1 s−1the rate itself
Units of k, first orders−1independent of concentration units
Units of k, second orderL mol−1 s−1general form: (mol L−1)1−n s−1

Integrated rate laws

WhatFormulaSymbols and notes
Zero order[R] = [R]₀ − kta straight line of concentration against time
Zero order half-lifet1/2 = [R]₀/2kproportional to the starting concentration
First orderk = (2.303/t)log([R]₀/[R])the most used equation in the chapter
First order half-lifet1/2 = 0.693/kindependent of the starting concentration
First order graphlog[R] against t is a straight lineslope = −k/2.303
After n half-lives[R] = [R]₀/2nthe quick route for whole numbers of half-lives

Temperature and catalysts

WhatFormulaSymbols and notes
Arrhenius equationk = Ae−Ea/RTA is the frequency factor
Logarithmic formlog k = log A − Ea/2.303RTa straight line of log k against 1/T
Two temperatureslog(k₂/k₁) = (Ea/2.303R)(1/T₁ − 1/T₂)the standard numerical
Rule of thumbrate roughly doubles for every 10 K risenear room temperature
Catalystlowers Ea, speeds both directions equallyit does not change ΔH or the position of equilibrium

How to remember this chapter

“Order from experiment, molecularity from the step”

Telling them apart

Order is measured and can be zero or fractional. Molecularity is counted from a single step and is always a whole number.

“First order forgets where it started”

Half-life

For first order, t1/2 = 0.693/k, whatever the initial concentration. For zero order the half-life depends on it.

“A catalyst lowers the hill”

What a catalyst does

It provides a path with lower activation energy. Both forward and reverse rates rise by the same factor, so the equilibrium position does not move.

The mistakes that cost the most

  • Reading the order off the balanced equation. Order comes from experiment.
  • Using the first order half-life formula for a zero order reaction.
  • Forgetting the 2.303 when moving between natural and base ten logarithms.
  • Thinking a catalyst shifts the equilibrium. It only makes both directions faster.
  • Using °C in the Arrhenius equation, which needs kelvin.

Open Chemical Kinetics on its own page

10The Solid State

Solid state is geometry with chemistry attached. Unit cells, packing and density make up nearly all of it, and every numerical uses the same density formula, so it is one of the fastest chapters to secure.

Unit cells

WhatFormulaSymbols and notes
Simple cubicatoms per cell Z = 1corners only, 8 × ⅛
Body centred cubicZ = 2corners plus one at the centre
Face centred cubicZ = 4corners plus six face centres, each shared by two
Edge and radius, simple cubica = 2ratoms touch along the edge
Body centred cubica√3 = 4ratoms touch along the body diagonal
Face centred cubica√2 = 4ratoms touch along the face diagonal

Packing and density

WhatFormulaSymbols and notes
Packing efficiency, simple cubic52.4%the most wasteful arrangement
Packing efficiency, bcc68%better, and common in alkali metals
Packing efficiency, fcc and hcp74%the closest packing possible for equal spheres
Coordination numbersimple cubic 6, bcc 8, fcc and hcp 12nearest neighbours touching an atom
Density of a unit cellρ = ZM/(a³NA)a in metres gives density in kg m−3
Voids in close packing2 tetrahedral and 1 octahedral per atomfor N atoms, 2N and N respectively
Radius ratior+/r− decides the geometry0.225 to 0.414 tetrahedral, 0.414 to 0.732 octahedral

Defects

WhatFormulaSymbols and notes
Schottky defecta cation and an anion both missingdensity falls; seen in NaCl, KCl
Frenkel defectan ion moves to an interstitial sitedensity unchanged; seen in AgCl, ZnS
Metal excess defectgives colour, as F-centresfor example NaCl heated in sodium vapour turns yellow
n-type semiconductordoped with a group 15 elementextra electrons carry the current
p-type semiconductordoped with a group 13 elementholes carry the current

How to remember this chapter

“1, 2, 4”

Atoms per unit cell

Simple cubic 1, body centred 2, face centred 4. Almost every solid state numerical starts by picking the right Z.

“Edge, face, body: 2, √2, √3”

Relating a and r

Simple cubic touches along the edge, fcc along the face diagonal, bcc along the body diagonal. a = 2r, a√2 = 4r, a√3 = 4r.

“Schottky slims, Frenkel shuffles”

The two defects

Schottky removes a pair of ions, so density falls. Frenkel just moves an ion inside, so density is unchanged.

The mistakes that cost the most

  • Using Z = 1 for a face centred cell. Corners give 1, but the six faces add 3 more.
  • Forgetting to convert the edge length from picometres to metres before finding density.
  • Mixing up which defect changes the density. Only Schottky does.
  • Assuming fcc and hcp differ in packing efficiency. Both are 74%.
  • Using the radius ratio rule without checking that the cation is the smaller ion.

Open The Solid State on its own page

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