Exam 2 Study Guide
This guide covers chapters 4 through 6. You will balance reactions and predict how much product forms. You will track where a reaction's energy goes. Then you will look at where electrons live, and why that builds the periodic table. Read each chapter's "what to know" list first. Copy the equations out by hand. Try the worked examples with the answers covered, then run the ten self-test items. The cram sheet at the end gathers every equation and constant.
What this guide covers
- Chapter 4 · balancing equations, reaction types, oxidation numbers, mole ratios, limiting reactant and percent yield.
- Chapter 5 · heat and work, calorimetry, enthalpy, Hess's law, formation enthalpies and bond energies.
- Chapter 6 · light and photons, the Bohr model, quantum numbers, electron configurations, orbital diagrams and periodic trends.
Chapter 4Stoichiometry of Chemical Reactions
What to know
A balanced equation has the same count of each element on both sides. Atoms are never lost in a reaction, so the two sides must match. You balance by changing coefficients, the big numbers in front. Never change a subscript. Changing a subscript turns the substance into something else.
Three reaction types show up again and again. In a precipitation reaction two solutions mix and a solid drops out. In an acid-base reaction a proton moves, and the net result is water. In a redox reaction electrons move from one species to another.
A net ionic equation strips out the spectator ions. Those are the ions that start and end as free ions in solution. What is left is the real change. For lead nitrate and potassium iodide, the net change is Pb2+(aq) + 2 I−(aq) → PbI2(s).
Oxidation numbers are a bookkeeping charge. Assign them in order. A free element is 0. A monatomic ion takes its own charge. Fluorine is always −1. Oxygen is usually −2, but it is −1 in a peroxide. Hydrogen is +1 with nonmetals and −1 with metals. Then the rest must make the total equal the species charge.
A rise in oxidation number is oxidation. A drop is reduction. The two always happen together. The substance that gets oxidized is the reducing agent. The one that gets reduced is the oxidizing agent.
Coefficients are mole ratios. They convert moles of one substance into moles of another. Every mass problem walks the same road. Grams of the known go to moles. Moles cross over with the mole ratio. Moles of the wanted go back to grams.
The limiting reactant is the one that runs out first. Do not guess from the masses. Convert each reactant to moles, divide each by its coefficient, and the smallest result limits the reaction. That reactant sets the theoretical yield.
Theoretical yield is the most that could form. Actual yield is what you really collect. Percent yield is actual divided by theoretical, times 100. It is almost never 100, because product sticks to glassware and side reactions steal material.
- Balance with coefficients only. Subscripts are part of the substance.
- A net ionic equation shows only the ions that actually change.
- Oxidation is a rise in oxidation number. Reduction is a drop.
- Divide moles by the coefficient before comparing reactants.
- The limiting reactant sets the theoretical yield, every time.
- A percent yield above 100 means an error, or a product that is still wet.
Equations, units and symbols
| Equation | What each symbol means | Units |
|---|---|---|
| n = m ÷ M | n is moles, m is mass, M is molar mass | mol = g ÷ (g/mol) |
| nB = nA × (b ÷ a) | a and b are the coefficients of A and B in the balanced equation | mol; the ratio itself has no unit |
| mB = (mA ÷ MA) × (b ÷ a) × MB | the whole road map in one line | g in, g out |
| limiting test: n ÷ coefficient | n is each reactant's moles; the smallest value wins | mol, so the results compare fairly |
| percent yield = (actual ÷ theoretical) × 100% | actual is what you collect, theoretical is the maximum | both in the same unit; the answer is a percent |
| actual = percent yield × theoretical ÷ 100 | same equation, solved for the real haul | g or mol |
| sum of oxidation numbers = species charge | 0 for a neutral compound, or the ion's charge | whole numbers, sometimes fractional in averages |
Worked examples
Worked example 1 · balance a combustion, then run the mass
Balance C3H8 + O2 → CO2 + H2O. Then find the CO2 from 22.0 g of C3H8.
Step 1 · carbon and hydrogen firstThree C go to 3 CO2. Eight H go to 4 H2O.
Step 2 · oxygen lastThe right side now holds 6 + 4 = 10 O atoms. So the left needs 5 O2. The balanced equation is C3H8 + 5 O2 → 3 CO2 + 4 H2O.
Step 3 · grams to molesC3H8 is 3(12.011) + 8(1.008) = 44.097 g/mol. So 22.0 ÷ 44.097 = 0.4989 mol.
Step 4 · cross with the ratioThe ratio is 3 CO2 per 1 C3H8. So 0.4989 × 3 = 1.4967 mol of CO2.
Step 5 · moles to gramsCO2 is 44.01 g/mol. So 1.4967 × 44.01 = 65.87 g, which rounds to 65.9 g.
Check it a second wayCount atoms on both sides. Left: 3 C, 8 H, 10 O. Right: 3 C, 8 H, 10 O. They match.
Worked example 2 · limiting reactant and percent yield
N2(g) + 3 H2(g) → 2 NH3(g). You start with 28.0 g of N2 and 5.00 g of H2. You collect 24.5 g of NH3. Find the limiting reactant, the theoretical yield and the percent yield.
Step 1 · both reactants to molesN2 is 28.014 g/mol, so 28.0 ÷ 28.014 = 0.9995 mol. H2 is 2.016 g/mol, so 5.00 ÷ 2.016 = 2.4802 mol.
Step 2 · divide by the coefficientsN2: 0.9995 ÷ 1 = 0.9995. H2: 2.4802 ÷ 3 = 0.8267. The smaller value belongs to H2, so hydrogen limits.
Step 3 · theoretical yield in molesThe ratio is 2 NH3 per 3 H2. So 2.4802 × (2 ÷ 3) = 1.6535 mol of NH3.
Step 4 · moles to gramsNH3 is 17.03 g/mol. So 1.6535 × 17.03 = 28.16 g, or 28.2 g.
Step 5 · percent yield(24.5 ÷ 28.2) × 100 = 86.9%.
Check it a second wayNitrogen would need 3 × 0.9995 = 2.999 mol of H2 to react fully. Only 2.4802 mol is there, so hydrogen really does run out first.
Worked example 3 · oxidation numbers and who did what
Find the oxidation number of chromium in Cr2O72−. Then name the oxidized and reduced species in Zn(s) + 2 HCl(aq) → ZnCl2(aq) + H2(g).
Step 1 · assign the sure oneOxygen is −2 here, and there are seven of them. That is 7 × (−2) = −14.
Step 2 · use the totalThe ion's charge is 2−. So the two chromiums must add to −2 − (−14) = +12.
Step 3 · split it+12 ÷ 2 = +6 for each Cr.
Step 4 · track the zinc reactionZinc starts at 0 as a free element and ends at +2 in ZnCl2. That is a rise, so zinc is oxidized and acts as the reducing agent.
Step 5 · track the hydrogenHydrogen starts at +1 in HCl and ends at 0 in H2. That is a drop, so hydrogen is reduced, and HCl is the oxidizing agent. Chlorine stays at −1 throughout, so it is a spectator.
- Balancing by changing subscripts. H2O2 is not water. Only coefficients may move.
- Treating masses as mole ratios. Coefficients count particles, not grams.
- Calling the smaller mass the limiting reactant. A light reactant can still supply more moles.
- Skipping the divide-by-coefficient step. Comparing raw moles gives the wrong limiter whenever coefficients differ.
- Using the excess reactant for the yield. Only the limiter sets the theoretical yield.
- Confusing oxidation number with ionic charge. In SO42− sulfur is +6, even though the ion is 2−.
- Forgetting oxygen's exceptions. Oxygen is −1 in a peroxide and 0 as O2.
Chapter 4 self-test · ten items
Chapter 5Thermochemistry
What to know
Draw a box around the part you care about. That is the system. Everything outside is the surroundings. Every sign in this chapter is written from the system's point of view. Energy going in is positive. Energy leaving is negative.
The first law says energy is conserved. For the system, ΔE = q + w. Here q is heat and w is work. Heat is energy that moves because of a temperature difference. Work here is usually a gas pushing on its surroundings.
Specific heat tells you how stubborn a substance is about changing temperature. Water's value is 4.184 J/g·°C, which is unusually high. That is why water makes such a steady heat sink. The working equation is q = mcΔT, and ΔT is always final minus initial.
In a calorimeter, heat lost by the hot object equals heat gained by the cold one. So qhot = −qcold. The minus sign is not optional. One object cools while the other warms, so the two q values must carry opposite signs.
Enthalpy, H, is built for constant pressure. At constant pressure the enthalpy change is the heat: ΔH = qp. A negative ΔH means heat flows out, which is exothermic. A positive ΔH means heat flows in, which is endothermic.
Thermochemical equations scale. Double every coefficient and ΔH doubles. Reverse the equation and ΔH flips sign. Halve it and ΔH halves.
Hess's law says enthalpy does not care about the path. You get only three moves: flip a step, scale a step, and add steps up. Stack known reactions until they sum to your target, then sum their ΔH values the same way.
A standard enthalpy of formation, ΔHf°, is the enthalpy change for making one mole of a substance from its elements in their standard states. An element already in its standard state has ΔHf° = 0. With a table of these, any reaction is products minus reactants, each value times its coefficient.
Bond energies explain where the heat comes from. Breaking a bond costs energy. Forming one pays energy back. So ΔH is roughly the energy of bonds broken minus the energy of bonds formed. Tables list averages, so the result is an estimate.
- Exothermic means negative ΔH. Endothermic means positive ΔH.
- ΔT is final minus initial, so a cooling object gives a negative ΔT.
- Water's specific heat is 4.184 J/g·°C.
- Reverse a reaction and flip the sign of ΔH. Scale it and scale ΔH.
- ΔHf° is zero for any element in its standard state.
- Bond energies give broken minus formed, which is the opposite order from the formation equation.
Equations, units and symbols
| Equation | What each symbol means | Units |
|---|---|---|
| ΔE = q + w | ΔE is internal energy change, q is heat, w is work on the system | all in J or kJ |
| q = mcΔT | m is mass, c is specific heat, ΔT is the temperature change | J = g × (J/g·°C) × °C |
| ΔT = Tfinal − Tinitial | the order matters, and it sets the sign of q | °C, or K for the same size of step |
| c = q ÷ (mΔT) | same symbols, solved for specific heat | J/g·°C |
| qhot = −qcold | the calorimeter balance between two objects | J on both sides |
| ΔH = Hproducts − Hreactants = qp | qp is the heat measured at constant pressure | kJ, or kJ/mol per mole of reaction |
| ΔH°rxn = Σ nΔHf°(products) − Σ nΔHf°(reactants) | n is each coefficient from the balanced equation | kJ; each ΔHf° is in kJ/mol |
| ΔH ≈ ΣD(broken) − ΣD(formed) | D is a bond dissociation energy | kJ; each D is in kJ/mol |
Worked examples
Worked example 1 · heating water with q = mcΔT
You heat 150.0 g of water from 22.0 °C to 78.5 °C. How much heat went in?
Step 1 · find ΔTΔT = 78.5 − 22.0 = 56.5 °C. It is positive, because the water warmed up.
Step 2 · put the numbers inq = 150.0 g × 4.184 J/g·°C × 56.5 °C.
Step 3 · multiply150.0 × 4.184 = 627.6. Then 627.6 × 56.5 = 35,459 J.
Step 4 · round and convertThree significant figures give 3.55 × 104 J, or 35.5 kJ.
Check the unitsGrams cancel with grams, and °C cancels with °C. Joules survive, which is what heat should be.
Worked example 2 · identify a metal by calorimetry
A 55.0 g metal block at 99.5 °C drops into 100.0 g of water at 21.0 °C. The mixture settles at 24.8 °C. Find the metal's specific heat.
Step 1 · heat gained by the waterΔT = 24.8 − 21.0 = 3.8 °C. So q = 100.0 × 4.184 × 3.8 = 1,590 J. The water gained it, so the value is positive.
Step 2 · heat lost by the metalqmetal = −1,590 J. The metal cooled, so its q is negative.
Step 3 · find the metal's ΔTΔT = 24.8 − 99.5 = −74.7 °C. Also negative, since it cooled.
Step 4 · solve for cc = q ÷ (mΔT) = (−1,590) ÷ (55.0 × −74.7) = 1,590 ÷ 4,109 = 0.387 J/g·°C.
Check it a second wayThe two minus signs cancel, so c comes out positive. That is required, because specific heat is never negative. Copper's value is 0.385 J/g·°C, so the block is very likely copper.
Worked example 3 · Hess's law and formation enthalpies
First use Hess's law to find ΔH for C(s) + ½O2(g) → CO(g). Then use formation data for the burning of methane.
Step 1 · list the known stepsStep A: C(s) + O2(g) → CO2(g), ΔH = −393.5 kJ. Step B: CO(g) + ½O2(g) → CO2(g), ΔH = −283.0 kJ.
Step 2 · flip step BReversed, it reads CO2(g) → CO(g) + ½O2(g), with ΔH = +283.0 kJ.
Step 3 · add A and flipped BCO2 cancels. What is left is C(s) + ½O2(g) → CO(g). So ΔH = −393.5 + 283.0 = −110.5 kJ.
Step 4 · now the methane routeCH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l). Use ΔHf° values of −74.6 for CH4, −393.5 for CO2 and −285.8 for H2O(l). Oxygen is an element in its standard state, so it is 0.
Step 5 · products minus reactantsProducts: −393.5 + 2(−285.8) = −965.1 kJ. Reactants: −74.6 + 0 = −74.6 kJ.
Step 6 · subtractΔH° = −965.1 − (−74.6) = −890.5 kJ. It is negative, which fits a fuel that gives off heat.
Worked example 4 · a reaction enthalpy from bond energies
Estimate ΔH for H2(g) + Cl2(g) → 2 HCl(g). Use D values of 436 for H–H, 242 for Cl–Cl and 431 for H–Cl, all in kJ/mol.
Step 1 · count bonds brokenOne H–H and one Cl–Cl break. That costs 436 + 242 = 678 kJ.
Step 2 · count bonds formedTwo H–Cl bonds form. That pays back 2 × 431 = 862 kJ.
Step 3 · subtractΔH ≈ 678 − 862 = −184 kJ.
Check it a second wayΔHf° for HCl(g) is −92.3 kJ/mol, and two moles form. That gives 2(−92.3) − 0 = −184.6 kJ. The bond estimate lands within 0.6 kJ, so both routes agree.
- Writing ΔT as initial minus final. That flips every sign in the problem.
- Dropping the minus sign in qhot = −qcold. Without it, the specific heat comes out negative.
- Using the metal's mass with water's specific heat. Each object brings its own m and its own c.
- Mixing joules and kilojoules. q = mcΔT gives joules. Formation tables give kilojoules.
- Forgetting the coefficients in products minus reactants. Two moles of water means 2 × (−285.8).
- Reversing a Hess step without flipping the sign. Flip the arrow, flip the sign.
- Reading bond energies as formed minus broken. It is broken minus formed, so a reaction that makes stronger bonds comes out negative.
Chapter 5 self-test · ten items
Chapter 6Electronic Structure and Periodic Properties
What to know
Light is a wave. Wavelength, λ, is the distance between crests. Frequency, ν, is how many crests pass each second. They multiply to the speed of light: c = λν. So short waves have high frequency, and long waves have low frequency.
Light also arrives in packets called photons. One photon carries E = hν, which is the same as hc/λ. So blue light carries more energy per photon than red light. Energy comes in fixed sizes, which is what quantized means.
A glowing gas gives only certain colors. That line spectrum is proof that an atom's electron energies are quantized. Bohr explained hydrogen with a ladder of allowed levels, En = −(2.179 × 10−18 J)/n2. An electron falling down a rung emits a photon. One climbing up absorbs one.
Four quantum numbers name an orbital and its electron. The number n gives the shell, starting at 1. The number l gives the subshell shape, and it runs from 0 to n − 1. Here l = 0 is s, 1 is p, 2 is d and 3 is f. The number ml picks the orientation, and it runs from −l to +l. The number ms is the spin, either +½ or −½.
So an s subshell holds one orbital, p holds three, d holds five and f holds seven. Each orbital holds two electrons at most. That gives 2, 6, 10 and 14 electrons.
Electrons fill the cheapest seats first. The order is 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p. Notice 4s fills before 3d. Chromium and copper break the pattern to gain a half-filled or full d set.
Orbital diagrams use boxes and arrows. Pauli's rule caps each orbital at two electrons with opposite spins. Hund's rule spreads electrons across equal-energy orbitals before any of them pair up. An atom with unpaired electrons is paramagnetic, which a magnet can detect.
One idea drives every periodic trend: effective nuclear charge, the net pull an outer electron feels. Going across a period, that pull rises, so atoms shrink. Going down a group, a new shell is added, so atoms grow. Ionization energy and electronegativity follow the opposite map from radius. They rise across and fall down.
- c = λν, so wavelength and frequency move in opposite directions.
- Short wavelength means high energy per photon.
- Emission gives a negative ΔE for the electron. Absorption gives a positive one.
- l runs from 0 to n − 1, and ml runs from −l to +l.
- 4s fills before 3d, even though 3d has the smaller n.
- Radius falls across a period and grows down a group. Ionization energy does the reverse.
Equations, units and symbols
| Equation | What each symbol means | Units |
|---|---|---|
| c = λν | c is the speed of light, λ is wavelength, ν is frequency | m/s = m × s−1 |
| ν = c ÷ λ | same symbols, solved for frequency | s−1, also called Hz |
| E = hν | E is one photon's energy, h is Planck's constant | J = (J·s) × s−1 |
| E = hc ÷ λ | the same photon energy, written from wavelength | J; λ must be in meters |
| En = −(2.179 × 10−18 J) ÷ n2 | En is a hydrogen level's energy, n is the level number | J; the values are negative, approaching 0 as n grows |
| ΔE = Efinal − Einitial | the electron's energy change in a jump | J; negative for emission, positive for absorption |
| λ = hc ÷ |ΔE| | the wavelength of the photon emitted or absorbed | m; multiply by 109 for nanometers |
| orbitals in a subshell = 2l + 1 | l is the subshell number | a plain count; electrons = 2(2l + 1) |
Worked examples
Worked example 1 · from wavelength to frequency to energy
A sodium lamp glows at 589 nm. Find the frequency and the energy of one photon.
Step 1 · convert to meters589 nm × (1 m / 109 nm) = 5.89 × 10−7 m.
Step 2 · find the frequencyν = c ÷ λ = (2.998 × 108 m/s) ÷ (5.89 × 10−7 m) = 5.09 × 1014 s−1.
Step 3 · find the photon energyE = hν = (6.626 × 10−34 J·s) × (5.09 × 1014 s−1) = 3.37 × 10−19 J.
Check it a second wayUse E = hc/λ instead. hc = 1.986 × 10−25 J·m, and dividing by 5.89 × 10−7 m gives 3.37 × 10−19 J. The two routes agree.
Worked example 2 · a hydrogen line from the Bohr levels
An electron in hydrogen falls from n = 3 to n = 2. Find ΔE and the wavelength of the photon.
Step 1 · the two levelsE3 = −(2.179 × 10−18) ÷ 9 = −2.421 × 10−19 J. E2 = −(2.179 × 10−18) ÷ 4 = −5.448 × 10−19 J.
Step 2 · final minus initialΔE = (−5.448 × 10−19) − (−2.421 × 10−19) = −3.026 × 10−19 J. It is negative, so the atom gave energy away.
Step 3 · the photon carries the sizeThe photon's energy is 3.026 × 10−19 J, with no sign.
Step 4 · find the wavelengthλ = hc ÷ E = (1.986 × 10−25) ÷ (3.026 × 10−19) = 6.56 × 10−7 m, which is 656 nm.
Check it a second way656 nm is red light, and the 3 to 2 drop is hydrogen's famous red line. Falling levels always emit, and emitted photons always carry positive energy.
Worked example 3 · configuration, orbital diagram and magnetism
Write the electron configuration of iron, Z = 26. Then count its unpaired electrons.
Step 1 · fill in order1s2 2s2 2p6 3s2 3p6 4s2 3d6. Add the superscripts: 2 + 2 + 6 + 2 + 6 + 2 + 6 = 26, which matches Z.
Step 2 · the short formArgon holds the first 18 electrons, so write [Ar] 4s2 3d6.
Step 3 · draw the 3d boxesThere are five d orbitals. Hund's rule puts one electron in each of the five first, all spinning the same way. The sixth electron must then pair up in the first box.
Step 4 · count the singlesOne box holds a pair. The other four hold one electron each. So iron has 4 unpaired electrons and is paramagnetic.
Check it a second wayIron sticks to a magnet, which fits unpaired electrons. If the sixth electron had gone in alone to a sixth orbital, that would break Pauli's rule, since d has only five orbitals.
- Leaving wavelength in nanometers. Convert to meters before using c = λν or E = hc/λ.
- Thinking long waves carry more energy. Energy rises as wavelength shrinks.
- Reversing ΔE. It is final minus initial, so a fall gives a negative value.
- Dropping the minus sign on En. Bound levels are negative, and level 1 is the most negative.
- Letting l equal n. For n = 3, l can only be 0, 1 or 2.
- Pairing electrons too early. Hund's rule fills every equal orbital singly first.
- Expecting trends with no exceptions. Ionization energy dips at boron and at oxygen in period 2.
Chapter 6 self-test · ten items
One pageCram sheet
Everything from chapters 4 to 6 in one place. Cover the right column and rebuild each row from memory.
Equations
| Name | Equation | Units |
|---|---|---|
| Mass to moles | n = m ÷ M | mol |
| Mole ratio | nB = nA × (b ÷ a) | mol |
| Limiting test | smallest value of n ÷ coefficient | mol |
| Percent yield | (actual ÷ theoretical) × 100% | percent |
| First law | ΔE = q + w | J or kJ |
| Heat and temperature | q = mcΔT, with ΔT = Tf − Ti | J |
| Calorimeter balance | qhot = −qcold | J |
| Enthalpy as heat | ΔH = qp | kJ |
| Formation route | ΔH°rxn = ΣnΔHf°(prod) − ΣnΔHf°(react) | kJ |
| Bond energy route | ΔH ≈ ΣD(broken) − ΣD(formed) | kJ |
| Wave relation | c = λν | m/s |
| Photon energy | E = hν = hc ÷ λ | J |
| Hydrogen levels | En = −(2.179 × 10−18 J) ÷ n2 | J |
| Jump energy | ΔE = Efinal − Einitial | J |
| Orbital count | 2l + 1 orbitals, holding 2(2l + 1) electrons | plain counts |
Constants and values
| Quantity | Value | Note |
|---|---|---|
| Speed of light | c = 2.998 × 108 m/s | in a vacuum |
| Planck's constant | h = 6.626 × 10−34 J·s | hc = 1.986 × 10−25 J·m |
| Hydrogen constant | 2.179 × 10−18 J | the size of the n = 1 level |
| Avogadro's number | NA = 6.022 × 1023 /mol | per-bond energy = (kJ/mol × 1000) ÷ NA |
| Specific heat of water | c = 4.184 J/g·°C | unusually high |
| Formation enthalpies | CH4 −74.6, CO −110.5, CO2 −393.5, H2O(l) −285.8, H2O(g) −241.8, SO2 −296.8, H2S −20.6, NH3 −45.9, HCl(g) −92.3 | all in kJ/mol; elements in standard state are 0 |
| Bond energies | H–H 436, C–C 347, C–H 415, C=C 614, N–H 390, C≡C 839, O–H 464, C–O 350, H–F 569, H–Cl 431, C–Cl 339, Cl–Cl 242, N≡N 945, O=O 498 | all in kJ/mol, and all averages |
| Filling order | 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p 7s 5f 6d 7p | Cr and Cu take 4s1 |
Rules that are not equations
| Topic | The rule |
|---|---|
| Balancing | Change coefficients only. Save oxygen and hydrogen for last. |
| Net ionic equations | Split every soluble strong electrolyte into ions, then cancel the spectators. |
| Oxidation numbers | Free element 0. F always −1. O usually −2, but −1 in peroxides. H is +1 with nonmetals, −1 with metals. The rest balance to the total. |
| Redox labels | Oxidation is a rise and the species is the reducing agent. Reduction is a drop and the species is the oxidizing agent. |
| Signs in thermochemistry | Into the system is positive. Out of the system is negative. |
| Scaling ΔH | Multiply the equation, multiply ΔH. Reverse the equation, flip the sign. |
| Quantum numbers | n = 1, 2, 3 and up. l = 0 to n − 1. ml = −l to +l. ms = +½ or −½. |
| Pauli and Hund | Two electrons per orbital with opposite spins. Fill equal orbitals singly before pairing. |
| Periodic trends | Across a period: radius down, ionization energy up, electronegativity up. Down a group: radius up, ionization energy down, electronegativity down. |
| Ion sizes | Cations are smaller than their atoms. Anions are larger. |