Chemistry · Exam 7

Exam 7 Study Guide

Exam 7 covers chapters 19, 20 and 21. Each one hands you a different piece of the table. Chapter 19 works the middle block, where d orbitals bring color and magnetism. Chapter 20 belongs to carbon, which builds millions of compounds out of one habit. Chapter 21 opens the nucleus itself. The three chapters share less math than the earlier units, so the work here is rules and patterns. Learn the rules, then run the self-tests until the answers come without hunting.

What this guide holds

Chapter 19: d-block configurations, coordination compounds, naming, isomers, crystal field splitting, color and magnetism.
Chapter 20: alkanes and isomers, alkenes and alkynes, aromatics, alcohols and ethers, the carbonyl families, amines, amides and polymers.
Chapter 21: nuclide symbols, binding energy, decay types, half-life and dating, nuclear equations, fission and fusion, radiation units and uses.
Every equation with its units, three worked examples per chapter, and the traps that cost the most points.
A 10-item self-test per chapter, and a one-page cram sheet at the end.

Chapter 19Transition Metals and Coordination Chemistry

What to know

The d-block runs from group 3 through group 12. A transition metal is a metal whose atoms or common ions carry a partly filled d subshell. That one trait gives the whole family its habits. These metals are hard and shiny, melt high, carry current well, show more than one charge, and form brightly colored compounds that a magnet often tugs.

Filling order and writing order are two different things. In a neutral atom the 4s orbital lies a little below 3d, so 4s takes electrons first. We still write 3d before 4s, because 3 comes before 4. Two metals in the first row break the pattern. Chromium is [Ar]3d⁵4s¹ and copper is [Ar]3d¹⁰4s¹, because a half filled or filled d set is worth the shift.

Here is the fact that trips up the most people. When a cation forms, the 4s electrons leave first, every time. So Fe²⁺ is [Ar]3d⁶, not [Ar]3d⁴4s². Once 3d holds electrons it drops below 4s, and the 4s orbital reaches further out anyway. Counting is then easy: d electrons in an ion equal the group number minus the charge.

A coordination compound is a central metal ion with a ring of attached partners. Each partner is a ligand, and every ligand carries at least one lone pair to share. So the ligand is a Lewis base and the metal ion is a Lewis acid. The bond that forms is a coordinate covalent bond, and both of its electrons came from the ligand. The atom holding that lone pair is the donor atom. The metal plus its ligands make the coordination sphere, written inside square brackets. Ions outside the brackets are counter ions.

Denticity and coordination number

Denticity is how many donor atoms one ligand uses. The word comes from teeth.
Monodentate ligands bite once: water, ammonia, chloride, cyanide, hydroxide.
Bidentate ligands bite twice with one body: ethylenediamine, written en, and the oxalate ion.
EDTA is the everyday champion, biting six times with two nitrogens and four oxygens.
Coordination number counts donor atoms, not ligands. [Co(en)₃]³⁺ has 3 ligands and a coordination number of 6.
Count 2 gives a line, count 4 gives a tetrahedron or a flat square, count 6 gives an octahedron.

Names look long and are built from small pieces in a fixed order. Name the cation, then the anion. Inside the brackets, list the ligands in alphabetical order and then the metal. Show how many of each ligand there is. Add an ending to the metal if the complex is negative. Finish with the metal's oxidation number in Roman numerals. The whole coordination sphere becomes one word with no spaces.

Naming details that get marked

A negative ligand gets an -o ending: chlorido, cyanido, hydroxido, oxalato. Older books write chloro and cyano.
Three neutral ligands have special names: water is aqua, ammonia is ammine with two letters m, and CO is carbonyl.
Use di, tri, tetra, penta and hexa. Switch to bis, tris or tetrakis with parentheses when the ligand name already holds a prefix.
Prefixes never affect alphabetical order. Sort by the ligand name itself.
A negative complex gives the metal an -ate ending, often on a Latin stem: ferrate, cuprate, argentate, aurate, plumbate, stannate.

Complexes offer more kinds of isomer than almost any other family. Structural isomers differ in which atoms are bonded to what. There are four types: ionization, hydrate, coordination and linkage. Stereoisomers have the same bonds and a different arrangement in space. Geometric isomers differ in which positions the ligands take, so neighboring is cis and across is trans. A tetrahedron cannot do this at all, because every corner touches every other corner.

Some complexes come in two forms that are mirror images and cannot be laid on top of each other. Such a complex is chiral, and the pair are enantiomers. They behave alike in almost every test and part ways only in polarized light, turning the plane in opposite directions by the same amount. The quick test is to hunt for a mirror plane. If one exists, the complex is not chiral.

Crystal field theory treats each ligand as a small point of negative charge and tracks how it pushes on the metal's d electrons. Two of the five d orbitals point straight down the axes at the incoming ligands, so they get pushed up hard. That pair is the e_g set. The other three point into the gaps, so they feel less, and they form the t_2g set. The gap between the two sets is the crystal field splitting energy, Δ. The average does not move, so the lower three drop by 0.4Δ and the upper two rise by 0.6Δ.

Now fill those split levels and you may have a choice. Climbing to the upper set costs Δ. Pairing two electrons in one orbital costs the pairing energy. Whichever is cheaper wins. A small Δ gives a high-spin complex with electrons spread out. A large Δ gives a low-spin complex with electrons stacked low. The choice only exists for d⁴ through d⁷. Tetrahedral complexes have a small gap, so they are high spin in practice.

What sets Δ is mostly the ligands, ranked in the spectrochemical series from weak field to strong field. A higher charge on the metal also raises Δ, and so does moving down a group, which is why 4d and 5d metals are nearly always low spin.

A d electron in the lower set can jump to the upper set if a photon hands it exactly Δ. That photon is swallowed, so that part of the light never reaches your eye. You see the complementary color of what was absorbed. This is the step most people get backward. A green solution is green because it is not absorbing green.

An electron spins, and a spinning charge acts like a tiny magnet. Pair two in one orbital and their magnets cancel. A sample with unpaired electrons is paramagnetic and gets pulled into a magnetic field. A sample with every electron paired is diamagnetic and is pushed out a little. The pull can be weighed and turned into a magnetic moment, which gives the unpaired count with no guessing.

A chelate is a complex in which one ligand bites more than once and closes a ring. Chelating ligands bind far more tightly than the same number of single-bite ligands. The bonds themselves are nearly the same, so something else is at work. Replacing six waters with three en molecules frees six waters while using only three ligands, so the number of loose particles goes up. That bonus is the chelate effect. Heme, chlorophyll and vitamin B₁₂ are all chelates of a flat nitrogen ring.

Equations, units and symbols

Equation	What each symbol means	Units
d electrons = group number − charge	Works for any d-block cation. Iron is group 8, so Fe³⁺ has 8 − 3 = 5.	A bare count of electrons.
x + Σ(ligand charges) = charge on the complex ion	x is the metal's oxidation number. Neutral ligands add 0, each chloride adds −1.	Charges in units of e.
Coordination number = total donor atoms	Multiply each ligand by its denticity, then add. Three bidentate ligands give 6.	A bare count of bonds.
t_2g drops 0.4Δ, e_g rises 0.6Δ	The average energy does not move, so three down by 0.4 balances two up by 0.6.	Same energy unit as Δ.
Δ = hc ÷ λ	h is Planck's constant and c is the speed of light. λ is the wavelength absorbed, in meters.	h = 6.626 × 10^-34 J·s, c = 2.998 × 10⁸ m/s, so hc = 1.987 × 10^-25 J·m. Δ in J per photon.
Δ per mole = (Δ per photon) × N_A	Scales one photon up to a mole of them, so the answer can sit next to a bond energy.	N_A = 6.022 × 10²³/mol. Divide by 1000 for kJ/mol.
μ = √(n(n + 2))	n is the number of unpaired electrons. A measured μ gives n with no guessing.	Bohr magnetons. n is a bare count.

Unpaired electrons n	0	1	2	3	4	5
μ in Bohr magnetons	0	1.73	2.83	3.87	4.90	5.92

Wavelength absorbed	Color absorbed	Color you see
400 to 430 nm	violet	yellow to green-yellow
430 to 490 nm	blue	orange
490 to 510 nm	blue-green	red to purple
530 to 570 nm	green to yellow-green	purple to violet
570 to 600 nm	yellow	violet to blue
600 to 700 nm	orange to red	blue to blue-green

Worked examples

Example 19.1 · Configurations, ions and unpaired electrons

Write the configuration of Fe, Fe²⁺ and Fe³⁺, then of Cr and Cr³⁺. Count the unpaired electrons in each ion.

Step 1 · place the neutral iron atom

Iron is element 26, in group 8. Argon covers 18 electrons, leaving 8 to place. Fill 4s with 2, then put 6 in 3d. So Fe is [Ar]3d⁶4s², and 18 + 6 + 2 = 26.

Step 2 · take the 4s pair off first

Fe²⁺ loses both 4s electrons, giving [Ar]3d⁶. Fe³⁺ loses the 4s pair plus one 3d electron, giving [Ar]3d⁵.

Step 3 · check with the shortcut

d electrons equal group number minus charge. For Fe²⁺ that is 8 − 2 = 6. For Fe³⁺ it is 8 − 3 = 5. Both match.

Step 4 · spread them by Hund's rule

Five d orbitals fill singly first. With 6 electrons, five go in alone and one gets a partner, leaving 4 unpaired. With 5 electrons, each orbital holds one, leaving 5 unpaired.

Step 5 · now chromium

Chromium is element 24, in group 6, and it is an exception. It is [Ar]3d⁵4s¹, not 3d⁴4s². For the ion, use the shortcut: 6 − 3 = 3, so Cr³⁺ is [Ar]3d³ with 3 unpaired.

Answer

Fe²⁺ is [Ar]3d⁶ with 4 unpaired, Fe³⁺ is [Ar]3d⁵ with 5 unpaired, and Cr³⁺ is [Ar]3d³ with 3 unpaired. These counts hold for a free ion. Ligands can change the iron(II) count, as the self-test shows.

Example 19.2 · Pull a compound apart, then name it

Work out the coordination number, the charge on the complex ion and the metal's oxidation number for K₃[Fe(CN)₆] and for [Co(NH₃)₄Cl₂]NO₃. Then name both.

Step 1 · split inside from outside

In the first compound, three K⁺ sit outside the brackets as counter ions. In the second, one NO₃^- sits outside.

Step 2 · balance the outside

Each compound overall is neutral. Three K⁺ means the iron complex is 3−. One nitrate means the cobalt complex is 1+.

Step 3 · solve for each metal

For iron: x + 6(−1) = −3, so x − 6 = −3 and x = +3. For cobalt: x + 4(0) + 2(−1) = +1, so x − 2 = +1 and x = +3.

Step 4 · count donor atoms

Cyanide is monodentate, so six of them give a coordination number of 6. Ammonia and chloride are both monodentate, so 4 + 2 = 6 again. Both complexes are octahedral.

Step 5 · build the names

The iron complex is negative, so iron takes the Latin stem and the -ate ending: potassium hexacyanidoferrate(III). The cobalt complex is positive, so cobalt keeps its name. Sort ammine before chlorido, since a beats c: tetraamminedichloridocobalt(III) nitrate.

Answer and one extra

Both are coordination number 6 with the metal at +3. Iron is group 8, so Fe³⁺ here is d⁵. Cobalt is group 9, so Co³⁺ is d⁶. The cobalt one also has two stereoisomers, with the chlorides either 90° apart, which is cis, or 180° apart, which is trans.

Example 19.3 · From wavelength to Δ, and back again

A complex absorbs most strongly at 600 nm. Find Δ in kJ/mol and name the color of the solution. Then work backward from Δ = 250 kJ/mol.

Step 1 · put the wavelength in meters

600 nm × (1 m ÷ 10⁹ nm) = 6.00 × 10^-7 m.

Step 2 · energy of one photon

Δ = hc ÷ λ, and hc = (6.626 × 10^-34 J·s)(2.998 × 10⁸ m/s) = 1.987 × 10^-25 J·m. Divide: (1.987 × 10^-25) ÷ (6.00 × 10^-7) = 3.31 × 10^-19 J per photon.

Step 3 · scale to a mole

(3.31 × 10^-19 J)(6.022 × 10²³/mol) = 1.99 × 10⁵ J/mol, which is 199 kJ/mol.

Step 4 · name the color

600 nm is orange light. Take orange out of white light and the solution looks blue.

Step 5 · now go the other way

250 kJ/mol is 2.50 × 10⁵ J/mol. Per photon: (2.50 × 10⁵) ÷ (6.022 × 10²³) = 4.15 × 10^-19 J. Then λ = hc ÷ Δ = (1.987 × 10^-25) ÷ (4.15 × 10^-19) = 4.79 × 10^-7 m, which is 479 nm.

Answer

Δ = 199 kJ/mol at 600 nm, and the solution looks blue. The bigger gap of 250 kJ/mol absorbs at 479 nm, which is blue light, so that solution looks orange. A bigger gap always means a shorter wavelength. Swap in a stronger field ligand and the band slides toward the blue end, so the color you see slides too.

The traps people fall into

Watch for these

Taking 3d electrons off first. The 4s pair always leaves first, so Fe²⁺ is [Ar]3d⁶.
Forgetting chromium and copper. Both shift one electron to reach a half filled or filled d set.
Counting ligands for the coordination number. Count donor atoms. Three bidentate ligands give 6.
Ignoring the counter ions. They set the charge on the complex, which sets the metal's oxidation number.
Alphabetizing by the prefix. Tetraammine files under a, not under t.
Writing diethylenediamine. The ligand name already holds a prefix, so it is bis(ethylenediamine).
Leaving off the -ate ending. A negative complex needs it, often on a Latin stem such as ferrate or cuprate.
Looking for cis and trans in a tetrahedron. There is no far side, so neither one exists.
Naming the color absorbed. You see the complementary color, so a green solution is not absorbing green.
Thinking high spin means more electrons. It means the same electrons are spread out, because Δ is small.
Calling the chelate effect stronger bonding. The bonds are about the same. The gain is in the number of loose particles.

Self-test · Chapter 19

1. Write the configurations of Cu, Cu⁺ and Cu²⁺, and give the unpaired count for each ion.

[Ar]3d¹⁰4s¹, then [Ar]3d¹⁰ with 0 unpaired, then [Ar]3d⁹ with 1 unpaired. Copper is element 29 in group 11, and it is the second exception, taking a filled d set over a filled 4s. The 4s electron goes first, then a 3d one. Use the shortcut to check: 11 − 2 = 9 d electrons in Cu²⁺.

2. How many d electrons does Mn²⁺ have, and how many are unpaired?

5 d electrons, all 5 unpaired. Manganese is group 7, so 7 − 2 = 5. Five electrons spread one to each of the five d orbitals by Hund's rule, so nothing is paired. That half filled set is why Mn²⁺ gives such a strong magnetic signal.

3. For [Cr(en)₂Cl₂]Cl, find the coordination number, the charge on the complex and the oxidation number of chromium.

6, then 1+, then +3. Each en is bidentate, so two of them give 4 donor atoms, and the two chloride ligands add 2 more, for 6 in all. One chloride sits outside as a counter ion, so the complex must be 1+. Then x + 2(0) + 2(−1) = +1 gives x = +3.

4. Name [Co(NH₃)₆]Cl₃.

Hexaamminecobalt(III) chloride. Three chlorides outside make the complex 3+, and the six ammonias are neutral, so cobalt is +3. Ammine takes two letters m. The complex is the cation, so cobalt keeps its plain name and no -ate ending appears.

5. Write the formula for sodium tetrahydroxidozincate(II).

Na₂[Zn(OH)₄]. Zincate tells you the complex is an anion. Four hydroxides add −4 and zinc is +2, so the complex ion is 2−. Each sodium is +1, so two of them balance it. This is the same hydroxo complex that makes zinc hydroxide amphoteric.

6. Which has optical isomers, cis-[Co(en)₂Cl₂]⁺ or the trans form? How many stereoisomers are there in all?

Only the cis form, and there are three stereoisomers in all. With the chlorides opposite, the two en rings lie in one plane through the metal, and that plane is a mirror, so trans is not chiral. With the chlorides at 90° no mirror plane survives, so cis exists as two enantiomers. Three forms in total: trans, plus the two mirror images of cis.

7. Cisplatin treats cancer and its trans partner does not. Why?

The two chlorides must be neighbors. In cis-[Pt(NH₃)₂Cl₂] the chlorides sit at one corner of a flat square, close enough to bind two neighboring sites on a strand of DNA. In the trans form they point in opposite directions and cannot reach the same strand. Same atoms, same bonds, and only the angle differs.

8. A complex absorbs at 450 nm. Find Δ in kJ/mol and name the color seen.

266 kJ/mol, and the solution looks orange. First 450 nm = 4.50 × 10^-7 m. Then Δ = (1.987 × 10^-25 J·m) ÷ (4.50 × 10^-7 m) = 4.42 × 10^-19 J per photon. Scale up: (4.42 × 10^-19)(6.022 × 10²³) = 2.66 × 10⁵ J/mol = 266 kJ/mol. Blue light was removed, so you see its complement.

9. A complex shows μ = 3.9 Bohr magnetons. How many unpaired electrons does it hold?

Three. Use μ = √(n(n + 2)). For n = 3 that is √(3 × 5) = √15 = 3.87, which rounds to the measured value. The neighbors are 2.83 for n = 2 and 4.90 for n = 4, so the three values are far enough apart to leave no doubt.

10. Iron(II) in water gives μ near 5.3, and in cyanide it gives μ = 0. Explain both.

Water is high spin with 4 unpaired, and cyanide is low spin with 0. Iron is group 8, so Fe²⁺ is d⁶ in both. For n = 4, μ = √(4 × 6) = √24 = 4.90, and iron(II) commonly reads a little above that. For n = 0, μ = 0. Water sits low in the spectrochemical series, so Δ is small and electrons spread into t_2g⁴e_g². Cyanide sits near the top, so Δ is large and all six pair up as t_2g⁶. Only the ligand changed.

Chapter 20Organic Chemistry

What to know

Carbon holds four valence electrons, so it makes four bonds every time, spread to the corners of a tetrahedron at about 109.5°. It also bonds to other carbons with ease. Chains, rings and branches all follow from that. The plainest carbon compounds hold carbon and hydrogen only, and those are the hydrocarbons.

An alkane has single bonds only, so each carbon carries all the hydrogen it can hold. That is what saturated means, and the formula is C_nH_2n+2. A ring closes two of those spots, so a cycloalkane is C_nH_2n. The alkanes form a homologous series, each member adding one CH₂ unit. Boiling points climb as the chain grows, because longer chains cling harder.

One molecule can be drawn three ways. A molecular formula counts atoms and stops. A condensed structural formula groups them carbon by carbon. A skeletal structure draws each bond as a line, where every corner and every free end is a carbon and the hydrogens are left out.

Structural isomers share a formula and wire their atoms differently, so they are different substances with different boiling points. The counts grow fast: C₄H₁₀ has 2, C₅H₁₂ has 3, C₆H₁₄ has 5, and C₁₀H₂₂ has 75.

The naming rules in order

Find the longest continuous carbon chain and let it set the base name.
Number that chain from the end nearest the first branch.
Name each branch as an alkyl group, so CH₃ is methyl and C₂H₅ is ethyl.
List the branches in alphabetical order with their numbers, using di, tri and tetra for repeats.
Those numbers are locants, and the right set is the lowest one the structure allows.

An alkene holds a carbon to carbon double bond, with the formula C_nH_2n. An alkyne holds a triple bond, at C_nH_2n-2. Both are unsaturated, which means they carry less hydrogen than the matching alkane. That missing hydrogen is the whole story, because the extra bonds are loose electron density and reagents go straight for it.

Shape follows the bonding. A doubly bonded carbon is flat, near 120°, and the two ends cannot spin. A triply bonded carbon is a straight line at 180°. That locked double bond is where cis and trans isomers come from, and they need two different groups on each carbon of the bond.

An addition reaction opens a multiple bond and attaches a piece to each carbon, so nothing is thrown away. Hydrogenation adds H₂ over a catalyst. Halogenation adds Br₂, and the fading orange color is a fast test for unsaturation. Markovnikov's rule sends an added hydrogen to the carbon that already carries more hydrogen. Alkanes cannot add, because they have no loose bond to open, so they substitute instead and throw off a second product as waste.

Benzene is C₆H₆, flat and regular, with 120° angles and six equal bonds of 139 pm. A plain single bond is 154 pm and a plain double bond is 134 pm, so benzene sits between them at every bond. Each ring carbon keeps one electron in a p orbital above and below the ring, and those six electrons spread over all six carbons at once. That spreading is delocalization, and it lowers the energy by a measurable amount. A flat ring that earns that bonus is aromatic, and it needs 4n + 2 delocalized electrons. Benzene has 6, with n = 1. Addition would break the ring apart and hand that bonus back, so benzene substitutes instead.

A functional group is the reactive piece hung on a carbon frame, and it sets the boiling point, the solubility and the reactions. The frame mostly comes along for the ride.

The families, their groups and their endings

Alcohol, R–OH, ends in -ol. Class it by the carbon holding the OH as primary, secondary or tertiary.
Ether, R–O–R'. No O–H bond, so it can accept a hydrogen bond but cannot donate one.
Aldehyde, –CHO, ends in -al. It sits at a chain end with a hydrogen on the carbonyl carbon.
Ketone, –CO–, ends in -one. It sits inside the chain with carbon on both sides.
Carboxylic acid, –COOH, ends in -oic acid. A weak acid with pK_a near 4 to 5.
Ester, R–COO–R'. Two words: the alcohol part first, then the acid part with -oate.
Amine, nitrogen with carbon groups. A weak base, because the lone pair can take a proton.
Amide, C(=O)N. Not basic, because the lone pair is shared with the carbonyl. This is the peptide link.

An O–H bond is very polar, so one molecule's hydrogen is drawn to the next molecule's oxygen. That hydrogen bond takes real heat to pull apart. Watch three molecules of nearly equal mass. Propane boils at −42.1 °C, dimethyl ether at −24.8 °C, and ethanol at 78.3 °C. Only ethanol has an O–H bond. Solubility follows the same logic, since the OH end is at home in water and the carbon chain is not.

These families are rungs on one ladder. A primary alcohol oxidizes to an aldehyde and then to a carboxylic acid. A secondary alcohol climbs one rung and stops at the ketone. A tertiary alcohol has no hydrogen left on that carbon, so it refuses.

A polymer is one very long molecule built from small repeating pieces, and the small starting molecule is a monomer. An addition polymer grows from alkenes, with the double bond opening and nothing leaving, so the repeat unit has exactly the monomer's formula. A condensation polymer drops a small molecule, usually water, at every link, so each monomer needs two reactive ends. A diacid with a diamine gives a polyamide such as nylon. A diacid with a diol gives a polyester such as PET. Amino acids join the same way, which is why a protein is a natural polyamide.

Equations, units and symbols

Equation	What each symbol means	Units
Alkane: C_nH_2n+2	n is the carbon count. This is the saturated ceiling for an open chain.	Bare atom counts.
Cycloalkane and alkene: C_nH_2n	One ring or one double bond costs the same two hydrogens.	Bare atom counts.
Alkyne: C_nH_2n-2	A triple bond costs four hydrogens, so it counts as two degrees.	Bare atom counts.
Degrees of unsaturation = (2C + 2 − H) ÷ 2	C and H are the atom counts from the formula. One degree is one ring or one double bond.	A bare count.
C_xH_y + O₂ → x CO₂ + (y ÷ 2) H₂O	A clean burn. Balance carbon, then hydrogen, then oxygen last. Double everything to clear a half.	Mole ratios from the coefficients.
Atom economy = (mass of wanted product ÷ total mass of reactants) × 100%	Use the molar masses from the balanced equation. Addition reaches 100%, and substitution cannot.	A percent, from g/mol over g/mol.
K_a = x² ÷ (c₀ − x)	For a carboxylic acid or, with K_b, for an amine. x is the ion made.	[ ] in mol/L. K_a has no units.
Degree of polymerization = (chain molar mass) ÷ (repeat unit molar mass)	Real samples hold chains of many lengths, so the mass quoted is always an average.	A bare count, from g/mol over g/mol.

Worked examples

Example 20.1 · Name a branched alkane, and check the formula

Name CH₃CH(CH₃)CH₂CH(CH₃)CH₃, then confirm its molecular formula.

Step 1 · find the longest chain

Count the carbons written in a row, skipping the ones in brackets. That gives 5, so the base name is pentane.

Step 2 · number it both ways

From the left the branches sit on carbons 2 and 4. From the right they sit on 2 and 4 as well. The chain is the same either way, so the locant set is 2,4.

Step 3 · name the branches

Each branch is one carbon, so each is a methyl group. Two of them means dimethyl.

Step 4 · write the name

2,4-dimethylpentane.

Step 5 · check the formula

Carbons: 5 in the chain plus 2 in the branches is 7. For an alkane, H = 2(7) + 2 = 16. Now add the written hydrogens: 3 + 1 + 3 + 2 + 1 + 3 + 3 = 16.

Answer

2,4-dimethylpentane, with the formula C₇H₁₆. The two counts agree, so nothing was dropped. If a hydrogen count comes out odd for an alkane, you lost an atom somewhere.

Example 20.2 · Degrees of unsaturation, then a clean burn

A hydrocarbon is C₇H₈. How unsaturated is it, and what structure fits? Then balance its combustion in oxygen.

Step 1 · find the saturated count

For 7 carbons a saturated alkane holds 2(7) + 2 = 16 hydrogens.

Step 2 · subtract and halve

16 − 8 = 8 hydrogen atoms missing, and 8 ÷ 2 = 4 degrees of unsaturation.

Step 3 · read it

Four degrees fits one ring plus three double bonds, which is a benzene ring. With one methyl group hung on it, that is toluene.

Step 4 · balance carbon and hydrogen

Seven carbons need 7 CO₂. Eight hydrogens need 4 H₂O.

Step 5 · count oxygen last

On the right, 7 × 2 = 14 oxygens in the carbon dioxide plus 4 in the water is 18. That is 9 O₂. So C₇H₈ + 9 O₂ → 7 CO₂ + 4 H₂O.

Answer and the check

Four degrees of unsaturation, and the equation above. Check it: carbon 7 and 7, hydrogen 8 and 8, oxygen 18 on each side. No half coefficient appeared here, so no doubling was needed. Note that the formula alone cannot prove it is toluene. Four degrees could also be two rings and two double bonds.

Example 20.3 · How long is a polymer chain?

A polystyrene sample averages 2.50 × 10⁵ g/mol. Its monomer is styrene, C₈H₈. Find the degree of polymerization. Then compare with a condensation polymer.

Step 1 · mass of the repeat unit

Carbon: 8 × 12.011 = 96.088 g/mol. Hydrogen: 8 × 1.008 = 8.064 g/mol. Together that is 104.15 g/mol.

Step 2 · nothing leaves in an addition polymer

The double bond opens and the carbons link up, so the repeat unit has the same formula as the monomer. That means 104.15 g/mol is both numbers at once.

Step 3 · divide

(2.50 × 10⁵) ÷ 104.15 = 2.40 × 10³, so about 2400 repeat units.

Step 4 · check it backward

(2400)(104.15) = 2.50 × 10⁵ g/mol, which matches the stated average.

Step 5 · now the condensation case

PET comes from terephthalic acid at 166.13 g/mol and ethylene glycol at 62.07 g/mol. Together that is 228.20 g/mol, but each link loses a water. Two links per unit means 228.20 − 2(18.02) = 192.16 g/mol for the repeat unit.

Answer

About 2400 repeat units of styrene. The difference between the two polymer types is the whole point. An addition polymer loses nothing, so the repeat unit equals the monomer. A condensation polymer drops water at every link, so the repeat unit is lighter than the monomers that built it.

The traps people fall into

Watch for these

Missing the longest chain. It may bend around a corner in the drawing, so count every path.
Numbering from the wrong end. The correct locant set is the lowest one available.
Alphabetizing by the prefix. Sort by the branch name, so dimethyl files under m.
Counting a triple bond as one degree. It costs four hydrogens, so it is two degrees.
Expecting cis and trans from any double bond. Each carbon of the bond needs two different groups.
Making benzene add like an alkene. Addition would cost the delocalization bonus, so the ring substitutes.
Mixing up aldehyde and ketone. An aldehyde sits at a chain end with a hydrogen on the carbonyl carbon.
Giving an ether a hydrogen bond to donate. It has no O–H, so it can only accept one.
Oxidizing a tertiary alcohol. That carbon has no hydrogen left, so nothing happens.
Calling an amide a base. The nitrogen's lone pair is shared with the carbonyl, so it is not available.
Using the monomer mass as a condensation repeat unit. Subtract the water lost at each link.

Self-test · Chapter 20

1. Give the molecular formula of the straight-chain alkane, alkene and alkyne with six carbons.

C₆H₁₄, C₆H₁₂ and C₆H₁₀. The alkane is C_nH_2n+2, so 2(6) + 2 = 14. The alkene is C_nH_2n, so 12. The alkyne is C_nH_2n-2, so 10. Each extra bond costs two hydrogens.

2. How unsaturated is C₅H₈, and what could it be?

Two degrees. The saturated count is 2(5) + 2 = 12, and 12 − 8 = 4 hydrogens are missing, so 4 ÷ 2 = 2 degrees. That could be two double bonds, one triple bond, or one ring plus one double bond. The formula cannot tell you which.

3. Balance the combustion of propane, C₃H₈.

C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O. Three carbons need 3 CO₂ and eight hydrogens need 4 H₂O. Oxygen on the right is 6 + 4 = 10 atoms, which is 5 O₂. No half coefficient appeared, so no doubling was needed.

4. How many structural isomers does C₅H₁₂ have? Name them.

Three. They are pentane, a straight chain of five; 2-methylbutane, a four chain with one branch; and 2,2-dimethylpropane, a three chain with two branches on the middle carbon. All three are C₅H₁₂, and all three sit in real bottles with different boiling points.

5. Ethanol boils at 78.3 °C and dimethyl ether at −24.8 °C. Both are C₂H₆O. Explain the gap.

Only ethanol has an O–H bond to donate a hydrogen bond. Both masses are 46.07 g/mol, from 2(12.011) + 6(1.008) + 15.999. So mass cannot explain a gap of 103.1 °C. The ether hides its oxygen in the middle with no hydrogen on it, so it can accept a hydrogen bond but never donate one. The group is doing the work, not the mass.

6. Hydrogenating cyclohexene releases 120 kJ/mol. Benzene plus 3 H₂ releases only 208 kJ/mol. Find the resonance energy.

152 kJ/mol. If benzene held three ordinary separate double bonds, it should release 3 × 120 = 360 kJ/mol. It releases 208, so 360 − 208 = 152 kJ/mol. Benzene started that much lower in energy than the drawing predicted, and giving up that bonus is why it will not add.

7. Find the pH and the percent ionization of 0.10 M ethanoic acid. K_a = 1.8 × 10^-5.

pH 2.87 and 1.3% ionized. Assume x is small: x² = (1.8 × 10^-5)(0.10) = 1.8 × 10^-6, so x = 1.3 × 10^-3 M. Then pH = −log(1.34 × 10^-3) = 2.87. The ionized share is (1.34 × 10^-3) ÷ 0.10 = 1.3%, under 5%, so dropping the x was fair. That pH matches a bottle of vinegar.

8. What does 1-propanol give on oxidation, and what does 2-propanol give? What about a tertiary alcohol?

Propanal then propanoic acid, propanone, and nothing. A primary alcohol climbs two rungs, through the aldehyde to the carboxylic acid. A secondary alcohol climbs one and stops at the ketone, because the ketone carbon has no hydrogen left to lose. A tertiary alcohol has no hydrogen on that carbon to begin with, so it refuses.

9. Ethanol and ethanoic acid are heated with an acid catalyst. Name the product and the reaction, then say how to reverse it.

Ethyl ethanoate, by esterification, reversed by hydrolysis. Water is the second product, so this is a condensation. The name takes the alcohol part first as an alkyl group, then the acid part with an -oate ending. Add extra water and the ester splits back apart. Boil it with sodium hydroxide instead and you get a salt plus the alcohol, which on a fat is saponification.

10. A nylon-6,6 sample averages 45,264 g/mol, with a repeat unit of 226.32 g/mol. Find the degree of polymerization and the water lost per chain.

200 repeat units, and about 400 water molecules. Divide: 45,264 ÷ 226.32 = 200.0. Each repeat unit forms two amide links, and each link drops one water, so one chain of 200 units releases about 400 waters. That is the difference from an addition polymer, which loses nothing at all.

Chapter 21Nuclear Chemistry

What to know

Protons and neutrons both live in the nucleus, and together they are called nucleons. A nuclide is one exact kind of nucleus. A nuclide symbol carries two numbers on the left. The top one is the mass number A, counting every nucleon. The bottom one is the atomic number Z, counting only protons. Subtract to get the neutrons, so N = A − Z.

Every proton shoves every other proton, and what holds them together is the strong nuclear force. It pulls any two nucleons together across a very short range. Neutrons add that pull without adding any shove, so they work like nuclear glue. Plot every nuclide with neutrons up and protons across, and the stable ones crowd into one narrow stripe, the band of stability. Light stable nuclei sit near a 1-to-1 ratio. Heavy ones need extra glue, so the band bends up toward 1.5 to 1. Nothing past bismuth is stable, because every nuclide with Z above 83 is radioactive.

Weigh a nucleus, then weigh its separate parts and add them up. The parts weigh more. The missing mass is the mass defect, and it left as energy when the nucleus formed. That energy is the binding energy, which is also the bill to pull the nucleus apart. Divide by the nucleon count to compare nuclei of very different sizes. The binding energy per nucleon peaks near iron-56 at about 8.79 MeV.

The five ways a nucleus decays

Alpha: a helium-4 nucleus leaves. A drops by 4 and Z drops by 2. Paper stops it.
Beta minus: a neutron turns into a proton and an electron leaves. A holds still and Z rises by 1. This is the fix for too many neutrons.
Positron emission: a proton turns into a neutron and a positron leaves. A holds still and Z falls by 1.
Electron capture: the nucleus swallows one of its own inner electrons. A holds still and Z falls by 1, just as for positron emission.
Gamma: a photon leaves an excited nucleus. Neither A nor Z changes, because only energy left.

A positron does not get far. It meets an ordinary electron and the two destroy each other, turning their mass into two gamma photons of 0.511 MeV fired in opposite directions. A PET scanner is built to catch both at once. One decay step is rarely enough for a heavy nucleus, so it runs through a decay series. Uranium-238 takes 14 steps to reach stable lead-206.

You cannot say when one nucleus will decay, since the event is random and the nucleus keeps no memory. Across a real sample the behavior is smooth. The half-life is the time for half the nuclei to decay, and each half-life halves whatever is left, not what you began with. So you pass 50 percent, then 25, then 12.5, and the curve bends toward zero without arriving.

Nuclear decay is first order. The rate depends on how many unstable nuclei are present and on nothing else. Heating the sample does not speed it up, and neither does crushing it, dissolving it or bonding it into a new compound. That is a sharp break from ordinary reaction rates.

Measure how much of a radioactive parent is left and you can solve for elapsed time. That is radiometric dating. Carbon-14 at 5,730 years is useful out to roughly 50,000 years, which is about nine half-lives. Rock needs a slower clock, so uranium-238 at 4.5 billion years and potassium-40 at 1.25 billion years take over.

A nuclear equation balances when two totals agree across the arrow. The mass numbers must match, and so must the atomic numbers, because charge is conserved. Notice what is missing from that list. Elements are not conserved, so atoms of one element really do become another. Give every particle a nuclide symbol, even the small ones, and finding an unknown becomes two subtraction problems. Transmutation is any change of one element into another, whether by decay or by firing a particle at a target. Neutrons make the best bullets, because a nucleus does not push them away.

Any change that moves nucleons toward the binding energy peak releases energy. A very heavy nucleus gets closer by splitting, which is fission. Very light nuclei get closer by joining, which is fusion. One fission of uranium-235 releases about 200 MeV, while burning one carbon atom releases about 4 eV. The nuclear event is roughly 50 million times larger.

Every fission throws out two or three fresh neutrons, and each can split another nucleus. That self-feeding loop is a chain reaction. The multiplication factor k is the number of fissions in one generation divided by the number in the one before. Below 1 the chain fades, at 1 it holds steady, and above 1 it climbs. The smallest amount of fuel that can hold the loop going is the critical mass. A reactor is a machine for parking k at exactly 1, using a moderator to slow the neutrons, control rods to soak up spares, and a coolant to carry the heat away.

Fusion pays better per unit of mass. Spread 17.6 MeV over the 5 nucleons of a deuterium and tritium pair and you get about 3.5 MeV each. Spread 200 MeV over 236 nucleons and you get about 0.85 MeV each. The obstacle is repulsion, since both nuclei are positive, so it needs temperatures near 100 million kelvin plus some way to hold the plasma together.

Three units that are easy to mix up

Activity says how busy the source is, in becquerels, where one becquerel is one decay per second.
Absorbed dose says how much energy tissue took in, in grays, where one gray is one joule per kilogram.
Dose equivalent says how much harm to expect, in sieverts, found by multiplying grays by a quality factor.
Beta and gamma rate a factor of 1, neutrons about 10, and alpha a factor of 20.

Where the source sits matters enormously. Alpha emitters outside the body are almost harmless, because dead skin stops the particles. Breathe them in and they become the worst case, since all their energy lands inside living tissue. Protection comes down to three words: time, distance and shielding. Distance is the cheapest, because intensity falls with the square of the distance.

Equations, units and symbols

Equation	What each symbol means	Units
N = A − Z	A is the mass number on top, Z is the atomic number below, N is the neutron count.	Bare counts of particles.
Δm = Σ(separate parts) − (nuclide mass)	The mass defect. Use Z hydrogen atoms and N neutrons, so the electrons cancel out.	Atomic mass units, u.
E = mc²	Turns the mass defect into the binding energy. m must be in kilograms.	c = 2.998 × 10⁸ m/s. 1 u = 1.6605 × 10^-27 kg. E in joules.
Binding energy = Δm × 931.5 MeV/u	The same calculation with the conversion folded in, which is faster by hand.	1 u is worth 931.5 MeV. 1 MeV = 1.602 × 10^-13 J.
Binding energy per nucleon = (binding energy) ÷ A	The fair way to compare nuclei of different sizes. It peaks near iron-56.	MeV per nucleon.
k = 0.693 ÷ t_1/2	The decay constant, since 0.693 is the natural log of 2.	Reciprocal time, matching the half-life's unit.
N = N₀(1/2)^t/t_1/2	Use this when the elapsed time is a tidy number of half-lives.	Any amount unit, as long as both sides match.
ln(N₀ ÷ N) = kt	Use this for everything else, including dating. N₀ is the starting amount and N is what is left.	The ratio is unitless, so t comes out in the unit k uses.
1 Ci = 3.7 × 10¹⁰ Bq	Activity, where 1 Bq is one decay per second.	Bq and Ci.
Dose equivalent = absorbed dose × quality factor	Grays times a factor gives sieverts. Alpha is 20, neutrons about 10, beta and gamma 1.	1 Gy = 1 J/kg. 1 rad = 0.01 Gy, 1 rem = 0.01 Sv.
Intensity ∝ 1 ÷ d²	Doubling the distance quarters the dose rate. Divide by the square of the distance ratio.	Any dose rate unit, such as µSv/h.

Worked examples

Example 21.1 · The binding energy of carbon-12

A carbon-12 atom weighs exactly 12.000000 u. A hydrogen-1 atom weighs 1.007825 u and a neutron weighs 1.008665 u. Find the binding energy per nucleon.

Step 1 · add up the loose parts

Carbon-12 has 6 protons and 6 neutrons. Six hydrogen atoms come to 6 × 1.007825 = 6.046950 u. Six neutrons come to 6 × 1.008665 = 6.051990 u. Together the parts total 12.098940 u. Using hydrogen atoms instead of bare protons makes the six electrons cancel out.

Step 2 · subtract

Δm = 12.098940 − 12.000000 = 0.098940 u of missing mass.

Step 3 · the fast route to MeV

One u is worth 931.5 MeV, so (0.098940 u)(931.5 MeV/u) = 92.16 MeV.

Step 4 · report it per nucleon

92.16 MeV ÷ 12 nucleons = 7.68 MeV per nucleon.

Step 5 · check it in joules

(0.098940 u)(1.6605 × 10^-27 kg/u) = 1.6425 × 10^-28 kg. Then E = mc² = (1.6425 × 10^-28)(2.998 × 10⁸)² = 1.476 × 10^-11 J. Divide by 1.602 × 10^-13 J/MeV and you get 92.1 MeV, which matches.

Answer

7.68 MeV per nucleon. The peak sits near iron-56 at about 8.79 MeV, so carbon is still climbing toward it. For a whole mole of carbon-12 that binding energy is around 8.9 × 10⁹ kJ, which is millions of times any chemical bond energy.

Example 21.2 · Balance three nuclear equations

Write the equation for the alpha decay of uranium-238 and the beta decay of carbon-14. Then fill the gap in ²³⁵₉₂U + ¹₀n → ¹⁴⁴₅₆Ba + ? + 2 ¹₀n.

Step 1 · alpha decay takes 4 and 2

238 − 4 = 234 for the mass number, and 92 − 2 = 90 for the atomic number. Element 90 is thorium, so ²³⁸₉₂U → ²³⁴₉₀Th + ⁴₂He.

Step 2 · beta decay holds A and raises Z

A stays at 14 and Z goes from 6 to 7. Element 7 is nitrogen, so ¹⁴₆C → ¹⁴₇N + ⁰_-1e. A neutron turned into a proton, which walks the nuclide down into the band of stability.

Step 3 · add the mass numbers in the fission line

The left side totals 235 + 1 = 236. The known products total 144 + 1 + 1 = 146. The gap is 236 − 146 = 90, so the missing particle has A = 90.

Step 4 · add the atomic numbers

The left side totals 92 + 0 = 92. The known products total 56 + 0 + 0 = 56. The gap is 92 − 56 = 36, so the missing particle has Z = 36.

Step 5 · name it and check

Element 36 is krypton, so the missing product is ⁹⁰₃₆Kr. Mass numbers on the right: 144 + 90 + 2 = 236. Atomic numbers: 56 + 36 = 92. Both totals match the left side.

Answer

Thorium-234, nitrogen-14 and krypton-90. Only two totals ever have to balance, mass number and atomic number. The elements themselves change, which is exactly what a nuclear equation is allowed to do.

Example 21.3 · Half-life two ways, then a date

Strontium-90 has a half-life of 28.8 years. How much of an 80.0 mg sample is left after 86.4 years? Then date a bone that holds 22.0% of its original carbon-14, where the half-life is 5,730 years.

Step 1 · count the half-lives

86.4 years ÷ 28.8 years = 3.00 half-lives exactly, so the tidy route works.

Step 2 · halve it three times

80.0 mg → 40.0 mg → 20.0 mg → 10.0 mg. The formula agrees: 80.0 × (1/2)³ = 80.0 × 0.125 = 10.0 mg.

Step 3 · the bone needs the log form

22.0% is not a tidy number of half-lives, so first find the decay constant. k = 0.693 ÷ 5,730 yr = 1.21 × 10^-4 yr^-1.

Step 4 · set up and solve

N₀ ÷ N = 1 ÷ 0.220 = 4.55, and ln(4.55) = 1.515. Then t = 1.515 ÷ (1.21 × 10^-4 yr^-1) = 1.25 × 10⁴ years.

Step 5 · check that it is sensible

12,500 years is 2.18 half-lives. Two half-lives would leave 25.0% and three would leave 12.5%, so 22.0% should land just past two. It does.

Answer

10.0 mg of strontium-90 after 86.4 years, and the bone is about 1.25 × 10⁴ years old. Use the halving route only when the time divides evenly into half-lives. Otherwise go to the log form every time.

The traps people fall into

Watch for these

Mixing the three radiation units. Becquerels measure the source, grays measure energy absorbed, sieverts measure expected harm.
Forgetting the quality factor. A gray of alpha counts for 20 times a gray of gamma.
Thinking heat or chemistry changes a decay rate. Nothing you can do to a sample changes its half-life.
Halving when the time is not a whole number of half-lives. Use ln(N₀ ÷ N) = kt instead.
Halving the original amount each time. Each half-life halves what is left at that moment.
Expecting the element to stay the same. Only A and Z have to balance, so elements change freely.
Swapping the Z change for beta and positron. Beta raises Z by 1, and positron emission and electron capture each lower it by 1.
Giving gamma emission a mass change. Neither A nor Z moves, because only energy left.
Thinking the parts of a nucleus weigh less. They weigh more, and the difference is the binding energy.
Calling fission the bigger payout. It is bigger per event and smaller per nucleon than fusion.
Treating alpha emitters as the harmless ones. Outside the body they are, and inhaled they are the worst case.

Self-test · Chapter 21

1. For ²³⁸₉₂U, give the protons, the neutrons and the nucleons.

92 protons, 146 neutrons, 238 nucleons. Z is the proton count, so 92. A is the nucleon count, so 238. Then N = A − Z = 238 − 92 = 146. The neutron-to-proton ratio is 146 ÷ 92 = 1.59, which is up near the top of the band of stability.

2. Write the alpha decay of radium-226, which has Z = 88.

²²⁶₈₈Ra → ²²²₈₆Rn + ⁴₂He. Take 4 off the mass number for 222 and 2 off the atomic number for 86. Element 86 is radon, which is a gas. That is why uranium ore leaks radon into basements.

3. Phosphorus-32 is a beta emitter. Write the equation.

³²₁₅P → ³²₁₆S + ⁰_-1e. A holds at 32 and Z climbs from 15 to 16, which is sulfur. A neutron became a proton and the electron left. Check the charge: 15 on the left, and 16 + (−1) = 15 on the right.

4. Fluorine-18 emits a positron. Write the equation, and say what happens to the positron.

¹⁸₉F → ¹⁸₈O + ⁰₊₁e, and the positron is annihilated. A holds at 18 while Z falls from 9 to 8, which is oxygen. The positron meets an ordinary electron and both vanish, turning their mass into two gamma photons of 0.511 MeV fired in opposite directions. A PET scanner catches both at once.

5. Fill the gap: ²⁷₁₃Al + ⁴₂He → ³⁰₁₅P + ?

A neutron, ¹₀n. Mass numbers on the left total 27 + 4 = 31, and the phosphorus accounts for 30, leaving A = 1. Atomic numbers on the left total 13 + 2 = 15, and the phosphorus accounts for all 15, leaving Z = 0. A particle with A = 1 and Z = 0 is a neutron.

6. A clinic gets 40.0 mg of iodine-131, whose half-life is 8.02 days. How much is left after 32.08 days?

2.50 mg. First 32.08 ÷ 8.02 = 4.00 half-lives. Halve four times: 40.0 → 20.0 → 10.0 → 5.00 → 2.50 mg. The formula agrees: 40.0 × (1/2)⁴ = 40.0 × 0.0625 = 2.50 mg.

7. Cobalt-60 has a half-life of 5.27 years. What fraction of a source is left after 10.0 years?

About 27%. This is not a whole number of half-lives, so use the log form. k = 0.693 ÷ 5.27 yr = 0.131 yr^-1. Then ln(N₀ ÷ N) = kt = (0.131)(10.0) = 1.31, so N₀ ÷ N = e^1.31 = 3.71 and N ÷ N₀ = 1 ÷ 3.71 = 0.27. Two half-lives would leave 25%, so 27% after a bit under two is right.

8. One fission of uranium-235 gives about 200 MeV. How much does a mole of fissions give, and how does it compare with burning a mole of carbon at 393.5 kJ/mol?

1.93 × 10¹⁰ kJ/mol, about 49 million times larger. First 200 MeV × (1.602 × 10^-13 J/MeV) = 3.20 × 10^-11 J per event. Scale up: (3.20 × 10^-11)(6.022 × 10²³) = 1.93 × 10¹³ J/mol, which is 1.93 × 10¹⁰ kJ/mol. Divide by 393.5 kJ and you get 4.9 × 10⁷.

9. A hand takes in 1.5 mGy from an alpha emitter and 4.0 mGy from a gamma source. Find the total dose equivalent.

34 mSv. Weight each dose by its quality factor. Alpha rates 20, so 1.5 mGy × 20 = 30 mSv. Gamma rates 1, so 4.0 mGy × 1 = 4.0 mSv. Add them: 30 + 4.0 = 34 mSv. The alpha dose was smaller in grays and counted for far more in sieverts.

10. A meter reads 60 µSv/h at 0.50 m from a small gamma source. What does it read at 2.0 m?

3.8 µSv/h. The distance ratio is 2.0 ÷ 0.50 = 4.0. Intensity falls as 1 ÷ d², so divide by 4.0² = 16. Then 60 ÷ 16 = 3.75, which is 3.8 µSv/h. Two steps back did more than a slab of lead would.

Cram sheet

One page for the last pass before the exam.

Constants worth knowing cold

h = 6.626 × 10^-34 J·s and c = 2.998 × 10⁸ m/s, so hc = 1.987 × 10^-25 J·m.
N_A = 6.022 × 10²³/mol, for turning one photon or one event into a mole of them.
1 u = 1.6605 × 10^-27 kg = 931.5 MeV. 1 MeV = 1.602 × 10^-13 J.
Binding energy per nucleon peaks near iron-56 at about 8.79 MeV.
Carbon-14: 5,730 years. Uranium-238: 4.5 billion years. Potassium-40: 1.25 billion years.
Quality factors: beta and gamma 1, neutrons about 10, alpha 20.
1 Ci = 3.7 × 10¹⁰ Bq. 1 Gy = 1 J/kg. 1 rad = 0.01 Gy and 1 rem = 0.01 Sv.
C–C single bond 154 pm, double bond 134 pm, and every benzene bond 139 pm.

Every equation on one screen

Use it for	Equation
d-block ions	d electrons = group number − charge, and the 4s electrons leave first
Metal oxidation number	x + Σ(ligand charges) = charge on the complex ion
Shape of a complex	coordination number = total donor atoms, not total ligands
Field splitting	t_2g drops 0.4Δ and e_g rises 0.6Δ
Color of a complex	Δ = hc ÷ λ, then times N_A for a mole, and you see the complement
Magnetism	μ = √(n(n + 2)) in Bohr magnetons
Hydrocarbon formulas	alkane C_nH_2n+2, alkene and ring C_nH_2n, alkyne C_nH_2n-2
How unsaturated	degrees = (2C + 2 − H) ÷ 2, with a triple bond worth 2
Burning anything	carbon first, then hydrogen, then oxygen last
Green chemistry	atom economy = wanted product mass ÷ total reactant mass × 100%
Chain length	degree of polymerization = chain molar mass ÷ repeat unit molar mass
Reading a nuclide	N = A − Z
Nuclear energy	Δm × 931.5 MeV/u, or E = mc² with m in kg
Decay rate	k = 0.693 ÷ t_1/2
Whole half-lives	N = N₀(1/2)^t/t_1/2
Any other time	ln(N₀ ÷ N) = kt
Harm from a dose	sieverts = grays × quality factor
Backing away	intensity ∝ 1 ÷ d²

Facts that show up every year

The 4s electrons leave first, so Fe²⁺ is [Ar]3d⁶. Chromium and copper are the two exceptions.
Count donor atoms for the coordination number, and watch for counter ions outside the brackets.
A small Δ gives high spin, a large Δ gives low spin, and the choice exists only for d⁴ through d⁷.
You see the complementary color of the light that was absorbed.
Longest chain, lowest locants, branches listed alphabetically.
Addition keeps every atom. Substitution makes waste. Benzene substitutes to keep its ring.
One O–H bond can move a boiling point by a hundred degrees.
Primary alcohol to aldehyde to acid. Secondary stops at the ketone. Tertiary refuses.
Nuclear equations balance on A and Z only, so elements change.
One half-life halves what is left, and nothing you do to a sample changes the rate.
Fission is bigger per event. Fusion is bigger per nucleon.

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