🧪 Contents · Interactive Lab Chapter 6 · Electronic Structure and Periodic Properties
Chemistry · Chapter 6

Electronic Structure and Periodic Properties

Light carries energy in small packets. An atom can take in or give off those packets only in fixed amounts. That one fact explains the red glow of a neon sign and the orange of a sodium streetlight. It also explains the shape of the periodic table. In this chapter you follow the electron. You will see where it can sit, how much energy each seat holds, and what happens when it moves. Then you will use that map to write the electron layout of any element. By the end the table will look less like a chart and more like a record of how electrons stack up.

By the end of this chapter you can

  • Link wavelength, frequency and photon energy using c = λν and E = hν.
  • Use the Bohr energy levels to predict the four visible lines of hydrogen.
  • Name the four quantum numbers and say what each one fixes.
  • Sketch an s, p, d or f orbital and count the nodes it carries.
  • Write the electron configuration of any element, including the common exceptions.
  • Draw an orbital diagram that obeys Pauli and Hund, then count unpaired electrons.
  • Predict radius, ionization energy, electron affinity and electronegativity across the table.

Section 1Light and quantized energy

Light is a wave you can measure

Light is a moving pair of electric and magnetic fields. In empty space it always travels at one speed. That speed is c = 2.998 × 108 m/s. Every wave has a wavelength, written λ. Wavelength is the distance from one crest to the next crest. Every wave also has a frequency, written ν. Frequency counts how many crests pass a fixed point each second. Its unit is the hertz, which is just s-1.

The speed of light ties those two numbers together.

c = λν

Because c never changes, λ and ν must trade off. A short wave has to come fast. A long wave has to come slowly. Blue light has a shorter wavelength than red light. So blue light must have the higher frequency. The amplitude is the height of the wave. Amplitude sets how bright the light looks. It does not set the color.

Visible light is a thin slice of the whole electromagnetic spectrum. Red light sits near 700 nm. Violet light sits near 400 nm. Radio waves can be meters long. X-rays are smaller than an atom. All of them are the same kind of wave at different sizes.

The parts of a light wave and the electromagnetic spectrum λ = one wavelength amplitude crest trough Frequency ν crests passing one point each second, in s -1 c = λν = 2.998 × 10 8 m/s radio microwave infrared ultraviolet X-ray gamma longer λ, lower energy → ← shorter λ, higher energy 400 nm 550 nm 700 nm the visible slice, opened up
One wave, two measurements. Wavelength is a distance and frequency is a count per second. Their product is always the speed of light. The gold band on the spectrum bar is the narrow range your eye can detect.

Energy comes in packets

For a long time light looked purely like a wave. Then two experiments broke that picture. Hot objects do not glow the way wave theory predicts. Light shining on a metal knocks electrons loose only above a threshold color. Max Planck and Albert Einstein fixed both problems with one idea. Light energy comes in countable pieces.

One piece of light energy is a photon. The energy of a photon depends only on frequency.

E = hν = hc / λ

Here h is Planck's constant, 6.626 × 10-34 J·s. The number is tiny, so one photon carries very little energy. Still, the rule is strict. A photon of 400 nm light carries almost twice the energy of a 700 nm photon.

Energy that comes in fixed pieces is quantized. Think about coins. You can pay 25 cents or 50 cents. You cannot pay 25.3 cents, because no coin that size exists. Light works the same way. Brighter light means more photons, not bigger ones.

Worked example · the energy in green laser light

A green laser pointer puts out light at 532 nm. Find its frequency, the energy of one photon, and the energy of a mole of those photons.

Step 1 · put the wavelength in meters

λ = 532 nm = 532 × 10-9 m = 5.32 × 10-7 m

Step 2 · use c = λν to get frequency

ν = c / λ = (2.998 × 108 m/s) ÷ (5.32 × 10-7 m) = 5.635 × 1014 s-1

Step 3 · use E = hν for one photon

E = (6.626 × 10-34 J·s)(5.635 × 1014 s-1) = 3.734 × 10-19 J

Step 4 · scale up to one mole

(3.734 × 10-19 J)(6.022 × 1023 /mol) = 2.249 × 105 J/mol = 225 kJ/mol

Check the units

J·s times s-1 leaves J, so step 3 is set up right. And 225 kJ/mol is the right size. It sits close to the strength of a chemical bond, which is why light can drive reactions.

Key points
  • Wavelength and frequency are locked together by c = λν.
  • Short wavelength means high frequency and high photon energy.
  • E = hν with h = 6.626 × 10-34 J·s gives the energy of one photon.
  • Amplitude controls brightness. Frequency controls color and energy.
  • Energy in packets is quantized. There is no half photon.
Interactive Change wavelength and frequency and see the energy change. Open full ↑

Section 2The Bohr model and line spectra

A glowing gas gives only certain colors

Put hydrogen gas in a glass tube and run current through it. The tube glows pink. Send that glow through a prism and the pink splits apart. You do not get a smooth rainbow. You get four sharp colored lines on a dark field. That pattern is a line spectrum.

Hydrogen's four visible lines sit at 656 nm, 486 nm, 434 nm and 410 nm. Every hydrogen sample in the universe gives the same four. Sodium gives a different set. Helium gives another. A line spectrum is a fingerprint for an element. That is how we know what stars are made of.

A smooth rainbow would mean an electron can hold any energy at all. Sharp lines mean something else. The electron can only hold certain fixed energies. When it drops from a higher one to a lower one, the leftover energy leaves as a single photon. One jump gives one color.

Bohr's ladder of allowed energies

In 1913 Niels Bohr built a model that got hydrogen right. He let the electron circle the nucleus only in certain orbits. Each orbit carries a fixed energy set by a whole number n, called the principal quantum number.

En = −(2.179 × 10-18 J) / n2

The energies are negative because the electron is trapped. Zero energy means a free electron, far from the nucleus. The lowest rung, n = 1, is the ground state. Any higher rung is an excited state. Notice how fast the rungs crowd together. The gap from n = 1 to n = 2 is huge. The gap from n = 5 to n = 6 is small.

To find the photon, subtract the two energies.

ΔE = Efinal − Einitial

A drop gives a negative ΔE. The atom lost that energy, so a photon of that size left. A climb gives a positive ΔE, which means the atom took a photon in. The four visible lines all end on n = 2. Jumps that end on n = 1 release far more energy, so they land in the ultraviolet.

Hydrogen energy levels and the four visible emission lines Allowed energies of the hydrogen electron n = ∞ 0 J (free) n = 6, 5 n = 4 n = 3 n = 2 n = 1 −0.242 −0.545 −2.179 units of 10 -18 J 122 nm ultraviolet, 2 → 1 656 nm (3 → 2) 486 nm (4 → 2) Rung spacing follows 1/n 2 , so high rungs crowd. What the prism shows 410 434 486 656 wavelength, nm 6 → 2 gives 410 nm. 5 → 2 gives 434 nm. 4 → 2 gives 486 nm. 3 → 2 gives 656 nm. Four jumps, four lines, no colors between.
One jump, one line. Every allowed energy is a rung on a ladder, and only the gaps between rungs can show up as light. The four visible hydrogen lines all come from jumps that land on n = 2.

Worked example · where the red line comes from

An excited hydrogen electron drops from n = 3 to n = 2. Find the photon's energy and wavelength.

Step 1 · find each energy

E3 = −(2.179 × 10-18 J)/9 = −2.421 × 10-19 J

E2 = −(2.179 × 10-18 J)/4 = −5.448 × 10-19 J

Step 2 · subtract, final minus initial

ΔE = (−5.448 × 10-19) − (−2.421 × 10-19) = −3.026 × 10-19 J

The sign is negative, so the atom gave off a photon of 3.026 × 10-19 J.

Step 3 · turn energy into wavelength

hc = (6.626 × 10-34 J·s)(2.998 × 108 m/s) = 1.986 × 10-25 J·m

λ = hc/E = (1.986 × 10-25 J·m) ÷ (3.026 × 10-19 J) = 6.56 × 10-7 m = 656 nm

Check it

656 nm is deep red, and red is the brightest hydrogen line by eye. The same steps with n = 4, 5 and 6 give 486, 434 and 410 nm.

Push an electron all the way off a ground-state hydrogen atom and you must supply the full 2.179 × 10-18 J. Multiply by Avogadro's number and that is 1312 kJ/mol. That number is the measured ionization energy of hydrogen. A model built to explain colored lines also predicts how tightly the atom holds its electron.
Key points
  • A line spectrum shows that electron energies are quantized.
  • En = −(2.179 × 10-18 J)/n2 for hydrogen, and the sign means bound.
  • ΔE = Efinal − Einitial. Negative means a photon came out.
  • The visible lines at 656, 486, 434 and 410 nm all end on n = 2.
  • Bohr's model nails hydrogen but fails for atoms with more electrons.
Interactive Jump an electron between levels and watch a spectral line appear. Open full ↑

Section 3Quantum numbers and orbital shapes

From orbits to clouds

Bohr's neat circles cannot be right. An electron is small and fast, and the Heisenberg uncertainty principle says we cannot pin down both its place and its motion. So we stop asking where the electron is. We ask where it is likely to be. Erwin Schrödinger wrote an equation whose answers give exactly that. Each answer is an orbital.

An orbital is a region around the nucleus where one electron is likely to be found. Picture a swarm of gnats around a porch light. No single gnat has a path you can trace. Still, the swarm has a clear shape and a clear center. The darker part of the cloud means higher probability density.

Four numbers name every orbital

Each orbital is labeled by three quantum numbers. A fourth one describes the electron sitting in it.

So s subshells hold 1 orbital, p hold 3, d hold 5 and f hold 7. Each orbital takes at most 2 electrons. A full shell therefore holds 2n2 electrons.

What the shapes look like

An s orbital is a ball centered on the nucleus. A p orbital is two lobes on opposite sides, like a long balloon pinched in the middle. The three p orbitals point along x, y and z. A d orbital usually has four lobes in a clover pattern, and the fifth d shape has two lobes with a ring around the middle. The seven f shapes are more involved still.

Most orbitals have places where the electron is never found. Those places are nodes. The count is fixed. Angular nodes equal l, so a p orbital has one flat nodal plane through the nucleus. Radial nodes equal n − l − 1, and those are thin shells of empty space. Add them and any orbital has n − 1 nodes in all. More nodes go with more energy.

Shapes and nodes of 1s, 2s, 2p and 3d orbitals 1s one ball, no nodes l = 0, one orbital radial node 2s bigger ball with a hollow shell inside nodal plane 2p two lobes, one flat node l = 1, three orbitals 3d four lobes, two flat nodes l = 2, five orbitals Shape comes from l. Node count comes from n and l. Shading shows where the electron is most likely to be. Gold dashes mark nodes, where it is never found.
Four orbital shapes side by side. The s cloud is a ball, the p cloud is two lobes split by a flat node, and this d cloud is a four-leaf pattern split by two nodes. Angular nodes equal l and radial nodes equal n − l − 1.

Worked example · counting the fourth shell

How many subshells, orbitals and electrons fit in the shell with n = 4?

Step 1 · list the allowed l values

l goes from 0 to n − 1, so l = 0, 1, 2, 3. That is four subshells: 4s, 4p, 4d and 4f.

Step 2 · count orbitals with 2l + 1

4s gives 1, 4p gives 3, 4d gives 5 and 4f gives 7. Total = 1 + 3 + 5 + 7 = 16 orbitals.

Step 3 · two electrons per orbital

2 × 16 = 32 electrons.

Check with the shortcut

2n2 = 2(4)2 = 2(16) = 32. The two routes agree.

Key points
  • An orbital is a probability cloud, not a path.
  • n sets size, l sets shape, ml sets direction, ms sets spin.
  • l must be less than n, and ml must lie between −l and +l.
  • s, p, d and f subshells hold 1, 3, 5 and 7 orbitals.
  • A shell holds 2n2 electrons, and every orbital carries n − 1 nodes.
Interactive Turn s, p, d and f orbitals in space. Open full ↑

Section 4Electron configurations

Fill the cheapest seats first

An atom at rest holds its electrons in the lowest energy arrangement it can. The Aufbau principle says to add electrons one at a time, always to the lowest open subshell. The written result is the electron configuration. Oxygen has 8 electrons, so oxygen is 1s2 2s2 2p4. The superscripts must add up to the atomic number.

The filling order is not simply 1s, 2s, 3s. Once several electrons crowd an atom, each one feels the pull of the nucleus plus the push of the others. That reshuffles the subshell energies. The useful order is this.

1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p 7s 5f 6d 7p

Look at 4s and 3d. The 4s subshell fills first, even though its n is larger. The diagonal chart below builds that order for you. Follow each gold arrow from its tail to its head, then start the next one.

The diagonal chart that gives the orbital filling order Read along each gold arrow, then drop to the next arrow 1s 2s2p 3s3p3d 4s4p4d4f 5s5p5d5f 6s6p6d 7s7p Order: 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p 7s 5f 6d 7p 4s comes before 3d, and 6s comes before 4f. The arrows put them in that order for you. Each row is one value of n. Each column is one shape: s, then p, then d, then f.
The diagonal chart. Writing the subshells in a grid and sweeping down the diagonals reproduces the measured filling order, including the swap that puts 4s ahead of 3d.

Shorthand, core and valence

Long configurations get clumsy, so we use the last noble gas as a stand-in. Calcium is 1s2 2s2 2p6 3s2 3p6 4s2, which we shorten to [Ar] 4s2. The electrons hidden in the bracket are the core electrons. The ones written out are the valence electrons, and those are the ones that do chemistry. Elements in the same group share a valence pattern, which is exactly why they behave alike.

The exceptions worth knowing

In the first row of transition metals, 4s and 3d sit very close in energy. Two elements break the simple pattern because of that.

ElementPattern predictsMeasuredWhy
Cr (24)[Ar] 4s2 3d4[Ar] 4s1 3d5Half-filled d set, six unpaired spins
Cu (29)[Ar] 4s2 3d9[Ar] 4s1 3d10Full d set of ten electrons
Mo (42)[Kr] 5s2 4d4[Kr] 5s1 4d5Same half-filled gain, one row down
Ag (47)[Kr] 5s2 4d9[Kr] 5s1 4d10Same full-set gain as copper

Moving one s electron into the d set costs almost nothing here. Spreading the electrons out lowers repulsion, so the swapped version wins. Chromium ends with six unpaired electrons, which is more than any other element in its row.

Ions follow one more rule. When an atom loses electrons, the highest n goes first. Iron is [Ar] 4s2 3d6. Fe2+ is [Ar] 3d6, not [Ar] 4s2 3d4. Fe3+ is [Ar] 3d5. Anions just add electrons to the next open spot, so Cl- is [Ne] 3s2 3p6, the same as argon.

Worked example · arsenic, element 33

Write the full and short configurations for arsenic.

Step 1 · fill in order until you reach 33

1s2 (2) → 2s2 (4) → 2p6 (10) → 3s2 (12) → 3p6 (18) → 4s2 (20) → 3d10 (30) → 4p3 (33)

Step 2 · write it out

1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p3

Step 3 · use the noble gas shorthand

Argon is element 18, so arsenic is [Ar] 4s2 3d10 4p3.

Check the sum

2 + 2 + 6 + 2 + 6 + 2 + 10 + 3 = 33. Arsenic has five valence electrons, which matches group 15.

Key points
  • Add electrons to the lowest open subshell, and the superscripts must total the atomic number.
  • The order 1s 2s 2p 3s 3p 4s 3d 4p 5s comes from the diagonal chart.
  • [Noble gas] shorthand hides the core and shows the valence electrons.
  • Cr is [Ar] 4s1 3d5 and Cu is [Ar] 4s1 3d10. Mo and Ag match them.
  • Cations lose the highest n electrons first, so Fe3+ is [Ar] 3d5.
Interactive Fill orbitals in order for any element. Open full ↑

Section 5Orbital diagrams and Hund's rule

Boxes and arrows

A configuration like 2p4 tells you the count. It does not tell you how those four electrons are spread over the three 2p orbitals. An orbital diagram shows that. Draw one box for each orbital and one arrow for each electron. An up arrow is spin +½ and a down arrow is spin −½. Two rules decide where the arrows go.

Pauli caps every box at two

The Pauli exclusion principle says no two electrons in one atom can share all four quantum numbers. An orbital already fixes n, l and ml. Only spin is left, and spin has two settings. So a box holds at most two electrons, and they must point opposite ways. Those two are a paired set. This single rule is why s holds 2, p holds 6, d holds 10 and f holds 14.

Hund spreads them out first

Orbitals in the same subshell have equal energy. We call them degenerate. Hund's rule says to put one electron in each degenerate box, all spinning the same way, before doubling up anywhere. Electrons repel each other, so two in one box cost extra energy. Keeping them in separate boxes with parallel spins keeps them farther apart and lowers the total energy.

Nitrogen shows the rule plainly. Its 2p3 electrons go one per box, all three arrows up. Oxygen has one more. That fourth electron must pair somewhere, so oxygen ends with two unpaired electrons.

Unpaired electrons are measurable. An atom or ion with unpaired electrons is paramagnetic, and a magnet pulls on it. With every electron paired it is diamagnetic, and a magnet pushes it away very weakly. Zinc, [Ar] 4s2 3d10, is diamagnetic. Iron, with four unpaired 3d electrons, is strongly drawn to a magnet.

Orbital diagrams for nitrogen and oxygen showing Pauli and Hund One box per orbital, one arrow per electron 1s2s2p N O 1s 2 2s 2 2p 3 3 unpaired, paramagnetic Hund: spread out, spins parallel 1s 2 2s 2 2p 4 2 unpaired, paramagnetic Pauli: the pair must be opposite Up arrow is spin +1/2. Down arrow is spin −1/2. A box never holds two arrows pointing the same way.
Nitrogen and oxygen, box by box. Nitrogen keeps its three 2p electrons apart with parallel spins, as Hund's rule requires. Oxygen's fourth 2p electron has to pair up, and Pauli forces it to take the opposite spin.

Worked example · unpaired electrons in sulfur and iron

How many unpaired electrons does a sulfur atom have? What about an iron atom?

Step 1 · write sulfur's configuration

Sulfur is element 16, so it is [Ne] 3s2 3p4.

Step 2 · place the 3p electrons

Three boxes, four electrons. One per box uses three. The fourth pairs with one of them.

Boxes read paired, single, single. That gives 2 unpaired electrons, so sulfur is paramagnetic.

Step 3 · do the same for iron

Iron is element 26, so it is [Ar] 4s2 3d6. The 4s pair is paired.

Five d boxes take one each, which uses five. The sixth pairs up.

Boxes read paired, single, single, single, single. That gives 4 unpaired electrons.

Why it matters

Those four unpaired spins are why iron sticks to a magnet. Zinc, with 3d10, has none and does not.

Key points
  • An orbital diagram shows one box per orbital and one arrow per electron.
  • Pauli: no two electrons share all four quantum numbers, so a box holds two at most.
  • Paired electrons in one box must have opposite spins.
  • Hund: fill degenerate boxes singly with parallel spins before pairing any.
  • Unpaired electrons mean paramagnetic. All paired means diamagnetic.
Interactive Place electrons in boxes and obey Pauli and Hund. Open full ↑

One idea runs all four trends

Every trend in this section comes from a tug of war. The nucleus pulls the outer electrons in. The inner electrons get in the way and push back. The net pull that an outer electron actually feels is the effective nuclear charge, written Zeff. The blocking by inner electrons is shielding.

Move right across a period and protons pile up fast. Shielding barely changes, because the new electrons go into the same shell. So Zeff climbs and the outer electrons get held tighter. Move down a group and the valence electrons start a brand new shell. They sit farther out, with a thick layer of core electrons in between. The grip loosens.

Atomic radius

The atomic radius shrinks left to right and grows top to bottom. Across period 2 it falls from 167 pm at lithium to 42 pm at fluorine. Down group 1 it rises from 167 pm at lithium to 298 pm at cesium. Adding a shell always wins over adding protons.

Ions break the pattern in a useful way. A cation lost electrons, so the remaining ones feel a stronger pull per electron. A sodium atom is near 190 pm, while Na+ is about 102 pm. An anion gained electrons, which adds repulsion. Cl- is about 181 pm, much wider than the chlorine atom it came from. In an isoelectronic set, every ion has the same electron count. Then the one with the most protons is smallest, so the order runs Na+ < Ne < F-.

Atomic radius across period 2 and down group 1, with trend directions Atomic radius in picometers Across period 2, left to right LiBeBC NOFNe 1671128767 56484238 Down group 1, top to bottom LiNaKRbCs 167190243265298 up and to the right Z eff , ionization energy, electron affinity and electronegativity all rise down and to the left Atomic radius rises Radii come from one calculated set, so they can be compared with each other.
Two slices of the table. Going right across period 2, radius drops by a factor of four. Going down group 1, each new shell pushes the radius out again. The box at the bottom gives the direction of all four trends.

Ionization energy

The ionization energy is the energy needed to pull the loosest electron off a gaseous atom. It rises left to right and falls top to bottom. Sodium needs 496 kJ/mol, while argon in the same period needs 1521 kJ/mol. Cesium, far down group 1, needs only 376 kJ/mol.

Two small dips break the rise, and both are worth knowing. Boron (801 kJ/mol) is lower than beryllium (899 kJ/mol), because boron's electron leaves a higher-energy 2p orbital. Oxygen (1314 kJ/mol) is lower than nitrogen (1402 kJ/mol), because oxygen already has a pair crowding one 2p orbital. Those dips are the orbital diagrams showing up in measured data.

Removing a second electron always costs more. Sodium's first ionization energy is 496 kJ/mol. Its second is 4562 kJ/mol, roughly nine times as much. The second electron has to come out of the neon-like core. That huge jump is why sodium forms Na+ and never Na2+.

Worked example · from kJ per mole to joules per atom

Hydrogen's ionization energy is 1312 kJ/mol. How much energy is that for a single atom?

Step 1 · convert kilojoules to joules

1312 kJ/mol = 1.312 × 106 J/mol

Step 2 · divide by Avogadro's number

(1.312 × 106 J/mol) ÷ (6.022 × 1023 /mol) = 2.179 × 10-18 J per atom

Step 3 · compare with section 2

That is exactly the Bohr ground-state energy, 2.179 × 10-18 J. Lifting the electron from n = 1 to n = ∞ costs the same energy as ionizing the atom, because those are the same event.

Electron affinity and electronegativity

The electron affinity is the energy change when a gaseous atom accepts an electron. A negative value means energy is released, so the atom wants the electron. Chlorine is the champion at −349 kJ/mol, with fluorine close behind at −328 kJ/mol. Beryllium and the noble gases have positive values, because their subshells are already full and a new electron has nowhere good to sit.

Finally, electronegativity measures how hard an atom pulls on electrons it is sharing inside a bond. The Pauling scale runs from 0.79 at cesium to 3.98 at fluorine. It rises toward the upper right, just like ionization energy. Oxygen is 3.44, nitrogen is 3.04, carbon is 2.55 and hydrogen is 2.20. You will use those four numbers constantly in the next chapter to decide whether a bond is polar.

All four trends point the same way for a reason. Each one asks a version of the question “how tightly does this atom hold electrons?” So if you remember only one thing, remember that Zeff rises toward fluorine and falls toward cesium. Then you can rebuild every trend on the spot.
Key points
  • Zeff rises across a period and barely changes down a group.
  • Radius falls across a period and rises down a group. Cations shrink, anions swell.
  • Ionization energy rises across and falls down, with dips at group 13 and group 16.
  • Electron affinity is most negative near chlorine and fluorine.
  • Electronegativity runs from 0.79 at cesium to 3.98 at fluorine.
Interactive Watch radius, ionization energy and electronegativity move across the table. Open full ↑

The chapter in one breath

  • Light is a wave with c = λν, and it delivers energy in photons of size E = hν.
  • Line spectra prove that an atom's electron energies are quantized, not continuous.
  • Bohr's levels, En = −(2.179 × 10-18 J)/n2, give hydrogen's lines at 656, 486, 434 and 410 nm.
  • Four quantum numbers name an orbital and the electron in it: n, l, ml and ms.
  • Shapes follow l: s is a ball, p is two lobes, d is mostly four lobes, f is more complex.
  • Configurations follow the diagonal order, with Cr and Cu taking 4s1 to gain a half or full d set.
  • Pauli caps an orbital at two opposite spins, and Hund spreads electrons out before pairing them.
  • Unpaired electrons make an atom paramagnetic, which is a property you can measure with a magnet.
  • Zeff drives every periodic trend, so radius, ionization energy and electronegativity all follow one map.

ReferenceWords worth knowing

These are the terms this chapter leans on most. Come back to the list whenever a problem uses a word you want to pin down.

amplitude
The height of a wave. It controls brightness, not color or photon energy.
Aufbau principle
The rule that electrons fill the lowest open subshell first.
core electrons
The inner electrons hidden inside the noble gas bracket. They rarely take part in bonding.
degenerate orbitals
Orbitals in the same subshell, which all have equal energy.
effective nuclear charge
Written Zeff. The net pull an outer electron feels after inner electrons shield it.
electron affinity
The energy change when a gaseous atom gains one electron. Negative means energy is released.
electron configuration
The list of subshells with electron counts, such as 1s2 2s2 2p4.
electronegativity
How strongly an atom pulls on electrons it shares in a bond. Fluorine leads at 3.98.
frequency
Written ν. The number of wave crests passing a point each second, in s-1.
ground state
The lowest energy arrangement of an atom's electrons.
Hund's rule
Fill degenerate orbitals singly with parallel spins before any orbital gets a pair.
ionization energy
The energy needed to remove the loosest electron from a gaseous atom.
isoelectronic
Having the same number of electrons. Na+, Ne and F- are an isoelectronic set.
node
A place inside an orbital where the electron is never found.
orbital
A region around a nucleus where one electron is likely to be. It holds two electrons at most.
paramagnetic
Having one or more unpaired electrons, so a magnet attracts the sample.
Pauli exclusion principle
No two electrons in an atom may share all four quantum numbers.
photon
One packet of light energy, with E = hν.
quantized
Allowed only in fixed amounts, like coins rather than a smooth stream.
wavelength
Written λ. The distance from one wave crest to the next.

Check yourself

1. A blue photon has a wavelength of 450 nm. What is its frequency?
6.66 × 1014 s-1. First convert: 450 nm = 4.50 × 10-7 m. Then ν = c/λ = (2.998 × 108 m/s) ÷ (4.50 × 10-7 m) = 6.66 × 1014 s-1.
2. What is the energy of one of those 450 nm photons?
4.41 × 10-19 J. E = hν = (6.626 × 10-34 J·s)(6.66 × 1014 s-1) = 4.41 × 10-19 J. Scaled to a mole that is 266 kJ/mol.
3. The 434 nm hydrogen line comes from a drop that ends on n = 2. Where did it start?
At n = 5. The photon energy is hc/λ = (1.986 × 10-25 J·m) ÷ (4.34 × 10-7 m) = 4.577 × 10-19 J. Divide by 2.179 × 10-18 J to get 0.210, and 0.210 = 1/4 − 1/25. So the start was n = 5.
4. How many orbitals and how many electrons fit in the n = 3 shell?
9 orbitals and 18 electrons. l = 0, 1, 2 gives 3s, 3p and 3d. Orbitals: 1 + 3 + 5 = 9. Electrons: 2 × 9 = 18, which matches 2n2 = 2(9) = 18.
5. Is the set n = 3, l = 3, ml = 0 allowed?
No. The rule is l ≤ n − 1, so with n = 3 the only choices are l = 0, 1 or 2. There is no 3f subshell.
6. Write chromium's electron configuration and count its unpaired electrons.
[Ar] 4s1 3d5, with 6 unpaired electrons. Check the count: 18 + 1 + 5 = 24, and chromium is element 24. One electron sits alone in 4s and one sits in each of the five 3d orbitals, so all six spins are parallel.
7. Copper is element 29. Why is it [Ar] 4s1 3d10 instead of [Ar] 4s2 3d9?
A full 3d set is worth the move. The 4s and 3d levels are nearly equal in energy. Shifting one 4s electron into 3d completes the ten-electron d set and lowers repulsion, so the swapped version is lower in energy.
8. What is the electron configuration of Fe3+?
[Ar] 3d5. Iron is [Ar] 4s2 3d6. Electrons leave the highest n first, so both 4s electrons go, then one 3d electron. That leaves five 3d electrons, all unpaired.
9. How many unpaired electrons does a phosphorus atom have?
Three. Phosphorus is [Ne] 3s2 3p3. Hund's rule puts one electron in each of the three 3p orbitals with parallel spins, so phosphorus is paramagnetic.
10. Rank Mg, Si and Ca from largest to smallest atomic radius.
Ca > Mg > Si. Calcium is one period below magnesium, so it has an extra shell. Silicon sits to the right of magnesium in the same period, where higher Zeff pulls the cloud in tighter.
11. Oxygen's first ionization energy is 1314 kJ/mol, below nitrogen's 1402 kJ/mol. Why?
Oxygen already has a crowded 2p orbital. Nitrogen's 2p3 has one electron in each orbital. Oxygen's 2p4 forces one pair to share a box, and that repulsion raises the electron's energy. The crowded electron is easier to remove.
12. Hydrogen's ionization energy is 1312 kJ/mol. How many joules is that per atom?
2.179 × 10-18 J per atom. 1312 kJ/mol = 1.312 × 106 J/mol. Divide by Avogadro's number: (1.312 × 106) ÷ (6.022 × 1023) = 2.179 × 10-18 J. That is the Bohr ground-state energy, which is the point.