Worked example 1: solution

Step 1 · Identify the unknown

c and λ are given; the frequency ν is the unknown.

Worked example 1: solution

c is meters per second, so put λ in meters: 600 nm × (1 m / 10⁹ nm) = 6.00×10⁻⁷ m.

Worked example 1: solution

ν = c/λ.

Worked example 1: solution

Worked example 1: the route on the map

Worked example 2: solution

Step 1 · Identify the unknown

c and ν are given; the wavelength λ is the unknown.

Worked example 2: solution

Put the frequency in s⁻¹: 100.0 MHz = 100.0 × 10⁶ s⁻¹ = 1.00×10⁸ s⁻¹.

Worked example 2: solution

λ = c/ν.

Worked example 2: solution

Worked example 2: the route on the map

Practice 1 · answer: A

B let λ and ν rise together; with c fixed, the longer wave has the lower frequency. C reversed the colors; ROY G BIV runs from red at 700 nm to violet at 400 nm. D read one speed as one frequency; each color has its own λ, so its own ν.

Practice 2 · answer: B

A left the wavelength in nanometers: 3.00×10⁸ ÷ 400. = 7.50×10⁵ s⁻¹, smaller by 10⁹. C multiplied c by λ: 3.00×10⁸ × 4.00×10⁻⁷ = 1.20×10², with units m²/s. D put c on the bottom: 4.00×10⁻⁷ ÷ 3.00×10⁸ = 1.33×10⁻¹⁵ s, a time.

Practice 3 · answer: C

A left the frequency in MHz: 3.00×10⁸ ÷ 315 = 9.52×10⁵ m, a wave nearly a thousand kilometers long. B put c on the bottom: 3.15×10⁸ ÷ 3.00×10⁸ = 1.05, with units m⁻¹. D used 10³ for mega: 3.00×10⁸ ÷ 3.15×10⁵ = 952 m. Mega means 10⁶.

Practice 4 · answer: D

A left the frequency in GHz: 3.00×10⁸ ÷ 5.00 = 6.00×10⁷ m, which is 6.00×10⁹ cm, a wave longer than the Earth. B put c on the bottom: 5.00×10⁹ ÷ 3.00×10⁸ = 16.7, with units m⁻¹. C stopped halfway: 3.00×10⁸ ÷ 5.00×10⁹ = 0.0600 is the wavelength in meters; the last hop, × (100 cm / 1 m), was never done.

Worked example 1: solution

Step 1 · Identify the given

ν = 5.00×10¹⁴ s⁻¹ is a frequency, already in s⁻¹. There is no wavelength to convert. The unknown is E.

Worked example 1: solution

A frequency is given, so E = hν gives the energy directly.

Worked example 1: solution

s and s⁻¹ cancel, leaving joules.

Worked example 1: solution

Worked example 1: the route on the map

Three lamps: answer B

A ranked energy rising with wavelength, the E = hλ slip; λ sits in the denominator of E = hc/λ. C took the heat lamp's warmth for photon energy: a heat lamp warms by sending many weak photons. D mixed up brightness and energy: wavelength fixes each photon's energy, and brightness counts photons.

Energy from a frequency: answer A

B divided instead of multiplying: 6.626×10⁻³⁴ / 6.52×10¹⁴ = 1.02×10⁻⁴⁸, in J·s², not joules. C divided by c as well: hν/c = 1.44×10⁻²⁷. D put the frequency where λ goes: hc/ν = 3.05×10⁻⁴⁰.

Worked example 2: solution

Step 1 · Identify the given

Convert the wavelength to meters so it matches c: 400 nm × (1 m / 10⁹ nm) = 400×10⁻⁹ m. The unknown is E.

Worked example 2: solution

A wavelength is given. Frequency and wavelength are tied by ν = c/λ, so E = hν becomes E = hc/λ.

Worked example 2: solution

J·s × m/s leaves J·m; dividing by m leaves J.

Worked example 2: solution

Worked example 2: the route on the map

A neon sign: answer C

A left λ in nanometers: hc/640. = 3.11×10⁻²⁸. B paired energy with wavelength: hλ = 4.24×10⁻⁴⁰. D flipped the factor: 640. nm × (10⁹ m / 1 nm) = 6.40×10¹¹ m, so hc/λ = 3.11×10⁻³⁷.

Wavelength from an energy: answer D

A skipped the last unit hop: 5.52×10⁻⁷ is the wavelength in meters, not nanometers. B flipped the conversion factor: 5.52×10⁻⁷ m × (1 m / 10⁹ nm) = 5.52×10⁻¹⁶, and meters do not cancel. C dropped c: ν = E/h = 5.43×10¹⁴ s⁻¹, then λ = 1/ν = 1.84×10⁻¹⁵, which is 1.84×10⁻⁶ after the hop; λ = c/ν needs c on top.

Wavelength from an energy: the route on the map

The energy gap: answer B

A stopped at E_B = 9.47×10⁻¹⁹ J and never subtracted E_A. C subtracted the wavelengths first: 630. − 210. = 420. nm, then hc/(4.20×10⁻⁷ m) = 4.73×10⁻¹⁹ J; subtract energies, not wavelengths. D left both λ in nm: hc/210. − hc/630. = 6.31×10⁻²⁸ J. E subtracted backwards: E_A − E_B = −6.31×10⁻¹⁹ J.

Worked example 1: solution

Step 1 · Identify nᵢ and n_f

Start at nᵢ = 3, end at n_f = 2. The final level is lower, so a photon leaves.

Worked example 1: solution

1/n_f² = 1/2² = 0.2500 and 1/nᵢ² = 1/3² = 0.1111.

Worked example 1: solution

Worked example 1: solution

Worked example 1: the jump on the ladder

Practice 1 · answer: D

Longest wavelength means least energy, which means the smallest drop.

B flipped the relation: 6 → 2 is the biggest drop in the set, so it has the most energy and the shortest wavelength. A and C are also bigger drops than 3 → 2.

Practice 1: the jumps on the ladder

Practice 2 · answer: B

Only jumps that land on n = 2 (the Balmer series) fall in the visible range.

A (2 → 1) is a Lyman jump: ultraviolet. C (4 → 3) and D (5 → 4) land on n = 3 or higher: infrared. Only B lands on n = 2.

Practice 2: the jumps on the ladder

Practice 3 · answer: A

Shortest wavelength means highest energy, which means the biggest (1/n_f² − 1/nᵢ²).

B, C, and D all land on n = 2 (visible): 3→2 = 3.03×10⁻¹⁹ J, 4→2 = 4.09×10⁻¹⁹ J, 5→2 = 4.58×10⁻¹⁹ J: each smaller than A, so each a longer wavelength.

Practice 3: the jumps on the ladder

Practice 4 · answer: B

A did not square n: −2.18×10⁻¹⁸ J × (1/2 − 1/6) = −7.27×10⁻¹⁹. C stopped at E_final: −5.45×10⁻¹⁹ J is the energy of n = 2, not the jump. D dropped the double negative: −5.45×10⁻¹⁹ − 6.06×10⁻²⁰ = −6.06×10⁻¹⁹. E reported the photon's energy, or subtracted initial − final: ΔE for emission is negative.

Practice 4: the jump on the ladder

Practice 5 · answer: D

A stopped at the photon's energy, one step short of λ. B used the n = 1 level's energy alone: 2.18×10⁻¹⁸ J drops the 1/7² term and gives 91.2 nm. C skipped the m to nm hop: 9.31×10⁻⁸ is meters. E flipped the factor: 9.31×10⁻⁸ m × (1 m / 10⁹ nm) = 9.31×10⁻¹⁷, and meters do not cancel.

Practice 5: the jump on the ladder

Practice 5: the route on the light map

Worked example 1: solution

Step 1 · Name the sublevel

A p sublevel always has 3 orbitals. The 3 in front is the shell number and does not change that count.

Worked example 1: solution

Worked example 1: solution

Worked example 2: solution

A common first attempt

Five is the orbital count, not the electron count. Every orbital holds two electrons, so the two numbers cannot be equal.

Worked example 2: solution

A d sublevel has 5 orbitals, two more than a p sublevel.

Worked example 2: solution

Worked example 2: solution

Practice 1 · answer: A

B doubled the orbitals: 7 × 2 = 14 is the electron count, but the question asks for orbitals. C gave n² = 5² = 25, the orbitals in the whole n = 5 shell. D gave 2n² = 50, the electron capacity of the whole n = 5 shell.

Practice 2 · answer: D

A counted orbitals, not electrons: 1 + 4 + 1 + 3 = 9 boxes. B stopped before the 3p: 2 + 8 + 2 = 12 leaves out the three 3p electrons. C skipped the 3s: 2 + 8 + 3 = 13 drops the full 3s pair the stem names.

Guided example: solution

Step 2 · Fill in the building-up order

The order starts 1s, 2s, 2p. Hydrogen's one electron enters 1s; helium's second fills it.

Guided example: solution

With 1s full, the next two electrons fill 2s.

Guided example: solution

Guided example: solution, continued

Step 3 · Fill sublevels to capacity

With 1s and 2s full, boron and carbon start 2p, which holds 6.

Guided example: solution, continued

Guided example: solution, continued

Worked example 1: solution

Step 2 · Fill in the building-up order

Start at the lowest sublevel and work up: 1s, then 2s, then 2p.

Worked example 1: solution

1s holds 2 and 2s holds 2. The remaining 4 electrons go into 2p, which holds up to 6.

Worked example 1: solution

The superscripts total 8, matching oxygen's 8 electrons.

Worked example 1: the route on the table

Worked example 2: solution

A common first attempt

Continuing straight from 3p into 3d:

Worked example 2: solution

The order places 4s before 3d. After 3p⁶, fill 4s, then 3d.

Worked example 2: solution

4s takes 2; the last 6 go into 3d, which holds up to 10. The superscripts total 26.

Worked example 2: solution

Worked example 2: the route on the table

Nickel: the route on the table

Practice 1 · answer: A

The superscripts add to 7, so the atomic number is 7: nitrogen. B, oxygen, has one electron too many: 2 + 2 + 4 = 8. C, phosphorus, shares the p³ ending but sits a period lower, at 15: 2 + 2 + 6 + 2 + 3 = 15. D, boron, matches only the 5 valence electrons, 2 + 3 = 5, not all 7.

Practice 2 · answer: D

After 3p⁶, 18 electrons are placed and 1 remains; it enters 4s. A put it in 3d: 2 + 2 + 6 + 2 + 6 + 1 = 19 counts right, but 4s is lower in energy than 3d. B wrote 3p⁷: 19 again, but a p sublevel holds only 6. C filled 4s to capacity: 2 + 2 + 6 + 2 + 6 + 2 = 20, calcium.

Worked example 3: solution

Step 2 · Fill in the building-up order

Fill through 3p⁶, reaching 18 electrons, then 4s before 3d.

Worked example 3: solution

4s takes 2; the last 2 go into 3d.

Worked example 3: solution

1s²2s²2p⁶3s²3p⁶ is argon, so it becomes [Ar]. The electrons still total 22.

Practice 3 · answer: C

Argon is the noble gas before bromine. After [Ar], 4s fills, then 3d to 10, then 4p. A used krypton, the noble gas after bromine: 36 + 2 + 10 + 5 = 53. B skipped 3d: 18 + 2 + 5 = 25. D filled 4p to capacity: 18 + 2 + 10 + 6 = 36, krypton itself.

Practice 4 · answer: B

The count passes, so only the order flags B: after 4s², 3d takes all 10 before any electron enters 4p. A is correct: Sc, 18 + 2 + 1 = 21, with 4s filled before 3d. C is correct: Cl, 10 + 2 + 5 = 17, and [Ne] is the noble gas before Cl. D is correct: F, 2 + 2 + 5 = 9; a last sublevel may be partly filled.

Guided example: solution

Step 2 · Count the charge

O²⁻ gained 2 electrons and F⁻ gained 1. Mg²⁺ lost 2.

Guided example: solution

The anions add in the filling order and finish 2p. Magnesium's highest level is n = 3, so its 3s pair leaves.

Guided example: solution

Practice 1 · answer: D

A added two electrons for the 2+ charge: 38 + 2 = 40, continuing into 4d. B removed only one: 38 − 1 = 37, which is Sr⁺. C removed nothing: [Kr]5s² is neutral strontium, 38 electrons.

Practice 2 · answer: B

A gained nothing: [Ne]3s²3p⁴ is neutral sulfur, 16 electrons. C removed two electrons for the 2− charge: 16 − 2 = 14. D put the gained pair in 4s: 10 + 2 + 4 + 2 = 18 counts right, but 3p still had room.

Worked example 1: solution

Step 1 · Write the neutral atom Step 2 · Count the charge

Neutral iron is [Ar]4s²3d⁶. Fe²⁺ has lost 2 electrons; Fe³⁺ has lost 3.

Worked example 1: solution

The highest level is n = 4, holding 4s². Empty it, both electrons, before any 3d leaves.

Worked example 1: solution

With 4s already empty, the next-highest occupied level is 3d. Take one electron from it.

Worked example 1: solution

Worked example 1: sublevel by sublevel

Practice 3 · answer: A

B took both electrons from 3d, the last filled: 18 + 2 + 6 = 26 counts right, but 4s (n = 4) empties first. C added two electrons for the 2+ charge: 28 + 2 = 30. D removed nothing: [Ar]4s²3d⁸ is neutral nickel, 28 electrons.

Practice 4 · answer: C

A read the 2 in Mn₂ as the charge: [Ar]3d⁵ is Mn²⁺, 18 + 5 = 23 electrons. B took all three from 3d: 18 + 2 + 2 = 22 counts right, but 4s (n = 4) empties first. D put the whole 6− on one Mn: Mn⁶⁺, 25 − 6 = 19 electrons, [Ar]3d¹. E removed nothing: that is neutral manganese, 25 electrons.

Worked example 1: solution

Step 2 · Fill the boxes lowest-energy first

1s takes a pair, 2s takes a pair. Three electrons remain for the three 2p boxes.

Worked example 1: solution

Step 2 · Fill the boxes lowest-energy first
Step 3 · Within a sublevel, singly first

Three boxes, three electrons: one in each, spins parallel; none pairs.

Worked example 1: solution

Step 2 · Fill the boxes lowest-energy first
Step 3 · Within a sublevel, singly first

The highest occupied shell is n = 2, holding 2s² and 2p³. The 1s pair is the core.

Practice 1 · answer: C

B counts the whole atom: 17 is every electron, core included. A reads only the last superscript: 3p⁵ gives 5, dropping the 3s pair in the same shell. D counts the core, 2 + 2 + 6 = 10, the electrons that stay out of the chemistry.

Practice 2 · answer: B

A read group 14 straight, skipping the 10 that groups 13 to 18 count past; counting the filled 5d¹⁰ as valence lands on the same 10 + 2 + 2 = 14. C read only the last superscript, 6p², and dropped the 6s pair. D read the period, 6: it numbers the highest shell, not its electrons.

Practice 3 · answer: A

B read the superscript of 2p⁴ as four lone arrows, but three boxes hold only three singles. C paired first, ↑↓ ↑↓ with one box empty, which breaks Hund's rule. D counted every valence electron as single, 2 + 4 = 6, but the 2s pair is matched.

Practice 4 · answer: D

The drawing pairs the two 2p electrons while two empty 2p boxes wait. That breaks Hund's rule. Spread them out, one per box with parallel spins, and both sit alone.

A named the wrong rule: Pauli limits a box to two opposite spins, and ↑↓ obeys it. B accepted the drawing and its count of 0. C counted every valence electron as single, 2 + 2 = 4, but the 2s pair is matched.

Worked example 3: solution

Step 2 · Fill the boxes lowest-energy first

4s takes 2 of the 5 before any 3d box opens, and the last 5 − 2 = 3 go to 3d.

Worked example 3: solution

Step 2 · Fill the boxes lowest-energy first

The five 3d boxes share one energy, so the three electrons spread out, one per box with parallel spins.

Worked example 3: solution

Step 2 · Fill the boxes lowest-energy first

Practice 5 · answer: D

A counted all 25 − 18 = 7 electrons past [Ar] as single, the group number too, but the 4s pair is matched. B paired before every box held one: ↑↓ ↑↓ ↑ leaves 5 − 2 × 2 = 1. C put all 7 into 3d before 4s: [Ar]3d⁷ pairs 2 of them, leaving 5 − 2 = 3.

Iron's ions, box by box

Iron's ions, box by box

Worked example 1: solution

Step 1 · Place the two elements

Sodium and sulfur share period 3, so their outer electrons occupy the same n = 3 shell. One trend decides it.

Worked example 1: solution

Across a period, atomic radius decreases: the nuclear charge grows while the shell stays the same.

Worked example 1: solution

Worked example 1: solution

Worked example 1: the path on the table

Worked example 2: solution

A common first attempt

Potassium holds 19 protons to lithium's 3, so its grip looks stronger. But those extra protons sit buried under two extra shells. Distance and shielding, not raw charge, set the pull on the outer electron.

Worked example 2: solution

Lithium and potassium share group 1. Their outer electrons sit in different shells: n = 2 for lithium, n = 4 for potassium.

Worked example 2: solution

Down a group, ionization energy decreases: each new shell holds the outer electron farther out, better shielded.

Worked example 2: solution

Worked example 2: the path on the table

Practice 1 · answer: C

A let more protons mean a bigger atom: fluorine's extra proton pulls the same shell in tighter. B carried the fluorine memory hook over to radius, the one trend that runs away from fluorine. D had the right atom for the wrong reason: both use the n = 2 shell, so the difference is nuclear charge, not shell height.

Practice 2 · answer: A

B let raw charge win: calcium's +20 nucleus sits under an extra shell, so its outer electrons are farther out and better shielded. C tied size to grip the wrong way: the larger atom holds its outer electrons more loosely. D had the right atom for the wrong reason: each neutral atom holds as many electrons as its nucleus has protons, so the count is no advantage. The difference is the shell: n = 3 against n = 4.

Practice 3 · answer: C

A ran the trend backwards: electronegativity increases, not decreases, across a period, so the right-hand atom wins. B confused the size trend: boron is the larger atom, but a larger atom pulls a shared pair less, not more. D reached for extra shells that are not there: both atoms bond through the n = 2 shell, and oxygen wins by nuclear charge, not shell count.

Worked example 3: two comparisons

Step 1 · Place the two elements

Magnesium is the corner. It shares group 2 with calcium and period 3 with chlorine, so two single comparisons cover all three.

Worked example 3: two comparisons

Worked example 3: the ranking

Step 3 · Apply it

Worked example 3: the ranking

Worked example 3: the path through the corner

Practice 4 · answer: B

A let raw charge win: sulfur's +16 nucleus sits under one more shell than fluorine's +9, so it holds its outer electrons more loosely. C had the right atom for the wrong reason: the electron count is not the cause. Fluorine's outer electrons sit in n = 2, closer in, and across a period the pull climbs. D ran the group trend backwards: moving down a group lowers the ionization energy.

Practice 4: the path through the corner

Practice 5 · answer: C

A reversed both trends: down group 1 the new n = 4 shell makes K the larger of K and Na, and across period 3 P's +15 draws the same shell in, so P is the smallest; Z > Y > X is the correct list read backwards. B ran the group trend backwards: K sits a row below Na and opens a new shell, so K is the larger. D let Z's stronger nucleus push electrons out: a stronger pull draws the same shell inward, so P is the smallest.

Worked example 1: solution

Step 1 · Count the electrons

The nucleus holds +12 in both. The atom carries 12 electrons; the ion carries 10, and its n = 3 shell is gone entirely.

Worked example 1: solution

Same element, fewer electrons: each remaining electron feels the +12 pull with less company, and the outermost shell dropped from n = 3 to n = 2.

Worked example 1: solution

Worked example 1: solution

Worked example 2: solution

Step 1 · Count the electrons

Every species holds 10 electrons: the series is isoelectronic. Electron count cannot separate them.

Worked example 2: solution

Same electron count, different nuclei: +8, +9, +11, +12, +13. The weakest nucleus holds its 10 electrons loosest, so oxygen's ion is the largest.

Worked example 2: solution

Worked example 2: solution

Practice 1 · answer: A

B ran the pull backwards: more protons draw the same 18 electrons inward, so +20 gives the smallest ion, not the largest. C ranked the parent atoms (K 227 > Ca 197 > S 104 > Cl 99 pm): the atoms' sizes are irrelevant once all four species hold the same 18 electrons. D repeated the equal-size trap: isoelectronic means equal electron count, and the differing nuclear charges still set different radii.

Practice 2 · answer: D

A read the highest charge as the largest ion: Sr²⁺ has the most protons of the 36-electron trio, +38, so it grips its 36 electrons tightest and is the smallest (118 pm). B carried over the parent atom: rubidium is a large atom, but Rb⁺ lost its whole n = 5 shell. C ranked the isoelectronic trio and stopped: Br⁻ beats Rb⁺ and Sr²⁺, but I⁻ sits a row lower with an extra shell, and an anion only adds to that size.