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 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

Practice 1 — 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.

Worked example 3 — solution

A common first attempt

That frequency belongs to radio waves, not ultraviolet light. The 250 went in as meters.

Worked example 3 — solution

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

Worked example 3 — solution

c is meters per second, so λ must be in meters: 250 nm × (1 m / 10⁹ nm) = 2.50×10⁻⁷ m.

Worked example 3 — frequency

Step 3 · Rearrange for the unknown

Solve for the unknown: ν = c/λ.

Worked example 3 — frequency

Practice 2 — answer: C

A left the wavelength in nanometers: 3.00×10⁸ ÷ 480 = 6.25×10⁵ s⁻¹, smaller by 10⁹. B put c on the bottom: 4.80×10⁻⁷ ÷ 3.00×10⁸ = 1.60×10⁻¹⁵ s, a time. D multiplied c by λ: 3.00×10⁸ × 4.80×10⁻⁷ = 1.44×10², with units m²/s.

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 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

Practice 1 — answer: C

A stopped at the frequency: ν = c/λ = 6.67×10¹⁴ s⁻¹, one step short of energy. B left the wavelength in nanometers: dividing by 450 instead of 450×10⁻⁹ m gives 4.42×10⁻²⁸ J, too small by 10⁹. D paired energy with wavelength: E = hλ = 2.98×10⁻⁴⁰ J·s·m, not joules.

Worked example 3 — solution

A common first attempt

650 is a bigger number than 325, so the red photon looks more energetic. But λ sits in the denominator of E = hc/λ, so a bigger λ makes a smaller E.

Worked example 3 — solution

Convert each wavelength to meters: 650 nm → 650×10⁻⁹ m, and 325 nm → 325×10⁻⁹ m. The unknown is which energy is larger.

Worked example 3 — solution

Both photons use E = hc/λ, with the same h and c. Only λ differs, and 325 nm is exactly half of 650 nm.

Worked example 3 — which carries more

Step 3 · Substitute and cancel units

Worked example 3 — which carries more

Step 3 · Substitute and cancel units

Worked example 3 — which carries more

Step 4 · Check the size

Practice 2 — answer: B

A reads λ backwards: the larger 600 nm sits in the denominator, so it gives the smaller energy, 3.31×10⁻¹⁹ J. C counts brightness as energy; brightness is the number of photons, not the energy of each. D squares the ratio: energy scales as 1/λ, so 600 / 200 = 3 times, not nine.

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 2 — solution

Step 1 · Identify nᵢ and n_f

Start at nᵢ = 4, end at n_f = 1.

Worked example 2 — solution

1/n_f² = 1/1² = 1.0000 and 1/nᵢ² = 1/4² = 0.0625.

Worked example 2 — solution

Worked example 2 — solution

Practice 1 — 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 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.

Worked example 1 — solution

Step 1 · n from the shell number

The digit in 3d is the shell: n = 3.

Worked example 1 — solution

Step 1 · n from the shell number
Step 2 · ℓ from the letter

d is the third subshell type, so ℓ = 2. Check the rule: ℓ ≤ n−1 = 2 ✓.

Worked example 1 — solution

Step 1 · n from the shell number
Step 2 · ℓ from the letter
Step 3 · mℓ from −ℓ to +ℓ Step 4 · mₛ = +½ or −½

mℓ may be any of −2, −1, 0, +1, +2. Choose one — say mℓ = +1 — and a spin, mₛ = +½.

Worked example 1 — solution

Step 1 · n from the shell number
Step 2 · ℓ from the letter
Step 3 · mℓ from −ℓ to +ℓ Step 4 · mₛ = +½ or −½

Worked example 2 — solution

Check ℓ against n

The rule is ℓ ≤ n−1. Here n = 2, so ℓ can only be 0 or 1 — the 2s and 2p subshells. There is no 2d.

Worked example 2 — solution

Check ℓ against n

To carry ℓ = 2 (a d subshell), an electron needs a shell that actually contains d — the first is n = 3. Change n to 3, or change ℓ to 0 or 1.

Practice 1 — answer: C

C fails ℓ ≤ n−1: with n = 1 the only allowed ℓ is 0, so there is no "1p." A is a legal 3d electron (ℓ = 2 needs n ≥ 3 ✓, mℓ = −2 is within −2…+2). B is a legal 2p electron. D is a legal 4s electron. Each of those obeys every rule.

Practice 2 — answer: A

An f subshell is ℓ = 3, which first appears at n = 4. A fits: ℓ = 3 ≤ n−1 = 3, and mℓ = −3 lies within −3…+3. B sets ℓ = 4, but ℓ can only reach n−1 = 3. C puts ℓ = 3 in n = 3, where ℓ stops at 2 — there is no 3f. D pushes mℓ = +4 past ℓ = 3.

Worked example 1 — solution

Step 2 · Nodal planes = ℓ

ℓ = 1, so the 3p orbital has one nodal plane — the flat surface through the nucleus that splits its two lobes.

Worked example 1 — solution

Step 2 · Nodal planes = ℓ

ℓ = 1, so the 3p orbital has one nodal plane — the flat surface through the nucleus that splits its two lobes.

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: B

A counted the 3 orbitals as electrons: a p sublevel has 3 orbitals, holding 3 × 2 = 6. C gave n² = 4² = 16, the orbitals in the whole n = 4 shell. D gave 2n² = 32, the electron capacity of the entire n = 4 shell, not one sublevel.

Worked example 3 — solution

Step 1 · Name the sublevel

The n = 4 shell holds 4 sublevels: 4s, 4p, 4d, 4f — with 1, 3, 5, and 7 orbitals.

Worked example 3 — solution

Double each sublevel's orbitals: 4s gives 2, 4p gives 6, 4d gives 10, 4f gives 14.

Worked example 3 — solution

Worked example 3 — solution

Practice 2 — answer: D

A used 2n = 2 × 3 = 6, forgetting to square the shell number. B added only 3s and 3p, 2 + 6 = 8, dropping the 3d sublevel. C gave n² = 3² = 9, the orbital count for the shell, not its electrons.

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 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

Practice 1 — answer: A

The superscripts add to 14, so the atomic number is 14: silicon. B, aluminum, has 13, one electron short: 2 + 2 + 6 + 2 + 1 = 13. C, carbon, shares the p² ending but sits a period higher, at 6: 2 + 2 + 2 = 6. D, beryllium, matches only the 4 valence electrons, not all 14.

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 2 — answer: A

After [Ar], 4s fills before 3d: 4s² then 3d³. B fills 3d before 4s; it sums to 23 (18 + 5) but breaks the order. C stops one electron short: 18 + 2 + 2 = 22, which is titanium. D skips 3d and fills 4p; it also sums to 23, but 3d is lower in energy than 4p.

Worked example — solution

Fe²⁺ · remove 2 electrons, highest n first

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

Worked example — solution

Fe²⁺ · remove 2 electrons, highest n first

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

Worked example — solution

Fe²⁺ · remove 2 electrons, highest n first

Worked example — chromium solution

Step 2 · What the order predicts

After [Ar] (18 electrons), fill 4s before 3d: 4s² then the last 4 into 3d.

Worked example — chromium solution

Step 2 · What the order predicts

Shift one 4s electron into 3d. Now 3d holds 5 — one electron in every d orbital, a half-filled subshell — and 4s holds 1.

Worked example — chromium solution

Step 2 · What the order predicts

Practice — answer: A

One 4s electron shifts into 3d, filling it to 10. B is the order's prediction (18 + 2 + 9 = 29) but not the observed ground state. C keeps 4s full and fills 3d — 18 + 2 + 10 = 30, one electron too many. D leaves 3d at 9 — 18 + 1 + 9 = 28, one short.

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 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

Practice 1 — 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 period 3 with chlorine and group 2 with calcium, 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

Practice 2 — answer: A

B ran the trend backwards: radius increases down a group, not up, so the lower halogen is larger. C picked iodine for the wrong reason: extra protons pull electrons in, never out — iodine is larger because it adds whole shells. D swapped in electronegativity: fluorine does grip electrons hardest, but that tight grip marks the smaller atom, not the larger.

Worked example 1 — solution

Step 2 · Draw the partly-filled sublevel

1s and 2s are full, their electrons quietly paired. The action is all in 2p: three boxes, four electrons.

Worked example 1 — solution

One electron goes into each of the three boxes first; the fourth has no empty box left and must pair up.

Worked example 1 — solution

Worked example 2 — solution

A common first attempt

Stripping three electrons from 3d and leaving 4s alone:

Worked example 2 — solution

A common first attempt

Take both 4s electrons, then one 3d electron: 2 + 1 = 3 removed.

Worked example 2 — solution

A common first attempt

Five electrons spread across the five 3d boxes, one apiece — not one is forced to pair.

Worked example 2 — solution

A common first attempt

Practice 1 — answer: C

Zn²⁺ has a completely filled 3d sublevel, so no electron is left unpaired. A, oxygen, has 2 unpaired 2p electrons. B, Fe³⁺ (3d⁵), has 5 unpaired electrons. D, sodium, has a lone 3s¹ electron. All three of those are paramagnetic.

Practice 2 — answer: C

A counts all six 3d electrons as if each were unpaired, ignoring the pair forced into the first box. B gives 5, the count for Fe³⁺ (3d⁵) — one electron fewer. D assumes the ion is fully paired, but six electrons cannot fill five boxes two at a time, so four stay single.