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: 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 2 · 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

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

Practice 1 · 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.

Practice 2 · 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: 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: D

A stopped at the photon's energy, one step short of λ. B did not square n: 2.18×10⁻¹⁸ J × (1/1 − 1/7) = 1.87×10⁻¹⁸ J gives 106 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.

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

Guided example: solution

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

n = 3 names shell 3. ℓ = 1 is the letter p, and ℓ ≤ n−1 = 2 ✓. So far, a 3p electron.

Guided example: solution

A p subshell allows mℓ = −1, 0, +1 only. mₛ = +½ is fine, but one broken rule forbids the set.

Guided example: solution

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 1 · answer: C

Practice 2 · answer: A

Statement 1 holds: f means ℓ = 3, and mℓ may be any whole number from −3 to +3. Statement 2 fails: ℓ = 1 is p (s = 0, p = 1), so n = 5, ℓ = 1 is a 5p electron. Statement 3 holds: n = 4 allows ℓ = 0, 1, 2, 3, the first shell with an f subshell.

B rejected statement 1, capping mℓ at ℓ − 1 = 2 as if mℓ followed the n − 1 rule. C accepted statement 2, pairing ℓ = 1 with s instead of p. D rejected statement 3, stopping n = 4 at d as if f first appeared in n = 5.

Practice 2 · answer: A

Practice 3 · answer: E

E passes the ℓ check at its limit, then fails |mℓ| ≤ ℓ. Reading mℓ's range from n (−3 to +3) misses the break. A caps |mℓ| at ℓ − 1 = 1, but a 4d orbital allows mℓ up to +2. B treats mₛ = −½ as illegal, but a 7s electron may spin down. C stops ℓ short of n − 1, but ℓ = 3 in shell 4 is a 4f electron. D treats a negative mℓ as illegal, but 5d allows −2 to +2.

Practice 3 · answer: E

Worked example 1: solution

Step 1 · Read n and ℓ

The digit gives n = 3; the letter p gives ℓ = 1.

Worked example 1: solution

Nodal planes = ℓ = 1. This is the flat surface splitting the two lobes. Every p orbital has exactly 1.

Worked example 1: solution

Worked example 1: solution

Worked example 2: solution

A common first attempt

No planes, so no nodes? Nodal planes are only one kind of node. The total is set by n, not by ℓ.

Worked example 2: solution

n = 3 and ℓ = 0: the s orbital truly has 0 nodal planes.

Worked example 2: solution

Worked example 2: solution

Practice 1 · answer: B

A stopped at the total, n − 1 = 6 − 1 = 5, and never took out the 2 planes. C dropped the minus one: n − ℓ = 6 − 2 = 4. D read n = 6 itself as the count.

Practice 1 · answer: B

Guided example: solution

Step 1 · Read n and ℓ

Both numbers are unknown. The counts supply them: the planes give ℓ, and the total gives n.

Guided example: solution

Two planes name the letter d. Planes plus radial nodes give the total, and the total is one less than n.

Guided example: solution

Practice 2 · answer: A

B read the total 6 as n, skipping the plus one: 6p has 6 − 1 − 1 = 4 radial nodes but only 6 − 1 = 5 nodes in all. C subtracted instead of adding: n − 1 = 6 gives n = 6 + 1 = 7, not 6 − 1 = 5, and 5p has 5 − 1 − 1 = 3 radial nodes. D stopped at the total and called all 6 nodes radial: 1 of them is the plane, so 6 − 1 = 5 are radial.

Practice 2 · answer: A

Practice 3 · answer: D

A drew the p shape (ℓ = 1) and counted electrons as orbitals: 5 orbitals × 2 = 10 electrons. B has the shape, but 10 is the electron capacity and 7 is n itself, not n − 1 = 6. C drew the p dumbbell and skipped the minus one: 7 nodes.

Practice 3 · answer: D

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.

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

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: 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.

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: 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 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

Practice 1 · 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 1 · Write the predicted form

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

Worked example 1: solution

Step 1 · Write the predicted form

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

Worked example 1: solution

Step 1 · Write the predicted form

Worked example 1: the route

Worked example 2: solution

Step 1 · Write the predicted form

After [Ar], the order gives 4s², then 9 electrons into 3d.

Worked example 2: solution

Step 1 · Write the predicted form

One 4s electron fills 3d to 10: a completely filled subshell.

Worked example 2: solution

Step 1 · Write the predicted form

Worked example 2: the route

Guided example: solution

Step 1 · Count the electrons

The sheet's superscripts total 26, iron's count. The building-up order gives 4s², then 6 electrons into 3d.

Guided example: solution

Step 1 · Count the electrons

Iron's predicted ending is 3d⁶, one past half full. Step 2 says no, so no electron moves. The sheet moved one anyway.

Guided example: solution

Step 1 · Count the electrons

Iron keeps the predicted form: 18 + 2 + 6 = 26 ✓. The sheet's 4s¹3d⁷ is not the ground state.

Guided example: solution

Step 1 · Count the electrons

Practice 1 · answer: C

A copied chromium: 18 + 1 + 5 = 24 electrons, one short of manganese. B over-applied the move: 18 + 1 + 6 = 25 counts right, but 3d⁶ is neither half full nor full, so the 4s electron gains nothing by moving. D moved both 4s electrons: 18 + 7 = 25, but 4s never drops below 4s¹, and 3d⁷ is no steady point.

Practice 1 · answer: C

Practice 2 · answer: D

A misread zinc: the plain order already fills its 3d, so nothing needs to move and Zn keeps 4s². B over-applied the rule: only a d subshell one short of half or full borrows; Zn, like Ni and V, keeps 4s². C gave copper the wrong reason: copper's move fills 3d to 3d¹⁰; the half-filled 3d⁵ is chromium's story.

Practice 2 · answer: D

Practice 3 · answer: A

A moved an electron titanium's 3d² never calls for. B is correct: picking it trusts the prediction, but 4s¹3d⁵ is chromium's real exception, 18 + 1 + 5 = 24 ✓. C is correct: picking it over-applies the rule; Ni's 3d⁸ is not one short of full, so 4s² stays, 18 + 2 + 8 = 28 ✓. D is correct: picking it trusts the prediction again; copper's move gives 4s¹3d¹⁰, 18 + 1 + 10 = 29 ✓. E is correct: picking it carries copper's 4s¹ on to zinc, whose plain order already fills 3d, 18 + 2 + 10 = 30 ✓.

Practice 3 · answer: A

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 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 2 · 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.

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

Two electrons in one orbital sit closest and repel most; an empty orbital is the cheaper seat, so pairing is the last resort. 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 1: the route

Worked example 1: the route

Guided example: solution

Step 1 · Write the configuration

Build neutral vanadium first. The 2+ ion then loses its two highest-n electrons: the 4s pair.

Guided example: solution

Step 1 · Write the configuration

3d has five boxes and three electrons. Each electron takes an empty box, so none pairs. Three lone arrows remain.

Guided example: solution

Step 1 · Write the configuration

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

Worked example 2: the route

Worked example 2: the route

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 1 · answer: C

Practice 2 · answer: B

A counted the full 3d too: 10 + 3 = 13, but ten electrons in five boxes sit in five pairs. C counted every electron past [Ar]: 2 + 10 + 3 = 15, but only the 4p electrons stand alone. D counted all five valence electrons, 4s²4p³: 2 + 3 = 5, but the 4s pair is matched.

Practice 2 · answer: B

Practice 3 · answer: A

B removed 3d first: [Ar]4s²3d⁵, 18 + 2 + 5 = 25, five singles, but the 4s pair leaves before any 3d electron. C stopped after the 4s pair: that is Ni²⁺, [Ar]3d⁸, 10 − 8 = 2 unpaired, but a 3+ ion lost three. D paired before every box had one: 3d⁷ as ↑↓ ↑↓ ↑↓ ↑ shows 7 − 6 = 1, but Hund's rule seats five singles before any pair. E counted every 3d electron, 28 − 3 − 18 = 7, but four of them sit in two pairs.

Practice 3 · answer: A