Actinides — electronic configuration, comparison with lanthanides; nuclear chemistry tie-in: Worked examples

The scenario matrix

Before solving, let's list every case class this topic can present. Think of it like the quadrants of a graph — we must visit each one.

Cell	Case class	What makes it tricky	Covered by
A	Regular config (5f fills normally)	none — the baseline	Ex 1
B	Irregular config (6d "tourist")	half/full-filled 5f stability	Ex 2
C	Oxidation state — early actinide	high states (+5, +6) possible	Ex 3
D	Oxidation state — late actinide	collapses back to +3	Ex 3
E	Balance an α-decay	conserve mass A and charge Z	Ex 4
F	Balance a β⁻-decay	neutron→proton, Z rises, A fixed	Ex 4
G	Half-life, whole number of periods	just halve repeatedly	Ex 5
H	Half-life, fractional number of periods	must use the exponential	Ex 6
I	Degenerate: $t = 0$ and $t \to \infty$	sanity limits of the decay law	Ex 7
J	Find $λ$ or $t_{1/2}$ from data (inverse)	solve for the constant	Ex 8
K	Real-world word problem (dating)	translate story → equation	Ex 9
L	Exam twist: multi-step decay chain	combine ΔZ, ΔA over several steps	Ex 10

We now walk cells A→L. Watch how the same two ideas — conservation (config, mass, charge) and the exponential decay law — solve everything.

Cell A — Regular configuration

Worked example Write the configuration of

Pu (Z = 94). Forecast: guess before reading. Pu is 4 past Ac. Do you think it keeps a 6d electron or not?

Steps:

Start from the noble-gas core. Pu comes after Rn (Z = 86), so write [Rn]. Why this step? The 86 inner electrons are chemically dead — collapsing them saves us writing 86 terms.
Fill 7s first. $[Rn] 7 s^{2}$ uses 2 electrons (now at Z = 88, equivalent of Ra). Why this step? 7s sits lowest among the valence orbitals, so it fills before 5f/6d.
Fill 5f. We have $94 - 88 = 6$ electrons left. Pu is a "regular" filler (no half/full-shell bonus to exploit), so all 6 go into 5f: $5 f^{6}$ . Why this step? Once past Am, the 5f–6d energy gap opens enough that 5f is clearly preferred, so no 6d tourist.

Answer: $[Rn] 5 f^{6} 7 s^{2}$ .

Verify: electron count $= 86 + 6 + 2 = 94 = Z$ ✓. Matches the general formula $[Rn] 5 f^{1 - 14} 6 d^{0 - 1} 7 s^{2}$ with $6 d^{0}$ .

Cell B — Irregular configuration (the 6d tourist)

Worked example Write the configuration of

Cm (Z = 96) and explain the anomaly. Forecast: naively you'd dump all electrons into 5f. Cm has $96 - 88 = 8$ f-region electrons → guess $5 f^{8} 7 s^{2}$ ? Let's see why that's wrong.

Steps:

Core + 7s. $[Rn] 7 s^{2}$ → accounts for Z = 88. Remaining $= 8$ electrons. Why this step? Same baseline as always.
Spot the stability bonus. A half-filled $5 f^{7}$ (7 electrons, one per orbital, all parallel spins) is unusually stable — exchange energy is maximised. So the atom parks 7 electrons in 5f and lets the 8th electron sit in 6d instead of pairing up in 5f. Why this step? Nature minimises energy; keeping $5 f^{7}$ intact + one lone $6 d^{1}$ beats a crowded $5 f^{8}$ .
Assemble. $[Rn] 5 f^{7} 6 d^{1} 7 s^{2}$ .

Answer: $[Rn] 5 f^{7} 6 d^{1} 7 s^{2}$ — a genuine 6d tourist, exactly like the parent table.

Verify: $86 + 7 + 1 + 2 = 96 = Z$ ✓. The 6d appears because of $5 f^{7}$ half-filled stability, not by accident. Contrast: No (Z=102) has full $5 f^{14}$ , so no tourist → $[Rn] 5 f^{14} 7 s^{2}$ .

Cells C & D — Oxidation states, early vs late actinide

Worked example Predict the

highest common oxidation state of (C) U and (D) Cf (Z = 98). Forecast: U is famous for one particular ion in yellow-cake chemistry. Cf is deep in the row — will it match U or behave like a lanthanide?

Steps (U, cell C):

Count removable valence electrons. U $= [Rn] 5 f^{3} 6 d^{1} 7 s^{2}$ . That's $3 + 1 + 2 = 6$ electrons at similar, easily-reached energies. Why this step? Oxidation state = how many electrons you can strip; early-actinide 5f is spatially extended and poorly shielded, so all 6 are accessible.
Read off the max. Removing all 6 → +6, realised as the uranyl ion $UO_{2}^{2 +}$ . Why this step? +6 is the empirically dominant high state for U (parent note).

Steps (Cf, cell D): 3. Locate in the row. Cf is well past Am. By now the 5f electrons have sunk deeper (rising $Z_{eff}$ pulls them in), so they resist removal. Why this step? Late-actinide 5f becomes lanthanide-like — buried and quiet. 4. Read off the state. Only the outer electrons come off → +3 dominant. Why this step? Same reason lanthanides love +3.

Answers: U → +6; Cf → +3.

Verify: Fits the trend "early = variable up to +6/+7, late = collapse to +3." Charge check on uranyl: U(+6) with two O(−2) = $+ 6 - 4 = + 2 =$ charge on $UO_{2}^{2 +}$ ✓.

Cells E & F — Balancing decays (α and β⁻)

Worked example Balance (E) the

α-decay of $_{94}^{242} Pu$ and (F) the β⁻-decay of $_{92}^{239} U$ . Forecast: in α, which two numbers drop and by how much? In β⁻, which number moves — the top or the bottom?

Steps (α, cell E):

Recall the α rule. An α particle is $_{2}^{4} He$ : it carries away mass 4, charge 2. Why this step? Mass number $A$ and atomic number $Z$ must both be conserved (nucleons and charge don't vanish).
Subtract. Daughter mass $= 242 - 4 = 238$ ; daughter charge $= 94 - 2 = 92$ → element 92 is U. $_{94}^{242} Pu ⟶_{92}^{238} U +_{2}^{4} He$

Steps (β⁻, cell F): 3. Recall the β⁻ rule. A neutron converts to a proton + emitted electron $_{- 1}^{0} e$ . So $A$ stays the same, $Z$ goes up by 1. Why this step? The nucleus gains a proton but total nucleon count is unchanged (n→p is a swap, not a loss). 4. Adjust. Mass $= 239$ (unchanged); charge $= 92 + 1 = 93$ → element 93 is Np. $_{92}^{239} U ⟶_{93}^{239} Np +_{- 1}^{0} e$

Verify (E): top $238 + 4 = 242$ ✓, bottom $92 + 2 = 94$ ✓. Verify (F): top $239 + 0 = 239$ ✓, bottom $93 + (- 1) = 92$ ✓. Both sides balance in $A$ and $Z$ .

Cell G — Half-life, whole number of periods

Cell H — Half-life, fractional number of periods

Worked example The same

$^{240} Pu$ : how much of the 64 g remains after 10000 yr? Forecast: 10000 yr is between 1 and 2 half-lives. So the answer must lie between $64/2 = 32$ and $64/4 = 16$ g. Guess a number in that window before computing.

Steps:

Why we can't just halve. $t / t_{1/2} = 10000/6500 = 1.538$ — not a whole number, so "halve twice" undershoots and "halve once" overshoots. We need the continuous law. Why this step? Decay happens every instant, not in discrete jumps; the exponential is the exact description.
Use the fractional-power form. $N = N_{0} (\frac{1}{2})^{t / t_{1/2}} = 64 (\frac{1}{2})^{1.538}$ . Why this step? This is just $N_{0} e^{- λ t}$ rewritten with $λ = ln 2/ t_{1/2}$ — the cleanest form for fractional periods.
Evaluate. $(\frac{1}{2})^{1.538} = 0.3443$ , so $N = 64 \times 0.3443 = 22.04$ g.

Answer: ≈ 22.0 g remain.

Verify: $16 < 22.0 < 32$ ✓ (inside our forecast window). Units: g in, g out ✓.

Cell I — Degenerate limits: $t = 0$ and $t \to \infty$

Worked example Sanity-check the decay law

$N = N_{0} e^{- λ t}$ at the two extreme times. Forecast: at $t = 0$ nothing has decayed; at $t = \infty$ everything has. Does the formula obey these?

Steps:

Set $t = 0$ . $N = N_{0} e^{- λ \cdot 0} = N_{0} e^{0} = N_{0} \cdot 1 = N_{0}$ . Why this step? The exponent vanishes, so the sample is untouched — the definition of "start." Look at the left edge of the curve in the figure: it starts at full height $N_{0}$ .
Let $t \to \infty$ . Since $λ > 0$ , $e^{- λ t} \to 0$ , so $N \to 0$ . Why this step? Decay never quite finishes (it's asymptotic), but the population drains toward zero — the curve hugs the axis on the right.
Degenerate stable case ( $λ = 0$ ). Then $N = N_{0} e^{0} = N_{0}$ for all $t$ : a nucleus that never decays stays put. Why this step? Confirms the model reduces sensibly when there's no radioactivity.

Answer: boundary values $N (0) = N_{0}$ and $N (\infty) = 0$ ; stable limit gives a flat line.

Verify: $e^{0} = 1$ and $lim_{t \to \infty} e^{- λ t} = 0$ for $λ > 0$ ✓ — matches the physical picture.

Cell J — Inverse problem: find $t_{1/2}$ from data

Cell K — Real-world word problem (radiometric dating)

Worked example A rock contains

$^{238} U$ ( $t_{1/2} = 4.5 \times 1 0^{9}$ yr). Analysis shows only 12.5% of the original uranium is left. How old is the rock? Forecast: 12.5% = one-eighth. An eighth is three halvings. Multiply three half-lives... roughly $13 \times 1 0^{9}$ yr? Let's confirm.

Steps:

Translate the story. "12.5% left" means $N / N_{0} = 0.125 = \frac{1}{8} = (\frac{1}{2})^{3}$ . Why this step? Turn the percentage into a fraction remaining — the only quantity the decay law cares about.
Count half-lives. $(\frac{1}{2})^{t / t_{1/2}} = (\frac{1}{2})^{3} \Rightarrow t / t_{1/2} = 3$ . Why this step? Same base ⇒ equal exponents (as in Cell J).
Multiply. $t = 3 \times 4.5 \times 1 0^{9} = 1.35 \times 1 0^{10}$ yr. Why this step? Three full half-lives have elapsed.

Answer: the rock is $1.35 \times 1 0^{10}$ yr (13.5 billion years) old.

Verify: $(\frac{1}{2})^{3} = 0.125 = 12.5%$ ✓. Units: (yr per half-life) × (number of half-lives) = yr ✓.

Cell L — Exam twist: a multi-step decay chain

Worked example

$_{92}^{238} U$ decays by a chain of 8 α and 6 β⁻ emissions to reach a stable lead isotope. Find the final $A$ and $Z$ . Forecast: α removes mass, β⁻ doesn't. Which one drives the mass number down, and by how much total?

Steps:

Account for mass ( $A$ ). Only α changes $A$ , each by $- 4$ . So $Δ A = 8 \times (- 4) = - 32$ . Final $A = 238 - 32 = 206$ . Why this step? β⁻ conserves $A$ (n→p swap), so it contributes nothing to the mass count.
Account for charge ( $Z$ ) — the α part. Each α gives $- 2$ : $8 \times (- 2) = - 16$ . Why this step? α carries off 2 protons each.
Account for charge ( $Z$ ) — the β⁻ part. Each β⁻ gives $+ 1$ : $6 \times (+ 1) = + 6$ . Why this step? β⁻ turns a neutron into a proton, raising $Z$ by 1.
Combine. $Z = 92 - 16 + 6 = 82$ → element 82 is Pb (lead).

Answer: $_{82}^{206} Pb$ — the stable end of the uranium series.

Verify: $A : 238 - 32 = 206$ ✓. $Z : 92 - 16 + 6 = 82$ ✓. Lead-206 is indeed the known terminus of the $^{238} U$ chain.

Recap of the matrix

Recall Which example hit which cell?

A regular config ::: Ex 1 (Pu) B irregular / 6d tourist ::: Ex 2 (Cm) C early-actinide high ox. state ::: Ex 3 (U → +6) D late-actinide +3 collapse ::: Ex 3 (Cf → +3) E α-decay balance ::: Ex 4 (Pu-242) F β⁻-decay balance ::: Ex 4 (U-239) G whole-number half-lives ::: Ex 5 (8 g) H fractional half-lives ::: Ex 6 (≈22 g) I degenerate t=0 and t=∞ ::: Ex 7 J inverse: find t½ ::: Ex 8 (24 h) K word problem / dating ::: Ex 9 (13.5 Gyr) L multi-step chain ::: Ex 10 (Pb-206)

Connections

3.3.09 Actinides — electronic configuration, comparison with lanthanides; nuclear chemistry tie-in (Hinglish)
Lanthanides — 4f filling & lanthanide contraction
Oxidation States of Transition & Inner-transition Metals
Radioactivity — alpha, beta, gamma decay
First-order Kinetics & Half-life
Nuclear Stability & Binding Energy
Effective Nuclear Charge & Shielding