Actinides — electronic configuration, comparison with lanthanides; nuclear chemistry tie-in: Question bank

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Figure — Actinides — electronic configuration, comparison with lanthanides; nuclear chemistry tie-in

Formula Deriving

$N = N_{0} e^{- λ t}$ and $t_{1/2} = \frac{ln 2}{λ}$ from scratch WHAT & WHY: each nucleus decays independently with per-nucleus probability $λ$ per unit time, so the number decaying per unit time is $λ$ times how many are present. That is the rate law: $\frac{d N}{d t} = - λ N$ (minus sign: $N$ shrinks). To solve, separate the variables — put all $N$ on one side, all $t$ on the other — then integrate both sides: $\int \frac{d N}{N} = - λ \int d t \Rightarrow ln N = - λ t + C$ At $t = 0$ we have $N = N_{0}$ , so $ln N_{0} = C$ . Substitute back and exponentiate both sides (undoing the $ln$ ): $ln N - ln N_{0} = - λ t \Rightarrow ln \frac{N}{N _{0}} = - λ t \Rightarrow N = N_{0} e^{- λ t}$ The exponential appears because the rate is proportional to the amount left — constant fractional loss per unit time is exactly what $e^{- λ t}$ describes. For the half-life, ask "when is $N = N_{0} /2$ ?": $\frac{1}{2} = e^{- λ t_{1/2}} \Rightarrow - ln 2 = - λ t_{1/2} \Rightarrow t_{1/2} = \frac{ln 2}{λ}$ Notice $N_{0}$ cancelled — half-life is a fixed property of the nuclide, not of sample size. The parallel to chemical first-order kinetics is in First-order Kinetics & Half-life.

Figure s02 shows the two energy pictures behind the configuration traps below: the small 5f–6d gap in early actinides and the level crossings that produce curium's and nobelium's 6d behaviour.

True or false — justify

The actinide series begins at actinium (Ac, Z = 89).

False — Ac is the reference element before the series; the 14 actinides run Th (90) to Lr (103), just as the lanthanides run Ce (58) to Lu (71) after La. See Lanthanides — 4f filling & lanthanide contraction.

Every actinide is radioactive.

True — all have unstable, overcrowded nuclei where proton–proton repulsion outweighs the strong force's reach, so all decay; this is a genuine "all", unlike lanthanides where only promethium is radioactive.

Actinide contraction is smaller than lanthanide contraction because 5f is farther out.

False — it is slightly larger; the diffuse 5f electrons (Figure s01) shield even worse than 4f, so

Z_{eff}

on the outer electrons climbs faster and pulls them in more per step.

Only elements beyond uranium (Z > 92) are radioactive.

False — Th, Pa and U are also radioactive; "transuranium" (Z > 92) just marks where elements are essentially all man-made and absent from nature in usable amounts, not where radioactivity begins.

All actinides show a +3 oxidation state.

True in effect — +3 is systematically favoured because removing the two 7s and one 6d/5f electron leaves a stable

M^{3 +}

core (a compact, well-screened 5f set) whose ionisation is energetically cheap across the whole row; early actinides additionally reach +4 to +7, but +3 is the common floor everyone shares.

The 6d electron in curium (Cm,

[Rn] 5 f^{7} 6 d^{1} 7 s^{2}

) proves 6d is lower in energy than 5f there.

False — the driver is the extra stability of the half-filled

5 f^{7}

(see the Cm crossing in Figure s02); the atom keeps seven electrons in 5f and parks the "spare" in 6d, so 6d hosting an electron is a consequence of 5f stability, not of 6d being lower.

Half-life depends on how much sample you start with.

False — in the derivation above

N_{0}

cancels, leaving

t_{1/2} = ln 2/ λ

with only the decay constant

λ

; it is a fixed property of the nuclide, so 100 g and 1 g of the same isotope both halve in the same time.

In alpha decay the mass number drops by 4 and the atomic number by 4.

False — the trap is the second "4": an alpha particle is

_{2}^{4} He

, so mass number falls by 4 but atomic number falls by only 2; both the top numbers and bottom numbers must balance across the arrow.

Spot the error

"Th is

[Rn] 5 f^{2} 7 s^{2}

because it is the first 5f element."

Wrong — thorium is

[Rn] 6 d^{2} 7 s^{2}

with no 5f electrons; at the very start of the row 6d still lies below 5f (left edge of Figure s02), so Th behaves almost like a d-block atom.

"Because 5f is buried deep like 4f, actinides also stick to +3 only."

Wrong premise — 5f is more spatially extended and higher in energy than 4f (close to 6d/7s), so its electrons are removable; that is exactly why actinides span +3 to +7 while lanthanides don't.

N = N_{0} e^{- λ t}

so after one half-life

N = N_{0} e^{- λ}

Error — you must plug in

t = t_{1/2} = ln 2/ λ

, giving

N = N_{0} e^{- l n 2} = N_{0} /2

; leaving

t = 1

silently invents units and gives the wrong fraction.

_{92}^{238} U \to_{92}^{234} Th +_{2}^{4} He

Charge doesn't balance — losing an alpha removes 2 protons, so the product is

_{90}^{234} Th

(Z = 90), not Z = 92; check that both sub- and superscripts sum equal on each side.

"Lanthanide contraction and shielding are unrelated ideas."

Wrong — the contraction is caused by poor f-shielding: each added f-electron shields badly, so

S

barely grows while

Z

keeps rising,

Z_{eff} = Z - S

climbs, and radii shrink. The two are one mechanism.

"Beta decay lowers the mass number, so it also moves you along the actinides."

Wrong — beta (β⁻) decay converts a neutron to a proton: mass number is unchanged, atomic number rises by 1. Only alpha changes A. Compare in Radioactivity — alpha, beta, gamma decay.

5 f^{14}

is unstable, so No (

[Rn] 5 f^{14} 7 s^{2}

) grabs a 6d electron."

Backwards — the filled

5 f^{14}

is especially stable (the No crossing in Figure s02), which is precisely why nobelium keeps 6d empty rather than promoting an electron into it.

Why questions

Why can actinides form more covalent complexes than lanthanides?

Their 5f orbitals reach farther out (Figure s01) and overlap better with ligand orbitals, so bonding is less purely ionic; buried 4f orbitals in lanthanides barely overlap, giving weak, ionic interactions.

Why do early actinides reach high oxidation states (+6, +7) but later ones don't?

Early on, 5f and 6d are close in energy and poorly held, so several electrons are removable; as Z rises the growing

Z_{eff}

pulls 5f deeper and tighter, so beyond Am the chemistry reverts toward +3. See Oxidation States of Transition & Inner-transition Metals.

Why is alpha the preferred decay for heavy actinides rather than shedding lone protons?

The alpha particle (

_{2}^{4} He

) is an exceptionally tightly bound cluster, so ejecting it releases the most energy per event — it is the cheapest route toward a more stable nucleus. See Nuclear Stability & Binding Energy.

Why is decay described by a first-order rate law,

d N / d t = - λ N

Each nucleus decays independently with the fixed per-nucleus probability

λ

, so the total rate is simply proportional to how many are present — the defining signature of first-order kinetics.

Why do 5f electrons shield worse than 4f electrons?

The 5f cloud is more diffuse and non-penetrating (its density peaks farther out, Figure s01), so it spends less time between the nucleus and the outer electrons and screens the nuclear charge less effectively.

Why is the energy gap between the filling f-orbital and the neighbouring d-orbital smaller in actinides than lanthanides?

The 5f orbitals are raised close to 6d in energy (Figure s02), whereas 4f sits well below 5d; this small 5f–6d gap is the single fact behind the irregular configs and variable oxidation states.

Edge cases

What is the configuration and oxidation-state behaviour at the very first member, Th?

[Rn] 6 d^{2} 7 s^{2}

(zero 5f electrons), and it strongly favours +4 by losing all four outer electrons — a d-block-like edge of the row.

What happens to oxidation-state variety at the tail end (No, Lr)?

It collapses to essentially +3 (with +2 notable for No); the deeply sunk, tightly held

5 f^{14}

core no longer offers electrons for higher states.

For a nuclide with an extremely long half-life, what does

λ

look like and why does the sample seem "stable"?

A huge

t_{1/2}

means a tiny

λ

(since

t_{1/2} = ln 2/ λ

), so the decay rate

λ N

is minute and almost nothing changes on a human timescale — yet it is still radioactive.

t = 0

what does the decay law give, and does it make physical sense?

N = N_{0} e^{0} = N_{0}

; at the start no time has passed, so the full initial amount remains — the formula correctly anchors to

N_{0}

t \to \infty

, does

N

ever reach exactly zero?

No —

e^{- λ t}

approaches but never equals 0, so mathematically some tiny amount always "remains"; physically the last individual nucleus decays at some finite random moment.

Lr is sometimes written

[Rn] 5 f^{14} 7 s^{2} 7 p^{1}

instead of

6 d^{1}

— why the ambiguity?

Relativistic effects on the heaviest atoms lower 7p until it competes with 6d for the last electron (the near-degenerate 6d/7p pair at the far right of Figure s02), so the ground-state assignment is genuinely debated rather than a textbook error.

Connections

Lanthanides — 4f filling & lanthanide contraction
Effective Nuclear Charge & Shielding
Oxidation States of Transition & Inner-transition Metals
Radioactivity — alpha, beta, gamma decay
First-order Kinetics & Half-life
Nuclear Stability & Binding Energy