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

What: state the template. $[\text{Rn}]5f^{1\text{–}14}6d^{0\text{–}1}7s^2$ Why the ranges: the $5f$ shell fills from 1 to 14 electrons; a lone $6d$ electron sometimes appears (0 or 1) because $5f$ and $6d$ sit at almost the same energy; $7s$ is filled first and stays at 2. Boundaries: lowest is Th, $Z=90$; highest is Lr, $Z=103$. That is $103-90+1=14$ elements — one for each $5f$ electron. ✓

What is $\alpha$ decay: the nucleus ejects a helium-4 cluster ${}_2^4\text{He}$ (2 protons + 2 neutrons). ${}_{92}^{238}\text{U}⟶{}_{90}^{234}\text{Th}+{}_2^4\text{He}$ Check (the two conservation laws):

• Top numbers (mass): $234+4=238$ ✓
• Bottom numbers (charge): $90+2=92$ ✓ Why Th and not something else: dropping $Z$ by 2 moves us two boxes left on the periodic table, from U (92) to Th (90).

What: $48=4\times 12$, so exactly 4 half-lives pass — perfect for the halving form. $N=N_0{\left(\frac{1}{2}\right)}^{t/t_{1/2}}=96{\left(\frac{1}{2}\right)}^{48/12}=96{\left(\frac{1}{2}\right)}^4=\frac{96}{16}$ $N=6\text{ g}$ Why the halving form here: when $t$ is a clean multiple of $t_{1/2}$, we avoid $\lambda$ entirely and just halve four times: $96\to 48\to 24\to 12\to 6$. Same answer, less arithmetic.

• Th: $[\text{Rn}]6d^{2}7s^2$
• U: $[\text{Rn}]5f^{3}6d^{1}7s^2$ Why Th skips $5f$: at the very start of the row, the $6d$ level lies slightly below $5f$, so the first two "extra" electrons prefer $6d$. Only from Pa onward does $5f$ start winning. Count check for U: $[\text{Rn}]$ = 86 electrons, plus $5f^3+6d^1+7s^2=3+1+2=6$, total $86+6=92=Z$. ✓

(a) What fraction is left: $25/100=1/4=(1/2{)}^2$, so 2 half-lives passed. (b) Why divide: 2 half-lives took 8 hr, so one takes $t_{1/2}=\frac{8}{2}=4\text{ hr}.$ (c) Why $\ln 2$ appears: the decay constant links to the half-life through $t_{1/2}=\ln 2/\lambda$, so $\lambda =\frac{\ln 2}{t_{1/2}}=\frac{0.6931}{4}\approx 0.1733{\text{ hr}}^{-1}.$ Meaning of $\lambda$: about a 17% chance per hour that any given nucleus decays.

What contraction is: ionic/atomic radius shrinks as we move right across the row. Why it happens: each new electron goes into an $f$-orbital. See Effective Nuclear Charge & Shielding — $f$-orbitals are diffuse and non-penetrating, so an $f$-electron is a poor shield. Each added proton is only weakly cancelled, so the net pull $Z_{\text{eff}}$ felt by outer electrons keeps rising → the cloud is pulled inward. Why bigger than lanthanides: $5f$ orbitals are even more diffuse than $4f$, i.e. even worse shields. Poorer shielding → $Z_{\text{eff}}$ climbs faster → each step shrinks a little more. Prediction: the late-actinide $M^{3+}$ is smaller — it has more protons plus all the extra poorly-shielding $5f$ electrons, so its outer cloud is squeezed harder.

Step 1 — $\alpha$ (mass $-4$, charge $-2$): ${}_{95}^{241}\text{Am}⟶{}_{93}^{237}\text{Np}+{}_2^4\text{He}$ Check: $237+4=241$ ✓, $93+2=95$ ✓. Step 2 — ${\beta }^{-}$ (mass unchanged, charge $+1$): a neutron becomes a proton, ejecting an electron ${}_{-1}^{0}e$. ${}_{93}^{237}\text{Np}⟶{}_{94}^{237}\text{Pu}+{}_{-1}^{0}e$ Check: mass $237=237$ ✓, charge $94+(-1)=93$ ✓. Final nuclide: ${}_{94}^{237}\text{Pu}$, so $\boxed{Z=94,A=237}$. Why this order matters for $Z$: $\alpha$ dropped $Z$ by 2 (95→93); ${\beta }^{-}$ raised it by 1 (93→94). Net $Z$ change $=-1$.

Find the oxidation state: oxygen is $-2$ each, and the whole ion is $+2$. Let U's state be $x$: $x+2(-2)=+2\Rightarrow x-4=2\Rightarrow x=+6.$ So U is $+6$ in ${\text{UO}}_2^{2+}$ (the uranyl ion). Why U can reach $+6$: in actinides the $5f$, $6d$ and $7s$ levels are all close in energy and the $5f$ electrons are spatially extended (poorly held). So several electrons ($7s^2$, then $6d^1$, then $5f$'s) can be removed at comparable cost — see Oxidation States of Transition & Inner-transition Metals. Why Nd (a lanthanide) can't: its $4f$ electrons are buried deep and tightly bound (well shielded from the outside, but experiencing strong pull). After losing the two $6s$ and one $5d/4f$ electron to reach $+3$, pulling out a core-like $4f$ electron costs far too much. Hence lanthanides park at $+3$.

Why we can use activity like amount: activity $\propto N$ (both fall as $e^{-\lambda t}$), so ratios of counts behave exactly like ratios of nuclei — see First-order Kinetics & Half-life. (a) $100/800=1/8=(1/2{)}^3$ → 3 half-lives in 15 hr. (b) $t_{1/2}=15/3=5$ hr. (c) A further 10 hr $=2$ more half-lives, starting from 100 counts/s: $A=100{\left(\frac{1}{2}\right)}^{10/5}=100{\left(\frac{1}{2}\right)}^2=\frac{100}{4}=25\text{ counts/s.}$ As a fraction of the original 800: $25/800=1/32=(1/2{)}^5$ (a total of 5 half-lives over the full 25 hr) ✓.

What each does in 6 yr:

• A: $6/2=3$ half-lives → survives $(1/2{)}^3=1/8$.
• B: $6/6=1$ half-life → survives $(1/2{)}^1=1/2$. Why we can just take ratios: both start from the same $N_0$, so $\frac{N_A}{N_B}=\frac{N_0(1/8)}{N_0(1/2)}=\frac{1/8}{1/2}=\frac{1}{8}\cdot \frac{2}{1}=\frac{1}{4}.$ $\boxed{N_A:N_B=1:4}$ Meaning: the faster-decaying isotope (shorter $t_{1/2}$) is far more depleted — after 6 years only a quarter as many A atoms survive as B atoms.

Cm — why the $6d$ tourist: a half-filled $5f^7$ set (one electron in each of the seven $f$-orbitals, all spins aligned) is an especially stable, low-energy arrangement (exchange stabilisation). Rather than pair up and spoil that half-filled shell, the incoming electron prefers the nearby, empty $6d$ orbital. No — why no $6d$: by $Z=102$, the $5f$ shell can be completely filled ($5f^{14}$). A full shell is even more stable than a half-filled one, and once $5f$ is full there's no half-filled bonus to protect — the electrons simply complete $5f$ and leave $6d$ empty. Count: $86+14+2=102=Z$ ✓. The unifying idea: the $5f$/$6d$ energies are so close that tiny extra stabilisations (half-filled at Cm, fully-filled at No) decide where the last electron lands — the same principle links the whole irregular row (Pa, U, Np, Cm all show a $6d^1$).

Alpha decay changes $A$ by ::: $-4$ and $Z$ by $-2$ Beta-minus decay changes $Z$ by ::: $+1$ (mass number unchanged) $N$ after $n$ whole half-lives ::: $N_0(1/2{)}^n$ Link between $\lambda$ and $t_{1/2}$ ::: $t_{1/2}=\ln 2/\lambda$ Oxidation state of U in ${\text{UO}}_2^{2+}$ ::: $+6$ Why Cm shows $6d^1$ ::: to keep the stable half-filled $5f^7$