HardChemistryHigh School (11–12)The d- and f-Block Elements › Lanthanoids and Actinoids (f-Block)NumericalConceptualA-LevelIBNEET
Americium has the configuration [Rn] 5f⁷ 7s² and its highest known oxidation state is +6…
Contributed by StemPy LabTry it on paper first. A hint, the key idea and the worked solution are here when you need them.
Americium has the configuration and its highest known oxidation state is . Find the number of electrons outside the radon core and the number that do not appear in the highest oxidation state. Compare it with uranium, , whose highest state is .
Need a reminder? The key ideaShow
The actinoids fill 5f. All are radioactive. In each half-life, half of the sample decays: . In a nuclear equation the mass numbers add up on both sides, and so do the atomic numbers. A beta emission raises the atomic number by one and leaves the mass number unchanged.
The worked solution
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Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.
Step 1. Americium. electrons outside the radon core.
Step 2. The unused ones. The highest state is , so electrons stay out of bonding.
Step 3. Uranium. electrons, and its highest state is also . None is left over.
Answer: electrons and unused in americium. Uranium has electrons, and all of them are used.
Check. Uranium sits early in the row, and its 5f electrons reach outward. In americium the nuclear charge has pulled the seven 5f electrons inward, so most of them stay out of bonding. That matches the actinoid contraction.
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