Question

How does `""^238U` undergo `alpha-"decay"` to give `""^234Th`? Is this a chemical reaction? How does uranium metal react with fluorine?

Answer 1

Well, it is a nuclear reaction, rather than a chemical reaction........

Explanation:

And what we know for chemical reactions is that `"mass"` and `"charge"` are conserved. But for this `"NUCULAR"` reaction, `"MASS"` may not be conserved.

And thus we may write the `alpha"-particle"` decay of `""^(238)U` (the `"92"` subscript described in the question is superfluous, why?).

`"(i)"` `""^238U rarr ""^234Th +""^4He`

And for `"(ii)"` we have a `"de jure"` chemical reaction, when fluorine gas, the most reactive element on the Periodic Table, oxidizes uranium metal.

`"(ii)"` `U(s) +3F_2(g) rarr UF_6(s)`

Historically, reaction (ii) is significant, in that the product `UF_6` is reasonably volatile. Thus this enables some means to separate the isotopomeric uranium products as their volatile fluorides. I do not think `""^238U` is fissile, but for better info you should consult a text.