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Title: No atom binds two extra electrons
Plain: Experiments find no stable atom carrying two extra electrons. Prove from the Schrodinger equation that $N_c(Z) \le Z + 1$ for every element.
Precise: With $H_{N,Z}$ and $N_c(Z)$ as in mathph.ionization.excess-charge-bound (static nucleus, nonrelativistic, spin $1/2$ electrons), prove $N_c(Z) \le Z + 1$ for all $Z \ge 1$, or exhibit $Z$ with $N_c(Z) \ge Z + 2$. A computer-assisted proof for small $Z$ combined with an asymptotic argument for large $Z$ would count.
Settled by: A proof that $E(N, Z) = E(N-1, Z)$ for every $N \ge Z + 2$ and every $Z$, i.e. that no ion with two or more extra electrons has a bound ground state.
Kind: well-posed
Answer type: proof
References: (empty)
Literature status: open
Status note: Lieb's bound $N_c < 2Z + 1$ gives this statement only for hydrogen ($Z = 1$); no stable doubly charged atomic negative ion has been observed.
Posed since: (empty)