Why is there no stable nucleus with two or more neutrons but no protons?

Question

Why is there no stable nucleus with two or more neutrons but no protons?

Answer 1

Two bound neutrons would not really be called a nucleus as there is no way for an atom to form, as the pair would have a neutral electric charge and be unable to bind with electrons in the usual way.

The terms for a two-neutron state is dineutron. This is a resonance state that is very short lived. It was reported as having been found in an experiment in 2012, although I'm not aware if this is widely accepted. It is very difficult to detect the events that indicate such a resonance state existed. The lifetime of such a state is about $10^{-22}s$ which puts it well below the accepted threshold for being called an element - that's at least $10^{-14}s$.

Answer 2

Without neutrons, the SEMF for the binding energy $E_B$ for a nucleus of multiple neutrons simplifies to$$E_B=-(a_A-a_V)N-a_SN^{2/3}+\delta_0[2|N],$$where the last term uses an Iverson bracket, i.e. is absent for odd $N$. In megaelectronvolts$$a_A-a_V\approx23.2-15.8=7.4>0,\,a_S\approx18.3,\,\delta_0\approx12N^{-1/2}$$according to the first least-squares fit summarized here. Regardless of the values you trust, $a_A>a_V$, and $\delta_0$ is too small for $N\ge2$ to achieve $E_B\ge0$ (well, at least some values suggest $N=2$ is only unachievable because we also have to worry about the surface term proportional to $a_S$). So in short, the asymmetry term is too large compared with the volume term.