For One For All

Does a $Ne$ nucleus exist?  Yes, the paired orbits all without opposite particles, each generates a weak $g$ field that sums to the resultant represented by the dotted arrow.

At the negative end of this weak field, a $g^{+}$ particle spins as the nucleus spins.  This is the start of the nucleus cyclic set ($g^{+}$, $T^{+}$, $p^{+}$), as the next layer to the nucleus.

All $n$ solutions can have a resultant weak field extending beyond the nucleus.