or,
In the first case, the last electron together with the two existing electron drives sn1, sn3 and p+ in a \(B\) orbit. In the second case, the last electron send both electrons along with sn1, sn3 and p+ in to its own \(B\) orbit. In the former case the electrons are shared, removing an electron from the configuration would have to bring the resulting configuration into an higher energy state, requiring energy input. In the latter case, removing the last electron collapses the \(B\) field and leave behind a positively charged sn(1,3,+). The change in energy in the \(B\) field is,
\(\Delta U=0-\cfrac{1}{2\mu_o}B^2\lt0\);
energy is released. The remaining positive charge on sn(1,3,+) will resist the change, but when the atom is bonded to others in a lattice by sharing electrons, the residue charge is effectively reduced, which makes it easier to remove the last electron. Sodium is more reactive bonded to a lattice.
The lattice of Sodium,
which the same as Lithium with a free electron and each atom is bonded to two others in the lattice.
