two proton orbits with two protons and two electrons attracted to each orbit by both the positive end of a weak \(E\) fields and the positive charge, proton, at the negative end of the weak \(E\) field. This weak field is produced by a spinning \(T^{+}\) particle.
And so, the bond energy of a single chemical covalent bond is...
\(U_{E}=-\cfrac{q}{4\pi\varepsilon_o}\left(\cfrac{1}{d}+\cfrac{1}{2r}-\cfrac{1}{\sqrt{d^2+4r^2}}\right)\)
where \(q=\left| e^{+} \right|=\left| p^{+} \right| \), that both particles have equal charge.
\(U_{wE}=-\cfrac{q_{T^{+}}}{4\pi\varepsilon_o}.\cfrac{1}{r}\)
where \(q_{T^{+}}\) is the equivalent charge that would establish the weak \(E\) field due to the spinning \(T^{+}\) particle on the next lower layer of the nucleus. It is expected that,
\(q_{T^{+}}\lt q^{+}\)
Although the particles are orbiting, they are static with respect to one another. The weak field holds the particles in place and the whole field rotates as the plane of the orbit of the \(T^{+}\) particle spins. In this case, bond energy is taken to be the energy required to remove of any one of the four particles.
\(U_{cov}=U_{E}+U_{wE}=-\cfrac{1}{4\pi\varepsilon_o}\left\{q\left(\cfrac{1}{d}+\cfrac{1}{2r}-\cfrac{1}{\sqrt{d^2+4r^2}}\right)-\cfrac{q_{T^{+}}}{r}\right\}\)
where \(U_{cov}\) is the covalent bond energy. Given that,
\(d\lt\lt r\)
\(U_{cov}\) is largely dependent on \(\frac{1}{d}\); that the separation \(d\) between the paired orbits determines bond energy and not the bond length taken as \(\small{2r}\).
\(U_{B}=0\)
For this reason, the bond is not radiating \(\small{EMW}\).
Given the positions of the negative charges, there might be measurable magnetic moment as the opposing fields do not cancel completely. In general, any forces changing the positions of the charge particles will result in a change in the magnetic moment of the bond.
Note: Chemical here implies electron and proton (the charge particles pair) interaction.
