which is just like the \(Ne\), Neon element. From this we can obtain a ethene, \(C_2H_4\) by unpairing two paired to obtain four unpaired orbits. To each of these unpaired orbits, a hydrogen atom is bonded to it. We can image that the horizontal bond that constitute part of the double bond being stretched when the two \(C\) nuclei cannot be concentric, and the vertical bond not effecting the spacial arrangement of the two bonded Hydrogen atoms on each side. As far as a single \(C\) nucleus is concerned three bonds (a double bond and two bonds with hydrogen) around it distribute themselves coplanar separated by \(120^o\).
This suggests a stronger double bond than a pair of single bonds filling the adjacent vertices of a tetrahedron being twisted into a parallel double bonds.
But why stop at double,
This also suggests a much stronger and stable triple bonds than three twisted single bonds from the corners of a tetrahedron. A single paired orbit is unpaired and one hydrogen atom bonds with each unpaired orbit. The resulting structure is linear.
The new interpretation of Quantum Numbers gives flexibility to bond formation as the number of orbits around a nucleus can be reduced by reducing \(n\). The remaining orbits are unpaired to provide for the number of radicals around the molecules.
For once not twisted! And among those paired, double and triple bonds, democracy! All equal. No need for Hybridization.
