Quantum Contradictions 8: Boron Mystery Resolved

With the simple model of the previous posts we get for Boron B with 5 electrons the following ground state energy E_B  in a configuration with 2 electrons in an innermost shell of radius d_1= 1/5 followed by 2 electrons in a next shell of radius d_2 = 1/3 and 1 electron in an outer shell of radius d_3=1:

• E_B = - 2 x 5/d_1 + 2 x 1/2 x 1/d_1^2 + 1/2d_1 - 2 x 3/(d_1+d_2)+ 2 x 1/2 x 1/d_2^2 + 1/2(d_1+d_2) - 1/(d_1+d_2+d_3) + 1/2 x 1/d_3^2 = - 24.0

to be compared with observed - 24.85. Again a good correspondence with a main difference coming from a too small first ionization energy of - 0.15 instead of observed - 0.3.

The same model with instead 3 electrons in a second and outer shell, gives larger energy.

The model thus suggests a 2 + 2 + 1 configuration with 3 shells rather than a standard 2 + 3 configuration with 2 shells, but its too small first ionization energy suggests that the single outer electron is about to enter the into the second shell. The next element in the table is Carbon C with 6 electrons and the question to be tackled in the next post is if the configuration is 2 + 2 + 2 or the standard 2 + 4?

PS1 Notice that the ionization energy of Boron is smaller than that of Beryllium in contradiction with the standard Aufbau principle, which signifies that Boron has an outer 5th electron subject to small attraction from the kernel which could possibly be represented as a 2 + 2 + 1 configuration.  With a slight decrease of the distance d_1 + d_2 + d_3 the 5th electron gets a correct ionization energy.