Beryllium atom pair energies

The critical choice was made of a CASSCF(4,4)/6-31G calculation the active space is thus the degenerate filled 2s + 2s and 2s 2s pair of MOs, and the degenerate empty 2px + 2px and 2px — 2px pair of MOs. CASSCF(4,4) was chosen because it corresponds to the CASSCF(2,2) calculation on one beryllium atom in the sense that we are doubling up the number of electrons and orbitals in our noninteracting system. This calculation gave an energy of 29.1709451 hartree. We can compare this with twice the energy of one beryllium atom, 2 x — 14.5854725 hartree = —29.1709450 hartree. 

An impressive success of the lEPA is its prediction of the ground state energy of the beryllium atom. The spin-orbital pair correlation energies obtained by... 

The increase to 900 kJ mol" in the case of the beryllium atom is due to the increase in effective nuclear charge, offset by interelectronic repulsion of the two 2s electrons. The electron most easily removed is one of the pair in the 2s orbital. In the case of the boron atom, in spite of an increase in nuclear charge there is a decrease in the first ionization energy to 799 kJ mol. This is because the electron removed is from a 2p orbital, which is higher in energy than the 2s level. 

Table 2.6 shows the electron affinities, for the addition of one electron to elements in Periods 2 and 3. Energy is evolved by many atoms when they accept electrons. In the cases in which energy is absorbed it will be noted that the new electron enters either a previously unoccupied orbital or a half-filled orbital thus in beryllium or magnesium the new electron enters the p orbital, and in nitrogen electron-pairing in the p orbitals is necessary. 

The fourth electron in an atom of beryllium. Be, must complete the pair in the 2s sublevel because this sublevel is of lower energy than the 2p sublevel. With the 2s sublevel filled, the 2p sublevel, which has three vacant orbitals of equal energy, can be occupied. One of the three p orbitals is occupied by a single electron in an atom of boron, B. Two of the three p orbitals are occupied by unpaired electrons in an atom of carbon, C. And all three p orbitals are occupied by unpaired electrons in an atom of nitrogen, N. Hund s rule applies here, as is shown in the orbital notations in Figure 3.5. 

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