According to molecular orbital theory, two atomic orbitals form two molecular orbitals analogous to waves combining constructively or destructively but how can a wave combine destructively and constructively at the same time to produce two molecular orbitals?
Why don't two atomic orbitals combine to form a molecular orbital that can be BMO or ABMO?
Answer
Well, one important thing to keep in mind is that atomic and molecular orbitals are not "real" things, they are mathematical constructions that are useful in helping understand and predict how electrons behave in atoms and molecules. They are like a set of rules that tell us all of the ways in which it is possible for the electrons with a particular set of quantum numbers to behave. For example, a H-atom in the ground state has only one single electron in a $1s$ orbital (quantum numbers: $n=1$, $l=0$, $m_l=0$), but we also know that if the H-atom absorbs some energy, the electron can end up in a higher electronic state, for example the $3\mathrm p_z$ orbital (quantum numbers: $n=3$, $l=1$, $m_l=0$).
Now, when you are dealing with molecular orbital theory, you are talking about the possible ways in which electrons in atomic orbitals can combine in a covalent interaction. So it's not that the two atomic orbitals combine "constructively and destructively at the same time", it's just that those are the two ways in which its possible for the two orbitals to combine: in the region where they overlap, they can either add their amplitudes, or subtract them. The "constructive" case where they add produces an molecular orbital that is lower in energy (that is to say, an MO that describes a state where the electrons will have lower energies) than the "destructive" case, so by the aufbau principle, electrons will occupy the lower energy "bonding" orbital first. In a case like $\ce{H2}$, where there are only two electrons, only the bonding orbital is occupied, but because the Pauli exclusion principle tells us we can only have two electrons in a given orbital, if you add another electron (as in $\ce{He2+}$), it will occupy the higher-lying "anti bonding" orbital.
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