Well, kudos to you for studying it for yourself. I don't want to discourage you. The problem if you haven't got a tutor is that you may not know what's the best order to take things, what you need to know before going on to X, if you don't grasp something in the textbook there's nobody to explain it to you. So by all means make use of this forum, though it would help us if you made clear your limited chemistry background, so we know where you're at.
Let's try and have an approach at this. You have heard of the Schrodinger equation. Its solutions for the hydrogen atom (one electron) are characterised by four quantum numbers, n, l, m
l and m
s.
States with the same n have the same energy, which is proportional to -1/n
2.
l (the orbital angular momentum quantum number) can take integral values from 0 to n-1. These are labelled s, p, d, f... So you get 1s, 2s, 2p, 3s, 3p, 3d, etc.
m
l indicates the orientation of the orbital angular momentum vector with the z axis, and can take values -l, -l+1, ... l-1, l.
m
s indicates the orientation of the electron spin vector with the z axis, and can take values -1/2 or +1/2.
One way of looking at this is to see that the energy increases with the number of nodal surfaces (surfaces where the value of the wavefunction is zero), which is equal to n-1.
n=1 has no nodes, only the spherical distribution (1s orbital) is possible.
n=2 has one node; 2s has a radial node; 2p has an angular node (separating the + and - lobes of the p orbital).
n=3 has 2 nodes; 3s has 2 radial nodes, 3p has one radial and one angular, and 3d has two angular nodes.
See e.g.
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_A_Molecular_Approach_(Tro)/07%3A_The_Quantum-Mechanical_Model_of_the_Atom/7.6%3A_The_Shape_of_Atomic_OrbitalsSo much for hydrogen. Heavier atoms have multiple electrons, and the complexity of their interactions means that exact solutions to the Schrodinger equation can't be found, but they can be approximated to 1s, 2s, 2p etc. And an important point that emerges is that levels of the same n don't all have the same energy. For example, consider a 2s and a 2p orbital. The 2s wavefunction has non-zero amplitude at the nucleus, while the 2p has a node at the nucleus. That means that a 2s electron has greater probability than a 2p of being found very close to the nucleus, so it is less shielded from the nucleus by the 1s electrons, and therefore it has lower energy. This is called the "penetration" effect.
For higher shells, where the difference between the shell energies is smaller (remember it goes roughly as 1/n
2), this splitting between the s, p, d... levels can mean that (for example) 4s is lower than 3d. This is reflected in the electron configurations of the elements, and the structure of the periodic table.
However, these things aren't fixed in stone. For example, though 4s is filled before 3d, in the transition elements it is usually the 4s electrons that are ionised first, leaving a d
n configuration. Thus Ti is 4s
23d
2, but Ti
2+ is 3d
2, although the isoelectronic Ca is 4s
2. It seems that although all the levels change in energy with increasing nuclear charge, some change more than others, so going from Ca to Ti
2+, 3d drops below 4s. When an electron is removed from an atom, there is an adjustment in the energy of
all the other electrons, so the configuration is not necessarily "the same with one less electron in the top level". Cr is 4s
1 3d
5 rather than 4s
2 3d
4, because of the special stability of a half-filled d shell. I don't expect you to remember all these, but it just illustrates the complexity of the situation where different orbitals are of similar energy, and a variety of effects are at work in determining the configuration.