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Now, in the other answer, there is some discussion about spin-orbitals, meaning that each electron would exist in its own orbital. For practical purposes, you don't need to worry about that - by the time those sorts of distinctions matter to you, there won't be any confusion about what people mean by "shells" and "sub-shells."
3. The orbitals are solutions (also called wave functions) to the (time-independent) Schrödinger equation. The solutions are related to Bessel functions and Legendre polynomials. The drawing of orbitals as you often find in text books is in a sense the Copenhagen interpretation of quantum mechanics. In this interpretation, the wave functions ...
s-orbitals can hold 2 electrons, the p-orbitals can hold 6 electrons. Thus, the second shell can have 8 electrons. The n=3 (third) shell has: The 3s orbital; The 3p orbitals; The 3d orbitals; s-orbitals can hold 2 electrons, p-orbitals can hold 6, and d-orbitals can hold 10, for a total of 18 electrons. Therefore, the formula $2n^2$ holds!
These orbitals have different shapes (e.g. electron density distributions in space) and energies (e.g. 1s is lower energy than 2s which is lower energy than 3s; 2s is lower energy than 2p). (image source) So for example, a hydrogen atom with one electron would be denoted as $\ce{1s^1}$ - it has one electron in its 1s orbital
Electron as standing wave: You can make standing waves with a slinky toy spring, and it takes more energy to get more nodes. Your diagram with n=8 is quite high in energy, but you can probably get n=2. You can say where the spring is going to move for each standing wave and make the analogy that the spring is the electric field and the movement ...
Since we usually fill electrons in the order of increasing energy, the next electron (in case of manganes) goes into the $\mathrm{4s}$-orbital. The same reason for effective nuclear charge makes the $\mathrm{3d}$-orbitals somewhat lower in energy than $\mathrm{4s}$-orbitals and hence, the unusual configuration of $\ce{Cr}$ and $\ce{Cu}$.
The electron in one p orbital does not interact with the electron in the other p orbital on the same carbon. In ethylene the pi bond is orthogonal (perpendicular) to the sigma skeleton as shown in the figure below. Since the orbitals are orthogonal there is no interaction between the sigma and pi electrons (wavefunctions).
H^ϕij =Eiϕij ∀j =1,..., k H ^ ϕ i j = E i ϕ i j ∀ j = 1,..., k. We call those orbitals "k-fold degenerate orbitals", i.e. k orbitals with the same energy value (due to the particular simmetry of the molecule/atom under study). Now: for the hydrogen atom there is no difference in the energy of any orbital with the same quantum number n n.
The amount of energy that must be gained is equal to the difference in energy between the orbitals. When the electron relaxes and returns to its ground state, it emits that energy. However, I am not sure I understand why "higher" orbitals are of greater energy, for a couple of reasons. When I think of the potential energy of the electron, as ...
There is no electron-electron repulsion, obviously. For Helium, however, if one of its electrons is promoted to 2s or 2p, there's still an electron remaining in the core 1s shell, so the other electron in the 2s or 2p orbitals will be shielded because of electron-electron repulsion.