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In a hydrogen atom, the presence of only one electron allows various orbitals' energy states to be dependent only on the principal quantum number and not on angular momentum. Orbital degeneracy and all that fun stuff. (This has been answered many times on SE)

In helium, would this still be true? How come? I can't seem to find a satisfactory explanation.

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    $\begingroup$ The gist is that with two electrons the spin must either be the same or opposite. If the electrons are in the same orbit then the spins must be opposite. $\endgroup$ – MaxW Apr 9 '16 at 5:26
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The energy of orbitals for hydrogen atom is dependent only on the principal quantum number: $$E_{n}(\rm{eV})=-\frac{13.6}{n^2}$$ For atoms with more than one electron (all the atoms except hydrogen atom and hydrogenoid ions), the energy of orbitals is dependent on the principal quantum number and the azimuthal quantum number according to the equation: $$E_{n,l}(\rm{eV})=-\frac{13.6Z^{*2}}{n^2}$$ Where $Z^{*}$ is the effective nuclear charge: $Z^{*}=Z-\sigma$.

where$Z$ is the atomic number and $\sigma (n,l)$ is the screen (shielding) constant and it depends on the principal quantum number and the azimuthal quantum number.

According to this, the atomic orbitals $2s$ and $2p$, that are at the same energy in the case of hydrogen atom, are no longer at the same energy for the atom of helium.

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One view of the situation is that there is one 1s orbital which may house two electrons. This, in turn means that the electron's spin is the property of the electron.

Another view of this situation is that there are two orbitals, a 1s spin up and a 1s spin down. This implies the electron's spin is the property of the orbital.

Since orbitals don't really exist, noone may say which is a correct interpretation. The first interpretation seems to dominate teaching.

There is a finite, greater than zero energy cost associated with aligning the electron's spins for or against the nuclear spins (indeed this is the basis for NMR). However, the amount of energy associated with this is trivially small compared with the change in energy from 1s to 2s, that for all pragmatic purposes we make the simplifying assumption that $$\ce{\Delta E_{1s(spin down)->2s(spin down)} \approx \Delta E_{1s(spin down)->2s(spin up)}}$$ which effectively means there is no impact on the angular momentum and the energy states.

Another relevant point is that transitioning between energy levels while simultaneously 'flipping' spin is a quantum mechanically forbidden process, which has the pragmatic effect of being 'unlikely' and hence 'slow'.

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I once had the same question as you and I did go to my chemistry teacher for an answer. This is his original response:

"In hydrogen, all orbitals with the same principal quantum number 'n' (1,2,3...) are degenerate, regardless of the orbital angular momentum quantum number 'l' (0,1...n-1 or s,p,d,). However, in atoms with more than one electron, orbitals with different values of l for a given value for n are not degenerate. Why is this? Surely the radial distribution functions are similar for hydrogen (in that there's still penetration of orbitals and so on). Or is it that orbitals with different values of l are degenerate for a value of n greater than would be occupied in that particular atom's ground state?

The answer is right there in your question. It's only the interaction of multiple electrons in an atom like He, Li, Be, etc. that makes the different angular momenta wave functions differ in energy. Consider this. For the one electron system, why should a p or d orbital differ in energy from an s? What makes them differ? In the multi-electron case, the p orbitals have different spatial extent, different angular components, so the electron density caused by an electron in that orbital will interact differently with the other electrons. In other words, you need to have more than one electron for the "shape" of the p and d and f orbitals to matter to the other electrons. In the H atom, there's only one electron, so there's no electron-electron repulsion to differentiate the s, p, and d orbitals."

Hope this clarifies your doubt.

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