Question

Chemistry question? For the hydrogen atom, list the following orbitals in order of increasing energy: 4f, 6s, 3d, 1s, 2p.

I know that the answer is 1s<2p<3d<4f<6s, but why does 6s have greater energy than 4f?

Answer 1

Well, in the hydrogen atom, there is only one electron. The principal cause for why the `np` orbital, for instance, is higher in energy than the `ns` orbital in many-electron atoms is the addition of a second electron.

Otherwise, with only one electron,

`E_(2s) = E_(2p)`

`E_(3s) = E_(3p) = E_(3d)`

`E_(4s) = E_(4p) = E_(4d)`

etc.

and the energies only vary according to the principal quantum number `n`, i.e. `E_1 < E_2 < E_3 < . . . < E_n`. Therefore,

`E_(6s) > E_(4f) > E_(3d) > E_(2p) > E_(1s)`

In regular atoms, this relative energy ordering is still the case, with some differences.

There are zero radial nodes, except in the `6s` (which has `n - l - 1 = 5` radial nodes). However, there is an increase in the number of angular nodes as well, and the energies now vary according to both `n` and `l`.