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?

1 Answer
Truong-Son N.
Nov 18, 2017

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#.

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