Question

What is the electron configuration for an excited atom of phosphorous, ready to emit a photon of energy and return to ground state? (configuration has to follow all rules except be in lowest energy)

Answer 1

Phosphorus has atomic number `15`.
As such its electronic configuration in ground state is
`1s^2 2s^2 2p^6 3s^2 3p^3`

The following electronic configurations could be excited states

1. `1s^2 2s^2 2p^6 3s^1 3p^4`
   Here one of `3s` electrons has been promoted to `3p` sub level

2. `1s^2 2s^2 2p^6 3s^2 3p^2 3d^1`
   Here one of `3p` electrons has been promoted to `3d` sub level

3. `1s^2 2s^2 2p^6 3s^2 3p^2 4s^1`
   Another possibility is that one electron from the `3p` promoted to the `4s` sub-level
   However, as explained below probability of transition listed at 2. is more than transition listed at 3.

The `3p` orbital has the following states for electron filling

`n = 3`, `l = 1`, `m_l = -1,0,1 and m_s = -1/2, +1/2`.

And the `4s` orbital has

`n = 4`, `l = 0`, `m_l = 0 and m_s = -1/2, +1/2`

And the `3d` orbital has

`n= 3`, `l = 2`, `m_l=-2,-1,0,1,2 and m_s =-1/2, +1/2`

For a transition from `3p` to `4s`, electron need to loose 1 quantum of angular momentum. This means absorbing a photon with spin `-1`.
For a transition from `3p` to `3d`, a photon of spin `+1` needs to be absorbed and there are all three `m_l` states.

There is only one `m_l` level for the `4s` while there are three in the `3p` state. This make the transition more probable - due to greater number of states available to end up in.