Thank you very much. I understand it now, but why do more repulsions necessarily mean higher energy? Let's take the 2s and 2p subshell. Also, 2p subshell is larger in size compared to the 2s subshell. The 2s subshell can accommodate 2 electrons and 2p can accommodate 6 electrons. But we can't deduce that 2p will have a higher energy because it has 1 angular node and 2s has 0. Even the larger volume of 2p needs to be taken into consideration. Even though 1 angular node is present, there is still a lot more space for the 6 electrons to occupy, thereby the repulsion between the electrons is not as much as we assumed initially (it MAY be less than the repulsions in the 2s subshell). Is this logic correct?
Why is a 3s orbital lower in energy than a 3p orbital in all atoms other than a hydrogen atom which only has a single electron?