Part 1: According to what I've learned so far, helium is in the first row of the periodic table, so its $n$ should always equal $1$. However, a question in my text asks about helium 2s. What is helium 2s? If helium 2s exists, why does the periodic table list helium in the first row instead of the second?

Part 2: I believe that He 1s is more stable than He 2s. I've reasoned thus:

  • The numerals '1' and '2' in 'He 1s' and 'He 2s' express the size of the orbital. The greater number represents a larger orbital.
  • The electrons of smaller orbitals exist closer to the nucleus than the electrons of larger orbitals do.
  • The intensity of the attractive force between electron and protons increases the nearer the protons are to the electron
  • Therefore, the attractive force between the protons and the electron will be greater in He 1s than in He 2s.
  • The greater the attractive force between the protons and the electron, the greater the stability of the electron.
  • Therefore the electrons in He 1s are more stable than the electrons in He 2s.

Is that correct?

Note the following is the original question:

For each pair of orbitals determine which is more stable and explain why:

(a) He 1s and He 2s;

(b) Kr 5p and Kr 5s;

(c) He 2s and He+ 2s


Part 1: The elements on the periodic table are arranged according to their ground-state electron configurations. He 2s is an excited state of He 1s. There are even He 3s and 4s depending on whether one gives enough energy for the atom to excite to such levels. In general, higher principal quantum number (n) indicates a higher energy state (less stable)

Part 2: Your reasoning is correct.


Part 1: The 2s is referring to the orbitals that are occupied by electrons. In the ground state the electrons in He are both in 1s orbitals, however it is possible to excite the electrons from the 1s orbital to a higher orbital like 2s. All orbitals are accessible to the electrons of an atom through excitation.

Part 2: The reasoning you are using is appropriate to the question you are answering.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.