Why isn't #"Be"^-#s electron configuration #1s^2 2s^3#?

1 Answer
Jun 14, 2016

Because there's no room.

Beryllium only has four electrons: two in the #1s# and two in the #2s# orbitals.

Every orbital can only contain two electrons, which must be of two different spins; there only exist two different spins for all electrons, and there only exists one #2s# orbital.

You can't place another electron in the #2s# orbital; there's only one #2s# orbital, and it's already full.


Recall the four quantum numbers:

  • The principal quantum number #n# is the energy level, basically corresponding to the row on the periodic table.

    #n# is an integer, and it can only be one of the following values at a time: #1, 2, 3, . . . , N#, where #N# is an arbitrarily large integer.

  • The angular momentum quantum number #l# can only be one of the following values at a time: #0, 1, 2, . . . , n-1#.

    i.e. if #n = 1#, then #l = 0# and only #0#.

  • The magnetic quantum number #m_l# takes on all values in the set #{0, pm1, pm2, . . . , pml}#.

    It basically tells you how many orbitals there are in a single subshell.

  • The magnetic spin quantum number #m_s# is the spin of the electron in the orbital.

    This can only be #pm1/2#, a binary restriction.

From the above, we have that...

  1. For a #2s# orbital, #n = 2#.

  2. By definition, for an #s# orbital, #l = 0#.

  3. By definition, since #l = 0#, #m_l = {0}#.

  4. The number of #m_l# values is equal to #2l + 1 = 2(0) + 1 = 1#, so can only be one #2s# orbital.

  5. For the same orbital, the only quantum number that can differ is #m_s#, which can only be #pm1/2#.

  6. Since the #2s# orbital already has the maximum two electrons, by the Pauli Exclusion Principle, it must go into the next orbitals, higher in energy (one of the three #2p# orbitals).

Hence, if #"Be"# gains one electron, it attains the configuration #1s^2 2s^2 2p^1#, not #1s^2 2s^3#. It simply isn't possible. Besides, #2 + 2 + 1 < 8#, so the octet "rule" doesn't matter here.