Why isn't #"Be"^#s electron configuration #1s^2 2s^3#?
1 Answer
Because there's no room.
Beryllium only has four electrons: two in the
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
You can't place another electron in the
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, . . . , n1# .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...

For a
#2s# orbital,#n = 2# . 
By definition, for an
#s# orbital,#l = 0# . 
By definition, since
#l = 0# ,#m_l = {0}# . 
The number of
#m_l# values is equal to#2l + 1 = 2(0) + 1 = 1# , so can only be one#2s# orbital. 
For the same orbital, the only quantum number that can differ is
#m_s# , which can only be#pm1/2# . 
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