Question

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

Answer 1

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.

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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.