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

Jun 14, 2016

Because there's no room.

Beryllium only has four electrons: two in the $1 s$ and two in the $2 s$ 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 $2 s$ orbital.

You can't place another electron in the $2 s$ orbital; there's only one $2 s$ 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 $\left\{0 , \pm 1 , \pm 2 , . . . , \pm l\right\}$.

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 $\pm \frac{1}{2}$, a binary restriction.

From the above, we have that...

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

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

3. By definition, since $l = 0$, ${m}_{l} = \left\{0\right\}$.

4. The number of ${m}_{l}$ values is equal to $2 l + 1 = 2 \left(0\right) + 1 = 1$, so can only be one $2 s$ orbital.

5. For the same orbital, the only quantum number that can differ is ${m}_{s}$, which can only be $\pm \frac{1}{2}$.

6. Since the $2 s$ 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 $2 p$ orbitals).

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