# Why do Lewis dot diagrams hold a maximum of 8 electrons in the outermost energy level?

Jan 4, 2016

The short answer is that there is a general "octet rule" where eight valence electrons tends to be stable for atoms without access to $d$ orbitals because they are occupying all allowed quantum states. For $n \le 2$, this holds.

The cool answer is derived from a bit of quantum mechanics (though you should have been taught this in General Chemistry).

THE QUANTUM NUMBERS

Let us consider the quantum numbers $n$, $l$, ${m}_{l}$, and ${m}_{s}$:

• $n$ - the principal quantum number, defining the energy level, taking on integer values $1 , 2 , . . . , n$, e.g. $2 s$ has $n = 2$.
• $l$ - the angular momentum quantum number, defining the orbital shape, taking on integer values $0 , 1 , . . . n$, e.g. $2 s$ has $l = 0$, while $2 p$ has $l = 1$.
• ${m}_{l}$ - the orbital angular momentum quantum number. In other words, projection of $l$, taking on the values $0 , \pm l$, e.g. ${m}_{l}$ for $l = 1$ is $0 , \pm 1$.
• ${m}_{s}$ - the magnetic quantum number, taking on values for the possible spins of the electron, e.g. $\pm \text{1/2}$.

TWO ELECTRONS MUST OCCUPY DIFFERENT QUANTUM STATES

Next, according to the Pauli Exclusion Principle, two electrons must occupy different quantum states. That is, no two electrons can share exactly the same four quantum numbers.

Now, if we look at the second energy level, $n = 2$, we have the $\setminus m a t h b f \left(2 s\right)$ and $\setminus m a t h b f \left(2 p\right)$ orbitals available as valence orbitals. Notice how many quantum numbers we have available:

• $l = 0 :$ ($2 s$ orbital)

• ${m}_{l} = 0$
• ${m}_{s} = + \text{1/2", -"1/2}$
• $l = 1 :$ ($2 p$ orbital)

• ${m}_{l} = - 1 , 0 , + 1$
• ${m}_{s} = + \text{1/2", -"1/2}$

ALLOWED QUANTUM STATES FOR THE ELECTRON

In the same subshell, $n$ and $l$ are automatically the same, and the only quantum numbers that can be different are ${m}_{l}$ and ${m}_{s}$; therefore, the two possible values of ${m}_{s}$ correspond to two distinct quantum states.

It also follows that since the orbital shapes are different for $l = 0$ vs. $l = 1$, and since different ${m}_{l}$ correspond to different vector projections of $l$, this corresponds to four distinct quantum states---one for ${m}_{l} = 0$ and three for ${m}_{l} = 0 , \pm 1$.

WHY EIGHT FOR THE 2S AND 2P ORBITALS?

That means in total, we have the four allowed states for the electron for different ${m}_{l}$, and two allowed states for the electron for different ${m}_{s}$ in each of four distinct ${m}_{l}$ states.

Overall, it follows that for $n = 2$ and $l = 0 , 1$, we have ${m}_{l} = 0$ and ${m}_{l} = 0 , \pm 1$ and ${m}_{s} = \pm \text{1/2}$, so we have eight allowed quantum states for the electron.

Therefore, the maximum number of electrons that the $2 s$ and $2 p$ valence orbitals can hold altogether is eight.