# Question #8e208

Jun 10, 2017

It is ONLY $1 {s}^{2} 2 {s}^{1}$.

#### Explanation:

This is the explanation of how to order electrons into shells, sub-shells and orbitals. The answer to the question is halfway down.

The coefficient (number before the letter, e.g the coefficients here are 1 and 2) denotes the principal quantum number, which is also known as the shell or energy level. This tells you the energy state, higher the number = more energy.

The letter denotes the sub-shell (you can imagine the shell/energy level as a big box, which contains smaller sub-shells). The number above the letter tells you how many electrons are in the sub-shell. Each sub-shell contains orbitals. The "s" sub-shell can only contain one orbital, whereas the p, d and f sub-shells contain more than one orbital.

The orbital is the smallest "container" of electrons, for any given orbital, the maximum number of electrons in it, is 2. For example, the "p" sub-shell contains 3 different "p orbitals". These are called ${p}_{x} {p}_{y} {p}_{z}$, the subscript letter denotes the axis in which the orbital lies in (e.g. in the x-axis, y-axis or z-axis). Therefore the "p" sub-shell (regardless of which shell it's in) can have a maximum of 6 electrons. "d" sub-shells have 5 orbitals, therefore can have a maximum of 10 electrons.

Answer: For lithium, you would order the electrons like so:

Lithium contains 3 electrons. The way we place these in energy levels is by starting from the first shell (the lowest energy level), and work our way up to the higher energy states (higher principal quantum numbers/energy levels/shells):

The first shell can only contain one "s" sub-shell. The 1s sub-shell has a maximum of 1 orbital, therefore can only hold 2 electrons. We can only put 2 electrons in here. Hence $1 {s}^{2}$

We have now assigned 2 of the 3 electrons, since the 1st shell is now full, we have to move up in energy to the second shell.
The second shell starts off with the "s" sub shell, which can hold 2 electrons. As we only have one electron, the 2s orbital is only going to contain 1 electron. Hence $2 {s}^{1}$.

Therefore combining to give $1 {s}^{2} 2 {s}^{1}$. (count the top numbers to make sure you've used up all the electrons!)

It will not be $1 {s}^{2} 3 {s}^{1}$, as the 3rd shell is of higher energy relative to the 2nd shell. In nature, objects will try and lose potential energy to get to a more stable energy state. You can see this when you drop something. When you drop something it loses potential energy, therefore the object moves to a lower, more stable energy state. Therefore the electron goes to the lower energy state/shell.

Hope this helped!

Alex