3.3. Electron Configuration¶
The four quantum numbers $n$, $l$, $m_l$, and $m_s$ enable us to label completely an electron in any orbital in any atom. In a sense, we can regard the set of four quantum numbers as the “address” of an electron in an atom, somewhat in the same way that a street address, city, state, and postal ZIP code specify the address of an individual.
Example. Write the four quantum numbers for an electron in a $3p$ orbital.
In $3p$ orbital, the principle quantum number $n = 3$, the angular momentum Quantum number $l = 1$, the magnetic quantum number could be $m_l = -1, 0, +1$, the electron spin quantum number could be $m_s = -1, +1$. So, all possibilities are:
| $n$ |
$l$ |
$m_l$ |
$m_s$ |
| $3$ |
$1$ |
$-1$ |
$+1/2$ |
| $3$ |
$1$ |
$-1$ |
$-1/2$ |
| $3$ |
$1$ |
$0$ |
$+1/2$ |
| $3$ |
$1$ |
$0$ |
$-1/2$ |
| $3$ |
$1$ |
$1$ |
$+1/2$ |
| $3$ |
$1$ |
$1$ |
$-1/2$ |
Electron configuration of hydrogen atom:
The electron in a ground-state hydrogen atom must be in the $1s$ orbital.
There are two electrons in a helium atom. The ground-state electron configuration is $1s^2$. Both electrons are in the $1s$ orbital. However, according to Pauli exclusion principle, the four quantum numbers should not be the same. Thus, in helium, $m_s = +1/2$ for one electron; $m_s = -1/2$ for the other. Electrons that have opposite spins are said to be paired.
Paramagnetic substances are those that contain net unpaired spins and are attracted by a magnet.
Diamagnetic substances do not contain net unpaired spins and are slightly repelled by a magnet. He is diamagnetic.
Consider the lithium atom (Z = 3), which has three electrons. The third electron cannot go into the $1s$ orbital because it would inevitably have the same set of four quantum numbers as one of the first two electrons. Therefore, this electron “enters” the next (energetically) higher orbital, which is the $2s$ orbital. The electron configuration is $1s^22S^1$
The ground-state electron configuration of beryllium is $1s^22s^2$.
The ground-state electron configuration of boron is $1s^22s^22p^1$.
Hund's rule, which states that the most stable arrangement of electrons in subshells is the one with the greatest number of parallel spins. Thus the ground-state electron configuration of carbon is $1s^22s^22p^2$.The two $p$ orbital electrons have different $m_l$ quantum numbers.
The ground-state electron configuration of nitrogen is $1s^22s^22p^3$.
The ground-state electron configuration of oxygen is $1s^22s^22p^4$.
The ground-state electron configuration of fluorine is $1s^22s^22p^5$.
The ground-state electron configuration of neon is $1s^22s^22p^6$.
General Rules¶
- Each shell or principal level of quantum number $n$ contains $n$ subshells. For example, if $n = 2$, then there are two subshells (two values of $l$) of angular momentum quantum numbers $0$ and $1$.
- Each subshell of quantum number $l$ contains ($2l + 1$) orbitals. For example, if $l = 1$, then there are three $p$ orbitals.
- No more than two electrons can be placed in each orbital. Therefore, the maximum number of electrons is simply twice the number of orbitals that are employed.
- A quick way to determine the maximum number of electrons that an atom can have in a principal level $n$ is to use the formula $2n^2$.
Example: What is the maximum number of electrons that can be present in the principal level for which $n = 3$?
Answer:
When $n=3$, $l$ could be 0, 1, 2, corresponding to $s$, $p$ and $d$ orbitals.
In $s$ orbital, $m_l$ must be 0. Thus, there is only one $s$ orbital.
When $l=1$, $m_l$ has three possible values ($-1$, 0, $=1$), thus there are three $p$ orbitals.
When $l=2$, $m_l$ has five possible values ($-2$, $-1$, 0, $=1$, $+2$), thus there are five $d$ orbitals.
The total number of orbitals is $1+3+5=9$. Each orbital can host two electrons with $m_s = \pm \frac{1}{2}$, respectively. Thus, the maximum number of electrons that can be present in the principal level $n=3$ is 18.
Example: An oxygen atom has a total of eight electrons. Write the four quantum numbers for each of the eight electrons in the ground state.
Answer:
(1, 0, $0$, $+\frac{1}{2}$)
(1, 0, $0$, $-\frac{1}{2}$)
(2, 0, $0$, $+\frac{1}{2}$)
(2, 0, $0$, $-\frac{1}{2}$)
(2, 1, $-1$, $+\frac{1}{2}$)
(2, 1, 0, $+\frac{1}{2}$)
(2, 1, $+1$, $+\frac{1}{2}$)
(2, 1, $+1$, $-\frac{1}{2}$)
Of course, the placement of the eighth electron in the orbital labeled $m_l = +1$ is completely arbitrary. It would be equally correct to assign it to $m_l = 0$ or $m_l = -1$.
