At some point, when start noticing the pattern of filling the orbitals, and their mapping on the periodic table, it may be easier to write the electron configuration simply based on the position of the element. So, instead of starting from 1s 2 …., we write the electron configuration of the last noble gas that comes before the element and then add the remaining electrons from the corresponding n level. For example, let’s say we want to write an electron configuration for sulfur. First, you need to write the inner electron configuration, that is the configuration of the noble gas that precedes it in the periodic table. In this case, it is Ne, and since it is in the last element in the second row, the last filled sublevel would be 2p 6 , and therefore, the electron configuration of the noble gas will be 1s 2 2s 2 2p 6 . Now, to this, we add 2 s electrons, and four p electrons since S is the fourth element in the p orbital area:
Therefore, the electron configuration will be 1s 2 2s 2 2p 6 3s 2 2p 4 .
In condensed electron configurations, we show the symbol of the noble gas that precedes the element in square brackets, and then add the remaining electrons just as we have been doing so far. So, for the sulfur, it will be S: [Ne]3s 2 2p 4 .
Another example is the electron configuration of Br. Since it is in the 4 th row, the inner configuration would be that of Ar, and after that, we add the electrons as we did for the d elements that we have discussed. Two electrons will go into the 4s orbital, 10 to the 3d orbitals, and 5 into the 4p sublevel – Br: [Ar]4s 2 3d 10 4p 5 .
There is one exception to keep in mind for the electron configuration of transition metals. That is the (n +1)s orbitals always fill before the nd orbitals.
For example, the 4s level fills before the 3d level, and therefore, the electron configuration of Ti is 1s ² 2s ² 2p ⁶ 3s ² 3p ⁶ 4s ² 3d ² or, the condensed configuration which will be [Ar]4s ² 3d ² .
As we get to higher values of n, other variations in the filling pattern occur because the energy sublevels become very close together (recall the figure on the energy levels above).
We mentioned that noble gases are characterized by great stability because of their completely filled orbitals. Now, similar to this, orbitals tend to get lower energy levels (high stability) when they are half-filled. For example, following what we have learned so far, we may expect the following electron configuration for Cr: [Ar]4s ² 3d 4 .
However, the electron configuration of Cr is [Ar]4s 1 3d 5 , and the reason for this is that the d orbital gets a half-filled configuration (remember d orbitals can have a maximum of 10 electrons). We can think of this as the electron jumping from the 4s level to the 3d level and compensating this energy uphill by a stabilization associated with half-filled orbitals:
Of course, in reality, it is not as though the electrons fill one by one, and then one of them jumps from 4s to 3d level. However, this is a visual representation to help better understand this behavior in the electron configuration of transition elements.
Let’s also discuss the electron configuration of Cu which stands before Zn, and one may expect it to have 9 electrons in the d sublevel. However, the correct electron configuration of Cu is [Ar]4s 1 3d 10 as this allows to attain filled d orbitals and a half-filled s orbital which is lower in energy (more stable) than the expected [Ar]4s 2 3d 9 configuration.
Starting from period 6, we have the inner transition elements which contain f orbitals. The first thing you need to remember here is that there are seven f orbitals because l = 3, so the possible ml values are −3, −2, −1, 0, +1, +2, and +3. Each orbital can accommodate two electrons and therefore, a total of 14 electrons can fill the given f sublevel.
Although the first element in this region is La, the filling of f orbitals starts from Cerium (Ce), and the elements together with it in the Period 6 are called lanthanides (or rare earth elements).
Notice the general filling order is 6s → 4f → 5d, however, ns goes first, then only the first of (n − 1)d, all (n − 2)f, and after that remainder of the (n − 1)d, and np. Therefore, the electron configuration of La is [Xe]6s ² 5d ¹ and not [Xe]6s ² 4f ¹ as the first d electron goes first.
The inner transition series in Period 7 start after actinium (Ac; Z = 89) and are called the actinides.
Hopefully, you won’t get tested on remembering all these exceptions as it should be the principle that is more important, but in any case, use the ptable.com website to practice filling electron configurations. Just be sure to click the “Electrons” tab in the upper area.
Valence electrons are the ones farthest away from the nuclei and therefore, we find them in the outermost two orbitals. For example, Li (1s 2 2s 1 ) has one valence electron, and it is the 2s orbital. Chlorine (1s ² 2s ² 2p ⁶ 3s ² 3p ⁵ ) has seven valence electrons in the 3s and 3p sublevels. If you ever forget, remember that you can easily double-check this based on the group number. The group number indicates the number of valence electrons in the atom.
The first thing you need to remember here is that cations are formed by losing an electron(s), and anions are forming by gaining an electron(s).
The charge of the ion is a result of an imbalance between the number of protons and electrons. If it is a cation, then the positive indicates how many more protons it has compared to the number of electrons. For anions, the charge tells how many extra electrons there are compared to the number of protons.
Recall, this pattern for the formation of anions and cations: Metals tend to lose electron(s) and become cations (positively charged ions).
Nonmetals tend to gain an electron(s) and become anions (negatively charged ions).
Notice that the number of protons is not changed, and the ions are charged because, unlike atoms, their number of protons and electrons is not equal.
Now, how do we determine the electron configuration of an ion? If it is, for example, a +1 charged cation, that means the atom has lost one electron. This electron is going to be from the outermost valence shell as these are the electrons farthest away from the nuclei and thus not as strongly attracted to it.
Let’s see an example on Na. It is in the first group, so it loses one electron to become Na + . The electron configuration of sodium is 1s ² 2s ² 2p ⁶ 3s ¹ , and the electron is removed from the energy level with the greatest n value – 3s. Therefore, the electron configuration of the Na + ion will be 1s ² 2s ² 2p ⁶ :
Notice that the ion has a configuration with a complete shell of p orbitals which is characteristic of noble gases. In fact, this is the electron configuration of Ne, and we say that Na + and Ne are isoelectronic (same electronic structure). The reason for this is that, remember, noble gases are very stable because of the low energy level of complete orbitals.
This pattern explains why the metals in the first group become +1, the ones in the second group become +2, and Al, for example, becomes +3. It takes removing one electron from a metal in the first group to obtain the electron configuration of the previous noble gas, it takes two for the group two metals, etc.
Na (1s 2 2s 2 2p 6 3s 1 ) ⟶ e − + Na + ([He]2s 2 2p 6 ) [isoelectronic with Ne ([He] 2s 2 2p 6 )]
Anions are formed when the atom gains as many electrons as necessary to attain the electron configuration of the next noble gas in the periodic table. For example, oxygen is in group 6, and therefore, it will need two electrons to attain the electron configuration of Ne:
Notice again that the two electrons go to the outermost valence orbital. Oxygen has 6 valence electrons – those in the 2s and 2p orbitals, however, since p sublevels are higher in energy, and they are the only ones capable of accepting additional electrons, the two electrons go to the 2p orbitals. The electron configuration of the oxide ion (O 2- ) is therefore, 1s ² 2s ² 2p ⁶ .
O (1s 2 2s 2 2p 4 ) + 2e − ⟶ + O 2- ([He]2s 2 2p 6 ) [isoelectronic with Ne ([He] 2s 2 2p 6 )]
Another common type of monoatomic anions are the halides. Halogens are in group 7, and therefore, they only need one electron to attain the electron configuration of the noble gas following them in the periodic table. For example, bromine takes one electron and becomes isoelectronic to t Kr:
Br([Ar] 4s 2 3d 10 4p 5 ) + e − ⟶ Br − ([Ar] 4s 2 3d 10 4p 6 ) [isoelectronic with Kr ([Ar] 4s 2 3d 10 4p 6 )]
In contrast to main-group ions, transition metal ions do not usually attain a noble gas configuration. This is because the ns level is the outermost level, and the (n-1)d is considered an inner level therefore, it will take too much energy to remove those electrons and achieve a noble gas configuration. Therefore, the cation of a transition metal is formed by removing first the electrons from the ns (highest principal quantum number) orbital and then from the (n -1)d orbitals.
For example, the electron configuration of Zn is 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 or [Ar]4s 2 3d 10 , and it loses the two electrons from the 4s orbital to become Zn 2+ [Ar]3d 10 . This is “not a bad” electron configuration considering the filled d orbitals.
Remember, when discussing the Bohr model of the hydrogen atom, we mentioned that absorbing light with sufficient energy moves the electron to a higher energy level, and when it falls back to a lower energy level, light is emitted as energy is lost.
Now, when the electron is in a higher energy level than it normally is, the atom or ion is said to be in the excited state. If the electrons are in the lowest possible energy levels, the atom is in the ground state. So, everything we have discussed today pertains to the electron configurations of atoms at the ground state. In general, unless mentioned otherwise, the term electron configuration refers to the atom in the ground state.
An example of switching from the ground to an excited state can be when the electron in a carbon atom jumps from the 2s to the 2p level. The electron configurations and orbital diagrams can be represented as:
These electron transitions from lower to higher energy orbitals are the basis of many spectroscopies for determining the molecular structures in inorganic and organic materials.
Check this 95-question, Multiple-Choice Quiz on the Electronic Structure of Atoms including questions on properties of light such as wavelength, frequency, energy, quantum numbers, atomic orbitals, electron configurations, and more.
