Electron Configurations of Atoms and Ions
Chapter 07
Electron Configurations
The electron configuration of an atom is a notation for describing the distribution of electrons in an atom or a molecule. The configuration provides information about the structure of the periodic table of the elements, chemical bonds, chemical formulas of compounds and geometries of molecules. Electron configurations implement the rules of the quantum numbers.
The periodic table is arranged into blocks and understanding these blocks is useful in determining the electron configurations of atoms.
Note: Elements 71 (Lutetium; Lu) and 103 (Lawrencium; Lr) has historically been categorized in the f-block. However, recent evidence and attention by the scientific community has increased the momentum in categorizing these elements to be d-block elements.
Additionally, the following diagram aids in ordering the assignment of electrons for ground-state atoms (from low to higher energy).
The aufbau principle states that in a ground state of an atom or ion, electrons first fill subshells of the lowest available energy followed by subshells of the next higher energy. The Mandelung rule (or the n + l rule) states that the order in which atomic orbitals are filled is based on the quantum numbers n and l and that orbitals are filled in the order of increasing n + l values. If two orbitals share the same n + l value, the orbital with the lower n value is filled first.
In a one-electron system (any hydrogen-like atom), the atomic orbital energies for a given shell are equal.
In a multi-electron system, the orbital energies of subshells are different. The quantum numbers n and l determine the energies of the orbital. Within a principle shell (of value n), the orbital energies increase with increasing l.
This change in energy ordering is caused by the effects of electron shielding and penetration.
Consider a lithium atom for example. An electron in the 1s orbital “feels” the full 3+ charge of the nucleus. An electron in the 2s orbital, resides further away from the nucleus most of the time. At large distance, the 2s electron will “feel” a 1+ charge from the nucleus since the two 1s electrons (each with a 1– charge) “cancel out”, or shield/screen, two units of positive charge. However, the 2s orbital penetrates the 1s orbital to a certain extent.
If the electron in the 2s orbital is within the region that a 1s orbital occupies, the 2s electrons feels the entire nuclear charge of 3+, though the probability of this occurring is relatively low. Therefore, a 2s electron feels either a nuclear charge anywhere from 1+ to 3+. This is referred to as effective nuclear charge, Z* or Zeff, which is the actual amount of positive (nuclear) charge an electron experiences in a multi-electron atom. The effective nuclear charge of a 2s electron in lithium is 1.279.
Effective nuclear charge tends to increase as you move right (across a period on the periodic table) due to the increase in the number of protons in the nucleus. Even though the number of electrons are increasing, they are being added to the same principle energy level (shell) and the shielding effect remains relatively constant.
Effective nuclear charge tends to increase slightly as you move down (a group on the periodic table) but this increase is much smaller than moving right across a period. This is due to the number of protons in a nucleus increasing. While the number of electrons and the extent of electron shielding increases, the increased positive charge at the nucleus has a slightly larger effect on the attraction to the outer electrons.
Writing Electron Configurations: A Guided Approach
This section guides the reader through writing electron configurations for ground state atoms from Z = 1 to Z = 10. Orbital (box) diagrams will accompany each presented electron configuration as well as the spdf notation.
Elements and substances with unpaired electrons are, strictly speaking, paramagnetic, and will experience an acttractive interaction with an external magnetic field. Those whose electrons are all paired are diamagnetic, and will experience a repulsive interaction with an external magnetic field.
Hydrogen, Z = 1
The hydrogen atom resides in the s-block on the periodic table and has one electron. The electron is placed in the lowest energy atomic orbital. The quantum numbers are:
- n = 1
- l = 0 ( s )
- ml = 0
Notice how there is only one orientation of an l = 0, or s, orbital. Note: It is typical to write a “spin-up” (ms = 1/2) electron first before drawing a spin-down (ms = –1/2) electron.
The corresponding electron configuration for hydrogen is therefore
\[1s^1\]
Realize that the first three quantum numbers (n, l, ml) indicate the available atomic orbital whereas, the inclusion of the fourth quantum number to the set describes a specific electron in a specific location. The quantum numbers for this electron is
- (1s “spin-up” valence electron) n = 1, l = 0, ml = 0, and ms = 1/2
The valence electrons are typically the electrons that are not the core electrons, here the one 1s electron. Valence electrons are commonly described as being the electrons that are able to participate in bonding. Other sources describe valence electrons as being the “outermost” electrons, a faulty but sometimes useful definition.
This element is paramagnetic.
Excited state configuration of hydrogen
An excited state of an atom is one where an electron is promoted to a higher energy state. We could envision an excited state hydrogen atom, denoted as H*, where the 1s electron is promoted to a 2p orbital (a 1s → 2p transition).
- n = 1, 2
- l = 0 ( s ), 1 ( p )
- ml = 0 (for l = 0) and ml = –1, 0, 1 (for l =)
and
\[2p^1\] The quantum numbers for this electron is
- (2p “spin-up” valence electron) n = 2, l = 1, ml = –1, and ms = 1/2
Helium, Z = 2
The helium atom resides in the s-block on the periodic table and has two electrons. The quantum numbers are:
- n = 1
- l = 0 ( s )
- ml = 0
The corresponding electron configuration is
\[1s^2\]
The quantum numbers for each electron are
- (1s “spin-up” valence electron) n = 1, l = 0, ml = 0, and ms = 1/2
- (1s “spin-down” valence electron) n = 1, l = 0, ml = 0, and ms = –1/2
This element is diamagnetic.
Lithium, Z = 3
The lithium atom resides in the s-block on the periodic table and has three electrons. The quantum numbers are:
- n = 1, 2
- l = 0 ( s ), 1 ( p )
- ml = 0 (for l = 0) and ml = –1, 0, 1 (for l = 1)
The aufbau principle states that the lowest energy atomic orbitals are filled first. Therefore, the 2s orbital receives the third electron and not a 2p orbital.
The corresponding electron configuration is
\[1s^22s^1\]
Electron configurations can be simplified using the noble gas notation where a portion of an electron configuration is abbreviated with the noble gas element symbol in square brackets.
A helium atom has an electron configuration of 1s2. Therefore, [He] can be used to replace the 1s2 portion of lithium’s electron configuration.
\[1s^22s^1 ~\longrightarrow~ [\mathrm{He}]~2s^1\]
The quantum numbers for each electron are
- (1s “spin-up” core electron) n = 1, l = 0, ml = 0, and ms = 1/2
- (1s “spin-down” core electron) n = 1, l = 0, ml = 0, and ms = –1/2
- (2s “spin-up” valence electron) n = 2, l = 0, ml = 0, and ms = 1/2
Core electrons are often referred to as electrons who are represented by the noble gas notation, here the two 1s electrons.
This element is paramagnetic.
Beryllium, Z = 4
The beryllium atom resides in the s-block on the periodic table and has four electrons. The quantum numbers are:
- n = 1, 2
- l = 0 ( s ), 1 ( p )
- ml = 0 (for l = 0) and ml = –1, 0, 1 (for l = 1)
The corresponding electron configuration is
\[1s^22s^2\]
or
\[[\mathrm{He}]~2s^2\]
The quantum numbers for each electron are
- (1s “spin-up” core electron) n = 1, l = 0, ml = 0, and ms = 1/2
- (1s “spin-down” core electron) n = 1, l = 0, ml = 0, and ms = –1/2
- (2s “spin-up” valence electron) n = 2, l = 0, ml = 0, and ms = 1/2
- (2s “spin-down” valence electron) n = 2, l = 0, ml = 0, and ms = –1/2
This element is diamagnetic.
Boron, Z = 5
The boron atom resides in the p-block on the periodic table and has five electrons. The quantum numbers are:
- n = 1, 2
- l = 0 ( s ), 1 ( p )
- ml = 0 (for l = 0) and ml = –1, 0, 1 (for l = 1)
The corresponding electron configuration is
\[1s^22s^22p^1\]
or
\[[\mathrm{He}]~2s^22p^1\]
The quantum numbers for each electron are
- (1s “spin-up” core electron) n = 1, l = 0, ml = 0, and ms = 1/2
- (1s “spin-down” core electron) n = 1, l = 0, ml = 0, and ms = –1/2
- (2s “spin-up” valence electron) n = 2, l = 0, ml = 0, and ms = 1/2
- (2s “spin-down” valence electron) n = 2, l = 0, ml = 0, and ms = –1/2
- (2p “spin-up” valence electron) n = 2, l = 1, ml = –1, and ms = 1/2
Note that an electron is placed in one of the three p orbitals. It does not matter which one it is placed in since the three p orbitals are degenerate (have the same energy). We also do not prescribe a specific ml quantum number to a specific p orbital orientation. I arbitrarily chose a value of –1 here.
This element is paramagnetic.
Carbon, Z = 6
The carbon atom resides in the p-block on the periodic table and has six electrons. The quantum numbers are:
- n = 1, 2
- l = 0 ( s ), 1 ( p )
- ml = 0 (for l = 0) and ml = –1, 0, 1 (for l = 1)
Notice how the 6th electron is placed in an empty 2p orbital. This is because unpaired electrons result in lower energy than paired electrons due to the fact that “like” charges repel. Hund’s Rule states that the lowest energy atomic state is the one that maximizes the total spin quantum number for the electrons in the open subshell (i.e. electrons should remain unpaired until it is impossible to do so).
Wrong energy diagram for carbon
The corresponding electron configuration is
\[1s^22s^22p^2\]
or
\[[\mathrm{He}]~2s^22p^2\]
The quantum numbers for each electron are
- (1s “spin-up” core electron) n = 1, l = 0, ml = 0, and ms = 1/2
- (1s “spin-down” core electron) n = 1, l = 0, ml = 0, and ms = –1/2
- (2s “spin-up” valence electron) n = 2, l = 0, ml = 0, and ms = 1/2
- (2s “spin-down” valence electron) n = 2, l = 0, ml = 0, and ms = –1/2
- (2p “spin-up” valence electron) n = 2, l = 1, ml = –1, and ms = 1/2
- (2p “spin-up” valence electron) n = 2, l = 1, ml = 0, and ms = 1/2
This element is paramagnetic.
Nitrogen, Z = 7
The nitrogen atom resides in the p-block on the periodic table and has seven electrons. The quantum numbers are:
- n = 1, 2
- l = 0 ( s ), 1 ( p )
- ml = 0 (for l = 0) and ml = –1, 0, 1 (for l = 1)
In keeping with Hund’s rule, the 7th electron is placed in an unoccupied p orbital.
The corresponding electron configuration is
\[1s^22s^22p^3\]
or
\[[\mathrm{He}]~2s^22p^3\]
The quantum numbers for each electron are
- (1s “spin-up” core electron) n = 1, l = 0, ml = 0, and ms = 1/2
- (1s “spin-down” core electron) n = 1, l = 0, ml = 0, and ms = –1/2
- (2s “spin-up” valence electron) n = 2, l = 0, ml = 0, and ms = 1/2
- (2s “spin-down” valence electron) n = 2, l = 0, ml = 0, and ms = –1/2
- (2p “spin-up” valence electron) n = 2, l = 1, ml = –1, and ms = 1/2
- (2p “spin-up” valence electron) n = 2, l = 1, ml = 0, and ms = 1/2
- (2p “spin-up” valence electron) n = 2, l = 1, ml = 1, and ms = 1/2
This element is paramagnetic.
Oxygen, Z = 8
The oxygen atom resides in the p-block on the periodic table and has eight electrons. The quantum numbers are:
- n = 1, 2
- l = 0 ( s ), 1 ( p )
- ml = 0 (for l = 0) and ml = –1, 0, 1 (for l = 1)
The 8th electron can now be paired with an unpaired electron. The corresponding electron configuration is
\[1s^22s^22p^4\]
or
\[[\mathrm{He}]~2s^22p^4\]
The quantum numbers for each electron are
- (1s “spin-up” core electron) n = 1, l = 0, ml = 0, and ms = 1/2
- (1s “spin-down” core electron) n = 1, l = 0, ml = 0, and ms = –1/2
- (2s “spin-up” valence electron) n = 2, l = 0, ml = 0, and ms = 1/2
- (2s “spin-down” valence electron) n = 2, l = 0, ml = 0, and ms = –1/2
- (2p “spin-up” valence electron) n = 2, l = 1, ml = –1, and ms = 1/2
- (2p “spin-up” valence electron) n = 2, l = 1, ml = 0, and ms = 1/2
- (2p “spin-up” valence electron) n = 2, l = 1, ml = 1, and ms = 1/2
- (2p “spin-down” valence electron) n = 2, l = 1, ml = –1, and ms = –1/2
This element is paramagnetic.
Fluorine, Z = 9
The fluorine atom resides in the p-block on the periodic table and has nine electrons. The quantum numbers are:
- n = 1, 2
- l = 0 ( s ), 1 ( p )
- ml = 0 (for l = 0) and ml = –1, 0, 1 (for l = 1)
The corresponding electron configuration is
\[1s^22s^22p^5\]
or
\[[\mathrm{He}]~2s^22p^5\]
The quantum numbers for each electron are
- (1s “spin-up” core electron) n = 1, l = 0, ml = 0, and ms = 1/2
- (1s “spin-down” core electron) n = 1, l = 0, ml = 0, and ms = –1/2
- (2s “spin-up” valence electron) n = 2, l = 0, ml = 0, and ms = 1/2
- (2s “spin-down” valence electron) n = 2, l = 0, ml = 0, and ms = –1/2
- (2p “spin-up” valence electron) n = 2, l = 1, ml = –1, and ms = 1/2
- (2p “spin-up” valence electron) n = 2, l = 1, ml = 0, and ms = 1/2
- (2p “spin-up” valence electron) n = 2, l = 1, ml = 1, and ms = 1/2 8 (2p “spin-down” valence electron) n = 2, l = 1, ml = –1, and ms = –1/2
- (2p “spin-down” valence electron) n = 2, l = 1, ml = 0, and ms = –1/2
This element is paramagnetic.
Neon, Z = 10
The neon atom resides in the p-block on the periodic table and has ten electrons. The quantum numbers are:
- n = 1, 2
- l = 0 ( s ), 1 ( p )
- ml = 0 (for l = 0) and ml = –1, 0, 1 (for l = 1)
The corresponding electron configuration is
\[1s^22s^22p^6\]
or
\[[\mathrm{He}]~2s^22p^6\]
The quantum numbers for each electron are
- (1s “spin-up” core electron) n = 1, l = 0, ml = 0, and ms = 1/2
- (1s “spin-down” core electron) n = 1, l = 0, ml = 0, and ms = –1/2
- (2s “spin-up” valence electron) n = 2, l = 0, ml = 0, and ms = 1/2
- (2s “spin-down” valence electron) n = 2, l = 0, ml = 0, and ms = –1/2
- (2p “spin-up” valence electron) n = 2, l = 1, ml = –1, and ms = 1/2
- (2p “spin-up” valence electron) n = 2, l = 1, ml = 0, and ms = 1/2
- (2p “spin-up” valence electron) n = 2, l = 1, ml = 1, and ms = 1/2
- (2p “spin-down” valence electron) n = 2, l = 1, ml = –1, and ms = –1/2
- (2p “spin-down” valence electron) n = 2, l = 1, ml = 0, and ms = –1/2
- (2p “spin-down” valence electron) n = 2, l = 1, ml = 1, and ms = –1/2
This element is diamagnetic.
Electron configurations for Z = 1 → 109
Valence electrons
Valence electrons of atoms reside in valence shell, a set of orbitals that are energetically accessible for accepting electrons to form chemical bonds. The valence orbitals of different groups of elements are outlined below. Note that the indicated orbitals are the outermost electron shells (largest n for a given element).
- Hydrogen and helium
- 1s
- s- and p-blocks (main group elements)
- ns
- np
- d-block (transition metals)
- ns
- (n-1)d
- np
- f-block (lanthanides and actinides)
- ns
- (n-2)f
- (n-1)d
- np
Be careful when determining the number of valence electrons in ions.
Fluoride (F–) has 8 valence electrons as all 8 electrons in the n = 2 shell are energetically accessible for bonding.
Sodium(1+) (Na+) has zero valence electrons as the electrons in the outermost shell (n = 2) are not energetically accessible for bonding. It would be very difficult share the electrons in the n = 2 shell with a neighboring atom as the electrons of negative charge are very highly attracted to the sodium cation with a +1 charge.
Examples of these definitions are given below.
Hydrogen and helium
The “outermost” electrons in hydrogen and helium equal the number of electrons they contain (1 and 2, respectively). These elements have no “core” electrons.
s- and p-block (main group elements)
A lithium atom ([He]2s1) has 1 valence electron (2 core). The largest principle quantum number, n, is 2. Therefore, the following electrons are counted as valence electrons:
- ns = 2s electrons = 1 electron
A nitrogen atom ([He]2s22p3) has 5 valence electrons (2 core). The largest principle quantum number, n, is 2. Therefore, the following electrons are counted as valence electrons:
- ns = 2s electrons = 2 electron
- np = 2p electrons = 5 electrons
A bromine atom ([Ar]4s23d104p5) has 7 valence electrons (28 core). The largest principle quantum number, n, is 4. The 3d electrons are core electrons. Therefore, the following electrons are counted as valence electrons:
- ns = 4s electrons = 2 electron
- np = 4p electrons = 5 electrons
d-block (transition metals)
An iron atom ([Ar]4s23d6) has 8 valence electrons. The largest principle quantum number, n, is 4. Therefore, the following electrons are counted as valence electrons:
- ns = 4s electrons = 2 electrons
- (n-1)d = 3d = 6 electrons
f-block (lanthanides and actinides)
A uranium ([Rn]4s24f4) atom has 6 valence electrons. The largest principle quantum number, n, is 6. Therefore, the following electrons are counted as valence electrons:
- ns = 6s electrons = 2 electrons
- (n-1)d = 5d = 0 electrons
- (n-2)f = 4f = 4 electrons
Electron configurations for transition metals
Transition metals can be tricky when it comes to electron configurations. A few d-block metals illustrate exceptions to the general rules laid out above.
Consider chromium (Z = 24). The lowest energy configuration of electrons leads to a half-filled set of d orbitals where only one electron resides in the 4s orbital.
\[[\mathrm{Ar}]~4s^13d^5 \quad \mathrm{not} \quad [\mathrm{Ar}]~4s^23d^4\]
Some made-up rules seen in textbooks will state that “half-filled d orbitals are more stable”, which sounds nice but ultimately doesn’t apply in all cases such as seen with the element, Nb.
It can be difficult to generalize the behavior of electron configurations for d-block, and certainly f-block elements.
Electron configurations of ions
When writing electron configurations of ions, electrons that are added (to create anions) to or removed from (to create cations) the electron shell of highest n and then maximum l.
Some examples are presented below.
Fluorine(1–), Z = 9
The fluoride anion, F–, has 10 electrons (2 core and 8 valence). The addition of an electron to a fluorine atom changes the electron configuration as shown below:
\[1s^22s^22p^5 ~\overset{+e^-}{\longrightarrow}~ 1s^22s^22p^6\]
or
\[[\mathrm{He}]~2s^22p^5 ~\overset{+e^-}{\longrightarrow}~ [\mathrm{He}]~2s^22p^6\]
The fluoride anion is isoelectronic with the neon atom (i.e. two particles with an equal number of electrons but a different number of protons).
This ion is diamagnetic.
Lithium(1+), Z = 3
The lithium cation, Li+ has 2 electrons (2 core). The removal of an electron from a lithium atom changes the electron configuration as shown below:
\[[\mathrm{He}]~2s^1 ~\overset{-e^-}{\longrightarrow}~ 1s^2\]
The quantum numbers for each electron are
- (1s “spin-up” core electron) n = 1, l = 0, ml = 0, and ms = 1/2
- (1s “spin-down” core electron) n = 1, l = 0, ml = 0, and ms = –1/2
This ion is diamagnetic.
Iron(2+), Z = 26
The iron dication, Fe2+ has 24 electrons (18 core and 6 valence). The removal of two electrons to an iron atom changes the electron configuration as shown below:
\[[\mathrm{Ar}]~4s^23d^6 ~\overset{-2e^-}{\longrightarrow}~ [\mathrm{Ar}]~3d^6\]
Note that the 4s electrons are removed and not the d electrons. This ion is paramagnetic.
Iron(3+), Z = 26
The iron trication, Fe3+ has 23 electrons (18 core and 5 valence). The removal of two electrons to an iron atom changes the electron configuration as shown below:
\[[\mathrm{Ar}]~4s^23d^6 ~\overset{-3e^-}{\longrightarrow}~ [\mathrm{Ar}]~3d^5\]
Note that the 4s electrons are removed first before the removal of a d electron. This ion is paramagnetic.