Aufbau’s Principle, Hund’s Rule, and Pauli’s Exclusion Principle are rules for writing electron configurations, so go over the main principles of electron configurations if you need to before we start discussing them.

**Aufbau’s Principle**

Aufbau (German *aufbauen, *“to build up”) principle tells us that electrons fill the orbitals in the order of increasing their energy level. Remember, the main energy level is given by the principal quantum number, *n*, and it increases down the periodic table.

Here is a chart on the energy levels for up n = 4 which include the d sublevel:

Consider also that within the **same principal level**, orbitals with a lower value of

*l*have lower energy (

*E*) and therefore, are filled first. So, for a given value of

*n*:

*E *(** s** orbital) <

*E*(

**orbital) <**

*p**E*(

**orbital) <**

*d**E*(

**orbital)**

*f*Notice that in the periodic table, it is opposite to what we show on the energy level diagram since the periods indicate the principal quantum number (*n*), which is the energy level, and they increase as we go down the periodic table. The general order for filling the electrons based on the energy levels and orbitals can be shown as in the picture below:

Let’s look at a few examples of following the Aufbau principle. For example, hydrogen has **one electron** and of course, we are going to put it in the **1s orbital** it has the lowest energy.

The electron configuration of He with orbital diagrams can be shown as:

Next, we have helium, He – the second element in the periodic table. Remember, the **number of electrons is equal to the atomic number** as the number of electrons must be equal to the number of electrons in a neutral atom. Therefore, He has **two electrons** and they **both go to the 1s level** because each orbital, regardless of if it is *s*, *p*, *d*, or *f*, can only accommodate a maximum of two electrons, and they **wouldn’t go to the 2s level without filling the 1s:**

The electron configuration of He with orbital diagrams can be shown as:

And this keeps going by adding one more electron to the next lowest energy orbital for every subsequent element in the periodic table.

**Pauli’s Exclusion Principle**

Notice that the arrow representing the second electron is pointing down, opposite to the first arrow which indicates that they must have **opposite spins**. This is according to the ** Pauli exclusion principle** which states that

**no two electrons**in an atom can have the

**same**. So, if the electrons are in the same orbital, they must have the same

*four*quantum numbers*n*,

*l*,

*m*

_{l }values, and therefore, the only one that can be different is the

*m*

_{s}which is the spin of the electron shown by the position of the arrow.

For example, for the electrons in a 2*p* orbital, *n* = 2, *l* = 1. and *m*_{l} = –1, 0, or +1 (doesn’t matter which one because for the same orbital it will be identical). Therefore, the *m*_{s} must be different (+1/2 or -1/2) so we don’t violate Pauli’s exclusion principle.

The principle applies to all the orbitals. For example, notice how the electrons in each p orbital of oxygen, fluorine, and neon are with opposite spins:

**Hund’s Rule**

To understand Hund’s tule, let’s first write the electron configuration of boron:

Now, the next election, which indicates carbon, is going to have the option of **pairing up** with the one in the p orbital or **going to the next empty p orbital**. And it turns out that the **electron goes** to the **next** (**empty**) *p* orbital rather than fitting in with the other electron.

This is the **Hund’s rule**, which states that **electrons will fill** all the degenerate orbitals (equal in energy) with **parallel spins (both arrows up or down) first** before pairing up in one orbital. We can also formulate it as the lowest energy configuration for an atom is the one having the **maximum number of unpaired electrons** within the **same energy sublevel.**

**Hund’s rule **is another demonstration of the same principle which is the **tendency to adopt the lowest energy** state possible. There is a stronger repulsive interaction between two electrons in the same orbital compared to when they occupy separate orbitals of equal energy.

Let’s show the application of Hund’s rule in explaining the electron configuration of carbon:

Notice that placing the electron unpaired in the 3s orbital is also incorrect because, it is important to mention, that **Hund’s rule applies to electrons in the same energy level**. We are not going to place one of the electrons in the 3s orbital just for the sake of keeping a maximum number of unpaired electrons. It is energetically more favorable to have the electrons in the same lower sublevel rather than separating them by placing one in a higher energy level.

** **We can also demonstrate this using the orbital diagram of carbon:

As expected, the next element, nitrogen will place the additional electron in the 3^{rd} *p* orbital which is empty in carbon.

Some exceptions occur in the electron configuration of d and f elements. First, recall that the **( n +1)s** orbitals always

**fill before**the

**For example, the 4s level fills before the 3**

*n*d orbitals.*d*level, and therefore, the electron configuration of Ti is 1

*s*

^{²}2

*s*

^{²}2

*p*

^{⁶}3

*s*

^{²}3

*p*

^{⁶}4

*s*

^{²}3

*d*

^{² }or, the condensed configuration which will be [Ar]4

*s*

^{²}3

*d*

^{²}.

Sometimes having half-filled orbitals compensate for the energy gain associated with placing the electron in a higher-level orbital.

For example, unlike the expected [Ar]4*s*^{²}3*d*^{4}, the **electron configuration of Cr is [Ar]4 s^{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 4

*s*level to the 3

*d*level and compensating this energy uphill by a stabilization associated with half-filled orbitals:

To summarize, remember that:

- The
**Aufbau principle**is about filling the orbitals from**lower to higher energy** - Pauli’s exclusion principle tells us to place the
**arrows**of electrons in the same orbital in**opposite directions**. **Hund’s rule**tells us to place the electrons in different orbitals of the same energy sublevel (s, p, d, f) rather than pairing them up –**more unpaired electrons**is better.

**Check Also**

- Atomic Orbitals
- Electron Configurations
- Electron Configurations of Ions
- Orbital Diagrams
- Hund’s Rule
- Pauli Exclusion Principle
- Quantum Numbers (
*n*,*l*,*m*_{l},*m*_{s}) - Bohr Model of the Hydrogen Atom
- Rydberg Formula
- The Photoelectric Effect
- Calculating The Energy of a Photon
- Ionization Energy
- Electron Affinity
**Energy, Wavelength, and Frequency Practice Problems**

#### Practice

What is the maximum number of electrons that can occupy one p orbital?

a.14

b. 2

c. 10

d.1

e. 6

The n = 2 shell can accommodate a maximum of ____ electrons.

A. 10

B. 8

C. 4

D. 6

E. 2

Which of the following is an incorrect orbital occupancy representation?

- 4d
^{3} - 3d
^{5} - 2s
^{3} - 1s
^{1} - 2p
^{2}

The maximum number of electrons that can be accommodated in 3s subshell is

- 10
- 2
- 6
- 1
- 8

The following electron configuration is incorrect:

- 1s
^{2}2s^{2} - 1s
^{2}2s^{3}2p^{3} - 1s
^{2}2s^{2}2p^{5} - 1s
^{2}2s^{2}2p^{4} - 1s
^{2}2s^{2}2p^{6}3s^{2}

Which electron configuration represents an excited state of the indicated atom?

**He**: 1s^{2}**Na**: 1s^{2}2s^{2 }2p^{6}3s^{2}3p^{2}3s^{1}**P**: 1s^{2}2s^{2}2p^{6}3s^{2 }3p^{3}**N**: 1s^{2}2s^{2 }2p^{2}3s^{1}**Ne**: 1s^{2}2s^{2}2p^{6}

Which of the following represents the ground state electron configuration of a transition element?

- 1s
^{2}2s^{2}2p^{6}3s^{2}3p^{5} - 1s
^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}4s^{1} - 1s
^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}4s^{2}4p^{4} - 1s
^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2} - 1s
^{2}2s^{2}2p^{3}

Which of the following is the correct ground-state electron configuration of V is?

- 1s
^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{3} - 1s
^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{5} - 1s
^{2}2s^{2}2p^{6}3s^{2}4s^{2}3d^{3} - 1s
^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{5} - 1s
^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{3}

Which of the following represents a possible excited-state electron configuration for an iron atom?

- [Ar]3d
^{7}4s^{2} - [Ar]3d
^{6}4s^{2} - [Ar]3d
^{7}4s^{1} - [Ar]3d
^{6}4s^{1} - [Kr]3d
^{7}4s^{1}

Each of the following is an accurate representation of ground-state electron configuration except:

**Fe**: [Ar]4s^{2}3d^{6}.**Ca**: [Ar]4s^{2}.**Se**: [Ar] 4s^{2}3d^{10}4p^{4}.**Ag**: [Kr] 5s^{2}4d^{9}.**Ti**: [Ar] 4s^{2}3d^{2}

How many electrons, in total, are present in p orbitals in a ground-state nickel atom?

- 8
- 12
- 6
- 24
- 3

Which one of the following represents electron configuration of an excited carbon atom?

- 1s
^{2}2s^{2}2p^{3} - 1s
^{2}2s^{2}2p^{1} - 1s
^{2}2s^{2}2p^{1}3s^{1} - 1s
^{2}2s^{2}3s^{1} - 1s
^{2}2s^{2}2p^{2}

Which of the following ground-state electron configurations is incorrect?

**Cl**: [Ne]3s^{2}3p^{5}**Ge**: [Ar] 4s^{2 }3d^{10}3p^{2}**Co**: [Ar]4s^{2}3d^{7}**Rb**: [Kr]5s^{1}**Mn**: [Ar] 4s^{2}3d^{5}

What is the ground-state electron configuration of chromium (Cr)?

- 1s
^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}4s^{2}4p^{6}4d^{10}5s^{2}5p^{4} - 1s
^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{1}3d^{5} - 1s
^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10} - 1s
^{2}2s^{2 }2p^{6}3s^{2}3p^{6}4s^{2}3d^{4} - 1s
^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{6}

What is the ground-state electron configuration of bromine (Br)?

- 1s
^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{1}3d^{10 }4p^{5} - [Kr] 4s
^{2}3s^{10}4p^{5} - [Ne] 3s
^{2}3p^{5} - [Ar] 4s
^{2}3d^{10}4p^{5} - 1s
^{2}2s^{2}2p^{5}3s^{2}3p^{6}4s^{2}3d^{10}4p^{5}

The ground state electron configuration of As is ________.

- 1s
^{2}2s^{2}3s^{2}3p^{6}4s^{2}3d^{10}4p^{2} - 1s
^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{1} - 1s
^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{3} - 1s
^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4d^{3} - [Kr]4s
^{2}3d^{10}4d^{3}

Which of the following elements has the same valence-shell electron configuration as lithium?

- calcium.
- sulfur.
- potassium.
- magnesium.
- argon.

How many unpaired electrons are found in the d orbital of nickel.

- 4 electrons
- 3 electrons
- 2 electrons
- 8 electrons
- none of these

The ground-state electron configuration of the element ________ is [Kr]5s^{2}4d^{10}

- Fe
- Cd
- Cr
- Mn
- Zn

The ground-state electron configuration of ________ is [Ar]4s^{2}3d^{7}.

- Ti
- Co
- Fe
- Cr
- Cs

Chlorine has the following ground-state configuration:

- [He]3s
^{2}3p^{2} - [Ne]3s
^{2}3p^{5} - [He]2s
^{2}2p^{5} - [Ne]2s
^{2}2p^{3} - [He]3s
^{2}3p^{5}

What is the maximum number of electrons that can be accommodated in each d-subshell?

- 10
- 5
- 6
- 3
- 2

What noble gas core should be used when writing the condensed electron configuration of Strontium (Sr)?

- [Xe]
- [Kr]
- [Ne]
- [Ar]
- [He]

# Electron Configuration of Ions

Which of the following is the ground state electron configuration of S^{2-}?

A) [Ne]

B) [Ne]3s^{2}3p^{6
}C) [Ne]3s^{2}3p^{4
}D) [Ne]3s^{2}3p^{2
}E) [Ne]4s^{2}

A cation of +3 indicates that an element has

A) lost three neutrons.

B) lost three protons.

C) lost three electrons.

D) gained three protons.

E) gained three electrons.

Indicate the ground state electron configuration for Cl⁻.

A) 1s^{2}2s^{2}2p^{6}3s^{2}3p^{4
}B) 1s^{2}2s^{2}2p^{6}3s^{2}3p^{5
}C) 1s^{2}2s^{2}2p^{6}3s^{2}3p^{6
}D) 1s^{2}2s^{2}2p^{6}3s^{1}3p^{6
}E) 1s^{2}2s^{2}2p^{6}3s^{3}3p^{5}

Which of the following is the ground state electron configuration of Ba^{2+ }?

A) [Kr]5s^{2}4d^{10}5p^{6}6s^{2
}B) [Kr]5s^{2}4d^{10}5p^{6
}C) [Kr]5s^{2}4d^{10}5p^{6}5d^{2
}D) [Kr]5s^{2}5p^{6
}E) [Kr]5s^{2}4d^{10}5p^{6}6s^{2}

Which of the following is the ground state electron configuration of Ti^{2⁺}.

A) [Ar]3d^{4
}B) [Ar]3d^{2
}C) [Ar]4s^{2
}D) [Ar]4s^{2}3d^{4
}E) [Ar]4s^{2}3d^{2}

Which of the following is the ground state electron configuration of Zn^{2⁺}?

A) [Ar]4s^{2}3d^{6
}B) [Ar]3d^{10
}C) [Ar]4s^{2}3d^{8
}D) [Ar]3d^{8
}E) [Ar]

Which of the following is the ground state electron configuration of Cr^{3⁺}?

A) [Ar]

B) [Ar]4s^{1}3d^{2
}C) [Ar]3d^{3
}D) [Ar]4s^{2}3d^{1
}E) [Ar]4s^{2}3d^{7}

How many valence electrons are present in the azide (N^{3-}) ion?

A) 3

B) 5

C) 8

D) 7

E) 2

Ca^{2+} has the following ground state electron configuration:

A) 1s^{2}2s^{2}2p^{6
}B) 1s^{2}2s^{2}2p^{6}3s^{2
}C) 1s^{2}2s^{2}2p^{6}3s^{2}3p^{2
}D) 1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{2
}E) 1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}

Rb+ has the following ground state electron configuration:

A) [Kr]5s^{2}4d^{6
}B) [Kr]4s^{2
}C) [Ar]4s^{2}4p^{4
}D) [Kr]5s^{1
}E) [Ar]4s^{2}3d^{10}4p^{6}

Which of the following is the ground state electron configuration of Fe^{3⁺}?

A) [Ar]3d^{5}** ^{
}**B) [Ar]4s

^{1}3d

^{3 }C) [Ar]

D) [Ar]4s

^{2}3d

^{9 }E) [Ar]4s

^{2}3d

^{1}

The correct ground state electron configuration for Br^{–} is:

A) [Ar]4s^{2}3d^{10}4p^{6
}B) [Ar]4s^{2}3d^{10}4p^{3
}C) [Ar]4s^{2}3d^{8}4p^{6
}D) [Ar]4s^{2}3d^{10}4p^{5
}E) [Ar]4s^{2}4p^{6}

How many electrons are present in the ground state of Mg^{2+} ion?

A) 2

B) 6

C) 10

D) 8

E) 12