2.6: Writing Electron Configurations

LEARNING OBJECTIVES

• Describe the locations of electrons in an atom using an electron configuration.

In the previous section, the following electron configuration was generated.  Recall that this notation can be used to explicitly state the location of individual electrons within the first six orbitals/orbital sets of an atom.

1s²2s²2p⁶3s²3p⁶4s²

When applying this pattern, two rules must be followed:

1. In order to maximize their stability, electrons will always prefer to occupy the lowest energy orbital that is available.  Therefore, the energy level/orbital combinations given within the electron configuration should be filled in the order shown above, beginning with the "1s" energy level/orbital combination.
2. The number of electrons assigned to each energy level/orbital combination should never exceed the maximum indicated by the superscript.  If an atom contains more electrons than what that energy level/orbital combination can hold, fill the next energy level/orbital combination in the configuration, until all electrons have been placed.

For example, consider an atom of sulfur.

In order to establish the location of sulfur's electrons, the total number of electrons in an atom of sulfur (S) must first be determined.  Recall that the number of protons and electrons in an atom must be equal, in order for that atom to have an overall neutral charge.  Furthermore, the number of protons present in an atom is defined by the element's atomic number.  Therefore, since sulfur has an atomic number of 16, an atom of sulfur contains 16 protons and 16 electrons. 

The location of each of sulfur's electrons can then be determined by applying the given electron configuration pattern until the superscripts total to 16.  Based on the rules given above, the "1s" energy level/orbital combination must be filled first.  However, all 16 electrons cannot be placed in the "1s" orbital, since the maximum number of electrons indicated by its superscript is 2.  Therefore, this energy level/orbital combination is completed by filling it to its maximum capacity, as follows:

1s²

Since sulfur still has 14 additional electrons that must be placed, the next energy level/orbital combination, the  "2s" orbital, must be filled.  However, all 14 electrons cannot be placed in the  "2s" orbital, since the maximum number of electrons indicated by its superscript is 2.  Therefore, this energy level/orbital combination is completed by filling it to its maximum capacity, as follows:

1s²2s²

Sulfur still has 12 additional electrons that must be placed, so the next energy level/orbital combination, the "2p" orbital, must be filled, but all 12 electrons cannot be placed in the "2p" orbital, since the maximum number of electrons indicated by its superscript is 6.  Therefore, this energy level/orbital combination is completed by filling it to its maximum capacity, as follows:

1s²2s²2p⁶

Sulfur still has 6 additional electrons that must be placed, so the next energy level/orbital combination, the "3s" orbital, must be filled.  However, all 6 electrons cannot be placed in the "3s" orbital, since the maximum number of electrons indicated by its superscript is 2.  Therefore, this energy level/orbital combination is completed by filling it to its maximum capacity, as follows:

1s²2s²2p⁶3s²

Sulfur still has 4 additional electrons that must be placed, so the next energy level/orbital combination, the "3p" orbital, must be filled.  In this case, all 4 electrons can be placed in the "3p" orbital, since the maximum number of electrons indicated by its superscript is 6.  Therefore, these 4 electrons are assigned to the "3p" orbital, as follows:

1s²2s²2p⁶3s²3p⁴

Since all of sulfur's electrons have been placed, this final entry represents sulfur's electron configuration.

Example \(\PageIndex{1}\)

Write the electron configuration for nitrogen.

Solution

In order to write an electron configuration for nitrogen (N), the total number of electrons in an atom of nitrogen must first be determined.  Since nitrogen has an atomic number of 7, an atom of nitrogen contains 7 protons and 7 electrons. 

The location of each of nitrogen's electrons is determined by applying the given electron configuration pattern until the superscripts total to 7.  Based on the rules given above, the "1s" energy level/orbital combination must be filled first.  However, all 7 electrons cannot be placed in the "1s" orbital, since the maximum number of electrons indicated by its superscript is 2.  Therefore, this energy level/orbital combination is completed by filling it to its maximum capacity, as follows:

1s²

Since nitrogen still has 5 additional electrons that must be placed, the next energy level/orbital combination, the "2s" orbital, must be filled.  However, all 5 electrons cannot be placed in the "2s" orbital, since the maximum number of electrons indicated by its superscript is 2.  Therefore, this energy level/orbital combination is completed by filling it to its maximum capacity, as follows:

1s²2s²

Nitrogen still has 3 additional electrons that must be placed, so the next energy level/orbital combination, the "2p" orbital, must be filled.  In this case, all 3 electrons can be placed in the "2p" orbital, since the maximum number of electrons indicated by its superscript is 6.  Therefore, these 3 electrons are assigned to the "2p" orbital, as follows:

1s²2s²2p³

Since all electrons have been placed, this final entry represents nitrogen's electron configuration.

Exercise \(\PageIndex{1}\)

Write the electron configuration for each of the following elements.

1. Neon
2. Calcium

Answer a

In order to write an electron configuration for neon (Ne), the total number of electrons in an atom of neon must first be determined.  Since neon has an atomic number of 10, an atom of neon contains 10 protons and 10 electrons. 

The location of each of neon's electrons is determined by applying the given electron configuration pattern until the superscripts total to 10.  The "1s" energy level/orbital combination must be filled first.  However, all 10 electrons cannot be placed in the "1s" orbital.  Therefore, this energy level/orbital combination is completed by filling it to its maximum capacity, as follows:

1s²

The "2s" orbital must be filled next.  The remaining 8 electrons cannot all be placed in the "2s" orbital, so it is filled to its maximum capacity, as follows:

1s²2s²

The "2p" orbital must be filled next.  In this case, all 6 electrons can be placed in the "2p" orbital, since the maximum number of electrons indicated by its superscript is 6.  Therefore, these 6 electrons are assigned to the "2p" orbital, as follows:

1s²2s²2p⁶

Since all electrons have been placed, this final entry represents neon's electron configuration.

Answer b

In order to write an electron configuration for calcium (Ca), the total number of electrons in an atom of calcium must first be determined.  Since calcium has an atomic number of 20, an atom of calcium contains 20 protons and 20 electrons. 

The "1s" energy level/orbital combination must be filled first.  However, all 20 electrons cannot be placed in the "1s" orbital.  Therefore, this energy level/orbital combination is completed by filling it to its maximum capacity, as follows:

1s²

The "2s" orbital must be filled next.  The remaining 18 electrons cannot all be placed in the "2s" orbital, so it is filled to its maximum capacity, as follows:

1s²2s²

The "2p" orbital must be filled next.  The remaining 16 electrons cannot all be placed in the "2p" orbital, so it is filled to its maximum capacity, as follows:

1s²2s²2p⁶

The "3s" orbital must be filled next.  The remaining 10 electrons cannot all be placed in the "3s" orbital, so it is filled to its maximum capacity, as follows:

1s²2s²2p⁶3s²

The "3p" orbital must be filled next.  The remaining 8 electrons cannot all be placed in the "3p" orbital, so it is filled to its maximum capacity, as follows:

1s²2s²2p⁶3s²3p⁶

The "4s" orbital must be filled next.  In this case, both of the remaining electrons can be placed in the "4s" orbital, since the maximum number of electrons indicated by its superscript is 2.  Therefore, these 2 electrons are assigned to the "4s" orbital, as follows:

1s²2s²2p⁶3s²3p⁶4s²

Since all electrons have been placed, this final entry represents calcium's electron configuration.  Note that calcium's electron configuration utilizes the complete six orbital/orbital set pattern.  The given pattern was deliberately designed to stop after the sixth orbital/orbital set, as the next orbital/orbital set would require the inclusion of d orbitals, which are not being discussed in this text.