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ChemWiki: The Dynamic Chemistry E-textbook > Inorganic Chemistry > Electronic Configurations

Electronic Configurations

The electron configuration of an atom is the representation of the arrangement of electrons that are distributed among the orbital shells and subshells. Commonly, the electron configuration is used to describe the orbitals of an atom in its ground state, but it can also be used to represent an atom that has ionized into a cation or anion by compensating with the loss of or gain of electrons in their subsequent orbitals. Many of the physical and chemical properties of elements can be correlated to their unique electron configurations. The valence electrons, electrons on the outer most shell, become the determining factor for the unique chemistry of the element.

Introduction

Before we begin assigning the electrons of an atom into orbitals, we must first familiarize ourselves with the basic concepts needed to become fluent in electron configurations. Every element on the periodic table consists of an atom which is composed of protons, neutrons, and electrons. In these situations, we are concerned with the electrons. Electrons exhibit a negative charge and are found around within the nucleus of the atom. Electron orbitals are the position of the electrons around the nucleus and is determined as the volume of space in which the electron can be found within 95% probability. The four different types of orbitals s,p,d, and f have different shapes and one orbital can hold a maximum of two electrons. The p, d, and f orbitals have different sublevels unlike the s orbital and thus can hold more electrons.

As stated, the electron configuration of each element is unique to its position on the periodic table. The energy level is determined by the period and amount of electrons by the atomic number of the element. Orbitals on different energy levels are similar to each other, but they occupy different areas in space. The 1s orbital and 2s orbital both have the characteristics of an s orbital (radial nodes, spherical volume probabilities, can only hold two electrons, etc.) but as they are found in different energy levels they occupy different spaces around the nucleus. Each orbital can be represented by specific blocks on the periodic table. The s-block is the region of the Alkali metals including Helium (groups 1 & 2), the d-block is the Transition metals (groups 3 to 12), the p-block are the main group elements from group 13 to 18, and the f-block are the Lanthanides and Actinides series. 

PeriodicTable2.jpg

Using the periodic table to determine the electron configurations of atoms is key, but also keep in mind that there are certain rules to follow when assigning electrons to different orbitals. The periodic table is an incredibly helpful tool in writing electron configurations. For more information on how electron configurations and the periodic table are linked, visit the Connecting Electrons to the Periodic Table module.

Rules for Assigning Electron Orbitals

Occupation of Orbitals

The first thing to keep in mind is that electrons fill orbitals in a way to minimize the energy of the atom. This would mean that the electrons in an atom would fill the principal energy levels in order of increasing energy (the electrons are getting farther from the nucleus). The order of levels filled would look like this: 

1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, and 7p

One way to remember this pattern, probably the easiest, is to refer to the periodic table and remember where each orbital block falls to logically deduce this pattern. Another way is to make a table like the one below and use vertical lines to determine which subshells correspond to each other. 

subshells.jpg

Pauli Exclusion Principle

The second major fact to keep in mind is the Pauli Exclusion Principle which states that no two electrons can have the same four quantum numbers. The first three (\(n\),\(l\), and \(m_l\)) may be similar but the fourth quantum number must be different.A single orbital can hold a maximum of two electrons, which must have opposing spins; otherwise they will have the same four quantum numbers and that is not allowed. One electron will be spin up (\(m_s =+1/2\)) and the other would spin down (\(m_s =-1/2\)). This tells us that each subshell has double the electrons per orbital. The s subshell has 1 orbital that can hold to 2 electrons, the p subshell has 3 orbitals that can hold up to 6 electrons, the d subshell has 5 orbitals that hold up to 10 electrons, and the f subshell has 7 orbitals with 14 electrons.

 

Example 1

We have the first three quantum numbers \(n=1\), \(l=0\), \(m_l=0\). Only two electrons can correspond to these, which would be either \(m_s = -1/2\) or \(m_s = +1/2\). As we already know from our studies of quantum numbers and electron orbitals, we can conclude that these four quantum numbers refer to 1s subshell. If only one of the \(m_s\) values are given then we would have 1s1 (denoting Hydrogen) if both are given we would have 1s2 (denoting Helium). Visually this would be represented as:

pauliexample.jpg

As you can see, the 1s subshell can hold only two electrons and when filled the electrons have opposite spins.

Hund's Rule

When assigning electrons in orbitals, each electron will first fill all the orbitals with similar energy (also referred to as degenerate) before pairing with another electron in a half-filled orbital. Atoms at ground states tend to have as many unpaired electrons as possible. When visualizing this processes, think about how electrons are exhibiting the same behavior as the same poles on a magnet would if they came into contact; as the negatively charged electrons fill orbitals they first try to get as far as possible from each other before having to pair up.

Example

If we look at the correct electron configuration of Nitrogen (Z = 7), a very important element in the biology of plants: 1s2 2s2 2p3

Nitrogenexample.jpg

We can clearly see that p orbitals are half filled as there are three electrons and three p orbitals. This is because Hund's Rule states that the three electrons in the 2p subshell will fill all the empty orbitals first before filling orbitals with electrons in them. If we look at the element after Nitrogen in the same period, Oxygen (Z = 8) its electron configuration is: 1s2 2s2 2p4

oxygenexample.jpg

Oxygen has one more electron than Nitrogen and as the orbitals are all half filled the electron must pair up.

The Aufbau Process

Aufbau comes from the German word "Aufbauen" which means "to build". When writing electron configurations, we are building up electron orbitals as we proceed from atom to atom. As we write the electron configuration for an atom, we will fill the orbitals in order of increasing atomic number.  However, there are some exceptions to this rule.

Example

If we follow the pattern across a period from B (Z=5) to Ne (Z=10) the number of electrons increase and the subshells are filled. Here we are focusing on the p subshell in which as we move towards Ne, the p subshell becomes filled.

  • B (Z=5) configuration: 1s2 2s2 2p1
  • C (Z=6) configuration:1s2 2s2 2p2
  • N (Z=7) configuration:1s2 2s2 2p3
  • O (Z=8) configuration:1s2 2s2 2p4
  • F (Z=9) configuration:1s2 2s2 2p5
  • Ne (Z=10) configuration:1s2 2s2 2p6

Exceptions

While Aufbau process is an accurate in determining the electron configuration of most elements, there are some notable exceptions that occur within the transition metals and heavier elements. The reason these exceptions occur is because some elements are more stable with less electrons in some subshells and more electrons within others.  A list of the exceptions to the Aufbau process can be found below.

 

Table 1: Exceptions to Electron Configuration Trends

Period 4:

Period 5:

Chromium: Z:24 [Ar] 3d54s1

Niobium: Z:41 [Kr] 5s1 4d4

Copper: Z:27 [Ar] 3d104s1

Molybdenum: Z:42 [Kr] 5s1 4d5

 

Ruthenium: Z:44 [Kr] 5s1 4d7

 

Rhodium:  Z:45 [Kr] 5s1 4d8

 

Palladium: Z:46 [Kr] 4d10

 

Silver: Z:47 [Kr] 5s1 4d10

 

 

Period 6:

Period 7:

Lanthanum: Z:57 [Xe] 6s2 5d1

Actinium: Z:89 [Rn] 7s2 6d1

Cerium: Z:58 [Xe] 6s2 4f1 5d1

Thorium: Z:90 [Rn] 7s2 6d2

Gadolinium: Z:64 [Xe] 6s2 4f7 5d1

Protactium: Z:91 [Rn] 7s2 5f2 6d1

Platinum: Z:78 [Xe] 6s1 4f14 5d9

Uranium: Z:92 [Rn] 7s2 5f3 6d1

Gold: Z:79 [Xe] 6s1 4f14 5d10

Neptunium: Z:93 [Rn] 7s2 5f4 6d1

 

Curium: Z:96 [Rn] 7s2 5f2 6d1

 

Lawrencium: Z:103 [Rn] 7s2 5f14 7p1

Writing Electron Configurations

When writing the electron configuration we first write the energy level (the period) then the subshell to be filled and the superscript, which is the number of electrons in that subshell. The total number of electrons as mentioned before is the atomic number, Z. Using the rules from above, we can now start writing the electron configurations for all the elements in the periodic table.

There are three main methods used to write electron configurations:

  1. orbital diagrams
  2. spdf notation
  3. noble gas notation

Each method has its own purpose and each has its own drawbacks.

Orbital Diagrams

As seen in some examples above, the orbital diagram is a visual way to reconstruct the electron configuration by showing each of the separate orbitals and the spins on the electrons. This is done by first determining the subshell (s,p,d, or f) then drawing in each electron according to the stated rules above.

Example: Aluminum

Write the electron configuration for aluminum.

SOLUTION

We known that aluminum is in the 3rd period and it has an atomic number of Z=13. If we look at the periodic table we can see that its in the p-block as it is in group 13. Now we shall look at the orbitals it will fill: 1s, 2s, 2p, 3s, 3p. We know that aluminum completely fills the 1s, 2s, 2p, and 3s orbitals because mathematically this would be 2+2+6+2=12.  The last electron is in the 3p orbital. Also another way of thinking about it is that as you move from each orbital block, the subshells become filled as you complete each section of the orbital in the period. The block that the atom is in (in the case for aluminum: 3p) is where we will count to get the number of electrons in the last subshell (for aluminum this would be one electron because its the first element in the period 3 p-block). From this we can construct the following:

Aluminum.jpg

Note that in the orbital diagram, the two opposing spins of the electron can be seen. This is why it is sometimes useful to think about electron configuration in terms of the diagram. But because it is the most time consuming method, it is more common to write or see electron configurations in the spdf notation and noble gas notation. Another example is the electron configuration of iridium:

Ir1.jpgIr2.jpgIr3.jpg

The electron configuration of iridium is much longer than aluminum. Though drawing out each orbital may prove to be helpful in determining unpaired electrons, its very time consuming and often not as practical as the spdf notation especially in the case of atoms with much longer configurations. The Hund's Rule is also present above as each electron fills up each 5d orbital before being forced to pair with another electron.

Electron Notation using spdf

The most common way to describe electron configurations is to write distributions in the spdf notation. Though, the distributions of electrons in each orbital may not be seen as in the diagram, the total number of electrons in each energy level is described by a superscript that follows the relating energy level. To write the electron configuration of an atom, we will first describe which energy level we are referring to and write the number of electrons in the energy level as its superscript like this: 1s2 This denotes a full s orbital and would refer to the electron configuration of helium. As usual, we will use the periodic table as a reference to accurately write the electron configurations of all atoms. 

Example: Yttrium

First we will start with a straight forward problem, finding the electron configuration of the element Yttrium. As always we will refer to our periodic table. The element Yttrium (symbolized as Y) is found in the fifth period and in group 3 making it a transition metal. In total it has thirty-nine electrons. Its electron configuration would be:

 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d1

This is a much simpler and efficient way to portray electron configuration of an atom. A logical way of thinking about it is that all we have to do is to fill orbitals as we move across a period and through orbital blocks. We know that the amount of elements in each block is the same as in the energy level it corresponds. For example, there are 2 elements in the s-block, and 10 elements in the d-block. As we move across we simply count how many elements fall in each block. We know that yttrium is the first element in the fourth period d-block, thus this corresponds to one electron in the that energy level. To check our answer we would just add all the superscripts to see if we get the atomic number. In this case 2+2+6+2+6+2+10+6+2+1= 39 and Z=39 thus we have the correct answer.

A slightly more complicated example is the electron configuration of bismuth (symbolized as Bi with Z = 83). Looking at our periodic table we can get the following electron configuration:

1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p65s2 4d10 5p6 6s2 4f14 5d10 6p3

The reason why this electron configuration would seem to be more complex is because we must go through the f-block, the Lanthanide series. Most students who first learn electron configurations often have trouble with configurations that must pass through the f-block because they often overlook this break in the table and will skip that energy level. Its important to remember that when passing the 5d and 6d energy levels that we must pass through the f-block Lanthaniod and Actiniod series. If we keep this in mind, this "complex" problem will seem like second nature.

Another way (but less commonly used) to write the spdf notation is the expanded notation format. Its basically the same concept except that each individual orbital is represented with a subscript. We know that in the p, d, and f orbitals have different sublevels. The p orbitals are px py pz  and if represented on the 2p energy with full orbitals would look like: 2px2 2py2 2pz2. If we look at the expanded notation for neon (Ne, Z=10) it would look like the following:

1s2 2s2 2px2 2py2 2pz2

The individual orbitals are represented here but the spins on the electrons are not; we assume opposite spins. When representing the configuration of an atom with half filled orbitals we would just write the two half filled orbitals. For carbon the expanded notation would look like the following:

1s2 2s2 2px1 2py1

As this is form of the spdf notation is not usually used, its not as important to dwell on this detail as it is to understand how to use the general spdf notation.  

Noble Gas Notation

This brings up an interesting point about elements and electron configurations. As the p subshell is filled in the above example about the Aufbau Principal (trend from boron to neon), it reaches the group commonly known as the noble gases. The noble gases have the most stable electron configurations (as all their subshells are filled) and are known for being relatively inert. We can conclude from this that all noble gases have their subshells filled and we can used them as a short hand way of writing electron configurations for subsequent atoms. This method of writing configurations is called the noble gas notation in which the noble gas in the period above the element that is being analyzed is used to denote the subshells that element has filled and after which the valence electrons (electrons filling orbitals in the outer most shells) are written. We will write this notation slightly different than the spdf notation because we must denote our reference noble gas being used. 

 

Example: Vanadium

What is the electronic configuration of vanadium (V, Z=23)?

SOLUTION

Vanadium is a the transition metals at the four period in the fifth group. The noble gas before it is argon, (Ar, Z=18) and knowing that vanadium has filled those orbitals before it, we will use argon as our reference noble gas. We denote the noble gas in the configuration as the symbol, E, in brackets: [E] Now, to find the valance electrons that follow, we will simply do some simple math by subtracting the atomic numbers: 23 - 18 = 5 Now instead of having 23 electrons to distribute in orbitals, we have 5. Now we have enough information to write the electron configuration:

Vanadium, V: [Ar] 4s2 3d3

This method streamlines the process of distributing electrons by showing the valence electron which are determinants in the chemical properties of atoms. Also, when determining the number of unpaired electrons in an atom, this method will allow us to quickly visualize the configurations of the valance electrons. In the example above, we clearly see that we have a full s orbital and three half filled d orbitals.

Outside links

References

  1. Petrucci, Ralph H et al. General Chemistry: Principles & Modern Applications Ninth Edition. , Upper Saddle River, NJ: Pearson Prentice Hall, 2007. Print
  2. Sherman, Alan, Sharon J. Sherman, and Leonard Russikoff. Basic Concepts of Chemistry Fifth Edition. Boston, MA: Houghton Mifflin Company, 1992. Print.
  3. IUPAC. Compendium of Chemical Terminology, 2nd ed. (the "Gold Book"). Compiled by A. D. McNaught and A. Wilkinson. Blackwell Scientific Publications, Oxford (1997). XML on-line corrected version: http://goldbook.iupac.org (2006-) created by M. Nic, J. Jirat, B. Kosata; updates compiled by A. Jenkins. ISBN 0-9678550-9-8.doi:10.1351/goldbook.
  4. Scerri, Eric R. "The Electron Configuration Model, Quantum Mechanics, and Reduction." The British Journal for the Philosophy of Science 42.3 (1991): 309 -325. 
  5. Ostrovsky, V.N. (2004). On recent discussion concerning quantum justification of the periodic table of the elements. Foundations of Chemistry, 75, 93-116.
  6. Meek, T.L., & Allen, L.C. (2002). Configuration irregularities: deviations from the madelung rule and inversion of orbital energy levels. ScienceDirect, 362(5), 362-364.

Problems

Remember, you can use any method you know unless specified. Try to use the one that is the most time efficient and simple for yourself. The answers will be in the noble gas abbreviation.

1. Find the electron configuration of the following:

  1. silicon
  2. tin
  3. lead

2. Scenario: You are currently studying the element iodine and wish to use its electron distributions to aid you in your work.

  1. Find the electron configuration of iodine
  2. How many unpaired electrons does iodine have?

3. Thought Questions:

  1. In your own words describe how to make an electron configuration and why its an important skill to have in the study of chemistry. 
  2. Describe the major concepts (Hunds, Pauli...etc.) and explain why they are a key part of our "tool kit" when describing electron configurations
  3. Why are we able to abbreviate electron configurations with a noble gas in the noble gas notation?

4. Identify the following elements:

  1. 1s2 2s2 2p6 3s2 3p6 4s2 3d6
  2. 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d7
  3. 1s2 2s2 2p6 3s2 3p4
  4. 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 4f14 5d10 6p4

5. Without referring to a Periodic Table or any other references, fill in the correct box in the Periodic Table with the letter of each question. (a)The element with electron configuration: 1s2 2s2 2p6 3s2 3p5; (b)A noble gases with f electrons; (c) a fifth-period element whose atoms have three unpaired p electrons; (d) First rowtransition metals having one 4s electron.

Answers

1. Find the electron configuration of the following:

a) silicon: [Ne] 3s2 3p2

b) tin: [Kr] 5s2 4d10 5p2

c) lead: [Xe] 6s2 4f14 5d10 6p2

2. Scenario: You are currently studying the element iodine and wish to use its electron distributions to aid you in your work.

a) Find the electron configuration of iodine

[Kr] 5s2 4d10 5p5

b) How many unpaired electrons does iodine have?

To find the answer we refer to part a) and look at the valence electrons. We see that iodine has 5 electrons in the p orbitals. We know that the full p orbitals will add up to 6. Using the Hund's rule and Pauli exclusion principals we can make a diagram like the following:

iodine_example.jpg

The answer is one.

3. Thought Questions:

a) In your own words describe how to make an electron configuration and why its an important skill to have in the study of chemistry. 

The first part of this question is straight forward and we can refer to the body of this text if the process cannot be described correctly. The second part is slightly more complicated. Because each individuals knowledge of chemistry differs, there are many answers to this question. The important aspect is that we realize that knowing electron configurations helps us determine the valence electrons on an atom. This is important because valence electrons contribute to the unique chemistry of each atom.

b) Describe the major concepts (Hunds, Pauli...etc.) and explain why they are a key part of our "tool kit" when describing electron configurations

This should also be a straight forward question, and if it seems a little difficult refer to the body of this text about these rules and how they relate to creating an electron configuration. Remember to make logical connections! We know that the main "tools" we have in writing electron configurations are orbital occupation, Pauli exclusion principle, Hund's rule, and the Aufbau Process. Orbitals are occupied in a specific order, thus we have to follow this order when assigning electrons. Pauli exclusion principle states that no two electrons can have all four quantum numbers the same. The fourth quantum number that refers to spin will denote one of two spin directions. This means that in one orbital there can only be two electrons and they mus have opposite spins. This is important when describing an electron configuration in terms of the orbital diagrams. Hund's rule states that electrons first occupy the similar energy orbitals that are empty before occupying those that are half full. This is especially helpful when determining unpaired electrons. Aufbau process denotes the method of "building up" each subshell before moving on to the next; we first fill the 2s orbitals before moving to the 2p orbitals.

c) Why are we able to abbreviate electron configurations with a noble gas in the noble gas notation?

We know that the noble gas has all of its orbitals filled thus it can be used as a "shorthand" or abbreviated method for writing all of the electron configurations after 1s.

4. Identify the following elements:

a) 1s2 2s2 2p6 3s2 3p6 4s2 3d6

The element is iron, Fe

b) 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d7

The element is Rhodium, Rh

c) 1s2 2s2 2p6 3s2 3p4

The element is Sulfur, S

d) 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 4f14 5d10 6p4

The element is Polonium, Po

5. Without referring to a Periodic Table or any other references, fill in the correct box in the Periodic Table with the letter of each question. (a) The element with electron configuration: 1s2 2s2 2p6 3s2 3p5; (b)A noble gases with f electrons; (c) a fifth-period element whose atoms have three unpaired p electrons; (d) First row transition metals having one 4s electron.

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