How to Write an Electron Configuration for an Atom of Any Element

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Being able to write electron configurations of an atom is necessary to ace a chemistry class, but writing them can seem hard, but they are actually easy, if you know the basics.

Steps

1. Find out how many electrons the atom has. On the periodic table, the atomic number is the number of protons of the atom, and thus equals the number of electrons in an atom with zero charge.
   
2. Refer to the image at the right for a list of orbitals that will hold the electrons.
   
   • The S orbital set (any number followed by an "S") contains a single orbital, and by Pauli's Exclusion Principle, a single orbital can hold a maximum of two electrons, so each S orbital set can hold two electrons.
     
   • The P orbital set contains three orbitals, and thus can hold a total of six elections.
     
   • The D orbital set contains five orbitals, so it can hold ten electrons.
     
   • The F orbital set contains seven orbitals, so it can hold fourteen electrons.
     
   
   
3. Put one electron into the lowest energy orbital available, starting with 1S (holds a maximum of two electrons). Be careful! Do not fill the orbitals in the order shown in the chart! Fill the orbitals in this order (the number following the orbital set is the maximum number of electrons it can hold):
   
   • 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 4f14 5d10 6p6 7s2 5f14 6d10.
     
   • Note: Energy level changes as you go up. For example, when you are about to go up to the 4th energy level, it becomes 4s first, then 3d. After that it follows the order once again. This only happens after the 3rd energy level!
     
   
   
4. Once you've put every electron into an orbital (according to the order), write the configuration as shown at the end of step 3, except, write the number of electrons that are in the orbital set instead of the numbers shown (they are shown as completely filled). So, an uncharged antimony atom's electron configuration would be 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p3. Notice that the number following 5p isn't 6, but three. That's because only three electrons are in the orbital set, so the orbital set is not completely occupied (it lacks three more electrons).
   
5. Be sure to read the "Tips" section for information on how to write the electron configuration of an ion or how to promote an electron in special cases.
   

Tips

• The number following the letter is actually a superscript, so don't make that mistake on a test.
  
• To find the atomic number of the atom when it is in electron configuration form, just add up all of the numbers that follow the letters (S, P, D, and F).
  
• You can also figure out the electron configuration of an atom without remembering the order of the orbital sets. Just look at the periodic table. Have you ever wondered why it's shaped like that? Take the top right element (Helium) and shift it all the way to the left. The two left columns are the S orbitals sets. The middle 10 columns are the D orbital sets. and the right 6 columns are the P orbital sets. The two rows on the bottom are the F orbital sets. Now, all you need to do is go horizontally across the periodic table until you arrive at the atom you are calculating. So, if you were looking for scandium, you first go through the S orbital sets, which are hydrogen and helium. Write down 1s2 because you are on the first row and you passed through all of the first S orbital set without arriving at scandium. Now, go to the second row. Go through the 2S orbital, which are lithium and beryllium. Write down 2s2. Now, go through the 2P orbital set and continue through all of them until you arrive at your atom.
  
• Beware, after you go through the 6S orbital set, you go to the 4F orbital set, instead of 5D. Same thing goes for 7S (you go through 5F before going to 6D).
  
• Writing long electron configurations can be avoided by writing them in their noble gas configurations. Simply find the last symbol containing p6 (such as 3p6 or 5p6) and add up all of the numbers following the letters of every symbol before and including the p6 orbital set. Then, using the sum of the numbers, locate the element with the atomic number equal to the sum you just calculated. It should appear at the very right of the periodic table. That's called a noble gas. Now, just remove all of the symbols that you added up and put in the noble gas' symbol in brackets. So, for an antimony atom, the noble gas configuration would be [Kr] 5s2 4d10 5p3. Notice that you don't add up the 5s2 and 4d10 because they are after 4p6, which you do add.
  
• You can also write an element's electron configuration by just writing the valence configuration, which is the last S and P orbital set. So, the valence configuration of an antimony atom would be 5s2 5p3.
  
• When the atom is an ion, it means that the number of protons does not equal the number of electrons. The charge of the atom will them be displayed at the top right (usually) corner of the chemical symbol. So, an antimony atom with charge +2 has an electron configuration of 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p1. Notice that the 5p3 changed into a 5p1. Be careful when the configuration of an uncharged atom ends in anything but an S and P orbital set. When you take away electrons, you can only take them away from the valence orbitals (the S and P orbitals). So, if a configuration ends in 4s2 3d7, and the atom gains a charge of +2, then the configuration would change to end with 4s0 3d7. Notice that 3d7 does not change. Instead, the S orbital electrons are lost.
  
• There are circumstances when an electron needs to be "promoted." When an orbital set is one electron away from being half occupied or completely occupied, remove one electron from the nearest S or P orbital set and move it to the orbital set that needs the electron.
  
• Every atom desires to be stable, and the most stable configurations have full S and P (s2 and p6) orbital sets. The noble gases have this configuration, which is why they are non-reactive and are on the right side of the periodic table. So, if a configuration ends in 3p4, it only needs two more electrons to become stable (losing six, including the S orbital set's electrons, takes more energy, so losing four is easier). And if a configuration ends in 4d3, it only needs to lose three electrons to reach a stable state.
  

Warnings

• Be careful when calculating the configurations of ions when promoting electrons. They can be tricky.
  

Things You'll Need

• A periodic table
  
• Memorization of the orbital set order (or you can use the periodic table method).
  

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