Explain Me This: Calculating Oxidation States

2019-05-28

oxidation state of PH4+


As a chemist, I find that many students are intimidated by more specific chemical terms such as "oxidation state" (oxidation number). Do not be intimidated: understanding what it means and calculating oxidation states is actually quite straight forward! Let's begin:



Oxidation States in Monatomic Examples


An oxidation state is the number given to an element in a chemical combination which represents the number of electrons it has gained or lost. First, we will tackle monatomic (single atom) examples.



Example 1: Mg, Mg2+



Magnesium (Mg) is an alkaline earth metal (Group 2 of the Periodic Table). In elemental form, magnesium exists as Mg, or Mg0. The oxidation state is zero - meaning that it has not lost or gained any electrons. Mg is not the only element that has an oxidation state of zero in its elemental form. This is very useful to know, and we will use this knowledge later to solve questions concerning oxidation states. Mg2+ is a cation (a positively charged ion) with a 2+ positive charge. This is because Mg has lost two electrons (remember, one electron has a charge of -1). The element itself is missing two electrons, and therefore in this case the oxidation state is +2.


From this example, we see that every element exists in the zero-oxidation state in its elemental form.



Example 2: Cl, Cl-



Chlorine (Cl) is a halogen (Group 7 of the Periodic Table). We have learned already that in its elemental form, Cl must have an oxidation state of zero. The chloride has a charge of -1, for it has gained one electron (originally, Cl has 35 electrons, so Cl- must have 36). The element itself has gained one electron, and we are able to deduce that its oxidation state is -1.


From the above, we see that the oxidation number of the monatomic ion (that is, ions containing only one atom) equals the charge of the ion.



To test your understanding so far, what are the oxidation states of the following? The solutions are at the end of this post.


  1. Al3+
  2. O2-
  3. Mn7+


And a few bonus questions:


  1. In terms of electrons, what does an oxidation state of +3 mean?
  2. Why does elemental carbon have an oxidation state of zero?


So far so good. You should now have a good understanding of the basics. Let's move on to harder examples.



Oxidation States in Molecules and Compounds


For finding the oxidation numbers in molecules and compounds, we must find the oxidation number for each element in the species. Molecules and compounds are neutral species. That is the species has no overall charge. An important idea is that the sum of the oxidation numbers of all of the atoms in a neutral compound is zero:



Example 1: Find the oxidation numbers of X in XY, given that Y has an oxidation number of -3.



The oxidation numbers of each atom in the species XY must sum to zero. As the oxidation state of Y is -3, then X must have an oxidation state of +3: (+3) + (-3) = 0.


Recalling the periodic table, we know that X and Y are not real elements - we have arbitrarily used these letters to practice the sum rule. In the above example, we are told the oxidation state of Y. In real questions, however, you most likely will not be told the oxidation numbers of any of the elements in a given species - you will have to figure each state out for yourself! But do not panic, by knowing a handful of simples rules, you will be able to tackle these questions with ease.



Example 2: Find the oxidation numbers of the elements in XH (X is more electronegative).


For hydrogen in a molecule or compound, the oxidation number is +1, but it is -1 in when combined with less electronegative elements. We are told that X is more electronegative, so H must be +1. From what we learned in the previous example, we can deduce that X must be -1, since the overall species' oxidation state is zero.



In most cases, the oxidation number of H is +1. In some instances, though, the oxidation state is -1. This occurs when hydrogen is paired with a less electronegative element, such as sodium (Na). Depending on your course, you may or may nor be expected to recall the relative oxidation states of elements. It is most likely that you are not expected to, and in this case you can use the general rule - when H is with a metal its oxidation number is -1. When H is with a non metal element its oxidation number is +1. For example, in lithium hydride (LiH), H is paired with a metal, and so H has an oxidation number of -1.



Example 3: Find the oxidation numbers of the elements in YO (this is not a peroxide).


Oxygen's oxidation number follows the rule the oxidation number of O in compounds is usually -2, but it is -1 in peroxides. Given that YO is not a peroxide, O has oxidation number -2 and Y must be +2 (to sum to zero).



Oxygen is usually in the oxidation state -2. However, in peroxides, such as barium peroxide (Ba2O2) and hydrogen peroxide (H2O2), the oxidation number is -1. Do not worry too much about memorizing this rule, for you know that H is usually +1 and Group 1 elements are always +1 (as we will see in the next example), so we can always deduce the oxidation state of the oxygen present.



Example 4: What is the oxidation number of Li in LiOH?



The next rule you need to remember is that the oxidation number of a Group 1 element in a compound is +1. Lithium (Li) is a Group 1 element, so it has an oxidation state of +1. Easy! How about oxygen and hydrogen? Well, we have learned before that hydrogen is usually +1 and oxygen -2. This is certainly the case here, as the sum of these values is zero: (+1) + (-2) + (+1) = 0.



Example 5: Find the oxidation numbers of the elements in CaF2.




Group 2 elements also follow a rule: the oxidation number of a Group 2 element in a compound is +2. Calcium (Ca) must, then, have an oxidation number of +2. We have not learned about Group 7 elements yet, but this does not matter: we can use the sum rule to figure out that fluorine (F) must be -1: (+2) + 2(-1) = 0.



Example 6: Find the oxidation number of N in NCl3.



The oxidation number of a Group 7 element in a binary compound is -1. A binary compound is basically something that is made up of only two elements. Whilst this is true, Group 7 elements have an oxidation state of -1 99% of the time, regardless of whether they are in a binary compound or not. In this example, there are three chlorine (Cl) atoms, each with an oxidation number of -1. In total, they add to -3. For the molecule to be neutral overall, the N must be +3.



These are all the rules that you need to know for molecules and compounds! Below I have included some more examples to help you to solidify your understanding:



Example 7: Find the oxidation state of carbon in CO2


We know that oxygen has an oxidation state of -2. In this molecule, there are two oxygen atoms. Combined, they have an oxidation state of 2(-2) = -4. The molecule overall has an oxidation state of zero. So, carbon must have an oxidation state of +4: (+4) + (-4) = 0.



Example 8: What are the oxidation states of each element in KMnO4?


Potassium (K) is an Alkali Metal (Group 1) and has an oxidation state of +1. We know that oxygen is -2. The compound overall is neutral. Therefore, manganese (Mn) must be +7: (+1) + (+7) + 4(-2) = 0.



Example 9: Find the oxidation number of carbon in CH4


Hydrogen (H) is +1. There are four hydrogens, and their combined oxidation number is +4. Therefore, carbon (C) must have an oxidation number of -4.



Example 10: What is the oxidation state of N in N2O5?


Here, oxygen's oxidation state is -2. There are five oxygen atoms in total, and their combined oxidation number is -10. Therefore, each nitrogen (N) has an oxidation state of +5.



Example 11: The oxidation number of oxygen in sodium peroxide, Na2O2, is equal to X. Find X.



Be careful here, as this is one of the exceptions for oxygen (oxygen has a -1 oxygen state in peroxides). Although, even if you happen to forget this rule, you can still reach the answer by using what you know about Group 1 elements (their oxidation number is always +1): 2(+1) + 2(-1) = 0


Hopefully these examples all make sense. To make sure, have a go at the following by yourself (the answers are at the end of this post):


  1. What is the oxidation state of the central atom in NH3?
  2. What is the oxidation number of sulfur in the ionic compound Na2SO4?
  3. What is the oxidation number of each element in barium chloride (BaCl2)?
  4. What is the oxidation number of nitrogen in N2?



Oxidation States in Polyatomic Ions


Typically, the hardest oxidation number questions involve polyatomic ions (ions that contain more than one atom). This is because we need to take into consideration that the overall charge of the compound / molecule is no longer zero: the sum of the oxidation numbers in a polyatomic ion is equal to the charge of the ion. Let's look at some examples together:



Example 1: Find the oxidation number of sulfur in sulfite (SO32-).


We approach this question just like before. First we know that oxygen has an oxidation number of -2. There are three oxygen atoms with a combined number of -6. Now, we know that the anion has an overall charge of -2. This means that sulfur (S) has an oxidation state of +4:(+4) + 3(-2) = -2. (A common mistake would be to calculate the oxidation state as +6.)



Example 2: What is the oxidation number of carbon and oxygen in C2O42-?


Again, we know that oxygen is -2. There are four oxygens, with a combined number of -8. The overall anion has a charge of -2. Therefore, the two carbons must have a combined oxidation number of +6, each with an oxidation number of +3: 2(+3) + 4(-2) = -2



Example 3: Find the oxidation state of P in the ion PH4+


Hydrogen has an oxidation number of +1. The cation has a charge of +1. Therefore, P has an oxidation state of -3: (-3) + 4(+1) = +1.



Here are some for you to try (with answers at the end of this post):


  1. What is the oxidation number of each element in MnO4-?
  2. What is the oxidation number of each element in Cr2O72-?



If you got this far, congratulations! You should now have a strong grasp on what an oxidation state is and how to calculate oxidation numbers.



The Rules You Need to Know


Here is a summary of all the rules we have learned for calculating oxidation numbers. Go over these once in a while to make sure you have these concepts fresh in your mind!


  1. The oxidation number of a free element is always 0.
  2. The oxidation number of a monatomic ion equals the charge of the ion.
  3. The oxidation number of H is +1, but it is -1 in when combined with less electronegative elements.
  4. The oxidation number of O in compounds is usually -2, but it is -1 in peroxides.
  5. The oxidation number of a Group 1 element in a compound is +1.
  6. The oxidation number of a Group 2 element in a compound is +2.
  7. The oxidation number of a Group 7 element in a binary compound is -1.
  8. The sum of the oxidation numbers of all of the atoms in a neutral compound is 0.
  9. The sum of the oxidation numbers in a polyatomic ion is equal to the charge of the ion.



Answers


Oxidation States in Monatomic Examples


  1. +3
  2. -2
  3. +7


  1. The element is missing three electrons.
  2. Because elemental carbon is not missing or has not gained any electrons.



Oxidation States in Molecules and Compounds


  1. -3
  2. +6
  3. +2, -1
  4. 0



Oxidation States in Polyatomic Ions


  1. +7, -2
  2. +6, -2

If you liked this article, check out the tutor who wrote it below:

Alex

Imperial College London, Bachelor's degree in Chemistry

Alex graduated from Imperial College in 2015 with a Bachelor's Degree in Chemistry. He has more than four years of experience teaching standardized exams such as the SAT and ACT as well as subjects including chemistry, mathematics and physics. Alex's classes have helped many students excel at exams within very short time frames. His teaching style includes easy-to-understand and logical explanations of difficult concepts. He is also familiar with British boarding school and university applications. Alex's hobbies include weightlifting and chess.

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