How To Calculate Oxidation Numbers - Basic Introduction

Summary

Highlights

• Pure elements: Oxidation state is always zero! 🌟
• Zinc's oxidation state is determined easily by its charge when ionized to Zn²⁺. ⚡
• Compounds' total charge helps deduce individual states. Like algebra! 🧮
• Fluorine's consistent -1 oxidation status due to being super electronegative. 🌬️✨
• In peroxide, each oxygen atom carries an oxidation state of -1. 🍏
• Electronegativity comparison explains oxidation states and polarity in compounds. 💪
• Fe3O4 presented with a trick: average oxidation state leads to unique insights. 🔄

Key Takeaways

• Pure elements always have an oxidation state of zero! 🎉
• Monatomic ions have an oxidation state equal to their charge. 💡
• Fluorine in compounds always carries an oxidation state of -1. Always the most electronegative! 🧪
• Know your polyatomic ions! They can change the game when computing oxidation numbers. 🧩
• Electronegativity is key! It determines which element holds that negative charge in a compound. 📉
• Complex compounds may have elements with non-integer average oxidation states! 🤓
• Periodic table position often helps predict oxidation states based on electronegativity. 🔍

Overview

Chapters

• 00:00 - 00:30: Introduction to Oxidation Numbers The chapter introduces oxidation numbers, explaining how to determine them for elements. It states that the oxidation state of any pure element is zero. Examples provided include zinc, oxygen gas, and fluorine gas, all having oxidation states of zero when in pure form.
• 00:30 - 02:00: Oxidation States of Pure Elements and Ions This chapter discusses the concept of oxidation states in chemistry, particularly focusing on pure elements and ions. It explains that for pure elements, without any charges or compounds involved, the oxidation state is always zero. This is considered the first rule to remember in understanding oxidation states. Additionally, the chapter covers ions by explaining that the oxidation state of an ion, such as the zinc 2+ ion, corresponds directly to its charge.
• 02:00 - 04:30: Oxidation States in Compounds The chapter discusses oxidation states in compounds, focusing on different ions and their oxidation states. It mentions the oxidation states of specific ions like Fe^3+ which has an oxidation state of +3. The chapter also covers diatomic ions, using mercury as an example, explaining that in a diatomic mercury (Hg^2+) ion, each mercury has an oxidation state of +1, due to the presence of two mercury ions.
• 04:30 - 07:00: Polyatomic Ions and Average Oxidation States This chapter focuses on polyatomic ions and the concept of average oxidation states. It explains how to determine the oxidation state of individual atoms within a polyatomic ion, using mercury as an example, where each atom has a net positive charge. The oxidation state of mercury was found to be +1 by dividing the total charge by the number of atoms. The chapter also discusses the peroxide ion and hints at forming equations to find the oxidation states of oxygen atoms within such ions.
• 07:00 - 09:30: Electronegativity and Oxidation Numbers The chapter focuses on the concept of electronegativity and oxidation numbers. It explains how to determine the oxidation state of oxygen in different ions, such as the peroxide ion and the superoxide ion. The chapter highlights that in a peroxide ion, each oxygen atom carries an oxidation state of -1. For the superoxide ion, the oxidation state is calculated by dividing the total charge by two, resulting in each oxygen atom having a net charge of -0.5. Additionally, it explains that when two oxygen atoms combine, they have a combined net charge based on their individual oxidation states.
• 09:30 - 12:30: Practice Examples and Complex Scenarios The chapter titled 'Practice Examples and Complex Scenarios' explores the fundamental principles of oxidation states in elements and ions. It clarifies that pure elements exhibit an oxidation state of zero, while monoatomic ions possess an oxidation state equivalent to their charge. Additionally, the chapter delves into the role of fluorine in compounds, emphasizing its behavior when not in its pure elemental form.
• 12:30 - 15:30: Summary and Conclusion The chapter entitled "Summary and Conclusion" provides a detailed explanation of oxidation states related to fluorine and oxygen in chemical compounds. It is mentioned that fluorine, being the most electronegative element, consistently maintains a negative one oxidation state. Oxygen, typically exhibits a negative 2 oxidation state, with exceptions occurring when it is bonded to fluorine, or when part of compounds identified by specific names such as peroxide or superoxide. In peroxide compounds, oxygen's oxidation state adjusts to negative one, whereas in superoxide compounds, it is negative one-half.

How To Calculate Oxidation Numbers - Basic Introduction Transcription

• 00:00 - 00:30 in this video we're going to go over oxidation numbers and how to find them so let's say if we're given the element zinc what is the oxidation number of zinc now the first rule that you need to know is that the oxidation state of any pure element is always zero so the oxidation state of oxygen gas as a pure element is zero fluorine gas as the pure element is zero
• 00:30 - 01:00 even uh phosphorus as a pure element is zero so if there's no charges and it's only one pure element it's not a compound the oxidation state will always be zero so that's the first rule we need to keep in mind now the second thing is the oxidation state of ions the oxidation state of the zinc 2 plus ion is basically the charge
• 01:00 - 01:30 of what you see there it's positive 2. the oxidation state of the fe plus 3 ion is simply positive 3. now sometimes you might have diatomic ions for example the mercury 2 plus ion individually each mercury ion has an oxidation state of one because there's two of them so you need to write an equation two mercury
• 01:30 - 02:00 atoms has a net charge of positive two so if you divide both sides by two you can get the individual oxidation state of each mercury particle which is plus one so here's another example this is the peroxide ion to find the oxidation state of each oxygen atom in this ion you could write an equation as two oxygen atoms with a total charge of
• 02:00 - 02:30 negative two so individually each oxygen atom has a charge of minus one so that's the oxidation state of oxygen individually in the peroxide ion this is the superoxide ion so if you want to find the oxidation state you need to divide the total charge by two so each oxygen atom has a net charge of negative one half so two of them combined will have a net
• 02:30 - 03:00 charge of negative one so keep this in mind anytime you have a pure element the oxidation state will always be zero and if you have an ion let's say if it's a monoatomic ion the oxidation state is the same as that ion now let's talk about compounds whenever you have fluorine inside a compound when it's not a pure element
• 03:00 - 03:30 fluorine is always going to have a negative one oxidation state fluorine is the most electronegative element when oxygen is in a compound it's going to have a negative 2 oxidation state unless it's bonded to fluorine or unless you hear the name peroxide or superoxide whenever you hear the name peroxide oxygen has a negative one oxidation state if you hear the word superoxide it has a negative one-half oxidation state but if
• 03:30 - 04:00 you hear the word oxide then the oxidation state is negative two which is 99 of the time now hydrogen will have an oxidation state of plus one when bonded to a non-metal when bonded to a metal hydrogen will have an oxidation state of negative one and really the key is electronegativity hydrogen is more electronegative than most metals that's why it bears a
• 04:00 - 04:30 negative charge but hydrogen is usually less electronegative than most non-metals and so that's why it bears a positive charge so typically the element that's more electronegative is the one that usually carries the negative charge now let's work on some examples what is the oxidation state of magnesium and chlorine in this compound
• 04:30 - 05:00 by the way most halogens are usually negative one chlorine typically has a negative one charge like fluorine if we write an equation mg plus 2cl this whole compound is neutral so therefore the total charge is 0. now if chlorine has a negative one oxidation state that means magnesium
• 05:00 - 05:30 has to have a positive two oxidation state you can literally solve it and it makes sense magnesium is an alkaline earth metal which typically has a positive two charge go ahead and find the oxidation states of aluminum and fluorine in this example well we know that fluorine is negative one in a compound always and aluminum based on where it's located in a periodic table
• 05:30 - 06:00 it's typically positive three within an ionic compound and you could solve it too a out plus three f should add up to zero because the net charge is zero so each fluorine atom has an oxidation state of negative one so now we gotta add three to both sides so aluminum has an oxidation state of positive 3. here's another example
• 06:00 - 06:30 find the oxidation state of vanadium and oxygen in this compound so this is called vanadium oxide so whenever you hear the word oxide oxygen has a negative two charge so we got two vanadium atoms plus five oxygen atoms with a net charge of zero so each oxygen atom has an oxidation state of negative two five times negative two is negative ten
• 06:30 - 07:00 and then add ten to both sides so 2v is equal to 10. next divide both sides by 2. so 10 divided by 2 is 5. and so the oxidation state of vanadium is positive 5. now let's go over some examples containing polyatomic ions consider sulfate what is the oxidation state of sulfur and sulfate
• 07:00 - 07:30 we know oxygen is usually negative two so let's write an equation sulfur plus four oxygen atoms has a net charge of negative two so each oxygen atom has an oxidation state of minus two and four times negative two that's negative eight next we need to add eight to both sides
• 07:30 - 08:00 negative two plus eight is positive six so this is the oxidation state of sulfur and sulfate let's look at another example phosphate go ahead and find the oxidation state of phosphorus and phosphate so once again oxygen is still negative two so we got a phosphorus atom plus four oxygen atoms and the net charge is
• 08:00 - 08:30 negative three based on what we see here so it's going to be p plus four times negative two and four times negative two is negative eight and then add eight to both sides so negative three plus eight that's going to be positive five and that's the oxidation state of phosphorus
• 08:30 - 09:00 let's look at another example let's try nitrate and also chlorate as perchlorate go ahead and find the oxidation state of nitrogen and chlorine in these two polyatomic ions so we have a nitrogen three oxygen atoms and that's going to equal a net charge of negative one so o is negative two three times negative two is going to be negative six
• 09:00 - 09:30 and negative one plus six if we add six to both sides that's going to be positive 5. so that's the oxidation state of nitrogen and for chlorine in perchlorate it's going to be cl plus 4 oxygens equals a net charge of negative 1. so this is gonna be four times negative two which is a negative eight and then negative one plus eight that's going to give us an oxidation
• 09:30 - 10:00 state of positive seven so now you know how to find the oxidation states of elements within compounds and polyatomic ions now i want you to understand the concept of electronegativity and how it relates to oxidation numbers electronegativity increases towards fluorine on a periodic table so as you go up and to the right the
• 10:00 - 10:30 electronegativity increases so let me give you some values of common elements so let's say hydrogen is somewhere in the corner over there and then we have boron carbon nitrogen oxygen fluorine chlorine bromine iodine phosphorus and sulfur hydrogen has an electronegativity value of 2.1 for boron is 2.0 carbon is 2.5 and then 3.0 3.5
• 10:30 - 11:00 is the highest it's 4.0 phosphorus it's 2.1 it's the same as hydrogen so first 2.5 chlorine is 3.0 and this is 2.8 iodine is 2.5 so keep these values in mind so here's a question for you what is the oxidation state of oxygen and fluorine in oxygen difluoride
• 11:00 - 11:30 now oxygen has an electronegativity value of 3.5 fluorine is 4.0 so which one is more electronegative electronegativity is the ability of an atom to attract electrons to itself so fluorine is going to pull on the electrons in this molecule it's going to have a stronger pull than oxygen so fluorine is going to acquire a
• 11:30 - 12:00 partial negative charge whereas oxygen is therefore going to acquire a partial positive charge because fluorine pulls on the electron stronger than oxygen can so in this example oxygen will not have its typical charge of negative two the only time oxygen will have its oxidation state of negative two is if it's the most electronegative element in that compound
• 12:00 - 12:30 if it's not then it's going to have a positive oxidation state keep in mind any time fluorine is in a compound it has an oxidation state of negative one and the reason for that is because fluorine is the most electronegative element on a periodic table so now we can solve for oxygen so o plus two f should have a net charge of zero because there's no number here so fluorine is negative one two times negative one is negative two
• 12:30 - 13:00 so if we add two to both sides oxygen is going to equal positive two which makes sense because it's partially positive in this particular example now let's look at two other examples hydrochloric acid and sodium hydride chlorine has an electronegativity value of 3.0 hydrogen is 2.1 and sodium
• 13:00 - 13:30 it's like one point something i'm not sure what the exact number is it could be like 1.7 but i know it's less than two so in this example hydrogen bears a partial positive charge chlorine bears a negative charge because chlorine is more electronegative than hydrogen so therefore chlorine is going to have its oxidation state of negative one
• 13:30 - 14:00 which is typical of most halogens hydrogen is going to have an oxidation state of plus one as you mentioned before whenever hydrogen is bonded to a non-metal the oxidation state is usually positive one now what about in sodium hydride well we know that sodium is an alkali metal which always have a positive one charge so therefore sodium is going to have an oxidation state of plus one but hydrogen has an oxidation state of
• 14:00 - 14:30 negative one typically when hydrogen is bonded to a metal it usually has a negative one oxidation state and it makes sense because hydrogen is more electronegative than most metals so it usually bears the partial negative charge that's why it has a negative oxidation state sodium has the positive charge so it has a positive oxidation state and so you could use electronegativity to help you determine what the oxidation
• 14:30 - 15:00 state will be so let me give you another example bh3 what is the oxidation state of boron and hydrogen feel free to try that one now hydrogen has an electronegativity value of 2.1 and boron is 2.0 now is boron a metal or non-metal in this example
• 15:00 - 15:30 hydrogen is more electronegative so hydrogen bears the partial negative charge or boron bears the partial positive charge so therefore hydrogen it's going to have its oxidation state of negative one because it's more electronegative than boron so then this is gonna be b plus three h which is equal to zero so three times negative one is negative three
• 15:30 - 16:00 so boron is going to have an oxidation state of positive three in this example now let's consider these two examples sulfuric acid or rather hydrosulfuric acid and also sulfur dioxide now hydrogen has an en value of 2.1 sulfur is 2.5 and oxygen is 3.5
• 16:00 - 16:30 so in sulfur dioxide oxygen has the partial negative charge sulfur has the partial positive charge now in h2s hydrogen has the partial positive charge sulfur has the partial negative charge now in a periodic table when you have elements like nitrogen oxygen fluorine typically nitrogen has a negative three charge oxygen minus two fluorine negative one sulfur two sulfur usually has a negative two charge if
• 16:30 - 17:00 if sulfur is the more electronegative element so looking at h2s hydrogen is bonded to a non-metal that is more electronegative than itself so hydrogen is going to have the positive one oxidation state and there's two of them so sulfur in this example has its normal oxidation state of negative two so you can base your answer on a periodic table if sulfur is the more electronegative element
• 17:00 - 17:30 now in so2 you can't do that because sulfur doesn't have the partial negative charge so you can't base the charge on a periodic table you can do so however for oxygen because oxygen is the electronegative element in that compound so you can use the negative two charge for oxygen so oxygen is going to have an oxidation state of minus two and to find it for sulfur it's going to be s plus two oxygen atoms equals zero so that's two times negative two
• 17:30 - 18:00 which is negative 4 so sulfur is going to have a positive oxidation state of 4 due to the positive partial charge so elements that are less electronegative typically those are the ones you got to solve for the ones that are more electronegative you can find a charge based on a periodic table if they carry a negative charge now let's look at some other examples nh3 and no2 go ahead and find the oxidation state
• 18:00 - 18:30 of each element now in ammonia hydrogen has an en value of 2.1 but nitrogen is more electronegative it's 3.0 so therefore nitrogen should have its normal charge of negative three if we write an equation n plus three h is equal to zero hydrogen is going to have a positive one charge it's partially positive whereas nitrogen is partially negative
• 18:30 - 19:00 so typically when hydrogen's bonded to a non-metal it's usually plus one which means n has to be negative three so as you can see nitrogen is the electronegative element in this example and it has its periodic charge of negative three which you could find in a periodic table now in this case o is more electronegative so nitrogen is going to have a different oxidation state it's not going to be its natural oxidation state of negative 3.
• 19:00 - 19:30 so in this example it's going to be positive 4. typically when you have elements like nitrogen sulfur phosphorus if they carry an element that's more electronegative than itself those are the elements you gotta solve for the one that usually has a partial positive charge
• 19:30 - 20:00 now try these two examples methane and carbon dioxide in methane hydrogen has a positive one charge hydrogen is less electronegative than carbon so it's going to be partially positive carbon is going to be partially negative so solving for carbon we have c plus four h is equal to zero
• 20:00 - 20:30 so that's four times one so c is negative four so when carbon is bonded to hydrogen carbon has a negative oxidation state but when carbon is bonded to oxygen it's going to have a positive oxidation state when it's bonded to hydrogen it has a negative oxidation state so oxygen is negative two and there's two of them so carbon is going to have to be positive four in this example now sometimes
• 20:30 - 21:00 you might have elements that have an average oxidation state that's not a whole number let's try these two c3h8 and fe3o4 in this example hydrogen is less electronegative than carbon so it's going to be positive 1. so if we write the formula 3c plus 8h is equal to 0. so that's going to be 8 times 1
• 21:00 - 21:30 and if we subtract 8 from both sides 3c is equal to negative 8. so carbon on average has an oxidation state of negative 8 over 3. now let's do the same thing for fe3o4 so we got three iron atoms and four oxygen atoms with a net charge of zero so oxygen has an oxidation state of negative two
• 21:30 - 22:00 so four times negative two that's negative eight and if we add eight to both sides we get this so f e has an oxidation state of eight over three so eight over three is about two point sixty seven now keep in mind an individual iron atom cannot have a charge of 2.67
• 22:00 - 22:30 it's usually a whole number like positive 2 or positive 3. because electrons and protons they're they basically have numerical charges an electron has a charge of negative one a proton has a charge of positive one so a typical ion won't have a decimal charge so what does it mean that the average oxidation state is 2.67 so what is meant by that
• 22:30 - 23:00 in this compound there are three iron ions and four oxygen ions each oxygen has a charge of negative two so the total negative charge is negative eight in order for the compound to be electrically neutral the total positive charge has to be positive eight iron metal has two common oxidation
• 23:00 - 23:30 states positive two and positive three now they all can't be positive three because 3 plus 3 plus 3 is not and they can't all be positive 2 because 2 plus 2 plus 2 is 6. so some of them is positive 2 and some are positive 3. so the question is how many iron ions have a plus two charge and how many have a plus three charge in order to get up to eight two of them
• 23:30 - 24:00 has to have a positive three charge and one of them has to have a positive two charge if you average the numbers 2 3 and 3 and divided by 3 that's going to be 8 over 3 which averages out to 2.67 so whenever you get a decimal value what it really means is that that's the average oxidation state individually some were positive three and some or positive two so the individual ions should have a numerical oxidation state but when you have multiple of them
• 24:00 - 24:30 the average could be a decimal value because these they don't all have to be the same they can be different so hopefully this makes sense in terms of why some oxidation states have a decimal value now let's try polyatomic ions that have three different elements in it go ahead and find the oxidation state of every element in that polyatomic ion so oxygen has an oxidation state of
• 24:30 - 25:00 negative two hydrogen is positive one when it's bonded to non-metals usually so all we gotta do is find sulfur so h plus s plus three oxygen atoms has a net charge of negative one so hydrogen is one oxygen is negative two and so three times negative two that's negative six and then one plus negative six is negative five
• 25:00 - 25:30 so now let's add five to both sides negative one plus five is positive four so in this example sulfur has an oxidation state of positive 4. go ahead and try this one k2 cro4 find the oxidation state of every element in that example so we have two potassium atoms
• 25:30 - 26:00 a chromium atom and four oxygen atoms now we know oxygen is going to have an oxidation number of negative two potassium is an alkali metal which all of them have a positive one charge chromium is the transition metal and it has a variable charge so that's the one we got to solve for so this is gonna be two times one plus cr plus four times negative two and all of that is equal to zero
• 26:00 - 26:30 so four times negative two that's negative eight and two plus negative eight is negative six so therefore in this example chromium has an oxidation state of positive six try this one potassium bicarbonate
• 26:30 - 27:00 find the oxidation state of carbon in this example so we know oxygen is going to be negative two potassium and alkaline metal is plus one and hydrogen hydrogen is actually bonded to the oxygen in bicarbonate if you were to draw the lewis structure so therefore hydrogen's bonded to a non-metal do a covalent bond and so it's going to be plus one
• 27:00 - 27:30 so we have k plus h plus c plus three o and that's equal to zero so k is positive one hydrogen is one an oxygen is going to be three well oxygen is negative two but we gotta multiply that by three so one plus one is two three times negative two is negative six and then two plus negative six that's negative four
• 27:30 - 28:00 so in this example carbon is positive for potassium bicarbonate you could break it up into two ions k plus and hco3 minus so just by looking at k plus that tells you that k has an oxidation state of positive one now bicarbonate is basically the sum of the hydrogen ion and the carbonate ion
• 28:00 - 28:30 so therefore you can see that hydrogen in this example also has a positive one charge and then from this you could find the oxidation state of carbon you could say c plus three o has a net charge of negative two and then you have to add six to both sides so negative two plus six is positive four so if you understand the ions and all the polyatomic ions
• 28:30 - 29:00 you could break it down individually to see that hydrogen has a positive one charge in this example and the same is true for king so that's why it's good to know the polyatomic ion sheet now i have two more examples for you br cl3 and ibr5 find the oxidation state of every
• 29:00 - 29:30 element in this example so most halogens like fluorine chlorine bromine iodine they typically have a negative one charge but both bromine and chlorine can't be negative so which one is negative and which one is positive keep in mind bromine has an electronegativity value of 2.8 chlorine is 3.0 iodine is 2.5 so in this example
• 29:30 - 30:00 chlorine bears the partial negative charge bromine is partially positive so therefore chlorine is going to have its natural oxidation state of negative one bromine we need to calculate it so it's going to be br plus three cl and that's equal to zero so this is going to be three times negative one and so we can see that bromine has an oxidation state of positive three
• 30:00 - 30:30 now in the second example bromine is going to carry the partial negative charge iodine carries the partial positive charge if it's incorrectly usually the electro positive element is written first the electronegative element is written second so the one that you see on the right side is usually the one that carries the natural charge that can be found on the periodic table so in this case bromine is going to have its natural oxidation state of negative one
• 30:30 - 31:00 so iodine is going to have an oxidation state of positive 5 in this example so hopefully you understand the relationship between electronegativity and oxidation numbers so that's it for this video thanks for watching and have a good day
• 31:00 - 31:30 you