Déterminez le degré d'oxydation de l'élément souligné dans chaque formule, sans établir explicit

Déterminez le degré d'oxydation de l'élément souligné dans...

Key Concepts

Electronegativity Considerations

Electronegativity plays a key role in determining oxidation states because, in a bond, the more electronegative atom generally attracts the shared electrons more strongly. This difference forms the basis for assigning electrons in oxidation state calculations: electrons are assumed to be completely transferred to the more electronegative element, which directly influences the oxidation numbers assigned to the atoms in a compound or ion.

Charge Balance Principle

The charge balance principle states that the sum of the oxidation states of all atoms in a molecule or ion must equal the overall charge of that molecule or ion. In neutral compounds, this sum is zero, while in ions, the sum corresponds to the ion's net charge. This principle is crucial for deducing unknown oxidation states and ensuring the consistency of oxidation state assignments.

Oxidation State

The oxidation state (or oxidation number) is a formalism used in chemistry to describe the degree of oxidation or reduction of an atom within a molecule or ion. It represents the hypothetical charge an atom would have if all bonds it formed were purely ionic. This concept is central in understanding electron transfer processes, balancing redox reactions, and predicting chemical reactivity.

Rules for Assigning Oxidation States

Several standard rules are applied when determining oxidation states. These include assigning an oxidation state of zero to elements in their elemental form, using common oxidation states for certain atoms (for example, oxygen is typically assigned -2, and hydrogen is typically +1), and basing the oxidation state on electronegativity differences in bonds. These rules provide a systematic approach to calculating and predicting the oxidation numbers in various compounds and ions.

Transcript

00:01 In this problem, we are given ions and compounds, and we're asked to determine the oxidation number of each element involved in those ions and compounds.

00:09 We can look at page 200 in the text to follow some rules that give us an indication of how to solve for these oxidation numbers.

00:17 In our first compound, we can start up by saying that oxygen has an oxidation number of negative 2.

00:23 And this is because, through rule number 5 on page 200, we see that it commonly does have an oxidation number of negative.

00:31 And there are only a few exceptions to this rule.

00:34 We can write an equation that will relate our oxidation numbers to the amount of these ions that are involved in each compound.

00:42 So we're solving for x, which is going to be the oxidation number of our bromine, and there's only one bromine ion that's involved in this compound.

00:52 And then we have our oxidation number of oxygen, which is negative 2, and there are three oxygen ions involved here.

00:59 Now this is going to be equal to the overall compound charge, which in this case is negative 1.

01:05 In some cases, it can be larger than negative 1, it can be more negative than that, and sometimes it can just be 0, which indicates a neutral compound.

01:14 So we can go ahead and do our algebra here, and we find that x is equal to positive 5.

01:24 So we can say that bromine has an oxidation number of positive 5, and oxygen has an oxidation number of negative 2.

01:36 Similar to our last compound, we can assume that oxygen also has an oxidation number of negative 2 here.

01:43 So we can go ahead and write our equation to determine the oxidation number of carbon.

01:49 So x is the oxidation number of carbon.

01:52 There are two carbon ions that are involved here.

01:57 And then we're saying that the oxidation number of oxygen is negative 2.

02:01 There are four oxygen ions involved, and that is equal to the overall compound charge, which is negative 2 in this case.

02:10 So we can go ahead and start our algebra here.

02:17 And solving for x through this problem, we get that x is equal to positive 3.

02:29 So we say that carbon has an oxidation number of positive 3, and like we determined first, oxygen has an oxidation number of negative 2.

02:40 In this next part, we are dealing with an ion that has a charge of negative 1.

02:46 And looking at one of the rules on page 200, we see that the oxidation number is equal to the charge on the ion when we're looking at monotomic ions.

02:56 So in this case, determining the oxidation number is fairly easy because we can just say that it's equal to the charge, which is negative 1.

03:07 In this next compound we're working with, we're actually dealing with an exception to some of our oxidation.

03:12 Number rules.

03:14 So we're seeing that hydrogen is forming a binary compound with a metal.

03:19 And the metal is forming a positive ion.

03:22 So our hydrogen has to become a hydride ion, which means that it has a negative charge.

03:27 So we're going to say that our calcium here has a positive two oxidation number, because that's equal to its group number.

03:35 So we can go ahead and write our equation to determine what the charge of our hydride ion is going to be in this case.

03:43 So if we're saying that our calcium has a charge of positive 2, there's only one calcium ion involved here.

03:52 We're solving for our hydride ion charge, and there are two hydride ions involved.

03:57 And this is a neutral compound, so our overall charge is equal to 0.

04:02 When we do our algebra, we find that x is equal to negative 1...

Please give Ace some feedback