Assign oxidation states for all atoms in each of the following compounds. a. KMnO4 f. Fe3 O4 b.

Assign oxidation states for all atoms in each of the...

Key Concepts

Application in Complex or Polyatomic Structures

In compounds containing polyatomic ions or complex structures, oxidation states are assigned by considering both the typical oxidation states of constituent elements and the overall charge of the ion or molecule. This approach allows for the correct distribution of oxidation states even in molecules with multiple types of bonding and structural motifs.

Oxidation States

Oxidation states, or oxidation numbers, are theoretical charges assigned to atoms in a compound based on the assumption that electrons are completely transferred between atoms. They help in tracking electron flow in redox reactions and understanding the electron configuration in various chemical species.

Rules for Assigning Oxidation Numbers

There are specific guidelines for assigning oxidation numbers which include: free elements always have an oxidation state of zero; for a neutral compound, the sum of oxidation states of all atoms equals zero; for an ion, the sum equals the ion's charge; and certain elements have common oxidation states (for example, oxygen is usually -2 and hydrogen is typically +1).

Common Oxidation States of Specific Elements

Many elements exhibit characteristic oxidation states in compounds. For instance, alkali metals typically have an oxidation state of +1, alkaline earth metals +2, and oxygen is most commonly found as -2 except in peroxides or when bonded to fluorine. Recognizing these common states simplifies the process of assigning oxidation numbers in diverse chemical compounds.

Charge Balance and Sum Rule

The concept of charge balance is crucial, where the sum of assigned oxidation states must equal the overall electric charge of the molecule or ion. This rule provides a systematic method for verifying the oxidation numbers that have been assigned to individual atoms in a compound.

Transcript

00:01 So for this problem, we want to find oxidation states for all of the different elements in each of the compounds we're going to be looking at.

00:11 So let's start with a.

00:14 And in a, we see k mno4.

00:24 So we know a couple of these going into it.

00:30 We know that potassium has a charge of plus one, and oxygen has a charge of minus two.

00:43 And looking at the total charge, we can see the total charge is zero.

00:49 So if we look, we just talked to that.

00:52 The potassium has a charge of positive one.

00:56 This manganese, we aren't sure.

01:00 And the oxygen has a charge of negative 2 based on where they are in the periodic table and what they usually are.

01:08 And so in order to find all the oxidation states, we're going to set it equal to the charge of the compound itself, which is zero.

01:16 It's a chargeless neutral molecule that we're looking at here.

01:21 So we have one potassium.

01:27 You don't know what manganese is, what its oxidation state is.

01:32 And then we have four oxygens, which have an oxidation state of negative 2.

01:40 So if we were to solve for this, we find that x is equivalent to positive 7.

01:47 So we know that potassium has a charge of positive 1.

01:57 Mangonese has a charge of positive 7.

02:02 And then oxygen has a charge of negative 2.

02:08 Going on to b, we have a nickel and an auction that we're wanting to find.

02:18 The oxidation states of we aren't sure.

02:23 We have no rule for nickel, but for oxygen, we know oxygen exists in a state of negative 2, just like we saw in a.

02:31 So in order to find, we're again going to set it equal to zero, because this is a neutral.

02:36 We have 2, negative 2, and we don't know nickel yet.

02:44 So if we do this out, you find that x is equivalent to positive 4.

02:49 That makes sense.

02:50 So we have the oxidation state of nickel plus 4, and oxygen is negative 2.

03:02 Going on to c, we have quite a few elements.

03:08 And this problem, so we have sodium, we have four sodium, iron, then we have oxygen, in hydrogen, but we have six of each.

03:27 So it's still a neutral atom.

03:29 So we're going to do zero.

03:31 I can do this in different colors, so we're dealing with a lot now.

03:35 So zero.

03:37 Then if we look at, so sodium we know based on where it is and the rules that we have oxidation states positive one we want to find iron auction as we've talked about a couple times now negative two and hydrogen has a plus one state so we're looking at four sodiums we don't know iron then we have six negative two plus one.

04:16 This is just oxygen plus the hydrogen.

04:19 And if we've solved for x, we find that x is equivalent to positive, which means we can say sodium, iron, oxygen, and hydrogen, plus one, plus two, minus two, and plus one, respectively.

04:51 We want to move down, so we have more room.

04:53 There's quite a few.

04:55 So next we're going to do d.

04:58 D we have n and h, all of that.

05:08 We have two of them.

05:09 So we have another h, phosphorus, and then four oxygen.

05:17 So just like we saw above, it's still neutral.

05:22 Still zero.

05:26 So then we can do two times.

05:33 We need to go through here first and know that nitrogen has a charge of negative three, hydrogen plus one, another hydrogen plus one.

05:42 We don't know phosphorus, and oxygen minus two.

05:48 So you got two types.

05:51 We have eight hydrogen here, and then we have one more hydrogen here, and then we don't know the phosphorus, and then we have four oxygen.

06:10 So that means that x for d is going to be plus five.

06:20 So that means our nitrogen is minus three.

06:26 Hydrogen plus one, phosphorus, plus five, and then oxygen, minus two.

06:45 This is the two.

06:46 I didn't write that.

06:47 There we go.

06:48 So now we can go on to the next.

06:51 This is definitely a longer.

06:53 Problem.

06:55 So for e, we have four phosphorus and six oxygen.

07:06 We know oxygen to be negative two.

07:09 We want to find phosphorus.

07:11 And we did find phosphorus in d, but you cannot use the same oxidation state because it's going to differ depending on the compound that it's a part of.

07:20 So we can again set it equal to zero...

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