Obtain the oxidation number for the element noted in each of...

Obtain the oxidation number for the element noted in each of the following.
(a) $\mathrm{Sn}$ in $\mathrm{SnO}_{2}$
(b) $\mathrm{Ta}$ in $\mathrm{Ta}_{2} \mathrm{O}_{3}$
(c) Br in $\mathrm{KBrO}_{4}$
(d) Os in $\mathrm{K}_{2} \mathrm{O}_{\mathrm{s}} \mathrm{O}_{4}$

Key Concepts

Algebraic Determination of Oxidation States

When some oxidation numbers in a compound or ion are unknown, an algebraic approach is used by setting up an equation where the sum of the oxidation numbers of all atoms equals the overall charge. Known oxidation numbers are substituted into this equation, and the unknown value is solved for, allowing for the determination of oxidation states even in complex compounds.

Oxidation Number

The oxidation number (or oxidation state) is a value assigned to an element in a compound, reflecting the number of electrons lost or gained by an atom compared to its elemental form. It serves as a bookkeeping tool in redox reactions to track electron transfer and helps in understanding the reactivity of compounds.

Rules for Assigning Oxidation Numbers

There are several standard rules used to determine oxidation numbers: elements in their pure form are assigned an oxidation number of zero; common oxidation states are assigned to certain elements (for example, oxygen is usually -2 and hydrogen is usually +1); and the sum of oxidation numbers in a neutral compound is zero while in polyatomic ions it equals the charge of the ion. These rules provide a systematic way to deduce unknown oxidation numbers in a compound.

Transcript

00:01 All right, so this question is giving us four examples of regular polyatomic compounds and asking us about the oxidation number or oxidation state of these, or of specific atoms in these molecules.

00:20 So i'll start with my little example skeleton here, and we will start with our first example, which is sn, sn, now because these are straight up compounds, they're not ionic compounds, meaning they don't have a charge outside.

00:48 These are fully equal compounds.

00:52 You can figure out the oxidation number of one of the molecules by figuring out the oxidation number of the other molecules within, or of the other atoms within the molecules.

01:04 So this question is asking for the oxidation number of rsn.

01:10 And we can find this out because oxygen, the oxidation number for o, just one, is minus two.

01:18 Now because it's minus two and we have two of them, the overall oxidation number for the oxygens is going to be minus four.

01:27 And each of these, all these examples need to have complete zero when combined for the oxygen.

01:35 Oxidation number, which means we just need to make zero from our negative 4 from the oxygen.

01:42 So we'll have a positive.

01:48 That's going to be our oxidation number for our sn because that is going to make zero.

01:54 The 4 and our plus 4 here, plus 4 and our minus 4 here are going to combine to make a total of 0.

02:10 Which is exactly what we want.

02:12 0.

02:14 All right, so next we have our t .a.

02:19 T .a.

02:22 O3.

02:23 Now, just like in the example a, we're looking to make zero out of this.

02:29 Now, there's oxygen.

02:32 Again, each oxygen is minus 2 because an oxygen can accept two electrons.

02:37 So we have three of them.

02:38 That's going to be a minus 6...

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