What is the oxidation number of each underlined element? a) P 4 b) B O 3 3– c) As O 4 2– d) H S O 4 –

Welcome back, guys. In this video, we're gonna look at the first practice question that deals with oxidation and reduction reactions. So it says, What is the oxidation number off each underlined element. For the first one, we have phosphorus, and it's connected to copies of itself. Other phosphorus atoms and it has no charge. Remember, this fits with our general rule. So here the oxidation number would be equal to zero for the next one. We have boron, but here we don't have a real rule for it. So it's gonna be X. Okay, Now, what we're gonna do is we're gonna try to solve for by just using on algebraic approach. We're going to say here that oxygen is not a super oxide or a peroxide. Therefore, it's oxidation numbers Negative, too. Now, we're gonna solve where we're gonna say X plus the three oxygen's. Each one is minus two equals the charge of the compound minus three. So X minus six equals minus three plus six plus six X equals plus three. So the oxidation number of boron here would be plus three. But don't get fooled by questions like this. Don't automatically say Oh, of course, it's plus three because boron is in Group three A. That's not always true. It's all dependent on the charge of the compound and the different elements connected to boron. It's oxidation. Numbers could really change. It could become even plus three. That's why it's important to do the math and work it out the next one. We have arsenic connected to four Oxygen's, and it has a negative to charge Now. Arsenic is not by itself, so we can't say zero, since it's connected to a different element. That's why we have to solve for its new oxidation state. We don't have a rule for it. So it's X oxygen is not a super oxide or a peroxide here, so it b minus two Yeah, X plus four oxygen's. Each one is minus two equals negative. Two for the charge. X minus eight equals negative, too. At eight to both sides, X equals plus six. So the oxidation state of arsenic is plus six. Then finally, the last one we have H s 04 minus one sulfur. We don't have a road for, so it's X hydrogen is connected to nonmetal, so it's plus one oxygen here would be minus two. We're gonna say one plus X plus the four oxygen's. Each one is minus two equals the charge of the compound minus one one plus X minus eight equals minus one. Combine the negative eight and the plus one, and that gives us minus seven. Add seven to both sides. X equals plus six. So the oxidation number here would be plus six. Hopefully, you guys got the correct answer for these ones. And remember, all we're doing is we're using the rules that we talked about for oxidation and reduction reactions. As long as you follow through with those types of rules, you can solve the oxidation number for any type of compound and any type of element.

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20. Electrochemistry

Standard Reduction Potentials

General Chemistry

20. Electrochemistry

Standard Reduction Potentials

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Multiple Choice

What is the oxidation number of each underlined element?
a) P₄ b) BO₃^3– c) AsO₄^2–d) HSO₄^–

a) -3 b) +3 c) -2 d) -2

a) -2 b) 0 c) -6 d) -6

a) 0 b) +3 c) +6 d) +6

a) 0 b) +3 c) -6 d) -6

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Verified step by step guidance

Identify the oxidation number of phosphorus in P4: Since P4 is a molecule consisting of only phosphorus atoms, the oxidation number of each phosphorus atom is 0 because the molecule is in its elemental form.

Determine the oxidation number of boron in BO3^3-: The overall charge of the ion is -3. Oxygen typically has an oxidation number of -2. Set up the equation: x + 3(-2) = -3, where x is the oxidation number of boron.

Calculate the oxidation number of arsenic in AsO4^2-: The overall charge of the ion is -2. Oxygen has an oxidation number of -2. Set up the equation: x + 4(-2) = -2, where x is the oxidation number of arsenic.

Find the oxidation number of sulfur in HSO4^-: The overall charge of the ion is -1. Hydrogen has an oxidation number of +1, and oxygen has an oxidation number of -2. Set up the equation: 1 + x + 4(-2) = -1, where x is the oxidation number of sulfur.

Solve each equation to find the oxidation numbers: For BO3^3-, solve x - 6 = -3 to find x. For AsO4^2-, solve x - 8 = -2 to find x. For HSO4^-, solve 1 + x - 8 = -1 to find x.