Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 6: Problem 81

Assign oxidation states for all atoms in each of the following compounds. a. \(\mathrm{KMnO}_{4}\) f. \(\mathrm{Fe}_{3} \mathrm{O}_{4}\) b. \(\mathrm{NiO}_{2}\) g. \(\mathrm{XeOF}_{4}\) c. \(\mathrm{Na}_{4} \mathrm{Fe}(\mathrm{OH})_{6}\) h. \(\mathrm{SF}_{4}\) d. \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{HPO}_{4}\) i. \(\mathrm{CO}\) e. \(\mathrm{P}_{4} \mathrm{O}_{6}\) j. \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\)

Short Answer

Expert verified
Here are the oxidation states for all the atoms in the given compounds: a. KMnO4: K(+1), Mn(+7), O(-2) f. Fe3O4: Fe(+2), Fe(+3), Fe(+3), O(-2) b. NiO2: Ni(+4), O(-2) g. XeOF4: Xe(+6), O(-2), F(-1) c. Na4Fe(OH)6: Na(+1), Fe(+2), O(-2), H(+1) h. SF4: S(+4), F(-1) d. (NH4)2HPO4: N(-3), H(+1), P(+5), O(-2) i. CO: C(+2), O(-2) e. P4O6: P(+3), O(-2) j. C6H12O6: C(0), H(+1), O(-2)

Step by step solution

01

Assign Oxidation States for a: KMnO4

To determine the oxidation states, we need to assign the known oxidation states and calculate the unknown ones. In KMnO4: 1. Potassium (K) has an oxidation state of +1 2. Oxygen (O) usually has an oxidation state of -2 Let x be the oxidation state of Mn. The sum of the oxidation states of all atoms in the compound should equal to the net charge, which is 0 in this case. So, +1 + x + 4(-2) = 0. Solving the equation, we get x = +7. The oxidation states are: K(+1), Mn(+7), O(-2).
02

Assign Oxidation States for f: Fe3O4

To determine the oxidation states in Fe3O4: 1. Oxygen (O) usually has an oxidation state of -2 Let x be the average oxidation state of Iron (Fe). We have 3 atoms of Iron. The sum of the oxidation states of all atoms in the compound should equal to the net charge, which is 0 in this case. So, 3x + 4(-2) = 0. Solving the equation, we get x = +8/3. The average oxidation state of Iron is +8/3. This implies that two of the three iron atoms have an oxidation state of +3, and the third one has an oxidation state of +2. The oxidation states are: Fe(+2), Fe(+3), Fe(+3), O(-2).
03

Assign Oxidation States for b: NiO2

To determine the oxidation states in NiO2: 1. Oxygen (O) usually has an oxidation state of -2 Let x be the oxidation state of Nickel (Ni). The sum of the oxidation states of all atoms in the compound should equal to the net charge, which is 0 in this case. So, x + 2(-2) = 0. Solving the equation, we get x = +4. The oxidation states are: Ni(+4), O(-2).
04

Assign Oxidation States for g: XeOF4

To determine the oxidation states in XeOF4: 1. Oxygen (O) usually has an oxidation state of -2 2. Fluorine (F) almost always has an oxidation state of -1 Let x be the oxidation state of Xenon (Xe). The sum of the oxidation states of all atoms in the compound should equal to the net charge, which is 0 in this case. So, x - 2 + 4(-1) = 0. Solving the equation, we get x = +6. The oxidation states are: Xe(+6), O(-2), F(-1).
05

Assign Oxidation States for c: Na4Fe(OH)6

To determine the oxidation states in Na4Fe(OH)6: 1. Sodium (Na) has an oxidation state of +1 2. Oxygen (O) usually has an oxidation state of -2 3. Hydrogen (H) has an oxidation state of +1 when bound to non-metal (OH- is a non-metal) Let x be the oxidation state of Iron (Fe). The sum of the oxidation states of all atoms in the compound should equal to the net charge, which is 0 in this case. So, 4(+1) + x + 6[-2 + 1] = 0. Solving the equation, we get x = +2. The oxidation states are: Na(+1), Fe(+2), O(-2), H(+1).
06

Assign Oxidation States for h: SF4

To determine the oxidation states in SF4: 1. Fluorine (F) almost always has an oxidation state of -1 Let x be the oxidation state of Sulfur (S). The sum of the oxidation states of all atoms in the compound should equal to the net charge, which is 0 in this case. So, x + 4(-1) = 0. Solving the equation, we get x = +4. The oxidation states are: S(+4), F(-1).
07

Assign Oxidation States for d: (NH4)2HPO4

To determine the oxidation states in (NH4)2HPO4: 1. Hydrogen (H) has an oxidation state of +1 when bound to non-metal (NH4 and PO4 are non-metals) 2. Oxygen (O) usually has an oxidation state of -2 Let x be the oxidation state of Phosphorus (P) in HPO4-. So, x + (-2) * 4 + 1 = -1. Solving the equation, we get x = +5. The oxidation states are: N(-3), H(+1), P(+5), O(-2).
08

Assign Oxidation States for i: CO

To determine the oxidation states in CO: 1. Oxygen (O) usually has an oxidation state of -2 Let x be the oxidation state of Carbon (C). The sum of the oxidation states of all atoms in the compound should equal to the net charge, which is 0 in this case. So, x + (-2) = 0. Solving the equation, we get x = +2. The oxidation states are: C(+2), O(-2).
09

Assign Oxidation States for e: P4O6

To determine the oxidation states in P4O6: 1. Oxygen (O) usually has an oxidation state of -2 Let x be the oxidation state of Phosphorus (P). The sum of the oxidation states of all atoms in the compound should equal to the net charge, which is 0 in this case. So, 4x + 6(-2) = 0. Solving the equation, we get x = +3. The oxidation states are: P(+3), O(-2).
10

Assign Oxidation States for j: C6H12O6

To determine the oxidation states in C6H12O6: 1. Oxygen (O) usually has an oxidation state of -2 2. Hydrogen (H) has an oxidation state of +1 when bound to non-metal (C and O are non-metals) Let x be the oxidation state of Carbon (C) in the compound. The sum of the oxidation states of all atoms in the compound should equal to the net charge, which is 0 in this case. So, 6x + 12(+1) + 6(-2) = 0. Solving the equation, we get x = 0. The oxidation states are: C(0), H(+1), O(-2).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Write the balanced formula, complete ionic, and net ionic equations for each of the following acid-base reactions. a. \(\mathrm{HNO}_{3}(a q)+\mathrm{Al}(\mathrm{OH})_{3}(s) \rightarrow\) b. \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}(a q)+\mathrm{KOH}(a q) \rightarrow\) c. \(\mathrm{Ca}(\mathrm{OH})_{2}(a q)+\mathrm{HCl}(a q) \rightarrow\)

A mixture contains only \(\mathrm{NaCl}\) and \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\). A 1.45 -g sample of the mixture is dissolved in water and an excess of NaOH is added, producing a precipitate of \(\mathrm{Al}(\mathrm{OH})_{3}\). The precipitate is filtered, dried, and weighed. The mass of the precipitate is \(0.107 \mathrm{g} .\) What is the mass percent of \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) in the sample?

Separate samples of a solution of an unknown soluble ionic compound are treated with \(\mathrm{KCl}, \mathrm{Na}_{2} \mathrm{SO}_{4},\) and \(\mathrm{NaOH}\). A precipitate forms only when \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) is added. Which cations could be present in the unknown soluble ionic compound?

Many oxidation-reduction reactions can be balanced by inspection. Try to balance the following reactions by inspection. In each reaction, identify the substance reduced and the substance oxidized. a. \(\mathrm{Al}(s)+\mathrm{HCl}(a q) \rightarrow \mathrm{AlCl}_{3}(a q)+\mathrm{H}_{2}(g)\) b. \(\mathrm{CH}_{4}(g)+\mathrm{S}(s) \rightarrow \mathrm{CS}_{2}(l)+\mathrm{H}_{2} \mathrm{S}(g)\) c. \(C_{3} H_{8}(g)+O_{2}(g) \rightarrow C O_{2}(g)+H_{2} O(l)\) d. \(\mathrm{Cu}(s)+\mathrm{Ag}^{+}(a q) \rightarrow \mathrm{Ag}(s)+\mathrm{Cu}^{2+}(a q)\)

The concentration of a certain sodium hydroxide solution was determined by using the solution to titrate a sample of potassium hydrogen phthalate (abbreviated as KHP). KHP is an acid with one acidic hydrogen and a molar mass of \(204.22 \mathrm{g} / \mathrm{mol}\). In the titration, \(34.67 \mathrm{mL}\) of the sodium hydroxide solution was required to react with 0.1082 g KHP. Calculate the molarity of the sodium hydroxide.
See all solutions

Recommended explanations on Chemistry Textbooks

Physical Chemistry

Read Explanation

Organic Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Kinetics

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Inorganic Chemistry

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free