Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 4: Problem 52

Mssign oxidation numbers to each element in (a) \(\mathrm{HIO}_{3}\) (b) \(\mathrm{NaMnO}_{4}\) (c) \(\mathrm{SnO}_{2}\) (d) NOF (e) \(\mathrm{NaO}_{2}\)

Short Answer

Expert verified
Answer: The oxidation numbers of the elements in the given compounds are as follows: - HIO3: H (+1), I (+5), and O (-2) - NaMnO4: Na (+1), Mn (+7), and O (-2) - SnO2: Sn (+4) and O (-2) - NOF: N (+3), O (-2), and F (-1) - NaO2: Na (+1) and O (-1/2)

Step by step solution

01

HIO3: Assigning Oxidation Numbers for Hydrogen, Iodine, and Oxygen

1. Hydrogen generally has an oxidation number of +1. 2. Oxygen generally has an oxidation number of -2. 3. Knowing that the overall charge of the compound is 0, we will determine the oxidation number of Iodine. \(+1*x -2*3+y=0\), x being the amount of H-atoms and y being the amount of I-atoms. In this compound: x=1 and y=1 Hence, the oxidation number of Iodine is +5. So, the oxidation numbers of the elements are: H (+1), I (+5), and O (-2).
02

NaMnO4: Assigning Oxidation Numbers for Sodium, Manganese, and Oxygen

1. Sodium generally has an oxidation number of +1. 2. Oxygen generally has an oxidation number of -2. 3. Knowing that the overall charge of the compound is 0, we will determine the oxidation number of Manganese. \(+1-2*4+y=0\), y being the oxidation number for Mn. Hence, the oxidation number of Manganese is +7. So, the oxidation numbers of the elements are: Na (+1), Mn (+7), and O (-2).
03

SnO2: Assigning Oxidation Numbers for Tin and Oxygen

1. Oxygen generally has an oxidation number of -2. 2. Knowing that the overall charge of the compound is 0, we will determine the oxidation number of Tin. \(y-2*2=0\), y being the oxidation number for Sn. Hence, the oxidation number of Tin is +4. So, the oxidation numbers of the elements are: Sn (+4) and O (-2).
04

NOF: Assigning Oxidation Numbers for Nitrogen, Oxygen, and Fluorine

1. Oxygen generally has an oxidation number of -2. 2. Fluorine generally has an oxidation number of -1. 3. Knowing that the overall charge of the compound is 0, we will determine the oxidation number of Nitrogen. \(y-1-2=0\), y being the oxidation number for N. Hence, the oxidation number of Nitrogen is +3. So, the oxidation numbers of the elements are: N (+3), O (-2), and F (-1).
05

NaO2: Assigning Oxidation Numbers for Sodium and Oxygen

1. Sodium generally has an oxidation number of +1. 2. Knowing that the overall charge of the compound is 0, we will determine the oxidation number of Oxygen. \(+1+y*2=0\), y being oxidation number for O. Hence, the oxidation number of Oxygen is -1/2 (in this case, oxygen is in a peroxide form). So, the oxidation numbers of the elements are: Na (+1) and O (-1/2).

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

A reagent bottle is labeled \(0.450 \mathrm{M} \mathrm{K}_{2} \mathrm{CO}_{3}\). (a) How many moles of \(\mathrm{K}_{2} \mathrm{CO}_{3}\) are present in \(45.6 \mathrm{~mL}\) of this solution? (b) How many milliliters of this solution are required to furnish \(0.800 \mathrm{~mol}\) of \(\mathrm{K}_{2} \mathrm{CO}_{3} ?\) (c) Assuming no volume change, how many grams of \(\mathrm{K}_{2} \mathrm{CO}_{3}\) do you need to add to \(2.00 \mathrm{~L}\) of this solution to obtain a \(1.000 \mathrm{M}\) solution of \(\mathrm{K}_{2} \mathrm{CO}_{3} ?\) (d) If \(50.0 \mathrm{~mL}\) of this solution is added to enough water to make \(125 \mathrm{~mL}\) of solution, what is the molarity of the diluted solution?

A solution contains both iron(II) and iron(III) ions. A \(50.00-\mathrm{mL}\) sample of the solution is titrated with \(35.0 \mathrm{~mL}\) of \(0.0280 \mathrm{M} \mathrm{KMnO}_{4}\), which oxidizes \(\mathrm{Fe}^{2+}\) to \(\mathrm{Fe}^{3+} .\) The permanganate ion is reduced to manganese(II) ion. Another \(50.00-\mathrm{mL}\) sample of the solution is treated with zinc, which reduces all the \(\mathrm{Fe}^{3+}\) to \(\mathrm{Fe}^{2+}\). The resulting solution is again titrated with \(0.0280 \mathrm{M}\) \(\mathrm{KMnO}_{4} ;\) this time \(48.0 \mathrm{~mL}\) is required. What are the concentrations of \(\mathrm{Fe}^{2+}\) and \(\mathrm{Fe}^{3+}\) in the solution?

When solutions of aluminum sulfate and sodium hydroxide are mixed, a white gelatinous precipitate forms. (a) Write a balanced net ionic equation for the reaction. (b) What is the mass of the precipitate when \(2.76 \mathrm{~g}\) of aluminum sulfate in \(125 \mathrm{~mL}\) of solution is combined with \(85.0 \mathrm{~mL}\) of \(0.2500 \mathrm{M} \mathrm{NaOH}\) ? (c) What is the molarity of the ion in excess? (Ignore spectator ions and assume that volumes are additive.)

Laundry bleach is a solution of sodium hypochlorite (NaClO). To determine the hypochlorite (ClO \(^{-}\) ) content of bleach (which is responsible for its bleaching action), sulfide ion is added in basic solution. The balanced equation for the reaction is $$ \mathrm{ClO}^{-}(a q)+\mathrm{S}^{2-}(a q)+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{Cl}^{-}(a q)+\mathrm{S}(s)+2 \mathrm{OH}^{-}(a q) $$ The chloride ion resulting from the reduction of HClO is precipitated as \(\mathrm{AgCl}\). When \(50.0 \mathrm{~mL}\) of laundry bleach \(\left(d=1.02 \mathrm{~g} / \mathrm{cm}^{3}\right)\) is treated as described above, \(4.95 \mathrm{~g}\) of \(\mathrm{AgCl}\) is obtained. What is the mass percent of \(\mathrm{NaClO}\) in the bleach?

Decide whether a precipitate will form when the following solutions are mixed. If a precipitate forms, write a net ionic equation for the reaction. (a) potassium nitrate and magnesium sulfate (b) silver nitrate and potassium carbonate (c) ammonium carbonate and cobalt(III) chloride (d) sodium phosphate and barium hydroxide (e) barium nitrate and potassium hydroxide
See all solutions

Recommended explanations on Chemistry Textbooks

Organic Chemistry

Read Explanation

Chemical Reactions

Read Explanation

Making Measurements

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Inorganic Chemistry

Read Explanation

Physical 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