Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 4: Problem 49

Determine the oxidation number for the indicated element in each of the following substances: (a) \(\mathrm{S}\) in \(\mathrm{SO}_{3},\) (b) Ti in \(\mathrm{TiCl}_{4}\), (c) \(P\) in AgPF \(_{6}\), (d) \(\mathrm{N}\) in \(\mathrm{HNO}_{3}\) (e) \(\mathrm{S}\) in \(\mathrm{H}_{2} \mathrm{SO}_{3},(\mathbf{f}) \mathrm{O}\) in \(\mathrm{OF}_{2}\)

Short Answer

Expert verified
The oxidation numbers for the indicated elements in the given substances are: (a) S in SO3: +6, (b) Ti in TiCl4: +4, (c) P in AgPF6: +5, (d) N in HNO3: +5, (e) S in H2SO3: +4, (f) O in OF2: +2.

Step by step solution

01

(a) S in SO3

: We have the molecule SO3. Oxygen has an oxidation number of -2. There are three oxygen atoms, and we can express the sum of all oxidation numbers as \((\mathrm{oxidation\, number\, of\, S})+(3\cdot (-2))=0\). We solve for the oxidation number of S: \(x + (-6) = 0\) \(x = 6\) The oxidation number of S in SO3 is +6.
02

(b) Ti in TiCl4

: In this case, we have the molecule TiCl4. Chlorine has an oxidation number of -1, and there are four chlorine atoms. The sum of oxidation numbers can be expressed as \((\mathrm{oxidation\, number\, of\, Ti})+(4\cdot (-1))=0\). We solve for the oxidation number of Ti: \(x + (-4) = 0\) \(x = 4\) The oxidation number of Ti in TiCl4 is +4.
03

(c) P in AgPF6

: We have the molecule AgPF6. The oxidation number of Ag is +1 (since it's a monatomic ion), and F has an oxidation of -1. There are six fluorine atoms, and we solve for the oxidation number of P: \(+1 + (\mathrm{oxidation\, number\, of\, P})+(6\cdot (-1))=0\) Now, solve for the oxidation number of P: \(1 + x - 6 = 0\) \( x = 5\) The oxidation number of P in AgPF6 is +5.
04

(d) N in HNO3

: In the molecule HNO3, hydrogen has an oxidation number of +1, oxygen has an oxidation number of -2, and there are three oxygen atoms. We solve for the oxidation number of N: \((+1)+(\mathrm{oxidation\, number\, of\, N})+(3\cdot (-2))=0\) \(1+x-6=0\) \(x=5\) The oxidation number of N in HNO3 is +5.
05

(e) S in H2SO3

: For the molecule H2SO3, hydrogen has an oxidation number of +1, oxygen has an oxidation number of -2, and there are three oxygen atoms. We solve for the oxidation number of S: \((2\cdot (+1))+(\mathrm{oxidation\, number\, of\, S}) + (3\cdot (-2))=0\) \(2+x-6=0\) \(x=4\) The oxidation number of S in H2SO3 is +4.
06

(f) O in OF2

: In the molecule OF2, fluorine has an oxidation number of -1, and there are two fluorine atoms. Since OF2 is an exception to the rule that oxygen always has an oxidation state of -2, oxygen will have an oxidation number different from -2. We solve for the oxidation number of O: \((\mathrm{oxidation\, number\, of\, O})+(2\cdot (-1))=0\) \(x-2=0\) \(x=2\) The oxidation number of O in OF2 is +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 \(0.5895-\mathrm{g}\) sample of impure magnesium hydroxide is dissolved in \(100.0 \mathrm{~mL}\) of \(0.2050 \mathrm{MHCl}\) solution. The excess acid then needs \(19.85 \mathrm{~mL}\) of \(0.1020 \mathrm{M} \mathrm{NaOH}\) for neutralization. Calculate the percentage by mass of magnesium hydroxide in the sample, assuming that it is the only substance reacting with the HCl solution.

Determine the oxidation number for the indicated element in each of the following substances: (a) \(\mathrm{N}\) in \(\mathrm{N}_{2} \mathrm{H}_{4}\), (b) \(\mathrm{Nin} \mathrm{NO}_{2}\), (c) \(\mathrm{Mn}\) in \(\mathrm{MnCl}_{3}\) (d) Fe in \(\mathrm{FeSO}_{4^{\prime}}\) (e) \(\mathrm{Pt}\) in \(\mathrm{PtCl}_{4}\) (f) \(\mathrm{Cl}\) in \(\mathrm{NaClO}_{4}\).

(a) Calculate the molarity of a solution that contains 0.175 mol \(\mathrm{ZnCl}_{2}\) in exactly \(150 \mathrm{~mL}\) of solution. (b) How many moles of protons are present in \(35.0 \mathrm{~mL}\) of a \(4.50 \mathrm{M}\) solution of nitric acid? (c) How many milliliters of a $6.00 \mathrm{M} \mathrm{NaOH}\( solution are needed to provide 0.350 mol of \)\mathrm{NaOH} ?$

Separate samples of a solution of an unknown ionic compound are treated with dilute \(\mathrm{AgNO}_{3}, \mathrm{~Pb}\left(\mathrm{NO}_{3}\right)_{2},\) and \(\mathrm{BaCl}_{2}\) Precipitates form in all three cases. Which of the following could be the anion of the unknown salt: $\mathrm{Br}^{-}, \mathrm{CO}_{3}^{2-}, \mathrm{NO}_{3}^{-} ?$

Ignoring protolysis reactions, indicate the concentration of each ion or molecule present in the following solu- tions: (a) \(0.35 \mathrm{MK}_{3} \mathrm{PO}_{4},\) (b) $5 \times 10^{-4} \mathrm{MCuCl}_{2},(\mathbf{c}) 0.0184$ \(\mathrm{MCH}_{3} \mathrm{CH}_{2} \mathrm{OH}(\mathbf{d})\) a mixture of $35.0 \mathrm{~mL}\( of \)0.010 \mathrm{MNa}_{2} \mathrm{CO}_{3}$ and \(50.0 \mathrm{~mL}\) of \(0.200 \mathrm{MK}_{2} \mathrm{SO}_{4}\). Assume the volumes are additive.
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Analysis

Read Explanation

Kinetics

Read Explanation

Organic Chemistry

Read Explanation

Physical Chemistry

Read Explanation

Nuclear Chemistry

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