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Chapter 4: Problem 49

On the basis of oxidation number considerations, one of the following oxides would not react with molecular oxygen: \(\mathrm{NO}, \mathrm{N}_{2} \mathrm{O}, \mathrm{SO}_{2}, \mathrm{SO}_{3}, \mathrm{P}_{4} \mathrm{O}_{6}\). Which one is it? Why?

Short Answer

Expert verified
SO3 is the compound that won't react with molecular Oxygen. This is because Sulphur in SO3 has already reached its maximum oxidation state +6.

Step by step solution

01

Oxidation state of Nitrogen in NO

NO consists of Nitrogen and Oxygen. Oxygen has an oxidation state of -2. As the sum of the oxidation states in a neutrally charged molecule is 0, Nitrogen in NO will have an oxidation state of +2.
02

Oxidation state of Nitrogen in N2O

N2O is composed of two Nitrogen atoms and one Oxygen atom. As Oxygen has an oxidation number of -2, to balance this and achieve a total oxidation number of 0, each Nitrogen must have an oxidation state of +1.
03

Oxidation state of Sulphur in SO2 and SO3

Oxygen is present in SO2 and SO3 with oxidation state -2. For SO2, to make the total oxidation number 0, Sulphur will have an oxidation number of +4. For SO3, to balance the total oxidation number to 0, Sulphur needs to have an oxidation state of +6.
04

Oxidation state of Phosphorus in P4O6

Oxygen in P4O6 has an oxidation state of -2. With 6 Oxygen atoms in the compound, the total negative oxidation charge is -12. To balance this negative charge and make the total oxidation state 0, each Phosphorus in P4O6 must have an oxidation state of +3.
05

Finding the compound that won't react with oxygen

Since the element Sulphur in SO3 has already reached its maximum oxidation state +6, it will not be able to further react with molecular Oxygen.

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Most popular questions from this chapter

A \(35.2-\mathrm{mL}, 1.66 \mathrm{M} \mathrm{KMnO}_{4}\) solution is mixed with \(16.7 \mathrm{~mL}\) of \(0.892 \mathrm{M} \mathrm{KMnO}_{4}\) solution. Calculate the concentration of the final solution.

Acetic acid \(\left(\mathrm{CH}_{3} \mathrm{COOH}\right)\) is an important ingredient of vinegar. A sample of \(50.0 \mathrm{~mL}\) of a commercial vinegar is titrated against a \(1.00 \mathrm{M} \mathrm{NaOH}\) solution. What is the concentration (in \(M\) ) of acetic acid present in the vinegar if \(5.75 \mathrm{~mL}\) of the base were required for the titration?

A 0.8870 -g sample of a mixture of \(\mathrm{NaCl}\) and \(\mathrm{KCl}\) is dissolved in water, and the solution is then treated with an excess of \(\mathrm{AgNO}_{3}\) to yield \(1.913 \mathrm{~g}\) of \(\mathrm{AgCl}\). Calculate the percent by mass of each compound in the mixture.

Hydrogen halides (HF, HCl, HBr, HI) are highly reactive compounds that have many industrial and laboratory uses. (a) In the laboratory, HF and HCl can be generated by reacting \(\mathrm{CaF}_{2}\) and \(\mathrm{NaCl}\) with concentrated sulfuric acid. Write appropriate equations for the reactions. (Hint: These are not redox reactions.) (b) Why is it that HBr and HI cannot be prepared similarly, that is, by reacting \(\mathrm{NaBr}\) and \(\mathrm{NaI}\) with concentrated sulfuric acid? (Hint: \(\mathrm{H}_{2} \mathrm{SO}_{4}\) is a stronger oxidizing agent than both \(\mathrm{Br}_{2}\) and \(\mathrm{I}_{2} .\) ) (c) \(\mathrm{HBr}\) can be prepared by reacting phosphorus tribromide \(\left(\mathrm{PBr}_{3}\right)\) with water. Write an equation for this reaction.

How would you prepare \(60.0 \mathrm{~mL}\) of \(0.200 \mathrm{M} \mathrm{HNO}_{3}\) from a stock solution of \(4.00 M \mathrm{HNO}_{3} ?\)
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