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Chapter 4: Problem 48

Determine the oxidation number of sulfur in each of the following substances: (a) barium sulfate, \(\mathrm{BaSO}_{4},\) (b) sulfurous acid, \(\mathrm{H}_{2} \mathrm{SO}_{3},\) (c) strontium sulfide, \(\mathrm{SrS},\) (d) hydrogen sulfide, \(\mathrm{H}_{2} \mathrm{~S} .\) (e) Based on these compounds what is the range of oxidation numbers seen for sulfur? Is there any relationship between the range of accessible oxidation states and sulfur's position on the periodic table?

Short Answer

Expert verified
The oxidation numbers of sulfur in the given compounds are: (a) in barium sulfate (BaSO4), +6; (b) in sulfurous acid (H2SO3), +4; (c) in strontium sulfide (SrS), -2; and (d) in hydrogen sulfide (H2S), -2. The range of oxidation numbers for sulfur is from -2 to +6. This range is related to sulfur's position in group 16 on the periodic table, as it can gain, lose, or share electrons with other elements during chemical reactions due to the presence of lone pair electrons.

Step by step solution

01

Assign Oxidation Numbers to Barium Sulfate (BaSO4)

To assign oxidation numbers to the elements in barium sulfate, let's look at the author's list and their typical oxidation numbers. Barium (Ba) belongs to group 2 and generally forms +2 ions. Oxygen (O) typically has an oxidation number of -2. We can now write an equation and find the oxidation number of sulfur (S), represented by x: \(Ba^{+2} + S^{x} + 4O^{-2} = 0\) \(2 + x - 8 = 0\)
02

Calculate the Oxidation Number of Sulfur in BaSO4

Solving for x in the equation for the oxidation number of sulfur in BaSO4 gives us: \(x = 8 - 2\) \(x = +6\) Therefore, the oxidation number of sulfur in barium sulfate (BaSO4) is +6.
03

Assign Oxidation Numbers to Sulfurous Acid (H2SO3)

In sulfurous acid, hydrogen (H) has an oxidation number of +1 and oxygen (O) has an oxidation number of -2. We can now write an equation to find the oxidation number of sulfur (S), represented by x: \(2H^{+1} + S^{x} + 3O^{-2} = 0\) \(2 + x - 6 = 0\)
04

Calculate the Oxidation Number of Sulfur in H2SO3

Solving for x in the equation for the oxidation number of sulfur in H2SO3 gives us: \(x = 6 - 2\) \(x = +4\) Therefore, the oxidation number of sulfur in sulfurous acid (H2SO3) is +4.
05

Assign Oxidation Numbers to Strontium Sulfide (SrS)

Strontium (Sr) belongs to group 2 and generally forms +2 ions. We can now write an equation to find the oxidation number of sulfur (S), represented by x: \(Sr^{+2} + S^{x} = 0\) \(2 + x = 0\)
06

Calculate the Oxidation Number of Sulfur in SrS

Solving for x in the equation for the oxidation number of sulfur in SrS gives us: \(x = -2\) Therefore, the oxidation number of sulfur in strontium sulfide (SrS) is -2.
07

Assign Oxidation Numbers to Hydrogen Sulfide (H2S)

In hydrogen sulfide, hydrogen (H) has an oxidation number of +1. We can now write an equation to find the oxidation number of sulfur (S), represented by x: \(2H^{+1} + S^{x} = 0\) \(2 + x = 0\)
08

Calculate the Oxidation Number of Sulfur in H2S

Solving for x in the equation for the oxidation number of sulfur in H2S gives us: \(x = -2\) Therefore, the oxidation number of sulfur in hydrogen sulfide (H2S) is -2.
09

Determine the Range of Oxidation Numbers for Sulfur and its Relationship with the Periodic Table

Based on the compounds given, the oxidation numbers of sulfur are -2, +4, and +6. The range of oxidation numbers for sulfur is from -2 to +6. Sulfur belongs to group 16, also known as the oxygen family, and typically displays multiple oxidation states because of the presence of lone pair electrons. The range of accessible oxidation states for sulfur is directly related to its position on the periodic table and its ability to gain, lose, or share electrons with other elements during chemical reactions.

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Most popular questions from this chapter

An \(8.65-g\) sample of an unknown group 2 A metal hydroxide is dissolved in \(85.0 \mathrm{~mL}\) of water. An acid-base indicator is added and the resulting solution is titrated with \(2.50 \mathrm{M} \mathrm{HCl}(a q)\) solution. The indicator changes color signaling that the equivalence point has been reached after \(56.9 \mathrm{~mL}\) of the hydrochloric acid solution has been added. (a) What is the molar mass of the metal hydroxide? (b) What is the identity of the metal cation: \(\mathrm{Ca}^{2+}, \mathrm{Sr}^{2+}, \mathrm{Ba}^{2+} ?\)

Identify the precipitate (if any) that forms when the following solutions are mixed, and write a balanced equation for each reaction. (a) \(\mathrm{NaCH}_{3} \mathrm{COO}\) and \(\mathrm{HCl},(\mathrm{b}) \mathrm{KOH}\) and \(\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}\), (c) \(\mathrm{Na}_{2} \mathrm{~S}\) and \(\mathrm{CdSO}_{4}\).

Suppose you have a solution that might contain any or all of the following cations: \(\mathrm{Ni}^{2+}, \mathrm{Ag}^{+}, \mathrm{Sr}^{2+},\) and \(\mathrm{Mn}^{2+}\). Addition of HCl solution causes a precipitate to form. After filtering off the precipitate, \(\mathrm{H}_{2} \mathrm{SO}_{4}\) solution is added to the resulting solution and another precipitate forms. This is filtered off, and a solution of \(\mathrm{NaOH}\) is added to the resulting solution. No precipitate is observed. Which ions are present in each of the precipitates? Which of the four ions listed above must be absent from the original solution?

(a) Starting with solid sucrose, \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\), describe how you would prepare \(250 \mathrm{~mL}\) of a \(0.250 \mathrm{M}\) sucrose solution. (b) Describe how you would prepare \(350.0 \mathrm{~mL}\) of \(0.100 \mathrm{M}\) \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\) starting with \(3.00 \mathrm{~L}\) of \(1.50 \mathrm{M} \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11} .\)

(a) Which will have the highest concentration of potassium ion: \(0.20 \mathrm{M} \mathrm{KCl}, 0.15 \mathrm{M} \mathrm{K}_{2} \mathrm{CrO}_{4},\) or \(0.080 \mathrm{M} \mathrm{K}_{3} \mathrm{PO}_{4} ?(\mathbf{b})\) Which will contain the greater number of moles of potassium ion: \(30.0 \mathrm{~mL}\) of \(0.15 \mathrm{M} \mathrm{K}_{2} \mathrm{CrO}_{4}\) or \(25.0 \mathrm{~mL}\) of \(0.080 \mathrm{MK}_{3} \mathrm{PO}_{4} ?\)
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