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Chapter 22: Problem 53

What is the oxidation state of sulfur in the following compounds? (a) \(\mathrm{SF}_{4} ;\) (b) \(\mathrm{S}_{2} \mathrm{F}_{10} ;\) (c) \(\mathrm{H}_{2} \mathrm{S} ;\) (d) \(\mathrm{CaSO}_{3}\).

Short Answer

Expert verified
The oxidation states of sulfur in the given compounds are: (a) +4, (b) +5, (c) -2, and (d) +4.

Step by step solution

01

Determining oxidation states in SF4

Let's denote the oxidation state of sulfur as \(x\). Fluorine usually has an oxidation state of \(-1\) in its compounds. As there are four fluorine atoms, the total charge due to fluorine is \(-4\). The molecule is neutral, so the sum of oxidation states must be equal to zero, thus \(x - 4 = 0\). By solving this equation, we find \(x = +4\).
02

Determining oxidation states in S2F10

Since we have two atoms of sulfur, we will denote their total charge as \(2x\). Same as in step one, fluorine contributes with \(-10\), and the total charge of the molecule is zero. We therefore have \(2x - 10 = 0\). Solving this equation gives \(x = +5\).
03

Determining oxidation states in H2S

In this molecule, hydrogen has its usual oxidation state of \(+1\), contributing with \(+2\). Hence, \(2 + x = 0\), thus \(x = -2\).
04

Determining oxidation states in CaSO3

In this compound, calcium has an oxidation state of \(+2\) and oxygen has an usual oxidation state of \(-2\), contributing with \(-6\) (we have three oxygen atoms in the molecule). The total charge is zero, so we have \(2 + x - 6 = 0\). This gives us the oxidation state of sulfur as \(x = +4\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Sulfur Compounds
Sulfur is a versatile element that can form a variety of compounds by adopting different oxidation states. In the world of chemistry, sulfur is found in numerous compounds such as
  • sulfides (e.g., \(\rm {H}_{2} S\)
  • oxyacids (e.g., \(\rm {H}_{2} {SO}_{4}\)
  • fluorides (e.g., \(\rm {SF}_{4}\)
Each compound exhibits distinct chemical and physical properties, often determined by the specific oxidation state of sulfur.

By adjusting its oxidation state, sulfur can form compounds ranging from simple hydrogen sulfide \(\rm {H}_{2} S\) to more complex molecules such as disulfur decafluoride \(\rm {S}_{2} {F}_{10}\), or calcium sulfite \(\rm {CaSO}_{3}\). This adaptability allows sulfur to play essential roles in biological systems and industrial processes.

Understanding sulfur compounds' nature aids in various scientific endeavors, including environmental science and pharmaceuticals.
Chemical Problem-Solving
Chemical problem-solving involves a step-by-step approach to understand and find the solution to chemical exercises. Whether you're calculating oxidation numbers or predicting reaction outcomes, a structured technique helps simplify the task.

Start by identifying what is known and what is being asked. In our exercise, the task was to find the oxidation states of sulfur in different compounds.
  • List all known oxidation states of other elements within the compounds, such as fluorine usually being \(-1\) and oxygen being \(-2\).
  • Sum the contributions of these elements and set the equation according to the charge of the compound, whether it's neutral or charged.
  • Solve the equation for the unknown, which in this case, is sulfur's oxidation state.
A systematic approach ensures you remain organized and consistent in solving complex chemical problems. Practice regularly to gain confidence in this methodology.
Oxidation Number Calculation
Understanding how to calculate oxidation numbers is fundamental in chemistry. Oxidation numbers provide insight into electron transfer and chemical bonding. Here's how to calculate oxidation numbers, using sulfur compounds from our exercise as examples.

The oxidation state of \(\rm SF_{4}\) was found by assigning fluorine \(-1\) and balancing the total charge to zero. So, sulfur's oxidation state is \(+4\).
  • Write the known oxidation states of other atoms in the compound.
  • Use the fact that the sum of oxidation states must equal the charge of the compound.
  • For example, in \(\rm {S}_{2}{F}_{10}\), setting \(-10\) for ten fluorine atoms and solving \(2x - 10 = 0\) gives sulfur an oxidation state of \(+5\).
In \(\rm H_{2}S\), hydrogen contributes \(+2\) total (since each hydrogen is \(+1\)), giving sulfur an oxidation state of \(-2\).

Once you know how to systematically apply these rules, calculating oxidation numbers becomes a straightforward process. This skill is crucial for understanding redox reactions and predicting compound behavior.

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Most popular questions from this chapter

The trichloride ion, \(\mathrm{Cl}_{3}^{-1}\), is not very stable in aqueous solution. The equilibrium constant for the following dissociation reaction is 5.5 at \(25^{\circ} \mathrm{C}\) : $$\mathrm{Cl}_{3}^{-}(\mathrm{aq}) \rightleftharpoons \mathrm{Cl}^{-}(\mathrm{aq})+\mathrm{Cl}_{2}(\mathrm{aq})$$ (a) Draw a Lewis structure for the \(\mathrm{Cl}_{3}^{-}\) ion and predict the geometry. (b) Calculate the equilibrium concentration of \(\mathrm{Cl}_{3}^{-}\) if 0.0010 moles each of \(\mathrm{KCl}\) and \(\mathrm{Cl}_{2}\) are dissolved in water at \(25^{\circ} \mathrm{C}\) to make \(1.0 \mathrm{L}\) of solution.

Suppose that the sulfur present in seawater as \(\mathrm{SO}_{4}^{2-}\) \(\left(2650 \mathrm{mg} \mathrm{L}^{-1}\right)\) could be recovered as elemental sulfur. If this sulfur were then converted to \(\mathrm{H}_{2} \mathrm{SO}_{4},\) how many cubic kilometers of seawater would have to be processed to yield the average U.S. annual consumption of about 45 million tons of \(\mathrm{H}_{2} \mathrm{SO}_{4} ?\)

Use information from this chapter and previous chapters to write plausible chemical equations to represent the following: (a) the reaction of silver metal with \(\mathrm{HNO}_{3}(\mathrm{aq})\) (b) the complete combustion of the rocket fuel, unsymmetrical dimethylhydrazine, \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NNH}_{2}\) (c) the preparation of sodium triphosphate by heating a mixture of sodium dihydrogen phosphate and sodium hydrogen phosphate.

Draw plausible Lewis structures for (a) dimethylhydrazine, \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NNH}_{2}\) (b) nitryl chloride, \(\mathrm{ClNO}_{2}\) (Central atom is \(\mathrm{N}\) ) (c) phosphorous acid, a diprotic acid with the empirical formula \(\mathrm{H}_{3} \mathrm{PO}_{3}\).

The so-called pyroanions, \(\mathrm{X}_{2} \mathrm{O}_{7}^{n-},\) form a series of structurally similar polyatomic anions for the elements Si, P, and S. (a) Draw the Lewis structures of these anions, and predict the geometry of the anions. What is the maximum number of atoms that can lie in a plane? (b) Each pyroanion in part (a) corresponds to a pyroacid, \(\mathrm{X}_{2} \mathrm{O}_{7} \mathrm{H}_{n} .\) Compare each pyroacid to the acid containing only one atom of the element in its maximum oxidation state. From this comparison, suggest a strategy for the preparation of these pyroacids. (c) What is the chlorine analogue of the pyroanions? For which acid is this species the anhydride?
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