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Chapter 22: Problem 41

Write the chemical formula for each of the following compounds, and indicate the oxidation state of the group 6 \(\mathrm{A}\) element in each: (a) selenous acid, (b) potassium hydrogen sulfite, ( c) hydrogen telluride, (d) carbon disulfide, (e) calcium sulfate, (f) cadmium sulfide, (g) zinc telluride.

Short Answer

Expert verified
The chemical formulas and oxidation states of the group 6A elements in each compound are as follows: (a) Selenous Acid: H2SeO3, Se oxidation state = +4. (b) Potassium Hydrogen Sulfite: KHSO3, S oxidation state = +4. (c) Hydrogen Telluride: H2Te, Te oxidation state = -2. (d) Carbon Disulfide: CS2, S oxidation state = -2. (e) Calcium Sulfate: CaSO4, S oxidation state = +6. (f) Cadmium Sulfide: CdS, S oxidation state = -2. (g) Zinc Telluride: ZnTe, Te oxidation state = -2.

Step by step solution

01

Compound (a) - Selenous Acid

Selenous acid is H2SeO3. The group 6A element in selenous acid is selenium (Se). To find the oxidation state of selenium in selenous acid, we will first assign oxidation states to all elements in the formula: - Hydrogen (H) = +1 - Oxygen (O) = -2 Then, we set up an equation to find the oxidation state of selenium (Se): 2(+1) + x + 3(-2) = 0 Solving for x, the oxidation state of selenium in selenous acid is +4.
02

Compound (b) - Potassium Hydrogen Sulfite

Potassium Hydrogen Sulfite is KHSO3. The group 6A element in potassium hydrogen sulfite is sulfur (S). To find the oxidation state of sulfur in potassium hydrogen sulfite, we will first assign oxidation states to all elements in the formula: - Potassium (K) = +1 - Hydrogen (H) = +1 - Oxygen (O) = -2 Then, we set up an equation to find the oxidation state of sulfur (S): (+1) + (+1) + x + 3(-2) = 0 Solving for x, the oxidation state of sulfur in potassium hydrogen sulfite is +4.
03

Compound (c) - Hydrogen Telluride

Hydrogen Telluride is H2Te. The group 6A element in hydrogen telluride is tellurium (Te). To find the oxidation state of tellurium in hydrogen telluride, we will first assign oxidation states to all elements in the formula: - Hydrogen (H) = +1 Then, we set up an equation to find the oxidation state of tellurium (Te): 2(+1) + x = 0 Solving for x, the oxidation state of tellurium in hydrogen telluride is -2.
04

Compound (d) - Carbon Disulfide

Carbon Disulfide is CS2. The group 6A element in carbon disulfide is sulfur (S). To find the oxidation state of sulfur in carbon disulfide, we will first assign oxidation states to all elements in the formula: - Carbon (C) = +4 Then, we set up an equation to find the oxidation state of sulfur (S): (+4) + 2x = 0 Solving for x, the oxidation state of sulfur in carbon disulfide is -2.
05

Compound (e) - Calcium Sulfate

Calcium Sulfate is CaSO4. The group 6A element in calcium sulfate is sulfur (S). To find the oxidation state of sulfur in calcium sulfate, we will first assign oxidation states to all elements in the formula: - Calcium (Ca) = +2 - Oxygen (O) = -2 Then, we set up an equation to find the oxidation state of sulfur (S): (+2) + x + 4(-2) = 0 Solving for x, the oxidation state of sulfur in calcium sulfate is +6.
06

Compound (f) - Cadmium Sulfide

Cadmium Sulfide is CdS. The group 6A element in cadmium sulfide is sulfur (S). To find the oxidation state of sulfur in cadmium sulfide, we will first assign oxidation states to all elements in the formula: - Cadmium (Cd) = +2 Then, we set up an equation to find the oxidation state of sulfur (S): (+2) + x = 0 Solving for x, the oxidation state of sulfur in cadmium sulfide is -2.
07

Compound (g) - Zinc Telluride

Zinc Telluride is ZnTe. The group 6A element in zinc telluride is tellurium (Te). To find the oxidation state of tellurium in zinc telluride, we will first assign oxidation states to all elements in the formula: - Zinc (Zn) = +2 Then, we set up an equation to find the oxidation state of tellurium (Te): (+2) + x = 0 Solving for x, the oxidation state of tellurium in zinc telluride is -2.

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Most popular questions from this chapter

Manganese silicide has the empirical formula MnSi and melts at \(1280^{\circ} \mathrm{C}\) . It is insoluble in water but does dissolve in aqueous HF. (a) What type of compound do you expect MnSi to be: metallic, molecular, covalent- network, or ionic? (b) Write a likely balanced chemical equation for the reaction of MnSi with concentrated aqueous HF.

Write a balanced equation for the reaction of each of the following compounds with water: (a) \(\mathrm{SO}_{2}(g),(\mathbf{b}) \mathrm{Cl}_{2} \mathrm{O}_{7}(g)\) (c) \(\mathrm{Na}_{2} \mathrm{O}_{2}(s),\) (d) \(\mathrm{BaC}_{2}(s),\) (e) \(\mathrm{RbO}_{2}(s)\) (f) \(\mathrm{Mg}_{3} \mathrm{N}_{2}(s)\) , (g) \(\mathrm{NaH}(s) .\)

(a) What is the oxidation state of \(P\) in \(P O_{4}^{3-}\) and of \(N\) in \(N O_{3}^{-} ?(\mathbf{b})\) Why doesn't \(N\) form a stable \(N O_{4}^{3-}\) analogous to P?

Explain each of the following observations: (a) At room temperature I \(_{2}\) is a solid, Br \(_{2}\) is a liquid, and \(C l_{2}\) and \(F_{2}\) are both gases. (b) \(F_{2}\) cannot be prepared by electrolytic oxidation of aqueous \(\mathrm{F}^{-}\) solutions. (c) The boiling point of HF is much higher than those of the other hydrogen halides. (d) The halogens decrease in oxidizing power in the order \(\mathrm{F}_{2}>\mathrm{Cl}_{2}>\mathrm{Br}_{2}>\mathrm{I}_{2} .\)

Ultrapure germanium, like silicon, is used in semiconductors. Germanium of "ordinary" purity is prepared by the high-temperature reduction of \(\mathrm{GeO}_{2}\) with carbon. The Ge is converted to GeCl_ by treatment with \(\mathrm{Cl}_{2}\) and then purified by distillation; GeCl_ is then hydrolyzed in water to GeO \(_{2}\) and reduced to the elemental form with \(\mathrm{H}_{2}\) . The element is then zone refined. Write a balanced chemical equation for each of the chemical transformations in the course of forming ultrapure Ge from GeO \(_{2} .\)
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