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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 16 elements in each: (a) potassium peroxide, (b) potassium thiosulfate, \((\mathbf{c})\) selenium oxychloride, \((\mathbf{d})\) sodium telluride, (e) magnesium sulfite, (f) selenium hexafluoride.

Short Answer

Expert verified
(a) Potassium Peroxide: K2O2 \((Oxidation State: O = -1)\) (b) Potassium Thiosulfate: K2S2O3 \((Oxidation State: S = +2, O = -2)\) (c) Selenium Oxychloride: SeOCl2 \((Oxidation State: Se = +4)\) (d) Sodium Telluride: Na2Te \((Oxidation State: Te = -2)\) (e) Magnesium Sulfite: MgSO3 \((Oxidation State: S = +4)\) (f) Selenium Hexafluoride: SeF6 \((Oxidation State: Se = +6)\)

Step by step solution

01

(a) Potassium Peroxide

Potassium Peroxide is formed by the reaction of potassium and oxygen under specific conditions. The chemical formula for potassium peroxide is K2O2. The group 16 element in this compound is oxygen. Since there are two potassium ions carrying a +1 charge each, the peroxide ion (O2^2-) carries a total of -2 charge. Each oxygen atom in the peroxide ion has an oxidation state of -1.
02

(b) Potassium Thiosulfate

Potassium thiosulfate consists of potassium, sulfur, and oxygen. The chemical formula for potassium thiosulfate is K2S2O3. The group 16 element in this compound is oxygen and sulfur. In order to determine the oxidation state of sulfur, we use the algebraic sum method. Let x be the oxidation state of sulfur, then: 2(+1) + 2x + 3(-2) = 0 Solving this, we get the oxidation state of sulfur is +2. The oxidation state of oxygen remains -2.
03

(c) Selenium Oxychloride

Selenium oxychloride is a compound consisting of selenium, oxygen, and chlorine. The chemical formula for selenium oxychloride is SeOCl2. Here, selenium is the group 16 element. To find the oxidation state of selenium, let x be its oxidation state: x + (-2) + 2(-1) = 0 Solving this, we get the oxidation state of selenium is +4.
04

(d) Sodium Telluride

Sodium telluride is a compound formed from sodium and tellurium. The chemical formula for sodium telluride is Na2Te. The group 16 element in this compound is tellurium. Since there are two sodium ions carrying a +1 charge each, tellurium must carry a -2 charge to balance the charges. Thus, the oxidation state of tellurium is -2.
05

(e) Magnesium Sulfite

Magnesium sulfite is a compound composed of magnesium, sulfur, and oxygen. The chemical formula for magnesium sulfite is MgSO3. The group 16 elements in this compound are sulfur and oxygen. The oxidation state of magnesium is +2 and that of oxygen is -2. Let x be the oxidation state of sulfur, then: (+2) + x + 3(-2) = 0 Solving this, we get the oxidation state of sulfur is +4.
06

(f) Selenium Hexafluoride

Selenium hexafluoride is a compound containing selenium and fluorine. The chemical formula for selenium hexafluoride is SeF6. The group 16 element in this compound is selenium. The oxidation state of fluorine is always -1. Let x be the oxidation state of selenium: x + 6(-1) = 0 Solving this, we get the oxidation state of selenium to be +6.

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Most popular questions from this chapter

The maximum allowable concentration of \(\mathrm{H}_{2} \mathrm{~S}(g)\) in air is \(20 \mathrm{mg}\) per kilogram of air ( 20 ppm by mass). How many grams of FeS would be required to react with hydrochloric acid to produce this concentration at \(101.3 \mathrm{kPa}\) and \(25^{\circ} \mathrm{C}\) in an average room measuring $3.5 \mathrm{~m} \times 6.0 \mathrm{~m} \times 2.5 \mathrm{~m} ?\( (Under these conditions, the average molar mass of air is \)29.0 \mathrm{~g} / \mathrm{mol} .)$

Give the chemical formula for (a) fullerene, (b) potassium cyanide, (c) zinc carbide, (d) zinc acetylide, (e) carbon disulfide.

(a) Determine the number of calcium ions in the chemical formula of the mineral hardystonite, $\mathrm{Ca}_{x} \mathrm{Zn}\left(\mathrm{Si}_{2} \mathrm{O}_{7}\right)$. (b) Determine the number of hydroxide ions in the chemical formula of the mineral pyrophyllite, $\mathrm{Al}_{2}\left(\mathrm{Si}_{2} \mathrm{O}_{5}\right)_{2}(\mathrm{OH})_{x}$.

In aqueous solution, hydrogen sulfide reduces (a) \(\mathrm{Fe}^{3+}\) to \(\mathrm{Fe}^{2+},(\mathbf{b}) \mathrm{Br}_{2}\) to \(\mathrm{Br}^{-},(\mathbf{c}) \mathrm{MnO}_{4}^{-}\) to \(\mathrm{Mn}^{2+},(\mathbf{d}) \mathrm{HNO}_{3}\) to \(\mathrm{NO}_{2}\) In all cases, under appropriate conditions, the product is elemental sulfur. Write a balanced net ionic equation for each reaction.

Consider the elements \(\mathrm{N}, \mathrm{F}, \mathrm{Si}, \mathrm{Rb},\) Te, and Ir. From this list, select the element that \((\mathbf{a})\) is most electronegative, \((\mathbf{b})\) exhibits a maximum oxidation state of \(+6,(\mathbf{c})\) loses an electron most readily, (d) forms \(\pi\) bonds most readily, (e) is a transition metal, \((\mathbf{f})\) forms four covalent bonds to achieve octet.
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