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Chapter 20: Problem 40

Tin forms compounds in the \(+2\) and \(+4\) oxidation states. Therefore, when tin reacts with fluorine, two products are possible. Write balanced equations for the production of the two tin halide compounds and name them.

Short Answer

Expert verified
The balanced chemical equations for the reaction of tin with fluorine are: 1. \( Sn + \frac{1}{2} F_2 \rightarrow SnF_2 \) (Tin(II) fluoride) 2. \( Sn + 2 F_2 \rightarrow SnF_4 \) (Tin(IV) fluoride)

Step by step solution

01

Write unbalanced chemical equations for the two tin halide compounds

The chemical symbols for tin and fluorine are Sn and F, respectively. We will write unbalanced equations for each oxidation state of tin: 1. Sn in +2 oxidation state: Sn + F → SnF₂ 2. Sn in +4 oxidation state: Sn + F → SnF₄
02

Balance the chemical equations

Now we need to balance the number of each atom on both sides of the equations: 1. Sn in +2 oxidation state: Sn + 1/2 * F₂ → SnF₂ 2. Sn in +4 oxidation state: Sn + 2 * F₂ → SnF₄
03

Name the tin halide compounds

Naming the compounds, we can follow the guidelines for naming binary compounds: 1. Sn in +2 oxidation state: The compound SnF₂ is named tin(II) fluoride. 2. Sn in +4 oxidation state: The compound SnF₄ is named tin(IV) fluoride. #Summary# The balanced chemical equations for the reaction of tin with fluorine to form tin halides are: 1. Sn + 1/2 * F₂ → SnF₂ (tin(II) fluoride) 2. Sn + 2 * F₂ → SnF₄ (tin(IV) fluoride)

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Most popular questions from this chapter

Slaked lime, \(\mathrm{Ca}(\mathrm{OH})_{2}\), is used to soften hard water by removing calcium ions from hard water through the reaction \(\mathrm{Ca}(\mathrm{OH})_{2}(a q)+\mathrm{Ca}^{2+}(a q)+2 \mathrm{HCO}_{3}^{-}(a q) \rightarrow\) \(2 \mathrm{CaCO}_{3}(s)+2 \mathrm{H}_{2} \mathrm{O}(l)\) Although \(\mathrm{CaCO}_{3}(s)\) is considered insoluble, some of it does dissolve in aqueous solutions. Calculate the molar solubility of \(\mathrm{CaCO}_{3}\) in water \(\left(K_{\text {sp }}=8.7 \times 10^{-9}\right)\).

While selenic acid has the formula \(\mathrm{H}_{2} \mathrm{SeO}_{4}\) and thus is directly related to sulfuric acid, telluric acid is best visualized as \(\mathrm{H}_{6} \mathrm{TeO}_{6}\) or \(\mathrm{Te}(\mathrm{OH})_{6}\) a. What is the oxidation state of tellurium in \(\mathrm{Te}(\mathrm{OH})_{6}\) ? b. Despite its structural differences with sulfuric and selenic acid, telluric acid is a diprotic acid with \(\mathrm{p} K_{\mathrm{a}_{1}}=7.68\) and \(\mathrm{P} K_{a_{2}}=11.29 .\) Telluric acid can be prepared by hydrolysis of tellurium hexafluoride according to the equation $$ \mathrm{TeF}_{6}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Te}(\mathrm{OH})_{6}(a q)+6 \mathrm{HF}(a q) $$ Tellurium hexafluoride can be prepared by the reaction of elemental tellurium with fluorine gas: $$ \mathrm{Te}(s)+3 \mathrm{~F}_{2}(g) \longrightarrow \mathrm{TeF}_{6}(g) $$ If a cubic block of tellurium (density \(\left.=6.240 \mathrm{~g} / \mathrm{cm}^{3}\right)\) measuring \(0.545 \mathrm{~cm}\) on edge is allowed to react with \(2.34 \mathrm{~L}\) fluorine gas at \(1.06 \mathrm{~atm}\) and \(25^{\circ} \mathrm{C}\), what is the \(\mathrm{pH}\) of a solution of \(\mathrm{Te}(\mathrm{OH})_{6}\) formed by dissolving the isolated \(\mathrm{TeF}_{6}(g)\) in \(115 \mathrm{~mL}\) solution? Assume \(100 \%\) yield in all reactions.

Write a balanced equation describing the reduction of \(\mathrm{H}_{2} \mathrm{SeO}_{4}\) by \(\mathrm{SO}_{2}\) to produce selenium.

Besides the central atom, what are the differences between \(\mathrm{CO}_{2}\) and \(\mathrm{SiO}_{2} ?\)

The compound \(\mathrm{Pb}_{3} \mathrm{O}_{4}\) (red lead) contains a mixture of lead(II) and lead(IV) oxidation states. What is the mole ratio of lead(II) to lead(IV) in \(\mathrm{Pb}_{3} \mathrm{O}_{4}\) ?
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