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Chapter 22: Problem 69

Write the formulas for the following compounds, and indicate the oxidation state of the group 4 A element or of boron in each: (a) boric acid, (b) silicon tetrabromide, (c) lead(II) chloride, (d) sodium tetraborate decahydrate (borax), (e) boric oxide, (f) germanium dioxide.

Short Answer

Expert verified
The chemical formulas and oxidation states for the given compounds are: (a) Boric acid: H3BO3, with boron in +3 oxidation state (b) Silicon tetrabromide: SiBr4, with silicon in +4 oxidation state (c) Lead(II) chloride: PbCl2, with lead in +2 oxidation state (d) Sodium tetraborate decahydrate (Borax): Na2B4O7·10H2O, with boron in +3 oxidation state (e) Boric oxide: B2O3, with boron in +3 oxidation state (f) Germanium dioxide: GeO2, with germanium in +4 oxidation state

Step by step solution

01

(a) Boric Acid

For boric acid, first we need to know the elements present in it. Boric acid contains boron (B), hydrogen (H), and oxygen (O). The chemical formula for boric acid is H3BO3. Now, we need to determine the oxidation state of boron in this compound. Let the oxidation state of boron be x. For a neutral compound, the sum of the oxidation states of all elements must be equal to zero. We can use the following equation to find the oxidation state of boron: \(x + 3(+1) +3(-2) = 0\) \(x+ 3 - 6 = 0\) \(x = \)3 The oxidation state of boron in boric acid is +3.
02

(b) Silicon Tetrabromide

Silicon tetrabromide contains silicon (Si) and bromine (Br). The chemical formula for silicon tetrabromide is SiBr4. In this compound, the oxidation state of silicon will be: Let the oxidation state of silicon be x. \(x + 4(-1) = 0\) \(x - 4 = 0\) \(x = \)4 The oxidation state of silicon in silicon tetrabromide is +4.
03

(c) Lead(II) Chloride

Lead(II) chloride contains lead (Pb) and chlorine (Cl). The chemical formula for lead(II) chloride is PbCl2. Since it is lead(II), it means that the oxidation state of lead in this compound is +2.
04

(d) Sodium Tetraborate Decahydrate (Borax)

Sodium tetraborate decahydrate (borax) contains sodium (Na), boron (B), oxygen (O), and water molecules (H2O). The chemical formula for sodium tetraborate decahydrate is Na2B4O7·10H2O. Now, we need to determine the oxidation state of boron in this compound. Let the oxidation state of boron be x. \(2(+1) + 4x + 7(-2) = 0\) \(2 + 4x - 14 = 0\) \(4x = 12\) \(x = \)3 The oxidation state of boron in sodium tetraborate decahydrate (borax) is +3.
05

(e) Boric Oxide

Boric oxide contains boron (B) and oxygen (O). The chemical formula for boric oxide is B2O3. Now, we need to determine the oxidation state of boron in this compound. Let the oxidation state of boron be x. \(2x + 3(-2) = 0\) \(2x - 6 = 0\) \(x = \)3 The oxidation state of boron in boric oxide is +3.
06

(f) Germanium Dioxide

Germanium dioxide contains germanium (Ge) and oxygen (O). The chemical formula for germanium dioxide is GeO2. Let the oxidation state of germanium be x. \(x + 2(-2) = 0\) \(x - 4 = 0\) \(x = \)4 The oxidation state of germanium in germanium dioxide is +4.

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Most popular questions from this chapter

(a) What is the characteristic geometry about silicon in all silicate minerals? (b) Metasilicic acid has the empirical formula \(\mathrm{H}_{2} \mathrm{SiO}_{3} .\) Which of the structures shown in Figure 22.32 would you expect metasilicic acid to have?

The physical properties of \(\mathrm{D}_{2} \mathrm{O}\) differ from those of \(\mathrm{H}_{2} \mathrm{O}\) because (a) \(\mathrm{D}\) has a different electron configuration than O. (b) \(\mathrm{D}\) is radioactive. (c) \(\mathrm{D}\) forms stronger bonds with O than \(\mathrm{H}\) does.(\boldsymbol{d} ) D is much more massive than H .

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 \(\mathrm{F}_{2}>\mathrm{Cl}_{2}>\mathrm{Br}_{2}>\mathrm{I}_{2} .\)

Account for the following observations: (a) Phosphorus forms a pentachloride, but nitrogen does not. (b) \(\mathrm{H}_{3} \mathrm{PO}_{2}\) is a monoprotic acid. (c) Phosphonium salts, such as \(\mathrm{PH}_{4} \mathrm{Cl}\) can be formed under anhydrous conditions, but they cannot be made in aqueous solution. (d) White phosphorus is more reactive than red phosphorus.

Complete and balance the following equations:$$ \begin{array}{l}{\text { (a) } \mathrm{CO}_{2}(g)+\mathrm{OH}^{-}(a q) \longrightarrow} \\ {\text { (b) } \mathrm{NaHCO}_{3}(s)+\mathrm{H}^{+}(a q)} \\\ {\text { (c) } \mathrm{CaO}(s)+\mathrm{C}(s) \stackrel{\Delta}{\longrightarrow}}\end{array} $$ $$\begin{array}{l}{\text { (d) ~ } \mathrm{C}(s)+\mathrm{H}_{2} \mathrm{O}(g) \stackrel{\Delta}{\longrightarrow}} \\ {\text { (e) } \mathrm{CuO}(s)+\mathrm{CO}(g) \longrightarrow}\end{array} $$
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