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Chapter 22: Problem 70

Write the formulas for the following compounds, and indicate the oxidation state of the group 14 element or of boron in each: (a) stannous fluoride, (b) germane, (c) diborane, (e) tin selenide, (d) tin(II) sulfate, (f) zinc carbonate.

Short Answer

Expert verified
(a) Stannous fluoride: SnF2 (Sn oxidation state: +2) (b) Germane: GeH4 (Ge oxidation state: +4) (c) Diborane: B2H6 (B oxidation state: +3) (e) Tin selenide: SnSe (Sn oxidation state: +2) (d) Tin(II) sulfate: SnSO4 (Sn oxidation state: +2) (f) Zinc carbonate: ZnCO3 (C oxidation state: +4)

Step by step solution

01

(a) Stannous fluoride

Stannous fluoride is a compound containing tin (Sn) and fluoride (F). The term "stannous" indicates that tin has an oxidation state of +2. Therefore, Sn^2+ will combine with F^- to form the compound SnF2. The oxidation state of tin in this compound is +2.
02

(b) Germane

Germane is a compound containing germanium (Ge) and hydrogen (H). It is similar in structure to methane (CH4), but with germanium instead of carbon as the central atom. Each hydrogen atom has an oxidation state of -1, so germanium needs to have an oxidation state of +4 in order to balance the charges. The formula for this compound is GeH4.
03

(c) Diborane

Diborane is a compound containing boron (B) and hydrogen (H). The prefix "di" indicates that there are two boron atoms in the molecule. Each boron atom has an oxidation state of +3, and each hydrogen atom has an oxidation state of -1. The formula for diborane is B2H6.
04

(e) Tin selenide

Tin selenide is a compound containing tin (Sn) and selenium (Se). Selenium forms a selenide ion with a -2 charge (Se^(2-)), while tin can form ions with +2 or +4 oxidation states. Considering the +2 oxidation state for tin, we would have Sn^2+ and Se^(2-) ions. These ions combine in a 1:1 ratio, forming the compound SnSe. The oxidation state of tin in this compound is +2.
05

(d) Tin(II) sulfate

Tin(II) sulfate is a compound containing tin (Sn), sulfur (S), and oxygen (O). The (II) in the name indicates that tin has a +2 oxidation state. Sulfate is a polyatomic ion containing sulfur and oxygen with a -2 charge, represented as SO4^(2-). The tin ion (Sn^2+) combines with the sulfate ion (SO4^(2-)) to form the compound SnSO4. The oxidation state of tin in this compound is +2.
06

(f) Zinc carbonate

Zinc carbonate is a compound containing zinc (Zn), carbon (C), and oxygen (O). Zinc forms a +2 ion (Zn^2+), and carbonate is a polyatomic ion containing carbon and oxygen with a -2 charge, represented as CO3^(2-). The zinc ion (Zn^2+) combines with the carbonate ion (CO3^(2-)) to form the compound ZnCO3. The oxidation state of the carbon in this compound is +4.

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Most popular questions from this chapter

Complete and balance the following equations: (a) \(\mathrm{NaOCH}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow\) (b) \(\mathrm{CuO}(s)+\mathrm{HNO}_{3}(a q) \longrightarrow\) (c) \(\mathrm{WO}_{3}(s)+\mathrm{H}_{2}(g) \stackrel{\Delta}{\longrightarrow}\) (d) \(\mathrm{NH}_{2} \mathrm{OH}(l)+\mathrm{O}_{2}(g) \longrightarrow\) (e) $\mathrm{Al}_{4} \mathrm{C}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow$

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Which of the following statements are true? (a) Both nitrogen and phosphorus can form a pentafluoride compound. (b) Although CO is a well-known compound, SiO does not exist under ordinary conditions. (c) \(\mathrm{Cl}_{2}\) is easier to oxidize than \(\mathrm{I}_{2}\). (d) At room temperature, the stable form of oxygen is \(\mathrm{O}_{2}\), whereas that of sulfur is \(S_{8}\)
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