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Chapter 22: Problem 32

Write the chemical formula for each of the following, and indicate the oxidation state of the halogen or noble-gas atom in each: (a) krypton tetrafluoride, (b) hexafluoroantimonate ion, \((\mathbf{c})\) sodium hypoiodite, \((\mathbf{d})\) perbromic acid, (e) aluminum perchlorate, (f) iron(II) iodite.

Short Answer

Expert verified
(a) Krypton Tetrafluoride: KrF4, Krypton oxidation state: +4 (b) Hexafluoroantimonate Ion: SbF6^-, Antimony oxidation state: +5 (c) Sodium Hypoiodite: NaIO, Iodine oxidation state: +1 (d) Perbromic Acid: HBrO4, Bromine oxidation state: +7 (e) Aluminum Perchlorate: Al(ClO4)3, Chlorine oxidation state: +7 (f) Iron(II) Iodite: Fe(IO2)2, Iodine oxidation state: +3

Step by step solution

01

a) Krypton Tetrafluoride

Write down the symbols for Krypton (Kr) and Fluorine (F). Since it is "tetrafluoride," there will be four Fluorine atoms. The chemical formula for Krypton Tetrafluoride is KrF4. Krypton in KrF4 is in its highest possible oxidation state (+4) because it is bonded to four fluorine atoms with a -1 oxidation state each.
02

b) Hexafluoroantimonate Ion

Write down the symbol for antimony (Sb) and Fluorine (F). This ion has six Fluorine atoms, and since it is an anion, it will have a negative charge. The chemical formula for Hexafluoroantimonate Ion is SbF6^-. Here, antimony (Sb) has an oxidation state of +5 because it is bonded to six fluorine atoms, each having a -1 oxidation state.
03

c) Sodium Hypoiodite

Write down the symbols for Sodium (Na), Iodine (I), and Oxygen (O). Hypoiodite means the presence of the ion in the compound, so the chemical formula for Sodium Hypoiodite is NaIO. The oxidation state of iodine is +1 because it is bonded to an oxygen atom with an oxidation state of -2 and a sodium atom with an oxidation state of +1.
04

d) Perbromic Acid

Write down the symbols for Hydrogen (H), Bromine (Br) and Oxygen (O). The "per" in perbromic acid means it has the maximum possible amount of oxygen atoms, which is BrO4. So the compound is an acid, and we need the hydrogen, resulting in the chemical formula HBrO4. The oxidation state of bromine is +7 because it is bonded to four oxygen atoms with an oxidation state of -2 and one hydrogen atom with an oxidation state of +1.
05

e) Aluminum Perchlorate

Write down the symbols for Aluminum (Al), Chlorine (Cl), and Oxygen (O). The "per" in perchlorate indicates the maximum possible amount of oxygen atoms, meaning ClO4. The compound is a salt, and we need to balance the charge of the ions. Aluminum has an oxidation state of +3, while perchlorate anion (ClO4^-) has a charge of -1. So, the chemical formula for Aluminum Perchlorate is Al(ClO4)3. In this compound, chlorine has an oxidation state of +7.
06

f) Iron(II) Iodite

Write down the symbols for Iron (Fe) and Iodine (I). The compound consists of Iron(II), meaning it has an oxidation state of +2, and iodite is a polyatomic ion (IO2^-). Since the charge of iron is +2 and iodite anion has a charge of -1, we need two iodite ions to balance it. The chemical formula for Iron(II) Iodite is Fe(IO2)2. In this compound, iodine has an oxidation state of +3.

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Most popular questions from this chapter

Complete and balance the following equations: (a) \(\mathrm{NaH}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow\) (b) \(\mathrm{Fe}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow\) (c) \(\mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g) \longrightarrow\) (d) \(\mathrm{Na}(l)+\mathrm{H}_{2}(g) \longrightarrow\) (e) \(\mathrm{PbO}(s)+\mathrm{H}_{2}(g) \longrightarrow\)

Write a molecular formula for each compound, and indicate the oxidation state of the group 15 element in each formula: (a) hexafluoroantimonate(V) ion, (b) calcium phosphate, (c) potassium dihydrogen phosphate, (d) arsenic trioxide, (e) tetraphosphorus hexaoxide, (f) arsenic trifluoride.

Complete and balance the following equations: (a) $\mathrm{Mg}_{3} \mathrm{~N}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow$ (b) $\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{OH}(l)+\mathrm{O}_{2}(g) \longrightarrow$ (c) \(\mathrm{MnO}_{2}(s)+\mathrm{C}(s) \stackrel{\Delta}{\longrightarrow}\) (d) \(\mathrm{AlP}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow\) (e) \(\mathrm{Na}_{2} \mathrm{~S}(s)+\mathrm{HCl}(a q) \longrightarrow\)

Write a balanced equation for each of the following reactions. (You may have to guess at one or more of the reaction products, but you should be able to make a reasonable guess, based on your study of this chapter.) (a) Hydrogen selenide can be prepared by reaction of an aqueous acid solution on aluminum selenide. (b) Sodium thiosulfate is used to remove excess \(\mathrm{Cl}_{2}\) from chlorine-bleached fabrics. The thiosulfate ion forms \(\mathrm{SO}_{4}^{2-}\) and elemental sulfur, while \(\mathrm{Cl}_{2}\) is reduced to \(\mathrm{Cl}^{-}\).

(a) How does the structure of diborane $\left(\mathrm{B}_{2} \mathrm{H}_{6}\right)\( differ from that of ethane \)\left(\mathrm{C}_{2} \mathrm{H}_{6}\right) ?$ (b) Explain why diborane adopts the geometry that it does. (c) What is the significance of the statement that the hydrogen atoms in diborane are described as "hydridic"?
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