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Iodic acid has the molecular formula \(\mathrm{HIO}_{3}\). Write the formulas for the following: (a) the iodate anion, (b) the periodate anion, (c) the hypoiodite anion, (d) hypoiodous acid, \((\mathrm{e})\) periodic acid.

Short Answer

Expert verified
(a) \(\mathrm{IO}_{3}^{-}\) (b) \(\mathrm{IO}_{4}^{-}\) (c) \(\mathrm{IO}^{-}\) (d) \(\mathrm{HIO}\) (e) \(\mathrm{HIO}_{4}\)

Step by step solution

01

(a) The iodate anion formula

To find the formula for the iodate anion, we first need to consider the iodic acid \(\mathrm{HIO}_{3}\). When an iodic acid loses one hydrogen ion, it becomes the iodate anion. Hence, the formula for the iodate anion is \(\mathrm{IO}_{3}^{-}\).
02

(b) The periodate anion formula

The periodate anion results from the loss of a hydrogen ion from periodic acid. Periodic acid has one more oxygen atom compared to iodic acid. Thus, the formula for periodic acid is \(\mathrm{HIO}_{4}\). When it loses one hydrogen ion, it becomes the periodate anion. Hence, the formula for the periodate anion is \(\mathrm{IO}_{4}^{-}\).
03

(c) The hypoiodite anion formula

The hypoiodite anion is formed when one hydrogen ion is lost from hypoiodous acid. Hypoiodous acid has one less oxygen atom compared to iodic acid, which means its formula is \(\mathrm{HIO}\). When it loses one hydrogen ion, it becomes the hypoiodite anion. Therefore, the formula for the hypoiodite anion is \(\mathrm{IO}^{-}\).
04

(d) Hypoiodous acid formula

As mentioned in the previous step, hypoiodous acid has one less oxygen atom compared to iodic acid. So, the formula for hypoiodous acid is \(\mathrm{HIO}\).
05

(e) Periodic acid formula

Periodic acid has one more oxygen atom compared to iodic acid. Since the formula for iodic acid is \(\mathrm{HIO}_{3}\), the formula for periodic acid is \(\mathrm{HIO}_{4}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Iodate Anion
The iodate anion is a compound that comes from iodic acid (\(\mathrm{HIO}_3\)). It forms when iodic acid loses a hydrogen ion. This process leaves the iodate anion with the formula \(\mathrm{IO}_3^-\).

This anion is characterized by its three oxygen atoms bonded to a single iodine atom. These bonds create a pyramidal shape. It is a part of various oxidizing agents and is commonly used in chemistry labs.

Knowing that hydrogen ions are lost helps us determine how an anion forms from its parent acid.
Periodate Anion
One of the more complex iodine-derived anions is the periodate anion. This anion originates when periodic acid sheds a hydrogen ion. Periodic acid itself contains one more oxygen than iodic acid, giving it the formula \(\mathrm{HIO}_4\). Subsequently, the periodate anion is formulated as \(\mathrm{IO}_4^-\).

The additional oxygen atom differentiates the periodate from the iodate anion, creating a tetrahedral molecular geometry. This makes it very useful in chemical reactions that require strong oxidation processes. You'll often find this anion in studies involving oxidative reactions.
Hypoiodite Anion
The hypoiodite anion is a more simplified version of iodine’s compounds. It is created from hypoiodous acid, which has fewer oxygen atoms compared to iodic acid. The formula for hypoiodous acid is \(\mathrm{HIO}\).

When hypoiodous acid loses its hydrogen, it forms the hypoiodite anion, \(\mathrm{IO}^-\). This loss leads to a linear molecular shape due to only having one oxygen bound to the iodine. Hypoiodite anions are known for being intermediate oxidizers, participating in reactions where gentle oxidation is needed.
Hypoiodous Acid
Hypoiodous acid is an essential member of the iodine-related acids and anions' family. Unlike iodic acid, hypoiodous acid has only one oxygen, which simplifies its structure. Its chemical formula is \(\mathrm{HIO}\).

This acid serves as a precursor to hypoiodite anion when it loses a hydrogen ion. The simplicity of hypoiodous acid makes it an intriguing compound in both theoretical and applied chemistry. It plays a role in oxidation-reduction reactions, particularly where mild conditions are required.
Periodic Acid
Periodic acid is an oxygen-rich compound among iodine-containing acids. This acid is notable for having an additional oxygen atom compared to iodic acid, resulting in the formula \(\mathrm{HIO}_4\).

When periodic acid loses a hydrogen ion, it gives rise to the periodate anion \(\mathrm{IO}_4^-\). Periodic acid is often used in organic chemistry, where its oxidative power helps break complex molecular structures into simpler components.

The presence of four oxygen atoms, as opposed to three in iodic acid, allows periodic acid to be a potent oxidizing agent.

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Most popular questions from this chapter

Gallium (Ga) consists of two naturally occurring isotopes with masses of \(68.926\) and \(70.925\) amu. (a) How many protons and neutrons are in the nucleus of each isotope? Write the complete atomic symbol for each, showing the atomic number and mass number. (b) The average atomic mass of Ga is \(69.72\) amu. Calculate the abundance of each isotope.

Determine the molecular and empirical formulas of the following: (a) The organic solvent benzene, which has six carbon atoms and six hydrogen atoms; (b) the compound silicon tetrachloride, which has a silicon atom and four chlorine atoms and is used in the manufacture of computer chips; (c) the reactive substance diborane, which has two boron atoms and six hydrogen atoms; (d) the sugar called glucose, which has six carbon atoms, twelve hydrogen atoms, and six oxygen atoms.

Many ions and compounds have very similar names, and there is great potential for confusing them. Write the correct chemical formulas to distinguish between (a) calcium sulfide and calcium hydrogen sulfide, (b) hydrobromic acid and bromic acid, (c) aluminum nitride and aluminum nitrite, (d) iron(Il) oxide and iron(III) oxide, (e) ammonia and ammonium ion, (f) potassium sulfite and potassium bisulfite, \((g)\) mercurous chloride and mercuric chloride, \((\mathrm{h})\) chloric acid and perchloric acid.

Using a suitable reference such as the CRC Handbook of Chemistry and Physics or http://www.webelements.com, look up the following information for nickel: (a) the number of known isotopes, (b) the atomic masses (in amu) and the natural abundance of the five most abundant isotopes.

Locate each of the following elements in the periodic table; indicate whether it is a metal, metalloid, or nonmetal; and give the name of the element: (a) \(\mathrm{Ca}\), (b) \(\mathrm{Ti}\), (c) \(\mathrm{Ga},(\mathrm{d}) \mathrm{Th},(\mathrm{e}) \mathrm{Pt},(\mathrm{f}) \mathrm{Se},(\mathrm{g}) \mathrm{Kr}\).

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