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Chapter 4: Problem 79

Assign oxidation states for all atoms in each of the following compounds. a. \(\mathrm{KMnO}_{4}\) b. \(\mathrm{NiO}_{2}\) c. \(\mathrm{Na}_{4} \mathrm{Fe}(\mathrm{OH})_{6}\) d. \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{HPO}_{4}\) e. \(\mathrm{P}_{4} \mathrm{O}_{6}\) f. \(\mathrm{Fe}_{3} \mathrm{O}_{4}\) g. \(\mathrm{XeOF}_{4}\) h. \(\mathrm{SF}_{4}\) i. CO j. \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\)

Short Answer

Expert verified
The oxidation states for each atom in the given compounds are: a. \(\mathrm{KMnO}_{4}\): \(\mathrm{K} = +1\), \(\mathrm{Mn} = +7\), \(\mathrm{O} = -2\) b. \(\mathrm{NiO}_{2}\): \(\mathrm{Ni} = +4\), \(\mathrm{O} = -2\) c. \(\mathrm{Na}_{4} \mathrm{Fe}(\mathrm{OH})_{6}\): \(\mathrm{Na} = +1\), \(\mathrm{Fe} = +2\), \(\mathrm{O} = -2\), \(\mathrm{H} = +1\) d. \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{HPO}_{4}\): \(\mathrm{N} = -3\), \(\mathrm{H} = +1\), \(\mathrm{P} = +5\), \(\mathrm{O} = -2\) e. \(\mathrm{P}_{4} \mathrm{O}_{6}\): \(\mathrm{P} = +3\), \(\mathrm{O} = -2\) f. \(\mathrm{Fe}_{3} \mathrm{O}_{4}\): \(\mathrm{Fe} = +8/3\), \(\mathrm{O} = -2\) g. \(\mathrm{XeOF}_{4}\): \(\mathrm{Xe} = +6\), \(\mathrm{O} = -2\), \(\mathrm{F} = -1\) h. \(\mathrm{SF}_{4}\): \(\mathrm{S} = +4\), \(\mathrm{F} = -1\) i. CO: \(\mathrm{C} = +2\), \(\mathrm{O} = -2\) j. \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) (glucose): \(\mathrm{C} = 0\), \(\mathrm{H} = +1\), \(\mathrm{O} = -2\)

Step by step solution

01

Recall the rules for assigning oxidation states

Here are some of the rules to remember for assigning oxidation states: 1. The oxidation state of an element in its free state is 0. 2. The oxidation state of a monoatomic ion is equal to its charge. 3. Oxygen usually has an oxidation state of -2 in compounds. 4. Hydrogen usually has an oxidation state of +1 in compounds. 5. In most cases, the sum of oxidation states of all atoms in a compound or ion is equal to its overall charge. With these rules, we can find the oxidation states for the given compounds.
02

Assign oxidation states for \(\mathrm{KMnO}_{4}\) (a)

Applying the rules to determine oxidation states: \(\mathrm{K}\): as a group 1 element, has an oxidation state of +1. \(\mathrm{Mn}\): we have to find out. \(\mathrm{O}\): usually has an oxidation state of -2. In \(\mathrm{KMnO}_{4}\), the sum of oxidation states of all atoms = 0 (because it is a neutral compound). Therefore, \[+1 + x + 4(-2) = 0\] Upon solving for x (Manganese oxidation state), we get x = +7. The individual oxidation states for \(\mathrm{KMnO}_{4}\) are: \(\mathrm{K} = +1\) \(\mathrm{Mn} = +7\) \(\mathrm{O} = -2\)
03

Assign oxidation states for the other compounds (b-j)

In a similar way, we can assign oxidation states for the other compounds: b. \(\mathrm{NiO}_{2}\): \(\mathrm{Ni} = +4\) \(\mathrm{O} = -2\) c. \(\mathrm{Na}_{4} \mathrm{Fe}(\mathrm{OH})_{6}\): \(\mathrm{Na} = +1\) \(\mathrm{Fe} = +2\) \(\mathrm{O} = -2\) \(\mathrm{H} = +1\) d. \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{HPO}_{4}\): \(\mathrm{N} = -3\) \(\mathrm{H} = +1\) (in NH4+) \(\mathrm{H} = +1\) (in HPO4^2-) \(\mathrm{P} = +5\) \(\mathrm{O} = -2\) e. \(\mathrm{P}_{4} \mathrm{O}_{6}\): \(\mathrm{P} = +3\) \(\mathrm{O} = -2\) f. \(\mathrm{Fe}_{3} \mathrm{O}_{4}\): \(\mathrm{Fe} = +8/3\) (average oxidation state for the 3 iron atoms) \(\mathrm{O} = -2\) g. \(\mathrm{XeOF}_{4}\): \(\mathrm{Xe} = +6\) \(\mathrm{O} = -2\) \(\mathrm{F} = -1\) h. \(\mathrm{SF}_{4}\): \(\mathrm{S} = +4\) \(\mathrm{F} = -1\) i. CO: \(\mathrm{C} = +2\) \(\mathrm{O} = -2\) j. \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) (glucose): \(\mathrm{C} = 0\) (since it is in elemental form) \(\mathrm{H} = +1\) \(\mathrm{O} = -2\) By following these steps and rules, we have found the oxidation states of each atom in the given compounds.

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Key Concepts

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

Oxidation State Rules
Oxidation states, also known as oxidation numbers, play a critical role in understanding the electronic structure of atoms in compounds and are essential in the study of redox reactions in inorganic chemistry. Here are some foundational rules for determining oxidation states:

  • The oxidation state of any element in its pure form, such as O2, N2, or P4, is always zero.
  • For monoatomic ions, like Na+ or Cl-, the oxidation state equals the ionic charge.
  • Oxygen is typically assigned an oxidation state of -2, except in peroxides or when bonded to fluorine.
  • Hydrogen is usually given a +1 oxidation state when bonded to nonmetals and -1 when bonded to metals.
  • The sum of oxidation states in a neutral molecule must equal zero, while in an ion it should equal the net charge of the ion.
The precise assignment of oxidation states involves a systematic approach using these rules as guidelines. By examining the chemistry of the atoms and the molecular structure, one can unravel the electron distribution among atoms in a molecule.
Oxidation Number Assignment
Assigning oxidation numbers is not just a rote application of rules but also a process of understanding electron distribution in molecules and ions. It's a balancing act to ensure that the total charges equal the overall net charge of the compound. For example, in potassium permanganate (KMnO4), potassium has a +1 oxidation state as an alkali metal, and oxygen is assigned -2. Manganese's state is determined by considering that the compound is neutral, thus requiring the manganese to have an oxidation state of +7 to balance the charges.

To systematically assign oxidation numbers, follow these steps:
  • Start with known oxidation states based on the above rules.
  • Calculate the unknown oxidation states to balance the overall charge of the compound.
  • Apply this methodology consistently across various inorganic compounds for accuracy.
Students often benefit from written examples and visual aids when learning oxidation number assignment.
Inorganic Chemistry
Inorganic chemistry is the branch of chemistry that deals with inorganic compounds, which are generally defined as substances not containing a C-H bond. This includes a wide range of substances including metals, salts, and minerals, each with unique properties and reactions. Oxidation states play a vital part in this field as they are involved in predicting the reactivity and stability of inorganic compounds. Understanding the oxidation state of an element within a compound reveals much about its bonding and reactivity. It allows chemists to determine the likely products of chemical reactions and to balance redox equations, which is crucial for laboratory synthesis and industrial processes. Moreover, the concept of formal charges in coordination complexes, which are a significant part of inorganic chemistry, is also based on the principles of oxidation states.
Redox Reactions
Redox reactions, short for reduction-oxidation reactions, are processes where the oxidation states of atoms change due to the transfer of electrons. These fundamental chemical reactions are characterized by one substance being oxidized (losing electrons) and another being reduced (gaining electrons). The grasp of oxidation states is invaluable in identifying which species are undergoing oxidation or reduction. For example, when copper reacts with silver nitrate, copper's oxidation state increases as it loses electrons, while silver's decreases as it gains electrons, indicating that copper is oxidized and silver is reduced.

Oxidation state changes are at the heart of electricity generation in batteries, corrosion processes, and metabolic pathways in living organisms. The knowledge of oxidation states thus forms a cornerstone for understanding and controlling chemical processes in everyday life and advanced scientific research.

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Most popular questions from this chapter

A \(10.00-\mathrm{mL}\) sample of vinegar, an aqueous solution of acetic acid \(\left(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right)\), is titrated with \(0.5062 \mathrm{M} \mathrm{NaOH}\), and \(16.58 \mathrm{~mL}\) is required to reach the equivalence point. a. What is the molarity of the acetic acid? b. If the density of the vinegar is \(1.006 \mathrm{~g} / \mathrm{cm}^{3}\), what is the mass percent of acetic acid in the vinegar?

The units of parts per million (ppm) and parts per billion (ppb) are commonly used by environmental chemists. In general, 1 ppm means 1 part of solute for every \(10^{6}\) parts of solution. Mathematically, by mass: $$ \mathrm{ppm}=\frac{\mu \mathrm{g} \text { solute }}{\mathrm{g} \text { solution }}=\frac{\mathrm{mg} \text { solute }}{\mathrm{kg} \text { solution }} $$ In the case of very dilute aqueous solutions, a concentration of \(1.0\) ppm is equal to \(1.0 \mu \mathrm{g}\) of solute per \(1.0 \mathrm{~mL}\), which equals \(1.0 \mathrm{~g}\) solution. Parts per billion is defined in a similar fashion. Calculate the molarity of each of the following aqueous solutions. a. \(5.0 \mathrm{ppb} \mathrm{Hg}\) in \(\mathrm{H}_{2} \mathrm{O}\) b. \(1.0 \mathrm{ppb} \mathrm{CHCl}_{3}\) in \(\mathrm{H}_{2} \mathrm{O}\) c. \(10.0 \mathrm{ppm}\) As in \(\mathrm{H}_{2} \mathrm{O}\) d. \(0.10 \mathrm{ppm} \mathrm{DDT}\left(\mathrm{C}_{14} \mathrm{H}_{9} \mathrm{Cl}_{5}\right)\) in \(\mathrm{H}_{2} \mathrm{O}\)

Calculate the sodium ion concentration when \(70.0 \mathrm{~mL}\) of 3.0 \(M\) sodium carbonate is added to \(30.0 \mathrm{~mL}\) of \(1.0 \mathrm{M}\) sodium bicarbonate.

Assign oxidation states for all atoms in each of the following compounds. a. \(\mathrm{UO}_{2}^{2+}\) b. \(\mathrm{As}_{2} \mathrm{O}_{3}\) c. \(\mathrm{NaBiO}_{3}\) d. As \(_{4}\) e. \(\mathrm{HAsO}_{2}\) f. \(\mathrm{Mg}_{2} \mathrm{P}_{2} \mathrm{O}_{7}\) g. \(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\) h. \(\mathrm{Hg}_{2} \mathrm{Cl}_{2}\) i. \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\)

Give an example how each of the following insoluble ionic compounds could be produced using a precipitation reaction. Write the balanced formula equation for each reaction. a. \(\mathrm{Fe}(\mathrm{OH})_{3}(s)\) b. \(\mathrm{Hg}_{2} \mathrm{Cl}_{2}(s)\) c. \(\mathrm{PbSO}_{4}(s)\) d. \(\mathrm{BaCrO}_{4}(s)\)
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