Which compound amongst the following has the highest oxidation number of Mn? (a) \(\mathrm{KMnO}_{4}\) (b) \(\mathrm{K}_{2} \mathrm{MnO}_{4}\) (c) \(\mathrm{MnO}_{2}\) (d) \(\mathrm{Mn}_{2} \mathrm{O}_{3}\)

Short Answer

Expert verified
KMnO4 has the highest oxidation number of Mn, which is +7.

Step by step solution

01

Determine the oxidation numbers in KMnO4

To find the oxidation state of Mn in KMnO4, start with the fact that the compound is neutral. The oxidation state of K is +1 and O is -2. The compound has 4 oxygen atoms, for a total of -8. Let x be the oxidation state of Mn. As the overall charge is zero, the equation is: 1 (+1) + x + 4(-2) = 0. Solve for x to find the oxidation state of Mn.
02

Determine the oxidation numbers in K2MnO4

In K2MnO4, K has an oxidation state of +1, and O has -2. With 2 potassium atoms, the positive charge is +2, and with 4 oxygen atoms, the negative charge is -8. Let x represent the oxidation state of Mn. The equation is: 2(+1) + x + 4(-2) = 0. Solve for x to find the oxidation state of Mn.
03

Determine the oxidation numbers in MnO2

For MnO2, O has an oxidation state of -2. There are 2 oxygen atoms for a total of -4. If x is the oxidation state of Mn, we use the equation: x + 2(-2) = 0. Solving for x gives the oxidation state of Mn.
04

Determine the oxidation numbers in Mn2O3

In Mn2O3, O has an oxidation state of -2. With 3 oxygen atoms, the total negative charge is -6. Since there are 2 Mn atoms, let the sum of their oxidation states be 2x. The equation is: 2x + 3(-2) = 0. Solve for x, which gives the oxidation state of each Mn atom.
05

Compare the oxidation numbers

Compare the calculated oxidation numbers of Mn in each compound. The highest oxidation number of Mn will indicate the correct answer.

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Key Concepts

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

Oxidation State Determination
Understanding the oxidation state of elements within compounds is crucial in the field of chemistry. It's a key player in understanding reactions, such as redox processes, and helps to predict how elements will interact.

Oxidation states are typically represented by integers, which can be positive, negative, or zero, indicating the degree of oxidation of an atom in a chemical compound. Determining oxidation states is like solving a puzzle where the total charges must balance to match the overall charge of the molecule or ion. For neutral compounds, the sum of oxidation states must equal zero, while for ions, it must equal the ion's charge.

In the case of the textbook exercise to determine which manganese (Mn) compound has the highest oxidation number, one must first understand the common rule that elements in group I (like K) have an oxidation state of +1, and oxygen generally has an oxidation state of -2. Applying these rules, along with the charge balance principle, enables us to calculate the oxidation state of Mn in the given compounds.

For example, in \(\mathrm{KMnO}_{4}\), the potassium ion (K) has an oxidation state of +1, and each oxygen (O) has a state of -2. Because the compound is neutral overall, the sum of the oxidation states must be zero, leading to the calculation that reveals the oxidation state of Mn.
Chemical Nomenclature
Chemical nomenclature is a systematic way of naming chemical substances. It ensures that each compound has a unique name, allowing chemists to communicate clearly about specific substances without confusion. The International Union of Pure and Applied Chemistry (IUPAC) has developed the rules for naming compounds.

Names of inorganic compounds typically reflect both the type of atoms in the molecule and their quantities. For instance, \(\mathrm{KMnO}_{4}\) is known as potassium permanganate, not only indicating the presence of potassium (K), manganese (Mn), and oxygen (O), but also suggesting a particular oxidation state for Mn, which can be deduced from the 'permanganate' part of the name.

The necessity to understand the chemical nomenclature lies in its power to convey information about the molecular structure and can infer properties such as reactivity and interaction with other substances. It also plays an essential role in the prediction and balancing of redox equations.
Balancing Redox Equations
Balancing redox equations is crucial for understanding the transfer of electrons during a reaction. Redox equations express the conservation of charge and mass, and their balancing involves ensuring the same number of atoms for each element and the same total charge on both sides of the equation.

To balance redox equations, one must identify the oxidation changes by determining the oxidation states of the atoms before and after the reaction, as we did in the previous steps for manganese compounds. Once the changes in oxidation states are known, we can add the appropriate number of electrons to one side of each half-reaction to balance the charges. Then, the half-reactions are combined to cancel out the electrons and create a balanced full redox reaction.

This process may require using coefficients to balance the elements involved in the reaction. The coefficients must be the smallest possible integers that equalize the number of atoms on both sides of the equation and neutralize the charges. In contexts like this, command over balancing redox equations is essential to mastering chemistry concepts and problem-solving.

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Most popular questions from this chapter

When a manganous salt is fused with a mixture of \(\mathrm{KNO}_{3}\) and solid \(\mathrm{NaOH}\) the oxidation number of \(\mathrm{Mn}\) changes from \(+2\) to (a) \(+4\) (b) \(+3\) (c) \(+6\) (d) \(+7\)

When a piece of sodium metal is dropped in water, hydrogen gas evolved because (a) sodium is reduced and acts as an oxidising agent (b) water is oxidised and acts as a reducing agent (c) sodium loses electrons and is oxidised while water in reduced (d) water loses electrons and is oxidised to hydrogen.

Various oxidation states of few elements are mentioned. Which of the options is not correctly matched? (a) Phosphorus: \(+3\) to \(+5\) (b) Nitrogen: \(+1\) to \(+5\) (c) lodine: \(-1\) to \(+7\) (d) Chromium : \(-3\) to \(+6\)

The oxidation state of \(\mathrm{Fe}\) in \(\mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]\) is (a) \(+2\) (b) \(+3\) (c) \(+4\) (d) \(+6\)

If \(\mathrm{Mn}^{3+}\) ions are unstable in solution and undergo disproportionation to give \(\mathrm{Mn}^{2+}, \mathrm{MnO}_{2}\) and \(\mathrm{H}^{+}\) ions. What will be the balanced equation for the reaction? (a) \(3 \mathrm{Mn}^{3+}+4 \mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{MnO}_{2}+\mathrm{Mn}^{2+}+8 \mathrm{H}^{+}\) (b) \(\mathrm{Mn}^{3+}+4 \mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{MnO}_{2}+4 \mathrm{H}^{+}\) (c) \(\mathrm{Mn}+2 \mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{MnO}_{2}+4 \mathrm{H}^{+}\) (d) \(2 \mathrm{Mn}^{3+}+2 \mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{MnO}_{2}+\mathrm{Mn}^{2+}+4 \mathrm{H}^{+}\)

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