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Chapter 14: Problem 5

\(\mathrm{N}\) atoms can have a wide range of oxidation numbers. Assign oxidation numbers for the \(\mathrm{N}\) atom in each compound, all of which are known compounds. a) \(\mathrm{N}_{2} \mathrm{O}_{5}\) b) \(\mathrm{N}_{2} \mathrm{O}_{4}\) c) \(\mathrm{NO}_{2}\) d) \(\mathrm{NO}\) e) \(\mathrm{N}_{2} \mathrm{H}_{4}\) f) \(\mathrm{NH}_{3}\)

Short Answer

Expert verified
a) +5, b) +4, c) +4, d) +2, e) -2, f) -3

Step by step solution

01

Understand the concept of oxidation numbers

Oxidation numbers are a way of tracking electrons in molecules and ions. They represent the hypothetical charge an atom would have if all bonds were ionic.
02

Calculate oxidation number for N in N2O5

In the compound \(2o5\), the oxidation number of O is \(-2\). Let the oxidation number of N be \(+x\). The formula is \{2x + 5(-2) = 0\}, which simplifies to \{2x - 10 = 0\} so \{2x = 10\}, leading to \{x = +5\}. Therefore, the oxidation number for N in \(2o5\) is +5.
03

Calculate oxidation number for N in N2O4

For \(2o4\), the oxidation number of O remains \(-2\). Let the oxidation number of N be \(+x\). The equation is \{2x + 4(-2) = 0\}, which simplifies to \{2x - 8 = 0\} so \{2x = 8\}, leading to \{x = +4\}. Therefore, the oxidation number for N in \(2o4\) is +4.
04

Calculate oxidation number for N in NO2

In \(no2\), let the oxidation number of N be \(+x\), and the oxidation number of O is \(-2\). Thus, the equation is \{x + 2(-2) = 0\}, resulting in \{x - 4 = 0\}, so \{x = +4\}. Therefore, the oxidation number for N in \(no2\) is +4.
05

Calculate oxidation number for N in NO

Here, let the oxidation number of N be \(+x\) in \(no\), and the oxidation number of O is \(-2\). The equation is \{x + (-2) = 0\}, which gives \{x - 2 = 0\}, thus \{x = +2\}. Therefore, the oxidation number for N in \(no\) is +2.
06

Calculate oxidation number for N in N2H4

For \(2h4\), let the oxidation number of N be \(+x\), and for H, it is \(+1\). The equation is \{2x + 4(1) = 0\}, simplifying to \{2x + 4 = 0\}, which leads to \{2x = -4\}, so \{x = -2\}. Therefore, the oxidation number for N in \(2h4\) is -2.
07

Calculate oxidation number for N in NH3

In \(nh3\), let the oxidation number of N be \(+x\), and for H, it is \(+1\). The equation is \{x + 3(1) = 0\}, which simplifies to \{x + 3 = 0\}, thus \{x = -3\}. Therefore, the oxidation number for N in \(nh3\) is -3.

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Key Concepts

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

Nitrogen Compounds
Nitrogen is a highly versatile element and is central to an array of compounds, many of which are significant in both nature and industry. Due to its ability to form various oxidation states, nitrogen exhibits fascinating chemistry. Here are a few key aspects of nitrogen compounds:

  • Nitrogen often pairs with oxygen to form oxides, such as dinitrogen pentoxide (\(\mathrm{N}_2\mathrm{O}_5\)) and nitrogen dioxide (\(\mathrm{NO}_2\)). These compounds display nitrogen in oxidation states of +5 and +4, respectively.
  • Nitrogen can bond with hydrogen to form amines and ammonia (\(\mathrm{NH}_3\)), where it has an oxidation state of -3. This demonstrates nitrogen's capability of forming rather negative oxidation numbers.
  • Other compounds like hydrazine (\(\mathrm{N}_2\mathrm{H}_4\)) show nitrogen in a -2 state, indicating its rich and diverse electron sharing capabilities.
Understanding nitrogen compounds begins with recognizing these varied oxidation states, as they illuminate how nitrogen atoms engage in electron transfer and bonding.
Electron Tracking
Electron tracking is a powerful method used in chemistry to understand how electrons are distributed among atoms within compounds. This is crucial when balancing redox reactions and determining the structure of molecules.

  • Oxidation numbers are a handy tool for tracking electrons. These numbers represent the hypothetical charge of an atom if all bonds in the compound were completely ionic.
  • For instance, in compounds such as \(\mathrm{N}_2\mathrm{O}_5\) and \(\mathrm{NH}_3\), it is the differences in oxidation states of nitrogen that help chemists track electron movement. In \(\mathrm{N}_2\mathrm{O}_5\), each nitrogen atom has an oxidation number of +5, indicating it has "lost" electrons.
  • Determining these numbers involves considering the known charges of other atoms in a molecule (e.g., oxygen is typically -2 and hydrogen +1), and using algebraic equations to solve for the unknown oxidation states.
Through careful analysis of oxidation numbers, one gains insight into not just electron positioning but also how atoms within a molecule will interact chemically.
Redox Chemistry
Redox chemistry involves the transfer of electrons between chemical species, leading to changes in oxidation numbers. This chemistry is fundamental to processes ranging from metabolism in living organisms to industrial manufacturing.

  • Redox reactions are characterized by two parts: reduction, where a species gains electrons, and oxidation, where a species loses electrons.
  • The concept of oxidation numbers plays a critical role in identifying which species are oxidized and which are reduced. For instance, in nitrogen oxides like \(\mathrm{NO}\) and \(\mathrm{N}_2\mathrm{O}_4\), monitoring changes in the oxidation state of nitrogen highlights the direction and magnitude of electron flow.
  • This type of chemistry allows for applications such as energy generation in batteries, where understanding how electrons move between compounds is key to efficient operation.
Appreciating the nuances of redox chemistry thus requires an understanding of oxidation numbers, making it an essential aspect of mastering chemical reactions and transformations.

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Most popular questions from this chapter

Identify what is being oxidized and reduced in this redox reaction by assigning oxidation numbers to the atoms. $$ \mathrm{Sr}+\mathrm{SO}_{3} \rightarrow \mathrm{SrSO}_{3} $$

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