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In which of the following compounds the oxidation state of oxygen is other than \(-2 ?\) (a) \(\mathrm{H}_{2} \mathrm{O}_{2}\) (b) \(\mathrm{O}_{2}\) (c) \(\mathrm{O}_{2} \mathrm{~F}_{2}\) (d) \(\mathrm{H}_{2} \mathrm{O}\)

Short Answer

Expert verified
Compounds (a), (b), and (c) have oxygen not in the \(-2\) state.

Step by step solution

01

Understanding Oxidation States

Oxidation states are used to explain the degree of oxidation of an atom within a molecule. Typically, oxygen has an oxidation state of \(-2\), except in a few specific cases.
02

Analyzing Each Compound

We'll look at each compound individually to decide if the oxidation state of oxygen is different from \(-2\):(a) \(\mathrm{H}_2\mathrm{O}_2\): In hydrogen peroxide, each oxygen has an oxidation state of \(-1\).(b) \(\mathrm{O}_2\): Molecular oxygen is a diatomic molecule in which the oxidation state of oxygen is \(0\) because it is in its elemental form.(c) \(\mathrm{O}_2\mathrm{~F}_2\): In dioxygen difluoride, oxygen has an unusual oxidation state of \(+1\).(d) \(\mathrm{H}_2\mathrm{O}\): Water contains oxygen with the typical oxidation state of \(-2\).
03

Identifying the Odd Cases

Compounds (a), (b), and (c) all have oxygen in oxidation states other than \(-2\). Only in compound (d) does the oxygen have the typical oxidation state of \(-2\).

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Key Concepts

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

Understanding Oxygen Compounds
Oxygen is a versatile element, playing a key role in forming various compounds. It is often involved in creating oxides, peroxides, and even superoxides, though the oxidation states can differ. Typically, oxygen exhibits an oxidation state of \(-2\), prominently present in compounds like water (\(\text{H}_2\text{O}\)).
However, exceptions occur, for example, in hydrogen peroxide (\(\text{H}_2\text{O}_2\)), where each oxygen atom has an oxidation state of \(-1\).
These compounds highlight the diversity of oxygen's chemistry, proving it can bond in unusual ways depending on the elements involved. For instance, in the compound dioxygen difluoride (\(\text{O}_2\text{F}_2\)), oxygen shows an oxidation state of \(+1\) due to fluorine's higher electronegativity. Such variations in oxidation states are pivotal in many chemical reactions and processes.
Delving Into Oxidation State Exceptions
Chemistry often presents surprising twists, especially when it comes to oxidation states. While oxygen usually takes the role of \(-2\) because of its high electronegativity, there are notable exceptions. In
  • Peroxides, like \(\text{H}_2\text{O}_2\), the oxidation state of oxygen is \(-1\).
  • In molecular oxygen \(\text{O}_2\), its oxidation state is \(0\) as it stands in its elemental form.
  • Dioxygen difluoride \(\text{O}_2\text{F}_2\) is another fascinating case, where oxygen adopts a positive \(+1\).
These exceptions arise from specific bonding scenarios and the relative electronegativity of the atoms involved. Understanding these peculiar numbers can give insights into the behavior of elements in compounds, showing how adaptable their roles can be in chemical reactions.
The Role of Chemistry Education in Understanding Exceptions
Learning chemistry can sometimes seem like deciphering a complex puzzle. That's why understanding exceptions, such as those in oxidation states, is crucial in chemistry education. It aids in reinforcing the fundamental concepts while encouraging critical thinking. Chemistry classrooms should emphasize these peculiar cases to help students:
  • Recognize and predict oxidation states in different chemical environments.
  • Develop problem-solving skills by analyzing why exceptions occur.
  • Appreciate the nuanced nature of chemistry, moving beyond simple rules.
By using real-world examples and encouraging experimental approaches, educators can demystify these concepts and make learning more engaging. This concrete understanding empowers students to tackle complex problems with confidence, laying a solid foundation for advanced studies and practical applications.

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Most popular questions from this chapter

In a hydrogen-oxygen fuel cell, combustion of hydrogen occurs to [2004] (a) produce high purity water (b) generate heat (c) remove adsorbed oxygen from electrode surfaces (d) create potential difference between the two electrodes

When \(\mathrm{Cl}_{2}\) is passed through hot \(\mathrm{NaOH}\) solution, oxidation number of chlorine changes from (a) 0 to \(-1\) (b) 0 to \(+5\) (c) 0 to \(+7\) (d) \(-1\) to 0

Match the following $$ \begin{array}{ll} \text { Column-I } & \text { Column-II } \\ \hline \begin{array}{l} \text { (a) } 50 \text { \% solution of } \mathrm{H}_{2} \mathrm{SO}_{4} \\ \text { using Pt electrodes } \end{array} & \text { (p) } \mathrm{H}_{2} \text { is evolved at } \\ \text { cathode } \\ \text { (b) } \begin{array}{l} \text { Dilute solution } \mathrm{NaCl} \\ \text { using Pt electrodes } \end{array} & \text { (q) } \mathrm{O}_{2} \text { is evolved at } \\ \text { (c) } \begin{array}{l} \text { Dilute solution of } \mathrm{H}_{2} \mathrm{SO}_{4} \\ \text { using Cu electrodes } \end{array} & \text { (r) } \mathrm{Cl}_{2} \text { is evolved at } \\ \text { (d) } \begin{array}{c} \text { Concentrated solution of } \\ \text { LiCl using Pt electrodes. } \end{array} & \text { (s) } \mathrm{H}_{2} \mathrm{~S}_{2} \mathrm{O}_{8} \text { is } \\ \text { formed at anode } \\ & \text { (t) non-spontaneous } \\ \text { process } \end{array} $$

Given the standard reduction potentials \(\mathrm{Zn}^{2+} / \mathrm{Zn}=\) \(-0.74 \mathrm{~V}, \mathrm{Cl}_{2} / \mathrm{Cl}^{-}=1.36 \mathrm{~V}, \mathrm{H}^{+} / 1 / 2 \mathrm{H}_{2}=0 \mathrm{~V}\) and \(\mathrm{Fe}^{2+} / \mathrm{Fe}^{3+}\) \(=0.77 \mathrm{~V}\). The order of increasing strength as reducing agent is (a) \(\mathrm{Zn}, \mathrm{H}_{2}, \mathrm{Fe}^{2+}, \mathrm{Cl}\) (b) \(\mathrm{H}_{2}, \mathrm{Zn}, \mathrm{Fe}^{2+}, \mathrm{Cl}^{-}\) (c) \(\mathrm{Cl}^{-}, \mathrm{Fe}^{2+}, \mathrm{Zn}, \mathrm{H}_{2}\) (d) \(\mathrm{Cl}^{-}, \mathrm{Fe}^{2+}, \mathrm{H}_{2}, \mathrm{Zn}\)

For a spontaneous reaction the \(\Delta \mathrm{G}\), equilibrium constant \((\mathrm{K})\) and \(E_{\mathrm{ccll}}^{\circ}\) will be respectively (a) \(-\mathrm{ve},>1,+\mathrm{ve}\) (b) \(+\mathrm{ve},>1,-\mathrm{ve}\) (c) \(-\mathrm{ve},<1,-\mathrm{ve}\) (d) \(-v \mathrm{e},>1,-v e\)

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