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What changes in the oxidation numbers occur when copper reacts with concentrated nitric acid to give a blue solution and a brown acidic gas? (1) \(\mathrm{Cu}(0)\) to \(\mathrm{cu}(\mathrm{II})\) and \(\mathrm{N}(\mathrm{V})\) to \(\mathrm{N}\) (IV) (2) \(\mathrm{Cu}(0)\) to \(\mathrm{Cu}(\mathrm{II})\) and \(\mathrm{N}(\mathrm{V})\) to \(\mathrm{N}(\mathrm{III})\) (3) \(\mathrm{Cu}\) (I) to \(\mathrm{Cu}\) (II) and \(\mathrm{N}(\mathrm{V})\) to \(\mathrm{N}\) (IV) (4) \(\mathrm{Cu}\) (II) to \(\mathrm{Cu}\) (I) and \(\mathrm{N}(\mathrm{V})\) to \(\mathrm{N}\) (IV)

Short Answer

Expert verified
The correct option is (1).

Step by step solution

01

- Identify the Reactants and Products

The reaction involves copper (Cu) reacting with concentrated nitric acid (HNO₃). The products are a blue solution and a brown acidic gas. The blue solution indicates the presence of copper(II) ions (Cu²⁺) and the brown gas is nitrogen dioxide (NO₂).
02

- Determine Oxidation Numbers of Copper

Initially, copper is in its elemental state with an oxidation number of 0. When copper forms Cu²⁺, its oxidation number changes to +2. So, the change is from 0 to +2.
03

- Determine Oxidation Numbers of Nitrogen

In concentrated nitric acid, nitrogen is usually in the form of nitrate (NO₃⁻), where nitrogen has an oxidation number of +5. The product nitrogen dioxide (NO₂) has nitrogen with an oxidation number of +4. Thus, the change is from +5 to +4.
04

- Analyze the Given Options

From the identified changes: Cu(0) to Cu(II) is a change from 0 to +2, and N(V) to N(IV) is a change from +5 to +4. The correct changes match with option (1).

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Key Concepts

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

oxidation state changes
Understanding oxidation states helps us track electron transfers in a chemical reaction. An oxidation state is a hypothetical charge that an atom would have if all bonds to atoms of different elements were fully ionic. In a reaction, the oxidation state of an element can increase (oxidation) or decrease (reduction).

For example, in the reaction between copper and concentrated nitric acid, copper starts with an oxidation state of 0 and changes to +2. This indicates that copper has lost two electrons. Similarly, the nitrogen in nitric acid (HNO₃) changes from an oxidation state of +5 to +4. This means nitrogen gains one electron. Recognizing these changes is crucial to identify redox reactions.
copper reactions
Copper is a versatile metal that undergoes interesting chemical changes. When copper reacts with concentrated nitric acid, it forms a blue solution (copper(II) ions) and a brown acidic gas (nitrogen dioxide).

Copper, in its metallic state, has an oxidation number of 0. In the presence of nitric acid, copper is oxidized to form Cu²⁺, with an oxidation number of +2. This shift from 0 to +2 means copper loses two electrons during the reaction.

The visible change to a blue solution indicates the formation of copper(II) ions, which are soluble in water.
nitric acid reactions
Nitric acid is a powerful oxidizing agent often used in redox reactions. When concentrated nitric acid reacts with metals, it typically leads to the formation of nitrogen dioxide (NO₂) and other nitrogen oxides.

In the reaction with copper, the nitric acid provides nitrate ions (NO₃⁻), where nitrogen has an oxidation state of +5. Upon reaction, nitrogen dioxide (NO₂) is produced, where nitrogen's oxidation state is +4.

This reduction from +5 to +4 shows that nitrogen gains an electron during the reaction. The evolution of a brown gas further confirms the presence of nitrogen dioxide.
redox reactions
Redox reactions, short for reduction-oxidation reactions, involve the transfer of electrons between substances. One substance loses electrons (oxidation), while another gains electrons (reduction).

In our example, copper is oxidized from an oxidation state of 0 to +2, meaning it loses two electrons. Simultaneously, nitrogen in nitric acid is reduced from +5 to +4, gaining an electron.

Such reactions are vital in many chemical processes and industrial applications. They help understand how materials change and react, enabling us to predict and manipulate chemical behavior.

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Most popular questions from this chapter

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