Question

Change in oxidation state of \(\mathrm{Cr}\) during oxidation of isopropanol into acetone by \(\mathrm{CrO}_{3} / \mathrm{H}_{2} \mathrm{SO}_{4}\) : (A) 2 (B) 3 (C) 4 (D) 5

Short answer

The change in oxidation state of Chromium (Cr) during the oxidation of isopropanol into acetone by CrO₃ / H₂SO₄ is 0, as the oxidation state of Chromium remains +6 throughout the reaction. None of the given options are correct.

Step-by-step solution

Write the balanced chemical equation for the reaction

First, let's write the balanced chemical equation for the oxidation of isopropanol into acetone by CrO₃ / H₂SO₄. \( (CH_{3})_{2}COH + 2CrO_{3} + 3H_{2}SO_{4} \rightarrow (CH_{3})_{2}CO + 2Cr(SO_{4})_{3} + 4H_{2}O \)

Determine the initial oxidation state of Chromium

In order to find the initial oxidation state of Chromium (Cr) in the CrO₃ molecule, we need to use the oxidation numbers of the oxygen atoms in the molecule and calculate the oxidation number of the Chromium atom. Oxygen has an oxidation number of -2. Since there are three oxygen atoms in CrO₃, the total oxidation number for all oxygen atoms is -6 (-2 * 3 = -6). In order for the molecule to be neutral, the oxidation number of Chromium must be the opposite of the oxidation number of oxygen, so that they cancel each other. Therefore the initial oxidation state of Chromium in CrO₃ is +6.

Determine the final oxidation state of Chromium

In the product side of the balanced chemical equation, Chromium is found in the Cr(SO₄)₃ molecule. Each sulfate ion (SO₄²⁻) has an overall oxidation number of -2. Since there are three sulfate ions, the total oxidation number is -6 (-2 * 3 = -6). To maintain a neutral molecule, the oxidation state of Chromium must be equal to the positive value of the total oxidation number of sulfate ions. Hence, the final oxidation state of Chromium in Cr(SO₄)₃ is +6.

Calculate the change in oxidation state of Chromium

Now that we have the initial and final oxidation states of Chromium, we can calculate the change in oxidation state: Change in oxidation state = Final oxidation state - Initial oxidation state Change in oxidation state = +6 - (+6) = 0 So, there is no change in the oxidation state of Chromium during the oxidation of isopropanol to acetone by CrO₃ / H₂SO₄. None of the given options are correct.

Key concepts

Organic Chemistry

Organic chemistry focuses on the study of carbon-containing compounds. These compounds often include hydrocarbons and their derivatives. Central to organic chemistry is understanding how these compounds undergo reactions. In this exercise, we look at the conversion of isopropanol, a type of alcohol, into acetone, a ketone.
This transformation is a classic example demonstrating the oxidation of alcohols, a key reaction in organic chemistry.
Isopropanol (or 2-propanol) has the formula \( (CH_{3})_{2}CHOH \) and features three carbon atoms. The reaction involves:

• Substitution of the hydroxyl group \( (OH) \) with a carbonyl group \( (C=O) \) to form acetone \( (CH_{3})_{2}CO \).
• Removal of an alcohol hydrogen atom, typically facilitating the oxidation process to form ketones.

The process involves not only changes in molecular structure but also changes in electronic structure, affecting properties and reactivity. Every step in this transformation offers insight into how organic reactions can be harnessed in synthetic applications.

Chemical Equations

Chemical equations serve as a means to communicate the reactants and products in a chemical reaction. They show how atoms are rearranged, conserved, and transformed from reactants to products.
Writing a balanced chemical equation is crucial, ensuring the conservation of mass and charge. In the example given, isopropanol is oxidized by \( CrO_{3}/H_{2}SO_{4} \), a strong oxidizing agent:\[ (CH_{3})_{2}COH + 2CrO_{3} + 3H_{2}SO_{4} \rightarrow (CH_{3})_{2}CO + 2Cr(SO_{4})_{3} + 4H_{2}O \]

• All atoms present in the reactants must be accounted for as products, which is achieved by balancing the equation.
• For instance, the chromium, sulfur, oxygen, and hydrogen atoms are balanced on both sides of the equation to ensure matter is conserved.

Writing a balanced equation simplifies predicting the products of the reaction. It forms the basis for further analysis, such as changes in oxidation states. Balanced equations also assist in quantifying reactants and products, key for laboratory synthesis and industrial applications.

Oxidation-Reduction Reactions

An oxidation-reduction (or redox) reaction involves the transfer of electrons between chemical species. It is characterized by changes in oxidation states of the atoms involved.
In the given problem, determining changes in the oxidation state illustrates how electrons are transferred and shared among species:

• Chromium starts and ends with an oxidation state of +6, indicating no net change.
• The essence of the redox reaction here is seen in the alcohol to ketone transformation. The isopropanol loses electrons, undergoing oxidation to acetone.

Understanding oxidation-reduction is vital. It aids in grasping electron flow, as seen in biological systems and energy technology. The exercise on oxidation states allows students to track electron movement symbolically. Redox reactions play fundamental roles in processes like metabolism and combustion, making this concept critical across various scientific disciplines.