Problem 93 - A Complete Resource Book in Chemistry for JEE Main

Question

The equivalent weight of \(\mathrm{MnSO}_{4}\) is half of its molecular weight, when it is converted to (a) \(\mathrm{Mn}_{2} \mathrm{O}_{3}\) (b) \(\mathrm{MnO}_{2}\) (c) \(\mathrm{MnO}_{4}^{-}\) (d) \(\mathrm{MnO}_{4}^{2-}\)

Step-by-Step Solution

Answer

The equivalent weight is half the molecular weight when converting MnSO4 to Mn2O3.

Step 1: Determine the Molecular Weight of MnSO4

First, calculate the molecular weight of \( \mathrm{MnSO}_{4} \). The atomic weights are approximately as follows: \( \mathrm{Mn} = 54.94 \), \( \mathrm{S} = 32.07 \), and \( \mathrm{O} = 16.00 \times 4 = 64.00 \). Adding these, the molecular weight of \( \mathrm{MnSO}_{4} \) is \[ 54.94 + 32.07 + 64.00 = 151.01 \].

Step 2: Equivalent Weight Relationship

The equivalent weight (E) is half the molecular weight for the given transformation. Therefore, \( E = \frac{151.01}{2} = 75.505 \).

Step 3: Calculate Possible Oxidation State Changes

\( \mathrm{MnSO}_{4} \) contains manganese in an oxidation state of +2. For each option, calculate the change in oxidation state when converted to the new compound: (a) \( \mathrm{Mn}_{2} \mathrm{O}_{3} \): Manganese is +3. (b) \( \mathrm{MnO}_{2} \): Manganese is +4. (c) \( \mathrm{MnO}_{4}^{-} \): Manganese is +7. (d) \( \mathrm{MnO}_{4}^{2-} \): Manganese is +6.

Step 4: Determine the Correct Conversion

To find which conversion results in an equivalent weight half of the molecular weight, subtract the initial oxidation state of Mn (+2) from the final oxidation state:(a) \(+3 - +2 = +1\) (b) \(+4 - +2 = +2\) (c) \(+7 - +2 = +5\) (d) \(+6 - +2 = +4\)The conversion where the equivalent weight is exactly half of the molecular weight involves a change equivalent to losing exactly one electron, consistent with the equivalent weight definition as half the molecular weight.

Key Concepts

Oxidation State

You know that an oxidation state helps to track how electrons are shared in a compound. Manganese, in various compounds, can exhibit different oxidation states. This concept is vital for understanding redox reactions. In chemistry, the oxidation state, also known as oxidation number, indicates the degree of oxidation of an atom in a chemical compound. It represents the number of electrons gained or lost by an atom to form the compound.
In manganese sulfate (MnSO_{4}), manganese has an oxidation state of +2. This means it has lost two electrons compared to its elemental state. As manganese transitions to different compounds like Mn_{2}O_{3}, MnO_{2}, MnO_{4}^{-}, and MnO_{4}^{2-}, its oxidation state changes. For instance, in MnO_{2}, manganese is in a +4 oxidation state, meaning it has lost two more electrons compared to the +2 state found in MnSO_{4}. Understanding these changes in oxidation states is crucial for determining equivalent weight in redox reactions.

Manganese Sulfate

Manganese sulfate (MnSO_{4}) is a chemical compound that contains manganese in a +2 oxidation state. It is a common manganese salt comprised of manganese, sulfur, and oxygen. This compound is often used in agriculture to treat manganese-deficient crops and has applications in various industries.
The composition of manganese sulfate is essential for calculating molecular weight and understanding its chemical behavior during transformations. The structure consists of manganese ions, sulfate ions, and crystal water in its hydrated form. The formula weight of manganese sulfate must be calculated accurately before proceeding with equivalent weight computations or any chemical reaction that involves this compound.

• In MnSO_{4}, manganese is the central atom bonded to the sulfate group.
• The sulfate group (SO_{4}^{2-}) carries a -2 charge, balancing the +2 charge of manganese.

Studying manganese sulfate helps students comprehend the fundamental aspects of transition metal chemistry and redox processes.

Equivalent Weight Calculation

The concept of equivalent weight is key in stoichiometry and chemical reactions. Equivalent weight refers to the mass of a substance that will react with or supply one mole of hydrogen ion (H⁺) or electron in a redox reaction. This is crucial in analyzing redox reactions like that of manganese sulfate (MnSO_{4}).
Calculating equivalent weight depends on the change in the oxidation state of the central atom. For MnSO_{4} transitioning to various manganese compounds, equivalent weight calculations require understanding oxidation state changes.

• First, identify the initial and final oxidation states of manganese in both compounds.
• Subtract the initial oxidation state from the final oxidation state to determine the number of electrons involved.
• The equivalent weight is then calculated as the molecular weight divided by the change in oxidation state in terms of electrons.

For instance, if manganese changes from +2 to +3, the equivalent weight equals the molecular weight divided by one (because of one electron change). This skill simplifies the balancing of redox equations and helps predict reaction outcomes.

Molecular Weight Calculation

Molecular weight is the sum of the atomic weights of all atoms in a molecule. It is indispensable in chemistry for converting between moles and grams, preparing solutions, and understanding reaction stoichiometry.
To calculate the molecular weight of manganese sulfate (MnSO_{4}), consider the atomic weights of individual component elements:

• Manganese (Mn): 54.94 g/mol
• Sulfur (S): 32.07 g/mol
• Oxygen (O): 16.00 g/mol (4 oxygen atoms = 64.00 g/mol)

Add these atomic weights to find the molecular weight of MnSO_{4}:\[54.94 + 32.07 + 64.00 = 151.01 \text{ g/mol}\]
This calculation allows you to understand how manganese sulfate behaves in chemical processes, supports the calculation of equivalent weight, and aids in reaction stoichiometry. Molecular weight calculations are fundamental to making conversions in chemical equations and understanding the quantitative aspects of reactions.