Question

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

Short answer

The equivalent weight of \(\mathrm{MnSO}_{4}\) is half its molecular weight when converted to \(\mathrm{MnO}_{2}\).

Step-by-step solution

Understand the Concept of Equivalent Weight

Equivalent weight of a substance is its molar mass divided by the change in oxidation number during a reaction. For a reaction involving a change in oxidation state, the equivalent weight is calculated as the molar mass divided by the n-factor, which is the number of electrons lost or gained per molecule.

Identify the Oxidation States

Determine the oxidation states of manganese in the initial compound, \(\mathrm{MnSO}_{4}\), and in the products: \(\mathrm{MnO}\), \(\mathrm{MnO}_{4}^{2-}\), \(\mathrm{MnO}_{2}\), and \(\mathrm{MnO}_{4}^{-}\). Manganese in \(\mathrm{MnSO}_{4}\) is in the +2 oxidation state. The oxidation states in products are: \(\mathrm{MnO}\) is +2, \(\mathrm{MnO}_{4}^{2-}\) is +6, \(\mathrm{MnO}_{2}\) is +4, and \(\mathrm{MnO}_{4}^{-}\) is +7.

Calculate Change in Oxidation States

For each option, calculate the change in the oxidation state of manganese:- In \(\mathrm{MnO}\): Change is +2 to +2, so \(n = 0\).- In \(\mathrm{MnO}_{4}^{2-}\): Change is +2 to +6, so \(n = 4\).- In \(\mathrm{MnO}_{2}\): Change is +2 to +4, so \(n = 2\).- In \(\mathrm{MnO}_{4}^{-}\): Change is +2 to +7, so \(n = 5\).

Determine the Actual and Required Equivalent Weight

To have the equivalent weight equal to half the molecular weight, the compound must have an n-factor of 2. This gives us a check:- \(n\) for \(\mathrm{MnO}\) is 0, not valid.- \(n\) for \(\mathrm{MnO}_{4}^{2-}\) is 4, not valid.- \(n\) for \(\mathrm{MnO}_{2}\) is 2, valid.- \(n\) for \(\mathrm{MnO}_{4}^{-}\) is 5, not valid.

Key concepts

Oxidation State

Oxidation state, sometimes called oxidation number, is a concept in chemistry that helps us to keep track of electrons in a compound. It's the hypothetical charge that an atom would have if all bonds to it were ionic. Understanding this number is crucial because it affects how substances interact and transform in chemical reactions.
The oxidation state can vary based on the chemical environment. For manganese in the compound \(\mathrm{MnSO}_{4}\), it starts at an oxidation state of +2. This means manganese has 'given away' two electrons in this compound. This state can change depending on the reactions it undergoes, affecting the equivalent weight in calculations.

• For \(\mathrm{MnO}\), the oxidation state of manganese remains +2.
• In \(\mathrm{MnO}_{4}^{2-}\), it increases to +6.
• With \(\mathrm{MnO}_{2}\), it is +4.
• For \(\mathrm{MnO}_{4}^{-}\), it is +7.

Recognizing these shifts is key to solving problems involving redox reactions, like the one with \(\mathrm{MnSO}_{4}\). These changes inform us about electron transfers, which directly impact the calculation of equivalent weight.

Molecular Weight

Molecular weight, also known as molar mass, is a measure of the mass of a molecule. It sums up the weights of all atoms in a molecule. Calculating molecular weight is essential as it allows chemists to convert between moles, a count of particles, and grams, a mass measurement. The molecular weight of a compound like \(\mathrm{MnSO}_{4}\) involves adding the atomic weights of one manganese atom, one sulfur atom, and four oxygen atoms. These atomic weights are usually found on the periodic table.
Once determined, the molecular weight aids in various calculations, such as determining molarity or, as here, equivalent weight. Knowing the molecular weight is particularly crucial because when comparing reactions or building equations, chemists need to reference a standard mass unit to ensure accuracy.

n-factor

The "n-factor" is an important concept that directly influences the calculation of equivalent weight. It represents the number of electrons lost or gained by one molecule of the substance during a chemical reaction. The n-factor is vital because it equates changes in oxidation states to an electron basis. This directs us to use a consistent method of thought when considering electron transfers. For a metallurgical oxidation states question, like with manganese transitions, the n-factor is essential in figuring out the equivalent weight. For example, if manganese's oxidation state moves from +2 in \(\mathrm{MnSO}_{4}\) to +6 in \(\mathrm{MnO}_{4}^{2-}\), the n-factor would be 4 since four electrons are transferred.
However, if choosing which form of manganese results in equivalent weight as half the molecular weight, you need \(n = 2\). This tells us that only \(\mathrm{MnO}_{2}\), where manganese moves from an oxidation state of +2 to +4, fulfills the required condition, as here the n-factor exactly equals 2. In equivalent weight calculations, this precise matching of n-factor and molecular properties allows chemists to predict the behavior of substances accurately.