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Chapter 1: Problem 13

The equivalent weight of \(\mathrm{MnSO}_{4}\) is half of its molecular weight when it is converted to: [1988-1 Mark] (a) \(\mathrm{Mn}_{2} \mathrm{O}_{3}\) (b) \(\mathrm{MnO}_{2}\) (c) \(\mathrm{MnO}_{4}^{-}\) (d) \(\mathrm{MnO}_{4}^{2-}\)

Short Answer

Expert verified
The correct answer is (b) \(\mathrm{MnO}_{2}\).

Step by step solution

01

Understand the Concept of Equivalent Weight

The equivalent weight of a substance is defined as the molecular weight divided by the number of electrons lost or gained by one molecule in a reaction. In this problem, the equivalent weight is given as half the molecular weight.
02

Identify Redox Change in Each Option

Examine the oxidation states of manganese in the given compounds: \(\mathrm{MnSO}_{4}\), \(\mathrm{Mn}_{2}\mathrm{O}_{3}\), \(\mathrm{MnO}_{2}\), \(\mathrm{MnO}_{4}^{-}\), and \(\mathrm{MnO}_{4}^{2-}\). In \(\mathrm{MnSO}_{4}\), Mn is in the +2 oxidation state.
03

Calculate Oxidation State Change

Determine the change in the oxidation state for Mn as it converts from \(\mathrm{MnSO}_{4}\) (+2 state) to each of the products: (a) In \(\mathrm{Mn}_{2}\mathrm{O}_{3}\), Mn is +3.(b) In \(\mathrm{MnO}_{2}\), Mn is +4.(c) In \(\mathrm{MnO}_{4}^{-}\), Mn is +7.(d) In \(\mathrm{MnO}_{4}^{2-}\), Mn is +6.
04

Analyse the Required Electron Exchange

Calculate the number of electrons exchanged for each reaction:(a) \(\mathrm{MnSO}_{4} \to \mathrm{Mn}_{2}\mathrm{O}_{3}\): Change per Mn = 1 electron (from +2 to +3) (b) \(\mathrm{MnSO}_{4} \to \mathrm{MnO}_{2}\): Change per Mn = 2 electrons (from +2 to +4) (c) \(\mathrm{MnSO}_{4} \to \mathrm{MnO}_{4}^{-}\): Change per Mn = 5 electrons (from +2 to +7) (d) \(\mathrm{MnSO}_{4} \to \mathrm{MnO}_{4}^{2-}\): Change per Mn = 4 electrons (from +2 to +6)
05

Relate to Molecular Weight and Equivalent Weight

The equivalence is half the molecular weight, implying that the valency factor or number of electron changes should be 2 (since molecular weight/2 = equivalent weight).
06

Confirm the Correct Answer

Out of the options, (b) \(\mathrm{MnSO}_{4} \to \mathrm{MnO}_{2}\) shows a valency factor of 2 because there is a change of 2 electrons per Mn. Therefore, the equivalent weight is half the molecular weight for this conversion.

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Key Concepts

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

Redox Reactions
Redox reactions are chemical processes involving both reduction and oxidation.
These occur simultaneously, as one substance loses electrons (oxidation) and another gains electrons (reduction).
In the context of manganese compounds like MnSO鈧� and MnO鈧�, recognizing these changes is crucial to identify equivalent weight transformations.
Redox reactions can dramatically alter the chemical properties of the compounds involved.
  • Reduction corresponds to a decrease in oxidation state, due to gaining electrons.
  • Oxidation involves an increase in oxidation state, typically through losing electrons.
  • The overall charge must stay balanced, ensuring that the number of electrons lost equals those gained in a complete reaction.
Understanding these key principles of redox reactions allows you to determine complex chemical interactions and analyze equivalent weight changes.
Oxidation States
Oxidation states are numeric indicators of an atom's degree of oxidation or reduction in a compound.
These states help in understanding how many electrons are exchanged in reactions, especially in redox processes.
In manganese compounds, tracking oxidation states is essential to comprehend transformations between different compounds such as from MnSO鈧� to MnO鈧�.
  • The oxidation state reflects the hypothetical charge an atom might have if all bonds were purely ionic.
  • In MnSO鈧�, manganese has an oxidation state of +2, which implies it can accept or lose electrons to form other manganese compounds.
  • The calculation of oxidation states enables predictions of electron transfers in reactions, crucial for tasks like determining equivalent weights.
By mastering oxidation states, you gain insight into chemical behavior and reaction mechanics.
Electron Exchange
Electron exchange refers to the transfer of electrons between chemical species in redox reactions.
This is a fundamental concept in determining equivalent weight, as the weight is influenced by the number of electrons involved in the reaction.
For example, when MnSO鈧� is converted to MnO鈧�, two electrons are exchanged per manganese atom.
  • Identifying electron exchange helps determine the valency factor in equivalent weight calculations.
  • In the case of MnSO鈧� to MnO鈧�, a two-electron exchange signifies that the equivalent weight is half the molecular weight, meeting the problem's criteria.
  • Observing electron exchanges enables you to comprehend more complex chemical transformations and reactant-product relationships.
By understanding electron exchange, you can better predict the outcomes of chemical reactions and find values like equivalent weight efficiently.
Manganese Compounds
Manganese compounds are diverse, with various oxidation states and applications.
Each compound has unique characteristics, affecting both its reactivity and potential usages.
Compounds like MnSO鈧� and MnO鈧� are often studied to evaluate oxidation-reduction behavior and equivalent weight calculations.
  • MnSO鈧� features manganese in a +2 oxidation state, commonly used in fertilizers and as a precursor to other manganese compounds.
  • MnO鈧�, where manganese is in a +4 oxidation state, is used in batteries, pigments, and as an oxidizing agent.
  • The transformation from one manganese compound to another involves understanding redox mechanics, including electron exchange and oxidation state changes.
Gaining an understanding of manganese compounds enriches your grasp of elemental chemistry and allows for better application of concepts like equivalent weight.

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Most popular questions from this chapter

A sugar syrup of weight \(214.2 \mathrm{~g}\) contains \(34.2 \mathrm{~g}\) of sugar \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\). Calculate : (i) molal concentration and (ii) mole fraction of sugar in the syrup.

A solution of two components containing \(n_{1}\) moles of the \(1^{\text {st }}\) component and \(n_{2}\) moles of the \(2^{\text {nd }}\) component is prepared. \(\mathrm{M}_{1}\) and \(\mathrm{M}_{2}\) are the molecular weights of component 1 and 2 respectively. If \(d\) is the density of the solution in \(g \mathrm{~mL}^{-1}, \mathrm{C}_{2}\) is the molarity and \(x_{2}\) is the mole fraction of the \(2^{\text {nd }}\) component, then \(\mathrm{C}_{2}\) can be expressed as: (a) \(\mathrm{C}_{2}=\frac{1000 x_{2}}{\mathrm{M}_{1}+x_{2}\left(\mathrm{M}_{2}-\mathrm{M}_{1}\right)}\) (b) \(\mathrm{C}_{2}=\frac{d x_{2}}{\mathrm{M}_{2}+x_{2}\left(\mathrm{M}_{2}-\mathrm{M}_{1}\right)}\) (c) \(\mathrm{C}_{2}=\frac{1000 d x_{2}}{\mathrm{M}_{1}+x_{2}\left(\mathrm{M}_{2}-\mathrm{M}_{1}\right)}\) (d) \(\mathrm{C}_{2}=\frac{d x_{1}}{\mathrm{M}_{2}+x_{2}\left(\mathrm{M}_{2}-\mathrm{M}_{1}\right)}\)

A mixture of \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) (oxalic acid) and \(\mathrm{NaHC}_{2} \mathrm{O}_{4}\) weighing \(2.02 \mathrm{~g}\) was dissolved in water and solution made upto one litre. Ten millilitres of the solution required \(3.0 \mathrm{~mL}\). of \(0.1 \mathrm{~N}\) sodium hydroxide solution for complete neutralization. In another experiment, \(10.0 \mathrm{~mL}\). of the same solution, in hot dilute sulphuric acid medium. require \(4.0 \mathrm{~mL}\). of \(0.1\) N potassium permanganate solution for complete reaction. Calculate the amount of \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) and \(\mathrm{NaHC}_{2} \mathrm{O}_{4}\) in the mixture. [1990 - 5 Marks]

A solid mixture \((5.0 \mathrm{~g})\) consisting of lead nitrate and sodium nitrate was heated below \(600^{\circ} \mathrm{C}\) until the weight of the residue was constant. If the loss in weight is \(28.0\) per cent, find the amount of lead nitrate and sodium nitrate in the mixture. [1990-4 Marks]

(a) One litre of a sample of hard water contains \(1 \mathrm{mg}\) of \(\mathrm{CaCl}_{2}\) and 1 \(\mathrm{mg}\) of \(\mathrm{MgCl}_{2}\). Find the total hardness in terms of parts of \(\mathrm{CaCO}_{3}\) per \(10^{6}\) parts of water by weight. (b) A sample of hard water contains \(20 \mathrm{mg}\) of \(\mathrm{Ca}^{++}\)ions per litre. How many milli-equivalent of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) would be required to soften 1 litre of the sample? (c) \(1 \mathrm{~g}\) of \(\mathrm{Mg}\) is burnt in a closed vessel which contains \(0.5 \mathrm{~g}\) of \(\mathrm{O}_{2}\). (i) Which reactant is left in excess? (ii) Find the weight of the excess reactants? (iii) How may milliliters of \(0.5 \mathrm{~N} \mathrm{H}_{2} \mathrm{SO}_{4}\) will dissolve the residue in the vessel.
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