Question

Which of the following is not correct here a. For the reaction \(\mathrm{Cu}_{2} \mathrm{~S} \rightarrow \mathrm{CuO}+\mathrm{SO}_{2}\) \(\mathrm{E}_{\mathrm{Cu2s}}=\mathrm{M} / 8\) b. a \(\mathrm{Cu}_{2} \mathrm{O}+\mathrm{bNO}_{3}^{-}+14 \mathrm{H}^{+} \rightarrow \mathrm{Cu}^{2+}+\mathrm{NO}+7 \mathrm{H}_{2} \mathrm{O}\) Here \(a: b\) is \(3: 2\) c. \(\mathrm{FeS}_{2} \rightarrow \mathrm{Fe}_{2} \mathrm{O}_{3}+\mathrm{SO}_{2}\) Here \(\mathrm{E}=\frac{\mathrm{M}}{11}\) d. The value of \(n\) - factor of \(\mathrm{Fe}(\mathrm{SCN})_{2}\) when it oxidizes into \(\mathrm{Fe}^{+3}, \mathrm{SO}_{4}^{-2}\) and \((\mathrm{CN})_{2}\) is 12

Short answer

Option D is not correct.

Step-by-step solution

Understanding the Problem

The problem provides four statements about different chemical reactions and asks which statement is not correct. Each statement involves concepts like stoichiometry, electronic equilibrium, oxidation states, or reaction balancing.

Analyzing Option A

In option A, the reaction is \(\mathrm{Cu}_{2}\mathrm{S} \rightarrow \mathrm{CuO}+\mathrm{SO}_{2}\). The equivalent weight expression \(\mathrm{E}_{\mathrm{Cu2S}}=\mathrm{M}/8\) implies the molecular weight \(\mathrm{M}\) of \(\mathrm{Cu}_2\mathrm{S}\) involves 8 equivalents. However, based on the reaction half-equations, the calculation appears incorrect for this reaction.

Analyzing Option B

In option B, the reaction \(\mathrm{Cu}_{2}\mathrm{O}+\mathrm{bNO}_{3}^{-}+14 \mathrm{H}^{+} \rightarrow \mathrm{Cu}^{2+}+\mathrm{NO}+7 \mathrm{H}_{2}\mathrm{O}\) claims the ratio \(a : b\) is \(3 : 2\). Balancing the reaction shows that for every 3 Cu, 2 NO3- are needed, confirming this statement is likely correct.

Analyzing Option C

In option C, the reaction \(\mathrm{FeS}_{2} \rightarrow \mathrm{Fe}_{2}\mathrm{O}_{3}+\mathrm{SO}_{2}\) suggests \(\mathrm{E}=\frac{\mathrm{M}}{11}\). Reviewing the equivalent calculation, which focuses on the oxidation of sulfur and iron, indicates the expression is correct under standard stoichiometric considerations.

Analyzing Option D

Option D states that the n-factor of \(\mathrm{Fe}(mathrm{SCN})_{2}\) when oxidized to \(\mathrm{Fe}^{+3}\), \(\mathrm{SO}_{4}^{-2}\), and \((\mathrm{CN})_{2}\) is 12. Checking the processes of oxidation and changes in oxidation states reveals this value is incorrect because the overall electron transfer calculations do not account for such a high n-factor.

Key concepts

Stoichiometry

Stoichiometry is like a recipe for chemical reactions, ensuring that the quantities of substances react properly. It involves using balanced chemical equations to calculate the masses of reactants and products.
For example, if you are baking bread, you need a specific ratio of flour to water to yeast. In chemical reactions, stoichiometry gives us those specific ratios for molecules.
Basic stoichiometry involves understanding the coefficients in balanced chemical equations. These coefficients tell you how many moles of each substance are involved. By comparing them, you can figure out how much of one substance will react with or produce another substance.
In IIT JEE Physical Chemistry, mastering stoichiometry is crucial as it helps solve complex chemistry problems effortlessly.

Oxidation States

The concept of oxidation states is vital in understanding chemical reactions involving electron transfer. An oxidation state is a number assigned to elements in a compound, reflecting the degree of oxidation (loss of electrons) or reduction (gain of electrons).
A higher oxidation state indicates a molecule or atom is more oxidized, while a lower state suggests it has gained electrons and reduced its oxidation potential.
Oxidation states help chemists easily identify which elements are oxidized and which are reduced in a reaction. For example, in the conversion of \( ext{FeS}_2 \rightarrow ext{Fe}_2 ext{O}_3 + ext{SO}_2\), we analyze the oxidation states of Fe and S to understand what changes occur during the reaction.
Getting comfortable with calculating oxidation numbers will help you excel in reaction balancing and redox chemistry.

Reaction Balancing

Reaction balancing is the process of ensuring that the number of atoms for each element is the same on both sides of a chemical equation. It is crucial for maintaining the law of conservation of mass.
Each element involved must have equal atoms on both reactant and product sides. For instance, when balancing \( ext{Cu}_2 ext{O} + ext{NO}_3^- + ext{H}^+ ightarrow ext{Cu}^{2+} + ext{NO} + ext{H}_2 ext{O}\), coefficients are adjusted to balance the number of atoms across the equation.
Balancing reactions involves a step-by-step approach:

• List each element involved in the reaction.
• Count the number of atoms of each element.
• Adjust coefficients to ensure equal numbers of each type of atom on both sides.

Understanding reaction balancing principles will simplify tackling complex chemical problems.

Equivalent Weight

Equivalent weight is a concept used to quantify how much of a substance will interact with another in a reaction. It is calculated by dividing the molecular weight of a substance by the number of electrons lost or gained in a reaction.
In the context of IIT JEE, students might encounter problems like determining the equivalent weight of \(Cu_2S ightarrow CuO + SO_2\), where the equivalent weight calculation might seem off due to a miscalculated oxidation state.
Solving such problems involves careful analysis of how the number of electrons changes between reactants and products. Mastery will aid in understanding reaction stoichiometry and substance equivalency in reactions.

Chemical Reactions

Chemical reactions are processes where substances are transformed into different substances. Learning about chemical reactions involves understanding reactants, products, and the rearrangement of atoms.
Each reaction type, such as synthesis, decomposition, or redox reactions, follows unique principles. The reaction \(Fe(SCN)_2 ightarrow Fe^{3+}, SO_4^{2-}, (CN)_2\) demonstrates a redox reaction where electron transfer occurs.
Grasping the types and characteristics of chemical reactions is essential in predicting product formation and energy changes, crucial for both exams and practical applications.
With practice and understanding, chemical reactions become as routinely logical and predictable as mathematical equations.