Problem 146 - A Complete Resource Book in Chemistry for JEE Main

Question

Which of the following statement is/are correct? (a) \(\mathrm{CrI}_{3}+\mathrm{KOH}+\mathrm{Cl}_{2} \rightarrow \mathrm{K}_{2} \mathrm{CrO}_{4}+\mathrm{KCl}+\mathrm{KIO}_{4}+\mathrm{H}_{2} \mathrm{O}\) So, in balanced chemical reaction coefficient of \(\mathrm{KOH}\) is 64 . (b) If \(\mathrm{Zn}+\mathrm{KMnO}_{4}+\mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow \mathrm{ZnSO}_{4}+\mathrm{K}_{2} \mathrm{SO}_{4}+\) \(\mathrm{MnSO}_{4}+\mathrm{H}_{2}\); then equivalents of \(\mathrm{Zn}=\) equivalents of \(\mathrm{Zn}=\) equivalents of \(\mathrm{KMnO}_{4}+\) equivalents of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) (c) \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} \stackrel{\Delta}{\longrightarrow} \mathrm{N}_{2}+\mathrm{Cr}_{2} \mathrm{O}_{3}+\mathrm{H}_{2} \mathrm{O}, \mathrm{n}\)-factor of \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) is 12 . (d) \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-} \rightarrow \mathrm{Fe}^{3+}+\mathrm{CO}_{2}+\mathrm{NO}_{3}^{-}, \mathrm{n}\)-factor for \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}\) is \(60 .\)

Step-by-Step Solution

Answer

All statements (a), (b), (c), and (d) are incorrect.

Step 1: Analyzing Statement (a)

We need to balance the given chemical reaction: \[ \mathrm{CrI}_{3} + \mathrm{KOH} + \mathrm{Cl}_{2} \rightarrow \mathrm{K}_{2}\mathrm{CrO}_{4} + \mathrm{KCl} + \mathrm{KIO}_{4} + \mathrm{H}_{2} \mathrm{O} \]. Upon inspecting the products and reactants, note that \(\mathrm{KOH}\) is not actually providing any potassium for the formation of \(\mathrm{K}_{2}\mathrm{CrO}_{4}\), \(\mathrm{KCl}\), or \(\mathrm{KIO}_{4}\). Thus, this reaction is incorrectly proposed, and a coefficient of 64 for \(\mathrm{KOH}\) is incorrect in this context. This statement is false.

Step 2: Analyzing Statement (b)

Examine the chemical reaction: \( \mathrm{Zn} + \mathrm{KMnO}_{4} + \mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow \mathrm{ZnSO}_{4} + \mathrm{K}_{2} \mathrm{SO}_{4} + \mathrm{MnSO}_{4} + \mathrm{H}_{2} \). In this reaction, \(\mathrm{Zn}\) is oxidized while \(\mathrm{MnO}_{4}^{-}\) is reduced to \(\mathrm{Mn}^{2+}\). Equivalents of reactants in a redox reaction must be balanced; hence, equivalents of Zn should equal the equivalents of \(\mathrm{KMnO}_{4}\). However, the reaction doesn't involve \(\mathrm{H}_{2}\) formation under normal conditions. This statement is not balanced as described, thus false.

Step 3: Analyzing Statement (c)

The decomposition reaction given \( \left(\mathrm{NH}_{4}\right)_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} \stackrel{\Delta}{\longrightarrow} \mathrm{N}_{2} + \mathrm{Cr}_{2} \mathrm{O}_{3} + \mathrm{H}_{2} \mathrm{O}\) is a standard decomposition reaction where the \(n\)-factor denotes how many electrons are exchanged in the redox process. By calculating changes in oxidation states, the \(n\)-factor is found to be 6, indicating the statement's assertion of 12 is incorrect. Hence, this statement is false.

Step 4: Analyzing Statement (d)

The reaction \( \left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-} \rightarrow \mathrm{Fe}^{3+} + \mathrm{CO}_{2} + \mathrm{NO}_{3}^{-} \) involves a complex redox process. However, realistically, the \(n\)-factor for this process (assuming it is feasible) does not approach 60 as the transformation from cyanide to nitrate and other highly oxidized states is not straightforward due to complex chemical behavior. Thus, this statement is practically false.

Key Concepts

Balancing Chemical Equations

Balancing chemical equations is a crucial step in ensuring that a chemical reaction equation correctly respects the principle of the conservation of mass. This means that the number of atoms of each element must be the same on both the reactant and product sides of the equation. A balanced chemical equation not only allows one to understand the stoichiometry of a reaction but also ensures the proper calculation and understanding of involved substances.

When you attempt to balance a chemical equation, you might start by balancing elements that appear in the least amount on either side or those that appear in one compound on each side. Coefficients (whole numbers placed in front of compounds) are used to adjust the number of molecules or moles involved.

Take, for instance, the given unbalanced equation for statement (a): \[ \mathrm{CrI}_{3} + \mathrm{KOH} + \mathrm{Cl}_{2} \rightarrow \mathrm{K}_{2}\mathrm{CrO}_{4} + \mathrm{KCl} + \mathrm{KIO}_{4} + \mathrm{H}_{2} \mathrm{O} \]Attempting to balance this equation reveals that simply assigning a coefficient of 64 to \(\mathrm{KOH}\) does not accurately balance the equation or make chemical sense. Carefully matching the number of atoms for each element on both sides while considering the actual role of each compound ensures a correct biochemical depiction.

Oxidation States

Oxidation states, also known as oxidation numbers, denote the degree of oxidation of an atom in chemical compounds. They provide insight into the transfer of electrons in redox reactions. Assigning oxidation states requires understanding of certain rules:

• The oxidation state of an atom in its elemental form is always 0.
• In a compound, the sum of the oxidation states of all atoms equals the overall charge.
• The typical oxidation states for common elements can help streamline calculations. For example, hydrogen is usually +1 and oxygen is -2.

In statement (c), the oxidation state changes during the decomposition reaction of \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\): Cr changes in its states, leading to electrons being exchanged, indicative of oxidation and reduction processes. Different elements within the compound undergo changes, and carefully calculating these shifts confirms the redox nature of the reaction.

n-factor Calculation

The \(n\)-factor in chemistry describes the amount of exchange (usually in terms of electron exchange) within a reaction. Especially in redox reactions, \(n\)-factor is crucial for stoichiometry and dictates equivalence of moles in titrations and other chemical calculations.

To determine the \(n\)-factor, you need to analyze how many electrons are transferred per molecule during the chemical reaction. For example, when evaluating statement (c) and its decomposition reaction of \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\), by evaluating changes in oxidation states, you can find that the \(n\)-factor is calculated by totaling the number of electrons gained and lost. In this particular case, the correct \(n\)-factor was found to be 6, contrary to the incorrect initial assertion of 12.

The role of these factors extends far beyond mere academic exercise, impacting calculations involving titration where precision is vital. Proper calculation ensures balance and stoichiometric accuracy in any chemically calculated solution.

Chemical Reaction Analysis

Analyzing chemical reactions involves scrutinizing both reactants and products to ensure the reaction is chemically feasible and correctly represents the exchange processes happening at the molecular level.

Comprehensive chemical reaction analysis starts with understanding the nature of reactants, the actual exchange of particles, and the resulting products. For instance, in statement (b) involving \(\mathrm{Zn} \) and \( \mathrm{KMnO}_{4} \), understanding the flow of electrons and the role of each reactant is key.

• Identify redox couples and how oxidation and reduction complement each other.
• Ensure the reaction respects mass and charge conservation.
• Consider adjustments to coefficients to reflect accurate stoichiometry.

In practice, incorrectly predicting or describing the nature of the chemical reaction, such as asserting equal equivalents without backing stoichiometric balance, leads to mistakes. Proper analysis demands both theoretical understanding and precise quantitative expressions.