Problem 76 - A Complete Resource Book in Chemistry for JEE Main

Question

Which one of the following represents the reaction between fluorine and cold dilute \(\mathrm{NaOH} ?\) (a) \(2 \mathrm{~F}_{2}+2 \mathrm{NaOH} \longrightarrow 2 \mathrm{NaF}+\mathrm{OF}_{2}+\mathrm{H}_{2} \mathrm{O}\) (b) \(\mathrm{F}_{2}+2 \mathrm{NaOH} \longrightarrow \mathrm{NaF}+\mathrm{NaOF}+\mathrm{H}_{2} \mathrm{O}\) (c) \(3 \mathrm{~F}_{2}+\mathrm{NaOH} \longrightarrow 5 \mathrm{NaF}+\mathrm{NaFO}_{3}\) (d) \(2 \mathrm{~F}_{2}^{2}+4 \mathrm{NaOH} \longrightarrow 4 \mathrm{NaF}+2 \mathrm{H}_{2} \mathrm{O}\)

Step-by-Step Solution

Answer

The correct reaction is option (a): \(2 \mathrm{~F}_{2}+2 \mathrm{NaOH} \longrightarrow 2 \mathrm{NaF}+\mathrm{OF}_{2}+\mathrm{H}_{2} \mathrm{O}\).

Step 1: Understand the Reaction Context

We are dealing with a reaction between fluorine gas \(\mathrm{F}_2\) and cold, dilute sodium hydroxide \(\mathrm{NaOH}\). In these conditions, fluorine acts as a strong oxidizing agent.

Step 2: Identify Typical Reaction Products

Fluorine typically forms fluoride ions \(\mathrm{F}^-\) when reacting with alkali conditions, leading to the formation of \(\mathrm{NaF}\). Some reactions with fluorine might generate a compound such as \(\mathrm{OF}_2\) along with water.

Step 3: Checking Each Option

- **Option (a):** Produces \(\mathrm{NaF}\), \(\mathrm{OF}_2\), and \(\mathrm{H}_2\mathrm{O}\), balancing the equation as \(2 \mathrm{~F}_{2}+2 \mathrm{NaOH} \longrightarrow 2\mathrm{NaF}+\mathrm{OF}_{2}+\mathrm{H}_{2}\mathrm{O}\).- **Option (b):** Breakdown into products different from typical fluorine reactions.- **Option (c):** Does not balance correctly considering the stoichiometry of \(\mathrm{NaF}\).- **Option (d):** Incorrect stoichiometry and products are not typical for the fluorine reaction.

Step 4: Determine the Correct Equation

From analyzing all options, option (a) correctly features products of \(\mathrm{NaF}\), \(\mathrm{OF}_2\), and \(\mathrm{H}_2\mathrm{O}\) which are possible products for this reaction with fluorine under cold dilute conditions.

Step 5: Final Decision

The reaction represented in option (a) aligns with the expected products when fluorine reacts with cold dilute \(\mathrm{NaOH}\), forming fluoride, an oxygen fluoride compound, and water.

Key Concepts

Oxidizing Agent

An oxidizing agent is a substance that promotes oxidation by accepting electrons from another substance in a chemical reaction. Fluorine gas ($F_2$), in particular, is renowned for its powerful oxidizing abilities because it has a high affinity for electrons.

When fluorine comes in contact with sodium hydroxide ($NaOH$), it acts as an oxidizing agent. This means that it forces the oxidation of another element or compound, in this case, $NaOH$, and in doing so, it is itself reduced. The fluorine molecules $F_2$ gain electrons to form fluoride ions $F^-$.
- **Fluorine, with its high electronegativity, strives to gain electrons to achieve stability.** - ~~As an oxidizing agent, fluorine induces a transfer of electrons from the substance it contacts, leading to its reduction.~~ In these chemical scenarios: - Fluorine acts to oxidize sodium $Na$ while it itself undergoes reduction by gaining electrons and transforming into fluoride ions $F^-$. - Through this process, different substances are formed, including sodium fluoride $NaF$ and possibly other compounds.

Fluoride Ions

Fluoride ions $(F^-)$ are negatively charged ions formed when fluorine atoms gain one electron. This transformation is a result of fluorine's reaction where it behaves as an electron acceptor due to its high electronegativity.

In the reaction between fluorine gas $F_2$ and sodium hydroxide $NaOH$, the fluorine molecules attract electrons, forming fluoride ions. These ions then bond with positively charged sodium ions $Na^+$, resulting in the formation of sodium fluoride $NaF$, an essential product of this reaction.
Key characteristics and roles: - **Stable Configuration:** Fluoride ions achieve a stable electron arrangement, similar to that of noble gases, making them highly stable. - **Industrial Use:** Sodium fluoride is commonly used in water fluoridation to prevent dental cavities due to the protective capabilities of fluoride ions. It's important to note that while fluoride ions are vital in numerous applications, their creation and behavior in reactions are deeply tied to fluorine's chemical characteristics as a highly electronegative element.

Reaction Stoichiometry

Reaction stoichiometry involves the calculation and relationship of reactants and products in a chemical reaction. It requires balanced chemical equations to ensure that the conservation of mass is adhered to within the reaction.

In the specific reaction between fluorine \(F_2\) and sodium hydroxide \(NaOH\) under cold, dilute conditions, stoichiometry becomes critical to predict and verify the resulting products and their amounts. This exercise requires balancing not only the reactants and products but also considering their respective molar ratios.
For example:- In option (a), the equation is balanced as \( 2F_2 + 2NaOH \rightarrow 2NaF + OF_2 + H_2O \). This implies: - Two molecules of fluorine react with two units of sodium hydroxide. - Resulting in two sodium fluoride molecules, one oxygen difluoride \(OF_2\), and water.Crucial elements of stoichiometry include:- **Mole Ratios:** These ratios are used to interpret how many molecules of a reactant will yield specific products.- **Conservation Law:** Ensures atoms are neither created nor destroyed in the equation, hence each side of the equation reflects identical quantities of each element.