Problem 25 - Chemistry in 30 days

Question

Fluorine reacts with dilute \(\mathrm{NaOH}\) to produce (a) \(\mathrm{NaOF}, \mathrm{NaF}\) and \(\mathrm{H}_{2} \mathrm{O}\) (b) \(\mathrm{O}_{2}, \mathrm{NaF}\) and \(\mathrm{H}_{2} \mathrm{O}\) (c) \(\mathrm{OF}_{2}, \mathrm{NaF}\) and \(\mathrm{H}_{2} \mathrm{O}\) (d) \(\mathrm{F}_{2} \mathrm{O}, \mathrm{NaF}\) and \(\mathrm{H}_{2} \mathrm{O}\)

Step-by-Step Solution

Answer

(c)  ext{OF}_2, ext{NaF} and  ext{H}_2 ext{O}

Step 1: Identify the Reaction

The reaction of fluorine (_2) with dilute sodium hydroxide (NaOH) involves the disproportionation of fluorine, a highly electronegative element, leading to different products. When fluorine reacts with alkaline solutions like NaOH, it disproportionates into oxygen difluoride (OF_2) and sodium fluoride (NaF), along with water.

Step 2: Determine the Products

The balanced chemical equation for the reaction can be written as: \[ 2 ext{F}_2 + 2 ext{NaOH} ightarrow ext{OF}_2 + 2 ext{NaF} + ext{H}_2 ext{O} \] Here, fluorine forms oxygen difluoride (OF_2) and sodium fluoride (NaF), alongside water (H_2O) as products.

Step 3: Match the Products with the Options

From the reaction equation, the products are OF_2, NaF, and H_2O. Look at the given options to find the correct match. The products in option (c) are the same as those from the chemical equation, indicating that option (c) is correct.

Key Concepts

Disproportionation

Disproportionation is a special type of chemical reaction where a single substance both oxidizes and reduces itself, forming two different substances. This fascinating process often involves elements in unusual oxidation states or highly electronegative elements like fluorine.
In the context of our exercise, fluorine undergoes disproportionation when it reacts with sodium hydroxide (\(\text{NaOH}\)). Here, fluorine breaks the usual pattern, where instead of forming a single compound, it simultaneously forms two products:

• Oxygen difluoride (\(\text{OF}_2\))
• Sodium fluoride (\(\text{NaF}\))

This reaction showcases the unique reactivity of fluorine. It's different from many other reactions because it uses the same element to play two roles—both donor and acceptor of electrons. Such reactions can be puzzling for beginners, but grasping the idea of one element doing 'double duty' in a chemical process can enhance understanding of many complex systems.

Fluorine Reactions

Fluorine is the most electronegative and reactive of all the elements. It's known for its ability to react with almost every other element and many compounds, often quite dramatically. Its interactions with other substances, like in our textbook exercise, demonstrate fluorine's powerful reactivity.
When fluorine reacts with \(\text{NaOH}\), an alkaline solution, it forms unique compounds like oxygen difluoride (\(\text{OF}_2\)). This occurs because fluorine can force the oxygen and sodium atoms to part ways in the sodium hydroxide, allowing it to create both sodium fluoride and oxygen difluoride.

• Fluorine is quite versatile because of its electronegativity.
• Often forms a variety of compounds, challenging predictions.
Understanding fluorine's ability to engage in disproportionation and other reactions helps predict how it will interact with different chemicals, emphasizing its role as a highly active participant in many reactions. Seeing all these interactions in context can help pave the way for mastering chemistry fundamentals.

Balanced Chemical Equation

Writing a balanced chemical equation is the key to understanding any chemical reaction. This process involves ensuring the same number of each type of atom on both sides of the equation.
In the given reaction of fluorine with sodium hydroxide, balancing ensures that:

• Each element conserves its quantity before and after the reaction.
• No atoms are mysteriously gained or lost in the process.

For our example, the equation is:\[ 2 \text{F}_2 + 2 \text{NaOH} \rightarrow \text{OF}_2 + 2 \text{NaF} + \text{H}_2 \text{O} \]Each component here is meticulously balanced:

• 2 fluorine molecules form 1 \(\text{OF}_2\) and 2 \(\text{NaF}\)
• Each sodium hydroxide molecule is accounted for in products.

Balancing chemical equations correctly is crucial. It ensures that we adhere to the law of conservation of mass and enables us to accurately predict the outcomes of chemical reactions. Whether it's a classroom exercise or real-world application, being adept at balancing equations is foundational to chemistry.