Problem 20 - Chemistry in 30 days

Question

Oxidation number of fluorine in \(\mathrm{F}_{2} \mathrm{O}\) is (a) \(+1\) (b) \(+2\) (c) \(-1\) (d) \(-2\)

Step-by-Step Solution

Answer

The oxidation number of fluorine in \(\mathrm{F}_{2} \mathrm{O}\) is \(-1\).

Step 1: Understand the compound

In the compound \(\mathrm{F}_{2} \mathrm{O}\), there are two fluorine atoms (\(\mathrm{F}\)) and one oxygen atom (\(\mathrm{O}\)). We need to determine the oxidation number of the fluorine in this compound.

Step 2: Know the rules of oxidation numbers

Fluorine is the most electronegative element and usually has an oxidation number of \(-1\) when it is not combined with another more electronegative element. However, in this compound \(\mathrm{F}_{2}\mathrm{O}\), it is different.

Step 3: Consider common oxidation states

The typical oxidation state of oxygen is \(-2\). However, in compounds where oxygen is bonded to more electronegative elements such as fluorine, it can have a positive oxidation state. Here, oxygen is less electronegative than fluorine.

Step 4: Apply the sum of oxidation numbers rule

For a neutral compound like \(\mathrm{F}_{2}\mathrm{O}\), the sum of the oxidation numbers of all atoms must equal zero. This means the sum of the oxidation numbers of the two fluorine atoms and the oxygen atom must be zero.

Step 5: Use algebra to find the oxidation state

Assume the oxidation number of each fluorine is \(x\). We have two fluorine atoms, so their contribution is \(2x\). If we assume oxygen is \(+2\), the equation becomes: \[ 2x + (+2) = 0 \] Solving for \(x\), we find: \[ 2x = -2 \] \[ x = -1 \]

Step 6: Conclusion

The calculation shows that the oxidation number of each fluorine atom in \(\mathrm{F}_{2}\mathrm{O}\) is \(-1\). This confirms that even in a compound like \(\mathrm{F}_{2}\mathrm{O}\), fluorine retains its typical oxidation state of \(-1\).

Key Concepts

Fluorine and Its Unique Oxidation State

Fluorine is regarded as the most electronegative element on the periodic table. This means it has a strong tendency to attract electrons from other elements during chemical bonding. Because of its high electronegativity, fluorine almost always assumes an oxidation state of

• egative one (-1)

in compounds. This is true even though it can form compounds with oxygen, which is also quite electronegative. However, in these rare exceptions, such as in fluorine oxide (\(\mathrm{F}_{2}\mathrm{O}\)), fluorine still retains

• its oxidation state of \(-1\).

Understanding that fluorine typically remains unchanged in its oxidation state highlights its strong electron-withdrawing capacity and helps us make better predictions in chemical reactions.

Understanding Neutral Compound Oxidation

When dealing with chemical compounds, one key rule is that the sum of the oxidation states within a compound must equal zero. This is because most compounds are neutral, meaning they do not have an overall charge. In the case of \(\mathrm{F}_{2}\mathrm{O}\), this rule helps us determine the oxidation states of the constituent elements precisely. For the neutral compound rule, consider:

• The sum of oxidation numbers in \(\mathrm{F}_{2}\mathrm{O}\) must equal zero.
• The known state of oxygen\((+2)\)is crucial for determining the fluorine state.

Understanding this principle is essential for making accurate assessments of oxidation states in complex molecules and predicting chemical behavior.

Electronegativity's Role in Oxidation States

Electronegativity is a chemical property that describes an atom's ability to attract and hold onto electrons. It's crucial in determining the distribution of oxidation states in a molecule. Elements with high electronegativity, such as fluorine, often take on negative oxidation states because they attract electrons towards themselves.

• Fluorine has the highest electronegativity value of any element.
• This ensures it takes electrons, resulting in a \(-1\)oxidation state.

In \(\mathrm{F}_{2}\mathrm{O}\), fluorine's greater electronegativity compared to oxygen translates into fluorine retaining its typical negative reduction state. Learning about electronegativity not only explains oxidation numbers but also gives deeper insight into why elements bond the way they do.

Algebraic Determination of Oxidation States

When faced with the task of determining the oxidation states of elements within a compound, we can use algebra to find solutions accurately. This approach simplifies the complex rules of chemistry into manageable equations. Examining \(\mathrm{F}_{2}\mathrm{O}\), we follow these steps:

• Assume the oxidation number of each fluorine atom is \(x\).
• The total contribution of two fluorine atoms is \(2x\).
• Knowing oxygen is \(+2\), set up the equation \(2x + (+2) = 0\).

Solving this equation gives us the oxidation state of fluorine as \(x = -1\). This step-by-step mathematical method provides a clear pathway to accurately find oxidation states, ensuring accuracy in complex chemical problems.