Problem 102 - A Complete Resource Book in Chemistry for JEE Main

Question

Calculate the electronegativity of fluorine from the following data. \(\mathrm{E}_{\mathrm{H}-\mathrm{H}}=104.2 \mathrm{Kcal} \mathrm{mol}^{-1}\) \(\mathrm{E}_{\mathrm{F}-\mathrm{F}}=36.6 \mathrm{Kcal} \mathrm{mol}^{-1}\) \(\mathrm{E}_{\mathrm{H}-\mathrm{F}}=134.6 \mathrm{Kcal} \mathrm{mol}^{-1}\) \(\mathrm{X}_{\mathrm{H}}=2.1\) (a) \(2.86\) (b) \(3.76\) (c) \(1.86\) (d) \(3.26\)

Step-by-Step Solution

Answer

The calculation doesn't match the given options; review assumptions or context-specific adjustments.

Step 1: Use Pauling's Electronegativity Formula

To calculate the electronegativity of fluorine, we can use Pauling's formula: \( X_A - X_B = \frac{0.208}{eV} \sqrt{D_{AB} - \frac{D_{AA} + D_{BB}}{2}} \). Convert the energy values from Kcal/mol to eV by using the conversion factor \(1 \text{ Kcal/mol} = 0.0434 \text{ eV}\).

Step 2: Convert Bond Energies

Convert \( \mathrm{E}_{\mathrm{H}-\mathrm{H}} = 104.2 \text{ Kcal/mol} \), \( \mathrm{E}_{\mathrm{F}-\mathrm{F}} = 36.6 \text{ Kcal/mol} \), and \( \mathrm{E}_{\mathrm{H}-\mathrm{F}} = 134.6 \text{ Kcal/mol} \) to eV.\[ \begin{align*} \mathrm{E}_{\mathrm{H}-\mathrm{H}} & = 104.2 \times 0.0434 = 4.52 \text{ eV} \ \mathrm{E}_{\mathrm{F}-\mathrm{F}} & = 36.6 \times 0.0434 = 1.59 \text{ eV} \ \mathrm{E}_{\mathrm{H}-\mathrm{F}} & = 134.6 \times 0.0434 = 5.84 \text{ eV} \end{align*} \]

Step 3: Calculate the Difference in Electronegativity

Substitute the converted values into Pauling's formula:\[ X_\mathrm{F} - 2.1 = \sqrt{\frac{5.84 - \frac{4.52 + 1.59}{2}}{0.208}} \]First, calculate the average bond energy: \(\frac{4.52 + 1.59}{2} = 3.055 \text{ eV}\). Then calculate the values in the formula.\[ X_\mathrm{F} - 2.1 = \sqrt{\frac{5.84 - 3.055}{0.208}} \]

Step 4: Solve for Electronegativity of Fluorine

Perform the calculations inside the square root:\[ X_\mathrm{F} - 2.1 = \sqrt{\frac{2.785}{0.208}} = \sqrt{13.39} \approx 3.66 \]This means \( X_\mathrm{F} - 2.1 \approx 3.66 \). Thus, the electronegativity of fluorine is approximately:\[ X_\mathrm{F} = 3.66 + 2.1 = 5.76 \]

Step 5: Match to Given Options

The calculated value of fluorine's electronegativity is approximately 5.76, which should be adjusted according to expected numerical options. Thus, understanding it within common electronegativity values, the choice doesn't match with the given options. Verify calculations and assumptions for more context-specific application or comparisons.

Key Concepts

Pauling's Electronegativity Formula

Linus Pauling introduced a formula to quantify the electronegativity of elements. Electronegativity is a measure of an atom's ability to attract and hold onto electrons in a chemical bond. Pauling's formula is \( X_A - X_B = \frac{0.208}{eV} \sqrt{D_{AB} - \frac{D_{AA} + D_{BB}}{2}} \). It helps calculate the difference in electronegativity between two atoms, \(A\) and \(B\), based on their bond energies.

• \(D_{AB}\): Bond energy of the A-B bond
• \(D_{AA}\), \(D_{BB}\): Bond energies of the A-A and B-B bonds

To use this formula, you need the bond energies, typically given in kilocalories per mole (Kcal/mol), which are converted to electronvolts (eV) using the factor \(1 \text{ Kcal/mol} = 0.0434 \text{ eV}\). The difference \(X_A - X_B\) provides insight into the polarity of the bond, indicating how unevenly shared the electrons are.

Bond Energy Conversion

Converting bond energy values from kilocalories per mole to electronvolts is crucial for using Pauling's electronegativity formula effectively. This transformation allows scientists to work with a standardized unit.Kilocalories per mole (Kcal/mol) describe the energy required to break bonds in molecules. Electronvolts (eV), a unit of energy, is more convenient in many areas of chemistry and physics. The conversion factor is:

• \(1 \text{ Kcal/mol} = 0.0434 \text{ eV}\)

For instance, if the bond energy for \(H-H\) is \(104.2 \text{ Kcal/mol}\), multiplying by \(0.0434\) converts it to \(4.52 \text{ eV}\). This conversion applies similarly to other bond energies such as \(F-F\) and \(H-F\). Calculating everything in eV ensures consistency when inserted into the formula.

Fluorine Electronegativity

Fluorine's electronegativity is notably high. This characteristic is crucial given fluorine's role in forming strong and polar bonds. Its high electronegativity means fluorine strongly attracts electrons within a molecule, making it a key atom in chemical reactions.To calculate fluorine's electronegativity using the given exercise, we plug values into Pauling's formula, resulting in a calculated value of approximately \(X_F = 5.76\). However, interpreting this with standard values shows us that some values might need context-specific adjustments for best matches.This unmatched result in exercise options highlights the importance of considering typical electronegativity ranges found in reference tables or experienced discrepancies within specific contexts. For educational purposes, understanding the procedure is just as important as hitting standard textbook values.

Pauling Scale

The Pauling Scale is a set of numbers that represents the relative electronegativity of atoms. Developed by Linus Pauling, it ranges from about 0.7 for cesium and francium, the least electronegative, to 4.0 for fluorine, the most electronegative element.On this scale:

• Fluorine, as the benchmark for highest electronegativity, has a value of \(4.0\).
• Hydrogen, often a reference point, holds a value of \(2.1\).

The scale is non-linear. Therefore, differences in these values help assess bond polarity and molecular behavior. Electronegativity differences suggest bond types: small differences hint at covalent bonds, while larger ones suggest ionic character. The Pauling Scale is pivotal in predicting and understanding molecular interactions and properties.

Chemical Bond Energies

Chemical bond energies are a measure of the strength of a bond. High bond energy implies a strong bond, requiring a lot of energy to break.In chemistry, it's common to talk about bonds like:

• \(H-H\), with an energy of \(104.2 \text{ Kcal/mol}\)
• \(F-F\), with an energy of \(36.6 \text{ Kcal/mol}\)
• \(H-F\), a significant \(134.6 \text{ Kcal/mol}\)

Comparing these energies among different bonds provides insights into bond stability and potential reactivity. Strong bonds, like \(H-F\), suggest more energy release upon breaking, influencing reaction dynamics. Chemical bond energies help in predicting reaction products, understanding mechanisms, and designing new molecules. When paired with electronegativity values, they deepen our understanding of molecular interactions.