Aluminum fluoride, AlF3, exists as a gas at 1000 degrees C.
(a) The Al-F bond energy is observed to be 659 kJ/mole. Calculate the Al-Al bond energy if the F-F bond energy 154.8 kJ/mol.
Calculate the % ionic character of the Al-F bond.
To calculate the Al-Al bond energy, we can use the concept of average bond energy in a diatomic molecule. The Al-Al bond energy can be determined by subtracting twice the F-F bond energy from the Al-F bond energy since there are two Al-F bonds in AlF3.
Step 1: Determine the Al-Al bond energy using the given information.
Al-F bond energy = 659 kJ/mol
F-F bond energy = 154.8 kJ/mol
Al-Al bond energy = Al-F bond energy - (2 × F-F bond energy)
Al-Al bond energy = 659 kJ/mol - (2 × 154.8 kJ/mol)
Al-Al bond energy = 659 kJ/mol - 309.6 kJ/mol
Al-Al bond energy = 349.4 kJ/mol
Therefore, the Al-Al bond energy is 349.4 kJ/mol.
To calculate the % ionic character of the Al-F bond, we can use the Pauling's equation, which relates the % ionic character of a bond to the electronegativity difference between the bonded atoms.
Step 2: Determine the electronegativity difference between Al and F using the Pauling Electronegativity Scale.
The electronegativity of Al on the Pauling scale is approximately 1.61.
The electronegativity of F on the Pauling scale is approximately 3.98.
Electronegativity difference = difference in electronegativity values
Electronegativity difference = 3.98 - 1.61
Electronegativity difference = 2.37
Step 3: Calculate the % ionic character using Pauling's equation.
% ionic character = (1 - e^(-0.25 * electronegativity difference^2)) * 100
% ionic character = (1 - e^(-0.25 * (2.37)^2)) * 100
% ionic character = (1 - e^(-0.25 * 5.6169)) * 100
% ionic character = (1 - e^(-1.40422)) * 100
% ionic character ≈ (1 - 0.2447) * 100
% ionic character ≈ 75.53%
Therefore, the % ionic character of the Al-F bond is approximately 75.53%.