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A standard enthalpies of formation of gaseous XeF2,XeF4&XeF5 are -108,-216,&-294kJ mol^-1 respectively and the bond energy in F2 is 159kJ mol^-1. calculate the average Xe-F bond energy in each of this compounds and use of the value for XeF2 to obtain the value for the electronegativity of xenon on the pauling scale assuming the electronegativity of fluorine to be 4?
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A standard enthalpies of formation of gaseous XeF2,XeF4&XeF5 are -108,...
The average Xe-F bond energy for XeF2 is calculated by taking the standard enthalpy of formation of XeF2 and dividing by the number of Xe-F bonds in the compound. The standard enthalpy of formation of XeF2 is -108 kJ/mol, and there is one Xe-F bond in the compound, so the average Xe-F bond energy is -108 kJ/mol / 1 = -108 kJ/mol.
The average Xe-F bond energy for XeF4 is calculated in a similar way. The standard enthalpy of formation of XeF4 is -216 kJ/mol, and there are two Xe-F bonds in the compound, so the average Xe-F bond energy is -216 kJ/mol / 2 = -108 kJ/mol.
The average Xe-F bond energy for XeF5 is calculated in the same way. The standard enthalpy of formation of XeF5 is -294 kJ/mol, and there are three Xe-F bonds in the compound, so the average Xe-F bond energy is -294 kJ/mol / 3 = -98 kJ/mol.
To obtain the electronegativity of xenon on the Pauling scale, we can use the following equation:
electronegativity of Xe = (average Xe-F bond energy for XeF2 - bond energy of F2) / 2
Plugging in the values, we get:
electronegativity of Xe = (-108 kJ/mol - 159 kJ/mol) / 2 = (-267 kJ/mol) / 2 = -133.5 kJ/mol
The Pauling scale is based on the electron volts (eV) unit of energy, so we need to convert -133.5 kJ/mol to eV. One kJ/mol is equal to 96.485 eV, so -133.5 kJ/mol is equal to -12968.4 eV. The electronegativity of xenon on the Pauling scale is approximately -12.97 eV.
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A standard enthalpies of formation of gaseous XeF2,XeF4&XeF5 are -108,...
Calculating Average Xe-F Bond Energy
To find the average Xe-F bond energy for each compound, we can use the standard enthalpy of formation values along with the bond energy of F2. The reaction for the formation of each compound from its elements is:
- For XeF2:
\[ \text{Xe} + F_2 \rightarrow \text{XeF}_2 \]
Enthalpy change: ΔH = -108 kJ/mol
- For XeF4:
\[ \text{Xe} + 2F_2 \rightarrow \text{XeF}_4 \]
Enthalpy change: ΔH = -216 kJ/mol
- For XeF5:
\[ \text{Xe} + \frac{5}{2}F_2 \rightarrow \text{XeF}_5 \]
Enthalpy change: ΔH = -294 kJ/mol
Now, we can calculate the average bond energy.
Calculating Electronegativity of Xenon
Using the average bond energy for XeF2 to determine the electronegativity of xenon (χ_Xe):
\[
\chi_Xe = \chi_F + \left( \frac{E_{F-Xe}}{2} \right)
\]
Where:
- χ_F = 4 (electronegativity of fluorine)
- E_{F-Xe} = 51 kJ/mol (average Xe-F bond energy from XeF2)
Thus:
\[
\chi_Xe = 4 + \left( \frac{51}{2 \times 96.485} \right) \approx 4 + 0.264 \approx 4.264
\]
Therefore, the electronegativity of xenon on the Pauling scale is approximately 4.26.
To find the average Xe-F bond energy for each compound, we can use the standard enthalpy of formation values along with the bond energy of F2. The reaction for the formation of each compound from its elements is:
- For XeF2:
\[ \text{Xe} + F_2 \rightarrow \text{XeF}_2 \]
Enthalpy change: ΔH = -108 kJ/mol
- For XeF4:
\[ \text{Xe} + 2F_2 \rightarrow \text{XeF}_4 \]
Enthalpy change: ΔH = -216 kJ/mol
- For XeF5:
\[ \text{Xe} + \frac{5}{2}F_2 \rightarrow \text{XeF}_5 \]
Enthalpy change: ΔH = -294 kJ/mol
Now, we can calculate the average bond energy.
- For XeF2:
\[ \text{Bond Energy} = \Delta H + \text{Bond Energy of F}_2 \]
\[ = -108 + 159 = 51 \text{ kJ/mol} \] - For XeF4:
\[ \text{Bond Energy} = \Delta H + 2 \times \text{Bond Energy of F}_2 \]
\[ = -216 + 2 \times 159 = 102 \text{ kJ/mol} \] - For XeF5:
\[ \text{Bond Energy} = \Delta H + \frac{5}{2} \times \text{Bond Energy of F}_2 \]
\[ = -294 + \frac{5}{2} \times 159 = 165.5 \text{ kJ/mol} \]
Calculating Electronegativity of Xenon
Using the average bond energy for XeF2 to determine the electronegativity of xenon (χ_Xe):
\[
\chi_Xe = \chi_F + \left( \frac{E_{F-Xe}}{2} \right)
\]
Where:
- χ_F = 4 (electronegativity of fluorine)
- E_{F-Xe} = 51 kJ/mol (average Xe-F bond energy from XeF2)
Thus:
\[
\chi_Xe = 4 + \left( \frac{51}{2 \times 96.485} \right) \approx 4 + 0.264 \approx 4.264
\]
Therefore, the electronegativity of xenon on the Pauling scale is approximately 4.26.
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A standard enthalpies of formation of gaseous XeF2,XeF4&XeF5 are -108,-216,&-294kJ mol^-1 respectively and the bond energy in F2 is 159kJ mol^-1. calculate the average Xe-F bond energy in each of this compounds and use of the value for XeF2 to obtain the value for the electronegativity of xenon on the pauling scale assuming the electronegativity of fluorine to be 4?
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A standard enthalpies of formation of gaseous XeF2,XeF4&XeF5 are -108,-216,&-294kJ mol^-1 respectively and the bond energy in F2 is 159kJ mol^-1. calculate the average Xe-F bond energy in each of this compounds and use of the value for XeF2 to obtain the value for the electronegativity of xenon on the pauling scale assuming the electronegativity of fluorine to be 4? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A standard enthalpies of formation of gaseous XeF2,XeF4&XeF5 are -108,-216,&-294kJ mol^-1 respectively and the bond energy in F2 is 159kJ mol^-1. calculate the average Xe-F bond energy in each of this compounds and use of the value for XeF2 to obtain the value for the electronegativity of xenon on the pauling scale assuming the electronegativity of fluorine to be 4? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A standard enthalpies of formation of gaseous XeF2,XeF4&XeF5 are -108,-216,&-294kJ mol^-1 respectively and the bond energy in F2 is 159kJ mol^-1. calculate the average Xe-F bond energy in each of this compounds and use of the value for XeF2 to obtain the value for the electronegativity of xenon on the pauling scale assuming the electronegativity of fluorine to be 4?.
A standard enthalpies of formation of gaseous XeF2,XeF4&XeF5 are -108,-216,&-294kJ mol^-1 respectively and the bond energy in F2 is 159kJ mol^-1. calculate the average Xe-F bond energy in each of this compounds and use of the value for XeF2 to obtain the value for the electronegativity of xenon on the pauling scale assuming the electronegativity of fluorine to be 4? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A standard enthalpies of formation of gaseous XeF2,XeF4&XeF5 are -108,-216,&-294kJ mol^-1 respectively and the bond energy in F2 is 159kJ mol^-1. calculate the average Xe-F bond energy in each of this compounds and use of the value for XeF2 to obtain the value for the electronegativity of xenon on the pauling scale assuming the electronegativity of fluorine to be 4? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A standard enthalpies of formation of gaseous XeF2,XeF4&XeF5 are -108,-216,&-294kJ mol^-1 respectively and the bond energy in F2 is 159kJ mol^-1. calculate the average Xe-F bond energy in each of this compounds and use of the value for XeF2 to obtain the value for the electronegativity of xenon on the pauling scale assuming the electronegativity of fluorine to be 4?.
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