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A compound of Xe and F is found to have 53.5% of Xe. What is oxidation number of Xe in this compound ?
- a)–4
- b)0
- c)+4
- d)+6
Correct answer is option 'D'. Can you explain this answer?
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Most Upvoted Answer
A compound of Xe and F is found to have 53.5% of Xe. What is oxidation...
Explanation:
Given data:
- Percentage of Xe in the compound = 53.5%
- Let's assume the compound is XeFn
Calculation:
- Let's assume the total mass of the compound is 100g.
- Therefore, the mass of Xe in the compound = 53.5g
- Let the oxidation number of Xe be 'x'
- The oxidation number of F is -1
- Since the compound is neutral, the sum of oxidation numbers should be zero.
- Therefore, the equation becomes: x + (-1)n = 0
- From the given data, we can calculate the moles of Xe and F in the compound.
- Moles of Xe = 53.5g / molar mass of Xe
- Moles of F = (100 - 53.5)g / molar mass of F
- Now, using the mole ratio, we can determine the value of 'n' in XeFn
- Finally, we can solve for the oxidation number of Xe using the equation x + (-1)n = 0
Conclusion:
- Based on the calculations, the oxidation number of Xe in the compound is +6 (option D).
Given data:
- Percentage of Xe in the compound = 53.5%
- Let's assume the compound is XeFn
Calculation:
- Let's assume the total mass of the compound is 100g.
- Therefore, the mass of Xe in the compound = 53.5g
- Let the oxidation number of Xe be 'x'
- The oxidation number of F is -1
- Since the compound is neutral, the sum of oxidation numbers should be zero.
- Therefore, the equation becomes: x + (-1)n = 0
- From the given data, we can calculate the moles of Xe and F in the compound.
- Moles of Xe = 53.5g / molar mass of Xe
- Moles of F = (100 - 53.5)g / molar mass of F
- Now, using the mole ratio, we can determine the value of 'n' in XeFn
- Finally, we can solve for the oxidation number of Xe using the equation x + (-1)n = 0
Conclusion:
- Based on the calculations, the oxidation number of Xe in the compound is +6 (option D).
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Community Answer
A compound of Xe and F is found to have 53.5% of Xe. What is oxidation...
Xe = 53.5 % ∴ F = 46.5%
Relative number of atoms Xe
Simple ratio Xe = 1 and F = 6 ; Molecular formula is XeF6
O.N.of Xe is +6
Relative number of atoms Xe
Simple ratio Xe = 1 and F = 6 ; Molecular formula is XeF6
O.N.of Xe is +6
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A compound of Xe and F is found to have 53.5% of Xe. What is oxidation number of Xe in this compound ?a)–4b)0c)+4d)+6Correct answer is option 'D'. Can you explain this answer?
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