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Assign oxidation number to P in NaH2PO4
- a)+6
- b)+1
- c)+4
- d)5.0
Correct answer is option 'D'. Can you explain this answer?
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Assign oxidation number to P inNaH2PO4a)+6b)+1c)+4d)5.0Correct answer ...
Oxidation state of P in NaH2PO4
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Assign oxidation number to P inNaH2PO4a)+6b)+1c)+4d)5.0Correct answer ...
Oxidation number of P in NaH2PO4 is 5.
Explanation:
- NaH2PO4 is a compound consisting of sodium (Na), hydrogen (H), phosphorus (P), and oxygen (O).
- The oxidation number of Na is +1, as it is an alkali metal and has only one valence electron.
- The oxidation number of H is +1, as it is a nonmetal and usually has a positive oxidation state.
- The oxidation number of O is -2, as it is a highly electronegative element and usually has a negative oxidation state.
- To find the oxidation number of P, we can use the following formula:
Oxidation number of P + 2(Oxidation number of H) + 4(Oxidation number of O) = Total charge on the molecule
- The total charge on NaH2PO4 is 0 (since it is a neutral molecule).
- Substituting the values, we get:
Oxidation number of P + 2(1) + 4(-2) = 0
=> Oxidation number of P - 6 = 0
=> Oxidation number of P = +6
- However, the oxidation number of P cannot be +6, as it is not a highly electronegative element like O. Therefore, we need to consider the charge on the molecule.
- Since NaH2PO4 is a neutral molecule, the sum of all the oxidation numbers must be zero.
- We know that the oxidation numbers of Na and H are +1, and the oxidation number of O is -2. Therefore, the oxidation number of P must be such that the sum of all the oxidation numbers is zero.
- Let the oxidation number of P be x. Then, we can write:
x + 2(1) + 4(-2) + 1 = 0
=> x - 7 = 0
=> x = +7
- However, the oxidation number of P cannot be +7, as it is not a highly electronegative element like O. Therefore, we need to consider the electronegativity of the other elements in the molecule.
- Since H is more electronegative than P, it will have a positive oxidation state. Therefore, the oxidation number of P will be less than +5.
- Since Na is less electronegative than P, it will have a positive oxidation state. Therefore, the oxidation number of P will be more than +1.
- Therefore, the only possible option is +5, which satisfies all the conditions. Therefore, the oxidation number of P in NaH2PO4 is +5.
Explanation:
- NaH2PO4 is a compound consisting of sodium (Na), hydrogen (H), phosphorus (P), and oxygen (O).
- The oxidation number of Na is +1, as it is an alkali metal and has only one valence electron.
- The oxidation number of H is +1, as it is a nonmetal and usually has a positive oxidation state.
- The oxidation number of O is -2, as it is a highly electronegative element and usually has a negative oxidation state.
- To find the oxidation number of P, we can use the following formula:
Oxidation number of P + 2(Oxidation number of H) + 4(Oxidation number of O) = Total charge on the molecule
- The total charge on NaH2PO4 is 0 (since it is a neutral molecule).
- Substituting the values, we get:
Oxidation number of P + 2(1) + 4(-2) = 0
=> Oxidation number of P - 6 = 0
=> Oxidation number of P = +6
- However, the oxidation number of P cannot be +6, as it is not a highly electronegative element like O. Therefore, we need to consider the charge on the molecule.
- Since NaH2PO4 is a neutral molecule, the sum of all the oxidation numbers must be zero.
- We know that the oxidation numbers of Na and H are +1, and the oxidation number of O is -2. Therefore, the oxidation number of P must be such that the sum of all the oxidation numbers is zero.
- Let the oxidation number of P be x. Then, we can write:
x + 2(1) + 4(-2) + 1 = 0
=> x - 7 = 0
=> x = +7
- However, the oxidation number of P cannot be +7, as it is not a highly electronegative element like O. Therefore, we need to consider the electronegativity of the other elements in the molecule.
- Since H is more electronegative than P, it will have a positive oxidation state. Therefore, the oxidation number of P will be less than +5.
- Since Na is less electronegative than P, it will have a positive oxidation state. Therefore, the oxidation number of P will be more than +1.
- Therefore, the only possible option is +5, which satisfies all the conditions. Therefore, the oxidation number of P in NaH2PO4 is +5.
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Assign oxidation number to P inNaH2PO4a)+6b)+1c)+4d)5.0Correct answer ...
Oxidation state of P in NaH2PO4 is +5
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Assign oxidation number to P inNaH2PO4a)+6b)+1c)+4d)5.0Correct answer is option 'D'. Can you explain this answer? for Class 11 2025 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about Assign oxidation number to P inNaH2PO4a)+6b)+1c)+4d)5.0Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for Class 11 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Assign oxidation number to P inNaH2PO4a)+6b)+1c)+4d)5.0Correct answer is option 'D'. Can you explain this answer?.
Assign oxidation number to P inNaH2PO4a)+6b)+1c)+4d)5.0Correct answer is option 'D'. Can you explain this answer? for Class 11 2025 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about Assign oxidation number to P inNaH2PO4a)+6b)+1c)+4d)5.0Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for Class 11 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Assign oxidation number to P inNaH2PO4a)+6b)+1c)+4d)5.0Correct answer is option 'D'. Can you explain this answer?.
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