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Direction (Q. Nos. 27 and 28) This section contains 2 question. when worked out will result in an integer from 0 to 9 (both inclusive)
Q. In acidic medium,H2O2 changes Cr2O72- to CrO5 which has two (-O-O-) bonds. Oxidation state of Cr in CrO3 is
Correct answer is '6'. Can you explain this answer?
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Direction (Q. Nos. 27 and 28) This section contains 2question. when wo...
When H2O2 is added to an acidified solution of a dichromate, Cr2O72-, aO deep blue coloured complex, chromic peroxide, CrO5 [ or CrO(O2)2] is formed.
Cr2O72- +2H++4H2O2 ---> 2CrO(O2)2 +5H2O
This deep blue coloured complex.
Oxidation state of Cr is +6 due to the presence of two peroxide linkages which can be calculated as Cr peroxide normal
x+(-1)4+(-2) =0
x-6 =0
x=+6
Cr2O72- +2H++4H2O2 ---> 2CrO(O2)2 +5H2O
This deep blue coloured complex.
Oxidation state of Cr is +6 due to the presence of two peroxide linkages which can be calculated as Cr peroxide normal
x+(-1)4+(-2) =0
x-6 =0
x=+6
Most Upvoted Answer
Direction (Q. Nos. 27 and 28) This section contains 2question. when wo...
Explanation:
In acidic medium, hydrogen peroxide (H2O2) changes dichromate ion (Cr2O72-) to perchromate ion (CrO5). The perchromate ion contains two peroxide (-O-O-) bonds. We need to determine the oxidation state of chromium (Cr) in chromium trioxide (CrO3).
Step 1:
To determine the oxidation state of chromium in CrO3, we need to assign oxidation states to the other elements present in the compound.
Step 2:
Oxygen (O) is usually assigned an oxidation state of -2, unless it is in a peroxide compound where it is assigned an oxidation state of -1.
Step 3:
Since there are three oxygen atoms in CrO3, the total oxidation state contribution from oxygen is -6 (-2 × 3 = -6).
Step 4:
The overall oxidation state of the compound is 0 (since the compound is neutral).
Step 5:
To determine the oxidation state of chromium, we subtract the oxidation state contribution from oxygen (-6) from the overall oxidation state (0).
0 - (-6) = +6
Therefore, the oxidation state of chromium in CrO3 is +6.
Conclusion:
The oxidation state of chromium in CrO3 is +6.
In acidic medium, hydrogen peroxide (H2O2) changes dichromate ion (Cr2O72-) to perchromate ion (CrO5). The perchromate ion contains two peroxide (-O-O-) bonds. We need to determine the oxidation state of chromium (Cr) in chromium trioxide (CrO3).
Step 1:
To determine the oxidation state of chromium in CrO3, we need to assign oxidation states to the other elements present in the compound.
Step 2:
Oxygen (O) is usually assigned an oxidation state of -2, unless it is in a peroxide compound where it is assigned an oxidation state of -1.
Step 3:
Since there are three oxygen atoms in CrO3, the total oxidation state contribution from oxygen is -6 (-2 × 3 = -6).
Step 4:
The overall oxidation state of the compound is 0 (since the compound is neutral).
Step 5:
To determine the oxidation state of chromium, we subtract the oxidation state contribution from oxygen (-6) from the overall oxidation state (0).
0 - (-6) = +6
Therefore, the oxidation state of chromium in CrO3 is +6.
Conclusion:
The oxidation state of chromium in CrO3 is +6.
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Direction (Q. Nos. 27 and 28) This section contains 2question. when worked out will result in an integer from 0 to 9 (both inclusive)Q.In acidic medium,H2O2changes Cr2O72-to CrO5which has two (-O-O-) bonds. Oxidation state of Cr in CrO3isCorrect answer is '6'. Can you explain this answer?
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Direction (Q. Nos. 27 and 28) This section contains 2question. when worked out will result in an integer from 0 to 9 (both inclusive)Q.In acidic medium,H2O2changes Cr2O72-to CrO5which has two (-O-O-) bonds. Oxidation state of Cr in CrO3isCorrect answer is '6'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about Direction (Q. Nos. 27 and 28) This section contains 2question. when worked out will result in an integer from 0 to 9 (both inclusive)Q.In acidic medium,H2O2changes Cr2O72-to CrO5which has two (-O-O-) bonds. Oxidation state of Cr in CrO3isCorrect answer is '6'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Direction (Q. Nos. 27 and 28) This section contains 2question. when worked out will result in an integer from 0 to 9 (both inclusive)Q.In acidic medium,H2O2changes Cr2O72-to CrO5which has two (-O-O-) bonds. Oxidation state of Cr in CrO3isCorrect answer is '6'. Can you explain this answer?.
Direction (Q. Nos. 27 and 28) This section contains 2question. when worked out will result in an integer from 0 to 9 (both inclusive)Q.In acidic medium,H2O2changes Cr2O72-to CrO5which has two (-O-O-) bonds. Oxidation state of Cr in CrO3isCorrect answer is '6'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about Direction (Q. Nos. 27 and 28) This section contains 2question. when worked out will result in an integer from 0 to 9 (both inclusive)Q.In acidic medium,H2O2changes Cr2O72-to CrO5which has two (-O-O-) bonds. Oxidation state of Cr in CrO3isCorrect answer is '6'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Direction (Q. Nos. 27 and 28) This section contains 2question. when worked out will result in an integer from 0 to 9 (both inclusive)Q.In acidic medium,H2O2changes Cr2O72-to CrO5which has two (-O-O-) bonds. Oxidation state of Cr in CrO3isCorrect answer is '6'. Can you explain this answer?.
Solutions for Direction (Q. Nos. 27 and 28) This section contains 2question. when worked out will result in an integer from 0 to 9 (both inclusive)Q.In acidic medium,H2O2changes Cr2O72-to CrO5which has two (-O-O-) bonds. Oxidation state of Cr in CrO3isCorrect answer is '6'. Can you explain this answer? in English & in Hindi are available as part of our courses for Class 11.
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