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Valency and Oxidation number are differe...

Valency and Oxidation number are different for an element. Valency of carbon is generally 4, however, the oxidation state may be `-4,-2 0,+2,-1` etc. In the compounds containing carbon. hydrogen and oxygen the oxidation number of carbon can be calculated as
`"Oxidation number of carbon"=(2n_(O)-n_(H))/(n_(C)`
Where, `n_(H), n_(O) and n_(C)` are number of respective atoms
Which of the following compounds have zero oxidation state at carbon

A

`C_(6)H_(12)O_(6)`

B

`HCOOH`

C

HCHO

D

`CH_(4)`

Text Solution

Verified by Experts

The correct Answer is:
A, C
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The valancy of carbons is generally 4 , but its oxidation state may be -4, -2, 0, +2, -1 , etc. In the compounds containing C, H , and O , the oxidation number of C is calculated as Oxidation number of C= (2n_(O)-n_(H))/(n_(C )) Where n_(O), n_(H) and n_(C ) are the numbers of oxygen, hydrogen, and carbons, atoms, respectively. In which of the following compounds is the oxidation state of carbon is zero?

The valancy of carbons is generally 4 , but its oxidation state may be -4, -2, 0, +2, -1 , etc. In the compounds containing C, H , and O , the oxidation number of C is calculated as Oxidation number of C= (2n_(O)-n_(H))/(n_(C )) Where n_(O), n_(H) and n_(C ) are the numbers of oxygen, hydrogen, and carbons, atoms, respectively. In which of the following compounds is the oxidation state of C a fraction?

Knowledge Check

  • Valency and Oxidation number are different for an element. Valency of carbon is generally 4, however, the oxidation state may be -4,-2 0,+2,-1 etc. In the compounds containing carbon. hydrogen and oxygen the oxidation number of carbon can be calculated as "Oxidation number of carbon"=(2n_(O)-n_(H))/(n_(C) Where, n_(H), n_(O) and n_(C) are number of respective atoms Which of the following oxides of carbon has fractional oxidation state?

    A
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    B
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    C
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    D
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    A
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    B
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  • Valency and Oxidation number are different for an element. Valency of carbon is generally 4, however, the oxidation state may be -4,-2 0,+2,-1 etc. In the compounds containing carbon. hydrogen and oxygen the oxidation number of carbon can be calculated as "Oxidation number of carbon"=(2n_(O)-n_(H))/(n_(C) Where, n_(H), n_(O) and n_(C) are number of respective atoms Oxidation state of carbon in diamond is:

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    B
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    C
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    D
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