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How many electrons should X(2)H(4) liber...

How many electrons should `X_(2)H_(4)` liberate so that in the new compound X shows oxidation number of `-1//2`(E.N. X gt H)

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To solve the problem, we need to determine how many electrons the compound \(X_2H_4\) must liberate for the element \(X\) to achieve an oxidation state of \(-\frac{1}{2}\). ### Step-by-Step Solution: 1. **Identify the Oxidation States:** - In the compound \(X_2H_4\), we need to find the oxidation state of \(X\). - The oxidation state of hydrogen (H) is \(+1\). - The total contribution of hydrogen in \(X_2H_4\) is \(4 \times (+1) = +4\). ...
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2 mole of N_(2)H_(4) loses 16 mole of electron is being converted to a new compound X . Assuming that all of the N appears in the new compound. What is the oxidation state of 'N' in X ?

2 mole of N_2H_4 loses 16 mole of electron is beings converted to a new compound X. Assuming that all of the N appears in the new compound. What is the oxidation state of 'N' in X ?

Knowledge Check

  • In electrolysis of dilute H_(2)SO_(4) what is liberated at anode?

    A
    `H_2`
    B
    `SO_4^(2-)`
    C
    `SO_2`
    D
    `O_2`

  • One mole of N_(2)H_(4) loses ten moles of electrons to form a new compound A . Assuming that all the nitrogen appears in the new compound, what is the oxidation state of nitrogen in A ? (There is no change in the oxidation state of hydrogen.)

    A
    `-1`
    B
    `-3`
    C
    `+3`
    D
    `+5`

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    How many isomers of molecular formula C_(3)H_(9)N will liberate N_(2) gas on treatment with HNO_(2) ?

    How many atoms are present in one mole of H_(2)SO_(4) ?

    How many gram equivalents of H_(2)SO_(4) are present in 200 ml of (N)/(10) H_(2)SO_(4) solution?

    How many grams of H_(2)SO_(4) are present in 500ml of 0.2M H_(2)SO_(4) solution ? .

    How many grams of H_(2)SO_(4) are present in 400ml of 0.2M H_(2)SO_(4) solution?

    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 highest?

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