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One mole of hydrazine (N(2)H(4)) loses 1...

One mole of hydrazine `(N_(2)H_(4))` loses 10 moles of electrons in a reaction to form a new compound X. Assuming that all the nitrogen atoms in hydrazine appear in the new compound, what is the oxidation state of nitrogen in X ? (Note There is no change in the oxidation state of hydrogen in the reaction)

A

`-1`

B

`-3`

C

`+3`

D

`+5`

Text Solution

Verified by Experts

The correct Answer is:
C
Given, 1 mol of `H_(2)N - NH_(2)` (hydrazine), it loses 10 moles of electron to form a new compound that contains both the N-atoms with same oxidation number means :
`N_(2)H_(4) rarr 10 e^(-1) + X` (product)
(Oxidation number of N-atom in `N_(2)H_(4) = -2`)
`because` Oxidation number of both the N-atoms are same.
New total oxidaiton number of new compound (X)
`= 4 - 10 + x = 0` (due to 4 H -atoms)
`therefore x = +6`
`therefore` Each N-atom has oxidation state = +3
Alternate method
`underset((-2))(N_(2)H_(4)) rarr X`
Number of electrons post per N-atom
`therefore` New oxidaiton state of N in X
= -2 +5 + x = 0
x = +3
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