Math miscellaneousHard
Question
In which of the folowing compounds, nitrogen exhibits highest oxidation state ?
Options
A.N2H4
B.NH3
C.N3H
D.NH2OH
Solution
Let the oxidation state of nitrogen in the given compounds be x.]
2( x ) + ( +1 )4 = 0
2x = - 4
∴ x = - 2
x + ( +1 )3 = 0
∴ x = - 3
( x )3 + ( +1 ) = 0
3x = - 1
∴ x = - 1/3
x + ( + 1 )2 + ( - 2 ) + ( + 1 ) = 0
x + 2 - 2 + 1 = 0
x + 1 = 0
∴ x = - 1
Thus, oxidation state of nitrogen is highest in N3H.
2( x ) + ( +1 )4 = 0
2x = - 4
∴ x = - 2
x + ( +1 )3 = 0
∴ x = - 3
( x )3 + ( +1 ) = 0
3x = - 1
∴ x = - 1/3
x + ( + 1 )2 + ( - 2 ) + ( + 1 ) = 0
x + 2 - 2 + 1 = 0
x + 1 = 0
∴ x = - 1
Thus, oxidation state of nitrogen is highest in N3H.
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