Question

One mole of \[{{N}_{2}}{{H}_{4}}\] loses 10 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

Answer 1

Hint: Start by writing the chemical reaction for the given reaction. Then, solve this question by summing the oxidation number of each element and equating it to 10. Assign a random variable ‘x’ for calculating oxidation number of N.

Complete step by step answer:
According to the question,
We can write it in the form of an equation as –
\[{{N}_{2}}{{H}_{4}}\to A+10{{e}^{-}}\]
Given, there is no change in oxidation state of hydrogen. Oxidation state of hydrogen is +1.
Let the oxidation state of Nitrogen be ‘x’.
We can say that the oxidation state of compound A = +10
Therefore,
2(x) + 4(+1) = +10
2x = 10 – 4 = 6
x = +3
Therefore, the answer is – option (c) – The oxidation state of nitrogen in the compound ‘A’ is +3.

Additional Information:
The oxidation number of an atom or element is a number that indicates the total number of electrons lost or gained by it.

Note: There are certain rules for calculating oxidation number –
Oxidation number of any free element is zero.
Oxidation number for a monoatomic ion is equal to the net charge on it.
Hydrogen, in general, has an oxidation state equal to +1. However, it is -1 in forms of a compound with an element with lesser electronegativity.
Alkali metals have oxidation number = +1
Alkali earth metals have oxidation number = +2
Halogens have oxidation number = -1
Oxygen has an oxidation number = -2, but in case of peroxides, it is -1.