Question

The oxidation state of nitrogen is highest in:
A. ${{N}_{3}}H$
B. $N{{H}_{3}}$
C. $N{{H}_{2}}OH$
D. ${{N}_{2}}{{H}_{4}}$

Answer 1

Hint: Oxidation state is also known by oxidation number and oxidation state can be defined as the number of electrons lost in an atom. Any chemical reaction which includes the movement of electrons is called oxidation.

Complete answer:
The substance which donates electrons during the oxidation process is known as oxidized substance. To determine in which compound nitrogen have greatest oxidation state we have to calculate the oxidation state of nitrogen in each atom which can be calculated as follows:
A.${{N}_{3}}H$: Oxidation state of hydrogen is always one and there is no charge present on the compound, so oxidation state of nitrogen is given by $3x+1=0$which represents $x=-1/3$
B. $N{{H}_{3}}$: In this case 3 hydrogen atoms are there so oxidation state can be calculated as $x+3=0$i.e. $x=-3$
C. $N{{H}_{2}}OH$: Oxidation state of hydroxide ion is taken as -1 so in this compound oxidation state of nitrogen is given by $x+2-1=0$i.e. $x=-1$
D. ${{N}_{2}}{{H}_{4}}$: In this case 2 nitrogen atoms and 4 hydrogen atoms are there by which oxidation state of nitrogen will be $2x+4=0$; $x=-2$
Out of all this we can conclude that -1/3 is the highest oxidation state.

Thus option A is the correct answer.

Note:
Oxidation state can be zero, negative or positive. In a chemical reaction if there is an increase in oxidation state then this process is known as oxidation and when there is decrease in oxidation state then the term is referred as reduction.