Question

Match list – I (Compounds) with List –II (Oxidation states of Nitrogen) and select answers using the codes given below the lists.

List I	List II
a) $Na{N}_3$	(1)+5
b)${N}_2{H}_2$	(2)+2
c)$NO$	(3)$\dfrac{-1}{3}$
d)${N}_2{O}_5$	(4)-1

A. a – 3 b – 4 c – 2 d – 1
B. a – 4 b – 3 c – 2 d – 1
C. a – 3 b – 4 c – 1 d – 2
D. a – 4 b – 3 c – 1 d – 2

Answer 1

Oxidation number is also called an oxidation state and it is defined as the number of electrons gained or lost by an atom when it is going from its free state to the combined state with the atoms of other elements. Nitrogen can have many types of oxidation states depending upon the other atom which is attached with it to form the molecule.

Complete answer:
From your chemistry lessons you have learned about the oxidation state. Oxidation number also called an oxidation state is defined as the number of electrons gained or lost by an atom when it is going from its free state to the combined state with the atoms of other elements. The oxidation state of any atom can be negative, positive as well as zero. The substance which is present in its elemental state will have the oxidation state as zero.
If the oxidation number of an atom is increasing then it refers to the oxidation of the atom whereas if the oxidation number is decreasing then it refers to reduction of that atom. And the oxidation number of any compound will be equal to the overall charge that is present at the atom.
Now in the question we are asked to match the oxidation state of nitrogen in different molecules. So let us take each one by one:
a) $Na{{N}_{3}}$
Now to find the oxidation state of N we should know the Oxidation state of Na. And as we know that the oxidation state of Na is +1.Let us take the oxidation number of the N atom as 'x'.
$Oxidation\,state\,of\,Na+Oxidation\,state\,of\,N\times 3=0$
$\therefore +1+x\times 3=0$
$\therefore x=-\dfrac{1}{3}$
Therefore the oxidation state of N in $Na{{N}_{3}}$ is $-\dfrac{1}{3}$
b) ${{N}_{2}}{{H}_{2}}$
Here the oxidation state of hydrogen will be +1 and the oxidation state of N will be 'x'
$Oxidation\,state\,of\,N\times 2+Oxidation\,state\,of\,H\times 2=0$
$\therefore 2x+2=0$
$\therefore x=-1$
Therefore the oxidation state of N in${{N}_{2}}{{H}_{2}}$ is -1.
c) $NO$
As we know that there are variable oxidation states shown by the oxygen because the oxidation states of the p-block element varies. The oxidation state shown by oxygen in most of the cases are -2. Here the oxidation number of oxygen will be -2 and the oxidation state of N will be 'x'
$Oxidation\,state\,of\,N+Oxidation\,state\,of\,O=0$
$\therefore x-2=0$
$\therefore x=+2$
Therefore the oxidation state of N in NO is +2
d) ${{N}_{2}}{{O}_{5}}$
Here also the oxidation state of O will be -2 and the oxidation state of N will be 'x'.
$Oxidation\,state\,of\,N\times 2+Oxidation\,state\,of\,O\times 5=0$
$\therefore 2x-10=0$
$\therefore x=+5$
Therefore the oxidation state of N in ${{N}_{2}}{{O}_{5}}$ is +5.
Hence the correct match will be a – 3 b – 4 c – 2 d – 1

Thus the correct option will be (A).

Note:
The other oxidation states shown by oxygen is -2, -1, +1 and +2. In case oxygen is bonded to more electronegative atoms then the oxidation state shown by that oxygen will be positive because the atom with more electronegativity will tend to pull the electron towards themselves and thus acquires negative charge on them. The more electronegative atom other than oxygen is only fluorine.