Question

The oxidation state of nitrogen in:
a.) ${{N}_{3}}H$
b.) $N{{H}_{3}}$
c.) $N{{H}_{2}}OH$
d.) ${{N}_{2}}{{H}_{4}}$

Answer 1

Nitrogen compounds exhibit oxidation states ranging from −3 (ammonia or amines) to +5 (nitric acid).
The electronic configuration of nitrogen is:$1{{s}^{2}}2{{s}^{2}}2{{p}^{3}}$
Nitrogen has 5 outer valence electrons, and the three electrons in 2p orbital are unpaired. Nitrogen can gain three electrons (into 2p orbital) or lose all the 5 valence electrons (from 2s and 2p orbitals) to obtain stable configuration.

Complete answer:
We have to calculate the oxidation states of nitrogen in the given compounds.
The given compounds are
${{N}_{3}}H$- Hydrazoic acid.
$N{{H}_{3}}$- Ammonia
$N{{H}_{2}}OH$- Hydroxylamine
${{N}_{2}}{{H}_{4}}$- Hydrazine
The oxidation state of nitrogen in ${{N}_{3}}H$ is as follows.
There are three nitrogen atoms and one hydrogen atom. So,
3x+1 = 0
x = $-\dfrac{1}{3}$
The oxidation state of nitrogen in ${{N}_{3}}H$is $-\dfrac{1}{3}$.
The oxidation state of nitrogen in $N{{H}_{3}}$ is as follows.
There is one Nitrogen atom and three hydrogen atoms. So,
x+3(1) = 0
x+3 = 0
x = -3
The oxidation state of nitrogen in $N{{H}_{3}}$is -3.
The oxidation state of nitrogen in $N{{H}_{2}}OH$ is as follows.
The structure of $N{{H}_{2}}OH$ is

We know that there is one hydrogen atom, one oxygen atom and two types of hydrogen atoms.
One type of hydrogen atom (two hydrogens) directly attached to nitrogen and the second type of hydrogen attached to oxygen atom.
x + 3 (+1) – 2 = 0
x = -1
The oxidation state of nitrogen in $N{{H}_{2}}OH$is -1.
The oxidation state of nitrogen in ${{N}_{2}}{{H}_{4}}$is as follows.
There are two Nitrogen atoms and four hydrogen atoms.
2x + 4(1) = 0
2x = - 4
x = -2
The oxidation state of nitrogen in ${{N}_{2}}{{H}_{4}}$ is -2.

Note: The various oxidation states of nitrogen are due to the presence of 5 valence electrons. It can donate 5 electrons or it can accept three electrons from other atoms to form stable molecules.