Question

Find the oxidation number of nitrogen in each case:
$N$ in$N{H_3}$ , $N{H_2}OH$ , $Na{N_3}$ and $N{H_4}N{O_2}$ .
A. $- 3, - 1,\dfrac{{ - 1}}{3}, + 3,- 3$
B. $- 3, - 2,\dfrac{{ - 1}}{3}, + 1$
C. $- 3, - 1,1, + 2$
D. None of the above

Answer 1

Oxidation number is the charge on an atom which appears due to formation of bonds with the neighboring atoms or ligands. It is denoted by $0$, positive or negative sign.

Complete step by step answer:
Oxidation number is also termed as oxidation state is equal to the total number of electrons which an atom either gains or loses during the formation of a chemical bond with another atom or molecule.
In order to determine the oxidation number of any atom the number of bonds both ionic and covalent is to be determined first. The number of bonds decides the number of electrons shared by the considered atom in its ionic form.
The steps involved in assigning the oxidation number or oxidation state of an atom or ion in a compound:
i) Add up the constant oxidation state of other atoms/molecules/ions that are bonded with the central metal ion
ii) The total oxidation state of a molecule or ion is equal to the total charge on the molecule or ion.
$N{H_3}$ is a neutral molecule. Let $x$ is the oxidation number of $N$ in $N{H_3}$​. The three hydrogens attached to nitrogen carry unit positive charge. Thus the equation becomes:
$x + 3( + 1) = 0$
$x = - 3$.
$N{H_2}OH$ is a neutral molecule. Let $x$ is the oxidation number of $N$ in $N{H_2}OH$​. The $N$ is attached to two hydrogens carrying unit positive charge and one hydroxyl group carrying unit negative charge. Thus the equation becomes:
$x + 2( + 1) + ( - 1) = 0$
$x = - 1$.
$Na{N_3}$ is a neutral molecule. Let $x$ is the oxidation number of $N$ in $Na{N_3}$​. The $N$ is attached to two sodium ions carrying unit positive charge. Thus the equation becomes:
$+ 1 + 3x = 0$
$x = - \dfrac{1}{3}$
$N{H_4}N{O_2}$ is a neutral molecule. In this case two nitrogen atoms are present. One is as $N{H_4}^ +$ ion and the other is $N{O_2}^ -$ ion. $N{H_4}^ +$ ion is a positively charged molecule. Let $x$ is the oxidation number of $N$ in $N{H_4}^ +$ ion. In $N{H_4}^ +$ ion the $N$ is attached to four hydrogen ions carrying unit positive charge. Thus the equation becomes:
$x + 4( + 1) = + 1$
$x = - 3$
$N{O_2}^ -$ ion is a negatively charged molecule. Let $x$ is the oxidation number of $N$ in $N{O_2}^ -$ ion. In $N{O_2}^ -$ ion the $N$ is attached to two oxygen ion carrying two negative charges. Thus the equation becomes:
$x + 2( - 2) = - 1$
$x = + 3$ .
Hence option A is the correct answer, i.e. $- 3, - 1,\dfrac{{ - 1}}{3}, + 3, - 3$.

Note: In compounds the more electronegative atoms gain electrons from a less electronegative atom and possess negative oxidation states. The numerical value of the oxidation state is equal to change in the number of electrons. The elements in the periodic table vary in their oxidation number or state according to the number of valence shell electrons present. The octet rule must be satisfied during the formation of bonds.