Question

Which ordering of compounds is according to the decreasing order of the oxidation state of nitrogen?
A. $HN{{O}_{3}},NO,N{{H}_{4}}Cl,{{N}_{2}}$
B. $HN{{O}_{3}},NO,{{N}_{2}},N{{H}_{4}}Cl$
C. $HN{{O}_{3}},N{{H}_{4}}Cl,NO,{{N}_{2}}$
D. $NO,HN{{O}_{3}},N{{H}_{4}}Cl,{{N}_{2}}$

Answer 1

Hint: Oxidation state generally tells us loss of electrons of an atom in a chemical compound. It may also be called an oxidation number and it can be positive, negative or zero.

Complete step by step answer:
Oxidation state describes the degree of oxidation i.e. loss of electrons. Increase in oxidation state of an atom through a chemical reaction is known as an oxidation and a decrease in oxidation state is known as a reduction. Such reactions involve the formal transfer of electrons where gain in electrons being a reduction and loss of electrons being an oxidation. For pure elements the oxidation state is zero.
Nitrogen is the compound represented by N having atomic number 7. Its electronic configuration can be written as $1{{s}^{2}}2{{s}^{2}}2{{p}^{3}}$. At standard temperature and pressure nitrogen combines with other nitrogen atoms to form dinitrogen which is a colorless and odorless gas with the formula \[{{N}_{2}}\]. Dinitrogen forms about 78% of Earth's atmosphere and is the most abundant uncombined element. Nitrogen occurs in all organisms in amino acids, proteins, nucleic acids (DNA and RNA) and in the energy transfer molecule adenosine triphosphate.

Oxidation state of oxygen is -2 and for hydrogen it is +1 while chlorine is halogen which have -1 .Hence the oxidation state in different compounds is as follows:
$HN{{O}_{3}}=1+x+2(-3);1+x-6;x-5;x=5$
$NO=x-2;x=2$
${{N}_{2}}=0$
$N{{H}_{4}}Cl=x+4-2;x+2;x=-2$
\[HN{{O}_{3}}\] = + 5; \[NO\] = +2; \[{{N}_{2}}\]= 0; \[N{{H}_{4}}Cl\]= -3
So, the correct answer is “Option B”.

Note: All alkali metals, alkaline earth metals and halogens have an oxidation state of +1, +2 and -1, respectively in their compounds.