
Oxidation number of $N$ in ${N_3}H$ is:
A.$ + \dfrac{1}{3}$
B.$ - \dfrac{1}{3}$
C.$0$
D.$ + 3$
Answer
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Hint:Conceptually, the oxidation number , which can be positive, negative or zero, is that the hypothetical charge that an atom would have if all bonds to atom of various elements were $100\% $ ionic, with no covalent component. This is often never exactly true for real bonds.
Complete step by step answer:
We need to remember that the oxidation number, sometimes mentioned as oxidation state, describes the degree of oxidation (loss of electrons) of an atom during a compound. Antoine Lavoisier first used the term oxidation to indicate the reaction of a substance with oxygen.
Oxidation number of an element or compound is assigned by using the following rules.
Oxidation number for a free element equal to zero.
Oxidation number for monatomic ions all the time has the similar value as the net charge corresponding to the ion.
Hydrogen atoms have an oxidation state $ + 1$ . When a less electronegative atom is bonded then the oxidation number is $ - 1$ in some hydrides also.
Group $1$ elements (alkali metals) have an oxidation number $ + 1$ for their compounds and Group $2$ elements (alkaline earth metals) have an oxidation number $ + 2$ for their compounds.
If a compound contains halogens, the oxidation number of halogen is $ - 1$ .
For neutral compounds, oxidation number is the sum of all of the constituents atoms totals to $0$
Oxidation number is the sum of all of the constituent’s atoms totals to the net charge of the polyatomic ion when considering polyatomic ions.
Calculation of oxidation number of $N$ in ${N_3}H$.
${N_3}H$ is a neutral compound so the sum of all of the constituents atoms totals to $0$ . And the oxidation state of hydrogen in this compound is $ + 1$ .
Let $x$ be the oxidation number of nitrogen in ${N_3}H$ and this compound contains three nitrogen.
$3x + ( + 1) = 0$
$ \Rightarrow 3x = - 1$
$ \Rightarrow x = - \dfrac{1}{3}$
Oxidation number of $N$ in ${N_3}H$ is \[ - \dfrac{1}{3}\] .
Option B. $ - \dfrac{1}{3}$ is the correct answer.
Note:
We must have to know that the name of the compound \[{N_3}H\] is Hydrazoic acid (also known as azomite or hydrogen azide). \[{N_3}H\] is also sometimes written as \[H{N_3}\] . Nitrogen has various oxidation states ranging from $ - 3$ to $ + 5$ . Nitrogen outermost electron configuration is \[2{s^2}2{p^3}\] and contains five electrons. Nitrogen donates five electrons or accepts $3$ electrons to become a stable molecule.
Complete step by step answer:
We need to remember that the oxidation number, sometimes mentioned as oxidation state, describes the degree of oxidation (loss of electrons) of an atom during a compound. Antoine Lavoisier first used the term oxidation to indicate the reaction of a substance with oxygen.
Oxidation number of an element or compound is assigned by using the following rules.
Oxidation number for a free element equal to zero.
Oxidation number for monatomic ions all the time has the similar value as the net charge corresponding to the ion.
Hydrogen atoms have an oxidation state $ + 1$ . When a less electronegative atom is bonded then the oxidation number is $ - 1$ in some hydrides also.
Group $1$ elements (alkali metals) have an oxidation number $ + 1$ for their compounds and Group $2$ elements (alkaline earth metals) have an oxidation number $ + 2$ for their compounds.
If a compound contains halogens, the oxidation number of halogen is $ - 1$ .
For neutral compounds, oxidation number is the sum of all of the constituents atoms totals to $0$
Oxidation number is the sum of all of the constituent’s atoms totals to the net charge of the polyatomic ion when considering polyatomic ions.
Calculation of oxidation number of $N$ in ${N_3}H$.
${N_3}H$ is a neutral compound so the sum of all of the constituents atoms totals to $0$ . And the oxidation state of hydrogen in this compound is $ + 1$ .
Let $x$ be the oxidation number of nitrogen in ${N_3}H$ and this compound contains three nitrogen.
$3x + ( + 1) = 0$
$ \Rightarrow 3x = - 1$
$ \Rightarrow x = - \dfrac{1}{3}$
Oxidation number of $N$ in ${N_3}H$ is \[ - \dfrac{1}{3}\] .
Option B. $ - \dfrac{1}{3}$ is the correct answer.
Note:
We must have to know that the name of the compound \[{N_3}H\] is Hydrazoic acid (also known as azomite or hydrogen azide). \[{N_3}H\] is also sometimes written as \[H{N_3}\] . Nitrogen has various oxidation states ranging from $ - 3$ to $ + 5$ . Nitrogen outermost electron configuration is \[2{s^2}2{p^3}\] and contains five electrons. Nitrogen donates five electrons or accepts $3$ electrons to become a stable molecule.
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