Indicate the oxidation number of nitrogen in each of the following compounds.(a) NH3(b) N2H4(c) NF3(d) NH2OH(e)…
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Question “Indicate the oxidation number of nitrogen in each of the following compounds.(a) NH3(b) N2H4(c) NF3(d) NH2OH(e)…”
Indicate the oxidation number of nitrogen in each of the following compounds.
(a) NH3
(b) N2H4
(c) NF3
(d) NH2OH
(e) Fe(NO3)3
(f) HNO2
Answer
The charge assigned to an element in a compound’s compound is known as an oxidation number.
These are the rules for assigning an element’s oxidation number.
-NH3(b)-N2H4(c)-NF3(d)-NH2OH(e).png?x-oss-process=image/resize,w_560/format,webp)
is the oxidation number for an uncombined element/molecule.
2. Monoatomic ions have an oxidation number equal to their ionic charges.
3. The oxidation number of the higher electronegative element in a molecule is equal to its charge when it was a monoatomic element.
4. Hydrogen’s oxidation number
-NH3(b)-N2H4(c)-NF3(d)-NH2OH(e).png?x-oss-process=image/resize,w_560/format,webp)
, but it can be combined with metal to get
-NH3(b)-N2H4(c)-NF3(d)-NH2OH(e).png?x-oss-process=image/resize,w_560/format,webp)
.
5. The
-NH3(b)-N2H4(c)-NF3(d)-NH2OH(e).png?x-oss-process=image/resize,w_560/format,webp)
oxidation number for oxygen is
-NH3(b)-N2H4(c)-NF3(d)-NH2OH(e).png?x-oss-process=image/resize,w_560/format,webp)
when the gas forms an oxide; its oxidation number is #media_tag_4$ if it forms peroxide $55$
6. The sum of all elements in molecules should equal zero, and for ionic substances should equal its ionic charges. Below is the rule to determine the oxidation status of an element within a molecule.
-NH3(b)-N2H4(c)-NF3(d)-NH2OH(e).png?x-oss-process=image/resize,w_560/format,webp)
…… (1)
(a)
-NH3(b)-N2H4(c)-NF3(d)-NH2OH(e).png?x-oss-process=image/resize,w_560/format,webp)
is a mixture of nitrogen and hydrogen. Because there is no net charge, the molecule is neutral. The nitrogen is part of group V
-NH3(b)-N2H4(c)-NF3(d)-NH2OH(e).png?x-oss-process=image/resize,w_560/format,webp)
. The electron-negative nature of nitrogen makes it more reactive than hydrogen. Therefore, the nitrogen will be lower in oxidation.
-NH3(b)-N2H4(c)-NF3(d)-NH2OH(e).png?x-oss-process=image/resize,w_560/format,webp)
has three hydrogens. The total positive oxidation number is as follows:
-NH3(b)-N2H4(c)-NF3(d)-NH2OH(e).png?x-oss-process=image/resize,w_560/format,webp)
To find the total negative oxygenation number of ammonia, substitute
-NH3(b)-N2H4(c)-NF3(d)-NH2OH(e).png?x-oss-process=image/resize,w_560/format,webp)
and
-NH3(b)-N2H4(c)-NF3(d)-NH2OH(e).png?x-oss-process=image/resize,w_560/format,webp)
-NH3(b)-N2H4(c)-NF3(d)-NH2OH(e).png?x-oss-process=image/resize,w_560/format,webp)
-NH3(b)-N2H4(c)-NF3(d)-NH2OH(e).png?x-oss-process=image/resize,w_560/format,webp)
contains one nitrogen atom. Below is how the oxidation number for nitrogen is calculated:
-NH3(b)-N2H4(c)-NF3(d)-NH2OH(e).png?x-oss-process=image/resize,w_560/format,webp)
-NH3(b)-N2H4(c)-NF3(d)-NH2OH(e).png?x-oss-process=image/resize,w_560/format,webp)
has a nitrogen oxidation value of
-NH3(b)-N2H4(c)-NF3(d)-NH2OH(e).png?x-oss-process=image/resize,w_560/format,webp)
.
(b)
-NH3(b)-N2H4(c)-NF3(d)-NH2OH(e).png?x-oss-process=image/resize,w_560/format,webp)
is a mixture of nitrogen and hydrogen. Because there is no net charge, the molecule is neutral. The nitrogen is part of group V
-NH3(b)-N2H4(c)-NF3(d)-NH2OH(e).png?x-oss-process=image/resize,w_560/format,webp)
. The electron-negative nature of nitrogen makes it more effective than hydrogen. Therefore, the nitrogen will be less oxidized.
-NH3(b)-N2H4(c)-NF3(d)-NH2OH(e).png?x-oss-process=image/resize,w_560/format,webp)
has four hydrogens. The total positive oxidation number is as follows:
-NH3(b)-N2H4(c)-NF3(d)-NH2OH(e).png?x-oss-process=image/resize,w_560/format,webp)
To find the total negative oxygenation number for hydrazine, substitute
-NH3(b)-N2H4(c)-NF3(d)-NH2OH(e).png?x-oss-process=image/resize,w_560/format,webp)
and
-NH3(b)-N2H4(c)-NF3(d)-NH2OH(e).png?x-oss-process=image/resize,w_560/format,webp)
-NH3(b)-N2H4(c)-NF3(d)-NH2OH(e).png?x-oss-process=image/resize,w_560/format,webp)
-NH3(b)-N2H4(c)-NF3(d)-NH2OH(e).png?x-oss-process=image/resize,w_560/format,webp)
-NH3(b)-N2H4(c)-NF3(d)-NH2OH(e).png?x-oss-process=image/resize,w_560/format,webp)
contains two nitrogen atoms. Below is how the oxidation number for nitrogen is calculated:
-NH3(b)-N2H4(c)-NF3(d)-NH2OH(e).png?x-oss-process=image/resize,w_560/format,webp)
-NH3(b)-N2H4(c)-NF3(d)-NH2OH(e).png?x-oss-process=image/resize,w_560/format,webp)
has a nitrogen oxidation value of
-NH3(b)-N2H4(c)-NF3(d)-NH2OH(e).png?x-oss-process=image/resize,w_560/format,webp)
.
(c)
The Nitrogen trifluoride
-NH3(b)-N2H4(c)-NF3(d)-NH2OH(e).png?x-oss-process=image/resize,w_560/format,webp)
has nitrogen and fluorine. Because there is no net charge, the molecule is neutral. Fluorine belongs in group VII
-NH3(b)-N2H4(c)-NF3(d)-NH2OH(e).png?x-oss-process=image/resize,w_560/format,webp)
while nitrogen is in group V . The oxidation number for fluorine, on the other hand, is
-NH3(b)-N2H4(c)-NF3(d)-NH2OH(e).png?x-oss-process=image/resize,w_560/format,webp)
. The oxidation rate of nitrogen is lower than that of fluorine.
-NH3(b)-N2H4(c)-NF3(d)-NH2OH(e).png?x-oss-process=image/resize,w_560/format,webp)
has three fluorine molecules. The total amount of negative oxidation will be shown as follows:
-NH3(b)-N2H4(c)-NF3(d)-NH2OH(e).png?x-oss-process=image/resize,w_560/format,webp)
To find the total positive oxygenation number of #media_tag_40$, substitute
-NH3(b)-N2H4(c)-NF3(d)-NH2OH(e).png?x-oss-process=image/resize,w_560/format,webp)
and @media_tag_38$ in equation $media_39$, as shown below:
-NH3(b)-N2H4(c)-NF3(d)-NH2OH(e).png?x-oss-process=image/resize,w_560/format,webp)
-NH3(b)-N2H4(c)-NF3(d)-NH2OH(e).png?x-oss-process=image/resize,w_560/format,webp)
contains one nitrogen atom. Below is how the oxidation number for nitrogen is calculated:
-NH3(b)-N2H4(c)-NF3(d)-NH2OH(e).png?x-oss-process=image/resize,w_560/format,webp)
-NH3(b)-N2H4(c)-NF3(d)-NH2OH(e).png?x-oss-process=image/resize,w_560/format,webp)
oxidation number for nitrogen is
-NH3(b)-N2H4(c)-NF3(d)-NH2OH(e).png?x-oss-process=image/resize,w_560/format,webp)
.
(d)
-NH3(b)-N2H4(c)-NF3(d)-NH2OH(e).png?x-oss-process=image/resize,w_560/format,webp)
is composed of nitrogen, hydrogen and oxygen. Because there is no net charge, the molecule is neutral. The oxygen belongs in group VI
-NH3(b)-N2H4(c)-NF3(d)-NH2OH(e).png?x-oss-process=image/resize,w_560/format,webp)
.
-NH3(b)-N2H4(c)-NF3(d)-NH2OH(e).png?x-oss-process=image/resize,w_560/format,webp)
has three hydrogen atoms. The total positive oxidation number is as follows:
-NH3(b)-N2H4(c)-NF3(d)-NH2OH(e).png?x-oss-process=image/resize,w_560/format,webp)
To find the total negative oxygenation number of #media_tag_54$, substitute
-NH3(b)-N2H4(c)-NF3(d)-NH2OH(e).png?x-oss-process=image/resize,w_560/format,webp)
and @media_tag_52$ in equation $media_53$, as shown below:
-NH3(b)-N2H4(c)-NF3(d)-NH2OH(e).png?x-oss-process=image/resize,w_560/format,webp)
-NH3(b)-N2H4(c)-NF3(d)-NH2OH(e).png?x-oss-process=image/resize,w_560/format,webp)
contains one nitrogen atom, and one oxygen atom. The sum of the oxidation numbers of nitrogen and oxygen will equal the total negative oxidation amount. As shown below, the oxidation number for nitrogen is calculated.
-NH3(b)-N2H4(c)-NF3(d)-NH2OH(e).png?x-oss-process=image/resize,w_560/format,webp)
-NH3(b)-N2H4(c)-NF3(d)-NH2OH(e).png?x-oss-process=image/resize,w_560/format,webp)
oxidation number for nitrogen is
-NH3(b)-N2H4(c)-NF3(d)-NH2OH(e).png?x-oss-process=image/resize,w_560/format,webp)
.
(e)
Iron
-NH3(b)-N2H4(c)-NF3(d)-NH2OH(e).png?x-oss-process=image/resize,w_560/format,webp)
Nitrate $61$ Ionic compounds that contain metal ion or polyatomic anion are iron
-NH3(b)-N2H4(c)-NF3(d)-NH2OH(e).png?x-oss-process=image/resize,w_560/format,webp)
Iron is the metal, while nitrate is a polyatomic ion. Because there is no net charge in the molecule, the molecule is neutral.
-NH3(b)-N2H4(c)-NF3(d)-NH2OH(e).png?x-oss-process=image/resize,w_560/format,webp)
is the charge of the nitrate Ion. Iron will have a positive oxidation number.
The nitrate is an ion that contains nitrogen and oxygen. It is ionic because it has
-NH3(b)-N2H4(c)-NF3(d)-NH2OH(e).png?x-oss-process=image/resize,w_560/format,webp)
cost. The oxygen is part of group VI
-NH3(b)-N2H4(c)-NF3(d)-NH2OH(e).png?x-oss-process=image/resize,w_560/format,webp)
. The total negative oxidation amount of nitrate will be shown as follows:
-NH3(b)-N2H4(c)-NF3(d)-NH2OH(e).png?x-oss-process=image/resize,w_560/format,webp)
To find the total positive oxygenation number for nitrate ion, substitute
-NH3(b)-N2H4(c)-NF3(d)-NH2OH(e).png?x-oss-process=image/resize,w_560/format,webp)
and #media_tag_68$ in equation
-NH3(b)-N2H4(c)-NF3(d)-NH2OH(e).png?x-oss-process=image/resize,w_560/format,webp)
-NH3(b)-N2H4(c)-NF3(d)-NH2OH(e).png?x-oss-process=image/resize,w_560/format,webp)
Therefore,
-NH3(b)-N2H4(c)-NF3(d)-NH2OH(e).png?x-oss-process=image/resize,w_560/format,webp)
has an N oxidation number.
(f)
-NH3(b)-N2H4(c)-NF3(d)-NH2OH(e).png?x-oss-process=image/resize,w_560/format,webp)
is composed of nitrogen, hydrogen, and oxygen. Because there is no net charge, the molecule is neutral. The oxygen belongs in group VI
-NH3(b)-N2H4(c)-NF3(d)-NH2OH(e).png?x-oss-process=image/resize,w_560/format,webp)
.
-NH3(b)-N2H4(c)-NF3(d)-NH2OH(e).png?x-oss-process=image/resize,w_560/format,webp)
has one hydrogen atom. The total positive oxidation number is as follows:
-NH3(b)-N2H4(c)-NF3(d)-NH2OH(e).png?x-oss-process=image/resize,w_560/format,webp)
To find the total negative oxygenation number of #media_tag_81$, substitute
-NH3(b)-N2H4(c)-NF3(d)-NH2OH(e).png?x-oss-process=image/resize,w_560/format,webp)
and @media_tag_79$ in equation $media_80$, as shown below:
-NH3(b)-N2H4(c)-NF3(d)-NH2OH(e).png?x-oss-process=image/resize,w_560/format,webp)
-NH3(b)-N2H4(c)-NF3(d)-NH2OH(e).png?x-oss-process=image/resize,w_560/format,webp)
contains one nitrogen atom, and two oxygen atoms. The sum of the oxidation numbers of nitrogen and oxygen will equal the total negative oxidation amount. As shown below, the oxidation number for nitrogen is calculated.
-NH3(b)-N2H4(c)-NF3(d)-NH2OH(e).png?x-oss-process=image/resize,w_560/format,webp)
-NH3(b)-N2H4(c)-NF3(d)-NH2OH(e).png?x-oss-process=image/resize,w_560/format,webp)
oxidation value of nitrogen is
-NH3(b)-N2H4(c)-NF3(d)-NH2OH(e).png?x-oss-process=image/resize,w_560/format,webp)
Conclusion
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