The oxidation number of phosphorus in $P{{O}_{4}}^{3-}$, ${{P}_{4}}{{O}_{10}}$and ${{P}_{2}}{{O}_{7}}^{4-}$ is?
A. $+5$
B. $+3$
C. $-3$
D. $+2$
Answer
356.4k+ views
Hint: Take the oxidation number of phosphorus as “x” and try to solve for it using algebraic methods. Equate the oxidation numbers of atoms to the overall charge on the molecule.
Complete answer:
In all the compounds given above, the oxidation number of one oxygen atom is “$-2$”. This is because it is more electronegative than phosphorus and there are no peroxide linkages in the molecules given. That keeps only the oxidation number of phosphorus as unknown. We solve this by taking the oxidation number of one phosphorus atom as “x”. Below, the compounds are solved for one-by-one:
-$P{{O}_{4}}^{3-}$
There are four atoms of oxygen that makes the total oxidation number of oxygen to be “$-8$”. The overall charge of the molecules is $-3$, so the equation is as follows-
\[\begin{align}
& x-8=-3 \\
& \Rightarrow x=5 \\
\end{align}\]
So the oxidation number of phosphorus in this compound is “$+5$”.
- ${{P}_{4}}{{O}_{10}}$
There are ten oxygen atoms here which make their total oxidation number to be “$-20$”. The overall molecule is neutral and there are four phosphorus atoms, so the equation is as follows-
\[\begin{align}
& 4x-20=0 \\
& \Rightarrow x=5 \\
\end{align}\]
So the oxidation number of a phosphorus atom here is “$+5$”.
- ${{P}_{2}}{{O}_{7}}^{4-}$
There are seven oxygen atoms here, making their total oxidation number to be “$-14$”. The overall charge of the molecule is “$-4$” and there are two phosphorus atoms. The equation is as follows:
\[\begin{align}
& 2x-14=-4 \\
& \Rightarrow x=5 \\
\end{align}\]
Therefore the oxidation number of each phosphorus atom here is “$+5$”.
So, the correct answer is “Option A”.
Note: Do not forget to take the charge of a non-neutral molecule into consideration while forming the equations. If there are multiple phosphorus atoms, then multiply the variable with that number so as to get the oxidation number of one atom in solution.
Here the oxidation number of oxygen was always taken as “$-2$” because phosphorus is less electronegative, but if atoms like fluorine would have been present then the oxygen atom would have a positive oxidation state.
Complete answer:
In all the compounds given above, the oxidation number of one oxygen atom is “$-2$”. This is because it is more electronegative than phosphorus and there are no peroxide linkages in the molecules given. That keeps only the oxidation number of phosphorus as unknown. We solve this by taking the oxidation number of one phosphorus atom as “x”. Below, the compounds are solved for one-by-one:
-$P{{O}_{4}}^{3-}$
There are four atoms of oxygen that makes the total oxidation number of oxygen to be “$-8$”. The overall charge of the molecules is $-3$, so the equation is as follows-
\[\begin{align}
& x-8=-3 \\
& \Rightarrow x=5 \\
\end{align}\]
So the oxidation number of phosphorus in this compound is “$+5$”.
- ${{P}_{4}}{{O}_{10}}$
There are ten oxygen atoms here which make their total oxidation number to be “$-20$”. The overall molecule is neutral and there are four phosphorus atoms, so the equation is as follows-
\[\begin{align}
& 4x-20=0 \\
& \Rightarrow x=5 \\
\end{align}\]
So the oxidation number of a phosphorus atom here is “$+5$”.
- ${{P}_{2}}{{O}_{7}}^{4-}$
There are seven oxygen atoms here, making their total oxidation number to be “$-14$”. The overall charge of the molecule is “$-4$” and there are two phosphorus atoms. The equation is as follows:
\[\begin{align}
& 2x-14=-4 \\
& \Rightarrow x=5 \\
\end{align}\]
Therefore the oxidation number of each phosphorus atom here is “$+5$”.
So, the correct answer is “Option A”.
Note: Do not forget to take the charge of a non-neutral molecule into consideration while forming the equations. If there are multiple phosphorus atoms, then multiply the variable with that number so as to get the oxidation number of one atom in solution.
Here the oxidation number of oxygen was always taken as “$-2$” because phosphorus is less electronegative, but if atoms like fluorine would have been present then the oxygen atom would have a positive oxidation state.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are

Which of the following would not be a valid reason class 11 biology CBSE

Why should electric field lines never cross each other class 12 physics CBSE

An electrostatic field line is a continuous curve That class 12 physics CBSE

What is meant by monosporic development of female class 11 biology CBSE

Draw labelled diagram of the following i Gram seed class 11 biology CBSE

Trending doubts
The ray passing through the of the lens is not deviated class 10 physics CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Difference Between Plant Cell and Animal Cell

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

What is pollution? How many types of pollution? Define it

What is the nlx method How is it useful class 11 chemistry CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

What is the difference between anaerobic aerobic respiration class 10 biology CBSE
