Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# 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. $+2$

Last updated date: 17th Sep 2024
Total views: 441.3k
Views today: 8.41k
Verified
441.3k+ 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.

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.