Question
Question: The oxidation number of phosphorus in \(P{{O}_{4}}^{3-}\), \({{P}_{4}}{{O}_{10}}\)and \({{P}_{2}}{{O...
The oxidation number of phosphorus in PO43−, P4O10and P2O74− is?
A. +5
B. +3
C. −3
D. +2
Solution
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:
-PO43−
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-
x−8=−3⇒x=5
So the oxidation number of phosphorus in this compound is “+5”.
- P4O10
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-