# How do P-H bonds in phosphorus acids impart reducing properties?

My textbook states:

The oxoacids of Phosphorous which contain $$\ce{P-H}$$ bond have strong reducing properties. Thus, hypophosphorous acid is a good reducing agent as it contains two $$\ce{P-H}$$ bonds and reduces, for example, $$\ce{AgNO3}$$ to metallic silver.

These $$\ce{P-H}$$ bonds do not ionize to give $$\ce{H+}$$ and do not play any role in basicity.

On the basis of what's given in the text; how do these $$\ce{P-H}$$ bonds account for the reducing properties of phosphorus acids?

## 2 Answers

The $\ce{P-H}$ bond imparts an oxidation state of -1 per bond to the acid. Upon oxidation, these are converted into $\ce{P-OH}$ bonds, which imparts a +1 oxidation state per bond. You can clearly see this in the reaction: $$\ce{H3PO2 + 4AgNO3 + 2H2O -> 4Ag + H3PO4 + 4HNO3}$$ There are two $\ce{P-H}$ bonds in $\ce{H3PO2}$, both of which are converted into $\ce{P-OH}$ bonds in $\ce{H3PO4}$.

Moreover the $\ce{P-H}$ bond has an enthalpy of 322kJ, while the $\ce{P-O}$ bond has an enthalpy of 335kJ/mol. This slight increase in bond energies drives the reaction in the forward direction. Resonances may also be a factor that contributes as well.

• Answer seems convincing. So, all it has to do with the $\ce{P-H}$ bonds is with comparision with the product, right? – Reeshabh Ranjan Apr 25 '17 at 5:01
• Yes, that right @Reeshab Ranjan – Pritt says Reinstate Monica Apr 25 '17 at 10:50

First of all, the electronegativity of hydrogen (2.20) is greater than phosphorous (2.19), so in the P-H bond, the P atom actually gets a +1 oxidation state, whereas hydrogen gets a -1 oxidation state.

It is a well known fact that negative oxidation state hydrogens are highly reducing in nature; hence, the H atoms attached to the phosphorous atom contribute to the high reducing nature of phosphorous acids having P-H bonds.