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?


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.

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


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