8
$\begingroup$

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?

$\endgroup$
4
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 Balagopal Apr 25 '17 at 10:50
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.