Animated Solution for Chemistry - s and p-Block Elements: The oxidation states of 'P' in $H_4{P}_2{O}_7$, $H_4{P}_2{O}_5$ and $H_4{P}_2{O}_6$, respectively are

The problem asks us to determine the oxidation states of phosphorus in three different oxoacids: pyrophosphoric acid ($H_4{P}_2{O}_7$), pyrophosphorous acid ($H_4{P}_2{O}_5$), and hypophosphoric acid ($H_4{P}_2{O}_6$).

To solve this, we can rely on the fundamental principle of oxidation states. In any neutral molecule, the algebraic sum of the oxidation states of all its constituent atoms must be exactly zero.

For these compounds, we assign the standard oxidation state of $+1$ to hydrogen (since it is bonded to more electronegative non-metals) and $-2$ to oxygen (as there are no peroxide linkages here).

Let's start with the first molecule, $H_4{P}_2{O}_7$. We assume the oxidation state of the central phosphorus atom is $x$.

The molecule contains four hydrogen atoms, two phosphorus atoms, and seven oxygen atoms. Setting up our algebraic equation, we get:

Now, we simply solve for $x$. Expanding the terms gives us:

Simplifying further, we find $2x=10$, which beautifully yields:

Next, we move to $H_4{P}_2{O}_5$. Following the exact same logic, we set up the equation for this molecule.

With four hydrogens, two phosphorus atoms, and five oxygens, the equation is:

Expanding the terms, we get:

This simplifies to $2x=6$, giving us the oxidation state of phosphorus as:

Finally, let's tackle $H_4{P}_2{O}_6$. The setup remains consistent.

We have four hydrogens, two phosphorus atoms, and six oxygens. The equation becomes:

Expanding this gives:

Solving for $x$, we get $2x=8$, which means:

We have successfully found the oxidation states to be $+5$, $+3$, and $+4$. This perfectly matches option (c).

But what if we wanted to verify this visually? By drawing the Lewis structures of these acids, we can calculate the oxidation state by assigning bond electrons to the more electronegative atom.

For instance, in $H_4{P}_2{O}_6$, the $P-P$ bond contributes zero to the oxidation state because both atoms have identical electronegativity. The structural method is a powerful tool to double-check your algebraic results and avoid silly mistakes!