The correct statement(s) related to oxoacids of phosphorous is(are)
(A) Upon heating, H3PO3 undergoe

Q: Why doesn't the hydrogen attached directly to phosphorus ionize in water?

Ionization in water requires a highly polar bond. The electronegativity of phosphorus (2.19) and hydrogen (2.20) are almost identical. This makes the P-H bond non-polar, so water molecules cannot pull the hydrogen atom away as an $H^+$ ion.

Q: How can I quickly determine the basicity of any phosphorus oxoacid?

The most reliable way is to draw the structure and count the number of P-OH groups. A quick trick for phosphorus oxoacids of the formula $H_3PO_x$ is that the basicity is usually $(x - 1)$. For example, $H_3PO_4$ is tribasic (4-1=3), $H_3PO_3$ is dibasic (3-1=2), and $H_3PO_2$ is monobasic (2-1=1).

Q: What exactly is a disproportionation reaction?

It is a specific type of redox reaction where a single substance acts as both the oxidizing and reducing agent. The central element in an intermediate oxidation state simultaneously oxidizes to a higher state and reduces to a lower state.

Q: Why can't $H_3PO_4$ act as a reducing agent?

To act as a reducing agent, a molecule must be able to undergo oxidation (lose electrons). Phosphorus is in Group 15 and has 5 valence electrons. In $H_3PO_4$, its oxidation state is +5, meaning it has already 'lost' all 5 valence electrons. It cannot be oxidized any further.

Q: Is $H_3PO_2$ a reducing agent?

Yes, it is a very strong reducing agent! In hypophosphorous acid ($H_3PO_2$), the oxidation state of phosphorus is +1. It can easily be oxidized to +3 or +5, making it an excellent reducing agent.

Question

The correct statement(s) related to oxoacids of phosphorous is(are)

Options:

(A) Upon heating, $	ext{H}_3	ext{PO}_3$ undergoes disproportionation reaction to produce $	ext{H}_3	ext{PO}_4$ and $	ext{PH}_3$.

(B) While $	ext{H}_3	ext{PO}_3$ can act as reducing agent, $	ext{H}_3	ext{PO}_4$ cannot.

(C) $	ext{H}_3	ext{PO}_3$ is a monobasic acid.

(D) The H atom of P–H bond in $	ext{H}_3	ext{PO}_3$ is not ionizable in water.

Sigma Percentile · Accepted Answer

The chemistry of phosphorus oxoacids is a fascinating journey into molecular structure and redox behavior. To truly master this topic, we must look beyond the chemical formulas and visualize the actual 3D architecture of these molecules. Let's break down this classic JEE Advanced problem step-by-step.

### The Structural Blueprint

The key to unlocking the properties of any oxoacid lies in its Lewis structure. Let's examine orthophosphorous acid ($H_3PO_3$) and orthophosphoric acid ($H_3PO_4$).

In $H_3PO_3$, the central phosphorus atom is bonded to one oxygen atom via a double bond ($P=O$), two hydroxyl groups ($-OH$), and one hydrogen atom directly attached to it ($P-H$).

In contrast, $H_3PO_4$ features the central phosphorus atom bonded to one oxygen atom via a double bond ($P=O$) and three hydroxyl groups ($-OH$). There are no direct $P-H$ bonds in orthophosphoric acid. This structural difference is the root cause of their distinct chemical behaviors.

### Analyzing Disproportionation (Option A)

Disproportionation is a special type of redox reaction where an element in a specific oxidation state is simultaneously oxidized and reduced.

Let's calculate the oxidation state of phosphorus in $H_3PO_3$. Assuming hydrogen is $+1$ and oxygen is $-2$, we get: 
$$3(+1) + x + 3(-2) = 0 \implies x = +3$$

When $H_3PO_3$ is heated, it undergoes disproportionation to form orthophosphoric acid ($H_3PO_4$) and phosphine gas ($PH_3$). 
$$4H_3PO_3 \xrightarrow{\Delta} 3H_3PO_4 + PH_3$$

In $H_3PO_4$, the oxidation state of phosphorus is $+5$. In $PH_3$, it is $-3$. Since the $+3$ state splits into $+5$ (oxidation) and $-3$ (reduction), statement (A) is perfectly correct.

### The Redox Potential (Option B)

A reducing agent is a substance that reduces another species while getting oxidized itself. For a molecule to act as a reducing agent, its central atom must be capable of increasing its oxidation state.

Phosphorus belongs to Group 15, meaning its maximum possible oxidation state is $+5$. In $H_3PO_4$, phosphorus is already at this $+5$ zenith. It has no more valence electrons to lose, making it impossible to be oxidized further. Thus, $H_3PO_4$ **cannot** act as a reducing agent.

However, in $H_3PO_3$, phosphorus is in the $+3$ state. It has the potential to lose two more electrons to reach the $+5$ state. Because it can be oxidized, $H_3PO_3$ acts as a good reducing agent. Therefore, statement (B) is correct.

### Decoding Basicity (Option C)

The basicity of an acid is defined by the number of protons ($H^+$ ions) it can donate in an aqueous solution. In oxoacids, only the hydrogen atoms attached to highly electronegative atoms (like oxygen) are ionizable.

Looking back at our structural blueprint for $H_3PO_3$, we see exactly two $P-OH$ bonds. These two hydroxyl hydrogens are acidic and can be released in water.

Because it can donate two protons, $H_3PO_3$ is a **dibasic acid**, not a monobasic acid. This makes statement (C) incorrect.

### The Nature of the P-H Bond (Option D)

Finally, let's address the third hydrogen atom in $H_3PO_3$, which is bonded directly to the phosphorus atom ($P-H$).

The electronegativity of phosphorus is approximately $2.19$, and that of hydrogen is $2.20$. Because these values are nearly identical, the $P-H$ bond is essentially **non-polar**.

Water is a polar solvent and can only break polar bonds (like $O-H$) to release ions. The non-polar $P-H$ bond remains intact in water, meaning this specific hydrogen atom is **non-ionizable**. Thus, statement (D) is correct.

By systematically analyzing the molecular structure, we can confidently conclude that statements (A), (B), and (D) are the correct answers.

Animated Solution for Chemistry - s and p-Block Elements: The correct statement(s) related to oxoacids of phosphorous is(are)

Select Answer:

Visualized Solution

$Visualizing Oxoacids of Phosphorus$

$Analyzing Option A: Disproportionation$

$Analyzing Option B: Reducing Nature$

$Analyzing Option C: Basicity$

$Analyzing Option D: The P-H Bond$

$Final Conclusion$

The Sigma Insight: Group 15 Elements

The Structural Blueprint

Analyzing Disproportionation (Option A)

The Redox Potential (Option B)

Decoding Basicity (Option C)

The Nature of the P-H Bond (Option D)

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