Animated Solution for Chemistry - s and p-Block Elements: White phosphorus on reaction with concentrated NaOH solution in an inert atmosphere of ${\text{CO}}_2$ gives phosphine and compound (X). (X) on acidification with HCl gives compound (Y). The basicity of compound (Y) is

Unraveling the Mystery of Phosphorus Oxoacids: A Deep Dive into Basicity

Imagine you are in a chemistry lab, carefully handling highly reactive white phosphorus. When you drop it into a concentrated solution of sodium hydroxide under an inert carbon dioxide atmosphere, a fascinating chemical transformation occurs. This isn't just any reaction; it's a classic disproportionation reaction that frequently appears in competitive exams like JEE and NEET. In this article, we will walk through this reaction step-by-step, identify the mysterious compounds (X) and (Y), and finally uncover the secret behind the basicity of phosphorus oxoacids.

The Disproportionation Reaction

The journey begins with the reaction of white phosphorus ($P_4$) with an alkali. White phosphorus is highly reactive, and in a basic medium, it undergoes disproportionation. This means that the phosphorus atoms are simultaneously oxidized and reduced.

The balanced chemical equation for this process is:

Here, phosphorus (which starts with an oxidation state of $0$) is reduced to form phosphine gas ($P{H}_3$), where its oxidation state is $-3$. Simultaneously, it is oxidized to form sodium hypophosphite ($Na{H}_{2}P{O}_2$), where its oxidation state is $+1$.

According to our problem statement, the salt formed alongside phosphine is compound (X). Therefore, Compound (X) is sodium hypophosphite ($Na{H}_{2}P{O}_2$).

Acidification of Compound (X)

Now that we have identified compound (X), the next step is to acidify it. The problem states that (X) is treated with hydrochloric acid ($HCl$).

When a salt of a weak acid (like sodium hypophosphite) reacts with a strong acid (like $HCl$), the strong acid displaces the weak acid from its salt. The reaction proceeds as follows:

The sodium ion is replaced by a proton ($H^{+}$), yielding hypophosphorous acid ($H_{3}P{O}_2$) and sodium chloride. This newly formed acid is our Compound (Y).

Decoding the Basicity

We have successfully identified compound (Y) as hypophosphorous acid ($H_{3}P{O}_2$). The final hurdle is to determine its basicity.

This is where many students fall into a common trap. Looking at the molecular formula $H_{3}P{O}_2$, it is tempting to assume that because there are three hydrogen atoms, the basicity must be three. However, the basicity of an oxoacid is strictly defined by the number of ionizable hydrogen atoms.

To understand which hydrogens are ionizable, we must look at the molecular structure of $H_{3}P{O}_2$. In phosphorus oxoacids, a hydrogen atom is only acidic (ionizable) if it is bonded to a highly electronegative oxygen atom. Hydrogens bonded directly to the central phosphorus atom are not acidic due to the small electronegativity difference between phosphorus and hydrogen.

Let's visualize the structure of $H_{3}P{O}_2$: - The central phosphorus atom is $s{p}^3$ hybridized. - It forms one double bond with an oxygen atom ($P=O$). - It forms one single bond with a hydroxyl group ($P-OH$). - It forms two single bonds directly with hydrogen atoms ($P-H$).

Because there is only one $P-OH$ bond, only one hydrogen atom can be released as a proton ($H^{+}$) in an aqueous solution.

Therefore, hypophosphorous acid is a monobasic acid, and its basicity is exactly 1.

Reducing Character

While the two $P-H$ bonds do not contribute to the basicity, they play a crucial role in the chemical behavior of hypophosphorous acid. These direct $P-H$ bonds make $H_{3}P{O}_2$ a powerful reducing agent. In future problems, whenever you see a phosphorus oxoacid, always draw its structure. Count the $P-OH$ bonds to find the basicity, and count the $P-H$ bonds to gauge its reducing strength!