Problem 61 - Chemistry: The Central Science

Question

Account for the following observations: (a) Phosphorus forms a pentachloride, but nitrogen does not. (b) \(\mathrm{H}_{3} \mathrm{PO}_{2}\) is a monoprotic acid. (c) Phosphonium salts, such as \(\mathrm{PH}_{4} \mathrm{Cl}\), can be formed under anhydrous conditions, but they cannot be made in aqueous solution. (d) White phosphorus is more reactive than red phosphorus.

Step-by-Step Solution

Answer

Phosphorus forms PCl\(_5\) due to its availability of d orbitals, while nitrogen cannot form NCl\(_5\) as it lacks available d orbitals. \(\mathrm{H}_{3} \mathrm{PO}_{2}\) is monoprotic, as only one hydrogen atom is bonded to oxygen, releasing one H\(^+\) ion. Phosphonium salt PH\(_4\)Cl forms under anhydrous conditions due to the hydrolysis of PH\(_4\)Cl in aqueous environments. White phosphorus is more reactive than red phosphorus due to its tetrahedral structure, resulting in angular strain and less stability.

Step 1: (a) Phosphorus pentachloride and nitrogen pentachloride

Phosphorus can form PCl\(_5\) while nitrogen does not form NCl\(_5\). This is because of the difference in the availability of orbitals. As phosphorus (third period element) has d orbitals available, it can accommodate five bonds in its extended octet. Nitrogen, on the other hand, belongs to the second period and has no d orbitals available. It can make only three bonds in its normal octet with no room for an extended octet. Therefore, nitrogen cannot form NCl\(_5\).

Step 2: (b) Monoprotic nature of \(\mathrm{H}_{3} \mathrm{PO}_{2}\)

\(\mathrm{H}_{3} \mathrm{PO}_{2}\) (Hypophosphorous acid) has the structure: H-O-P(OH)-H. In this structure, only one hydrogen atom is bonded to an oxygen atom. When we consider the acidic strength of a compound, we focus on the number of hydrogen atoms that can be released as H\(^+\) ions. In the case of \(\mathrm{H}_{3} \mathrm{PO}_{2}\), only one H\(^+\) ion can be released because only one H is bonded to an oxygen atom (O-H bond). Thus, \(\mathrm{H}_{3} \mathrm{PO}_{2}\) acts as a monoprotic acid.

Step 3: (c) Anhydrous vs. aqueous conditions for forming PH\(_4\)Cl

Phosphonium salts, like PH\(_4\)Cl, are formed when PH\(_3\) (phosphine) reacts with a hydrogen halide, like HCl. The reaction proceeds as follows: PH\(_3\) + HCl → PH\(_4\)Cl. This reaction takes place under anhydrous (dry) conditions because aqueous conditions will cause the water to hydrolyze the PH\(_4\)Cl into its original reactants. In aqueous solution, PH\(_4\)Cl will break down into PH\(_3\) and HCl, making it impossible to produce in this environment.

Step 4: (d) Reactivity of white phosphorus and red phosphorus

White phosphorus (P\(_4\)) is more reactive than red phosphorus due to differences in their molecular structures. White phosphorus has a tetrahedral structure where each phosphorus atom is bonded to three other phosphorus atoms. This structure results in a high level of angular strain, making the molecule less stable and more reactive. Red phosphorus, on the other hand, has a polymeric chain structure with each phosphorus atom bonded to two other phosphorus atoms. This arrangement makes red phosphorus a more stable and less reactive form of phosphorus as compared to white phosphorus.