Question

Select the correct statement among the following about $P_4O_{10}$.

• A. It is an anhydride of $H_3PO_3$
• B. All the P atoms are $sp^3d$ hybridized
• C. There are 10(P - O - P) linkage in one molecule
• D. Total number of $sp^3$ hybridized atoms per molecules is 10

Answer 1

Answer: (D)

Total number of $sp^3$ hybridized atoms per molecules is 10

Answer 2

The question asks to select the correct statement about $P_4O_{10}$. Let's analyze each option based on the structure and properties of $P_4O_{10}$.

The structure of $P_4O_{10}$ is a cage-like structure derived from a tetrahedron of four phosphorus atoms.

• There are 4 phosphorus (P) atoms.

• There are 10 oxygen (O) atoms.

  • 6 oxygen atoms act as bridging atoms (P-O-P linkages) between the phosphorus atoms, forming the edges of the P-tetrahedron.
  • 4 oxygen atoms are terminal atoms, double-bonded to each phosphorus atom (P=O).

Let's evaluate each option:

A. It is an anhydride of $H_3PO_3$

An acid anhydride is formed by the removal of water from an acid. When $P_4O_{10}$ reacts with water, it forms orthophosphoric acid ($H_3PO_4$):

$P_4O_{10} + 6H_2O \rightarrow 4H_3PO_4$

The oxidation state of phosphorus in $P_4O_{10}$ is +5. The oxidation state of phosphorus in $H_3PO_4$ is also +5. The oxidation state of phosphorus in $H_3PO_3$ (phosphorous acid) is +3. Since the oxidation state of the central atom must be conserved when forming an anhydride, $P_4O_{10}$ cannot be the anhydride of $H_3PO_3$. Therefore, statement A is incorrect.

B. All the P atoms are $sp^3d$ hybridized

To determine the hybridization of phosphorus atoms, we use the steric number (number of sigma bonds + number of lone pairs).

Each phosphorus atom in $P_4O_{10}$ is bonded to four oxygen atoms: three single bonds (P-O) to bridging oxygens and one double bond (P=O) to a terminal oxygen.

Counting sigma bonds: Each P atom forms 3 P-O single bonds and 1 P=O double bond (which consists of one sigma bond and one pi bond). So, each P atom forms a total of 4 sigma bonds. There are no lone pairs on the phosphorus atom.

Steric number = 4 (sigma bonds) + 0 (lone pairs) = 4.

A steric number of 4 corresponds to $sp^3$ hybridization. Therefore, all P atoms are $sp^3$ hybridized, not $sp^3d$. Statement B is incorrect.

C. There are 10(P - O - P) linkage in one molecule

In the cage structure of $P_4O_{10}$, the four phosphorus atoms form a tetrahedron. The six edges of this tetrahedron are bridged by oxygen atoms.

Thus, there are 6 P-O-P linkages in one molecule of $P_4O_{10}$. Therefore, statement C is incorrect.

D. Total number of $sp^3$ hybridized atoms per molecules is 10

Let's count the $sp^3$ hybridized atoms:

1. Phosphorus (P) atoms: As determined in option B, all 4 P atoms are $sp^3$ hybridized (steric number = 4).

2. Oxygen (O) atoms: There are two types of oxygen atoms:

   • Bridging oxygen atoms (P-O-P): There are 6 such oxygen atoms. Each bridging oxygen atom forms two single bonds (to two P atoms) and has two lone pairs. Steric number = 2 (sigma bonds) + 2 (lone pairs) = 4. Therefore, these 6 bridging oxygen atoms are $sp^3$ hybridized.
   • Terminal oxygen atoms (P=O): There are 4 such oxygen atoms. Each terminal oxygen atom forms one double bond (one sigma, one pi) to a P atom and has two lone pairs. Steric number = 1 (sigma bond) + 2 (lone pairs) = 3. Therefore, these 4 terminal oxygen atoms are $sp^2$ hybridized.

Total number of $sp^3$ hybridized atoms = (Number of $sp^3$ P atoms) + (Number of $sp^3$ bridging O atoms)

Total $sp^3$ hybridized atoms = 4 (P atoms) + 6 (bridging O atoms) = 10.

Therefore, statement D is correct.

$P_4O_{10}$ is the anhydride of $H_3PO_4$, not $H_3PO_3$, as the oxidation state of P is +5 in $P_4O_{10}$ and $H_3PO_4$, but +3 in $H_3PO_3$. So A is incorrect.

Each P atom in $P_4O_{10}$ forms 4 sigma bonds (3 P-O single and 1 P=O sigma) and has no lone pairs, leading to a steric number of 4. Thus, P atoms are $sp^3$ hybridized, not $sp^3d$. So B is incorrect.

The $P_4O_{10}$ molecule has a cage structure where 4 P atoms form a tetrahedron, and 6 oxygen atoms bridge the 6 edges of this tetrahedron. Hence, there are 6 P-O-P linkages, not 10. So C is incorrect.

There are 4 P atoms, each $sp^3$ hybridized. There are 6 bridging oxygen atoms (P-O-P), each forming 2 sigma bonds and having 2 lone pairs, resulting in $sp^3$ hybridization. The 4 terminal oxygen atoms (P=O) are $sp^2$ hybridized. Therefore, the total number of $sp^3$ hybridized atoms is 4 (P) + 6 (bridging O) = 10. So D is correct.