Question

Select the correct statement for ${P_4}{O_{10}}$.
(This question has multiple correct options)
A.It has four $s{p^3}$ hybridised phosphorus atoms
B.It has high $s\%$ character in P-O bond than the ${P_4}{O_6}$
C.It has a cage like structure
D.It has ${p_\pi } - {d_\pi }$ bonding

Answer 1

${P_4}{O_{10}}$ is an oxide of phosphorus. Its chemical name is phosphorus pentoxide, because it is a dimer of ${P_2}{O_5}$ . In ${P_4}{O_{10}}$ , oxygen is in $- 2$ oxidation state and phosphorus is in $+ 5$ oxidation state.

Complete step by step answer:
Structure of ${P_4}{O_{10}}$ is shown below.

In ${P_4}{O_{10}}$ , all the phosphorus atoms are $s{p^3}$ hybridised with a tetrahedral structure. $s{p^3}d$ hybridisation is preferred for phosphorus atom over $s{p^3}$ hybridisation. But for ${P_4}{O_{10}}$ , $s{p^3}$ hybridisation is more stable than $s{p^3}d$. Hence ${P_4}{O_{10}}$has four $s{p^3}$ hybridised phosphorus atoms. Option A is correct.
${P_4}{O_6}$ is another oxide of phosphorus. Both ${P_4}{O_6}$ and ${P_4}{O_{10}}$ contains six P-O-P bond and $4$ six membered ring. In ${P_4}{O_6}$ , the atomic orbitals containing lone pairs have more s-character and less p-character. Hence they have shorter bond length. But the bonding orbitals have more p-character and less s-character and hence a longer bond length. But in ${P_4}{O_{10}}$ the bonding orbitals have more $s\%$ character compared to that in ${P_4}{O_6}$ . The bonding orbitals means the P-O bond. Hence ${P_4}{O_6}$ has a higher $s\%$ character in P-O bond than the ${P_4}{O_6}$. Option B is also correct.
From the structure given above, we can see that ${P_4}{O_{10}}$ has a cage like structure. Hence option C is correct.
The terminal P-O bonds in ${P_4}{O_{10}}$ are formed by ${p_\pi } - {d_\pi }$ bonding. p-orbitals of oxygen overlap with empty d-orbitals of phosphorus. The phosphorus atom donates its lone pair to this bonding forming a $\pi$ - back bonding. Hence option D is correct.
Options A,B,C and D correct for ${P_4}{O_{10}}$ .

Note:
Phosphorus can show $+ 3$ and $+ 5$ oxidation states. Hence it forms two oxides ${P_2}{O_3}$ and ${P_2}{O_5}$ with oxidation states $+ 3$ and $+ 5$ respectively. But their monomeric form is not stable. This is the reason why they exist as dimer - ${P_4}{O_6}$ and ${P_4}{O_{10}}$.