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 $s\mathrm{\%}$ 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

Hint: $P_4{O}_{10}$ 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 $s\mathrm{\%}$ 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 $s\mathrm{\%}$ 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 $P_4{O}_{10}$.