Question

In $P_4{O}_{10}$ the:
(A) Second bond in $P=O$ is formed by $p\pi -d\pi$ back bonding
(B) $P=O$ bond is formed by $p\pi -p\pi$ bonding
(C) $P=O$ bond is formed by $d\pi -d\pi$ bonding
(D) $P=O$ bond is formed by $d\pi -d\pi -3\sigma$ back bonding

Answer 1

Hint:$P_4{O}_{10}$ (phosphorus Pentoxide) is a chemical compound. This while crystalline solid is the anhydride of phosphoric acid. It is a powerful desiccant and dehydrating agent.

Complete step by step answer:
In $P_4{O}_{10}$, the terminal $P-O$ bonds are formed by $p\pi -d\pi$ back bonding. This bonding results in the formation of a coordination bond resulting from an overlap of the p-orbitals of oxygen with empty $d\pi$ orbitals of phosphorus. Here, phosphorus atoms donate one pair of electrons resulting in $\pi$ back bonding.

The molecules are held together in a hexagonal lattice by weak Vander Waals forces.
This structure contains $6P-O-P$ bonds where the hybridization of oxygen is $s{p}^3$. Another $4$ oxygens are attached to each phosphorus by $P=O$ where hybridization of oxygen is $s{p}^2$. All the $10$ oxygen contains $20$lone pairs ($2$ lone pairs each).
Preparation: $P_4+5O_2\to P_4{O}_{10}$.
Hence, option B is the answer.

Note:
$P_4{O}_{10}$ is called pentoxide because of the oxidation state of phosphorus in the compound. The name of phosphorus pentoxide itself suggests that the name is related to phosphorus. The naming of any compound is done using its empirical formula. Since, the empirical formula of $P_4{O}_{10}$ is $P_2{O}_5$, it is called phosphorus pentoxide.