Question

We write oxygen as \[{{O}_{2}}\] because \[(10-8\](atomic no. Of oxygen)) but why not write phosphorus as \[P_3(18-15\](atomic no. of phosphorus)) while we write it as \[P_4\]?

Answer 1

Hint: We know that Hybridization involves the mixing of atomic orbitals of an atom (generally a central atom) to generate a new set of atomic orbitals which is called the hybrid orbitals. Hence to find the hybridization of the central metal atom we need to first write the electronic configuration and then draw the orbitals in which the bonded atom will be filled.

Complete answer:
Let us firstly know what the electronic configuration ion of an atom really is before moving onto answering the given question. The electron configuration is a representation of the arrangement of electrons distributed among the shells and subshells in the orbitals of an atom; and is mostly used for describing the electronic arrangement in the orbitals of an atom in its ground state. The easiest way to solve this question is by calculating the hybridization of the compound given above. We generally consider the hybridization of the central element, but in this case, all the atoms will act the same, i.e. hybridization of all four phosphorus will be the same.
Due to the large atomic size of the phosphorus, it is unable to form pi bonds and it is tetratomic in which each p atom is linked with the \[3\] other p atoms by sigma bond. To form a triple bond between the two phosphorus there should be enough space in between them. As the atomic size is large it is not possible. They use d orbitals to make pi bonds.

Note:
Remember that the Hund’s maximum spin multiplicity implies only for the orbitals that have the same amount of energy. That means you need not to put electrons in the \[4p\] orbital of Krypton until the \[4s\] orbital is full.