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What is the percentage of ${\text{p - }}$ character in forming \[{\text{P - P}}\] bonds in ${\text{P_4}}$ molecule?

Answer
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Hint: Since Phosphorus can form three bonds, it can form a ${\text{P_4}}$ white phosphorus tetrahedron, while sulphur can only form two. As a result, sulphur just forms rings and chains. Red phosphorus is a cross-linked, polymeric chain of atoms that is the most stable allotrope of phosphorus.

Complete answer:
Phosphorus is a derivative of the ${\text{P_4}}$ molecule and is one of the most common allotropes of phosphorus. It is made up of a network of phosphorus atoms that is amorphous (non-crystalline). Phosphorus is found in a polymeric chain of tetrahedrally shaped ${\text{P_4}}$ molecules, one of which has one of its ${\text{P - P}}$ bonds broken to allow the tetrahedrons to be connected.
In a tetrahedral structure, each phosphorus atom in ${\text{P_4}}$ is connected to three other phosphorus atoms. When one of these bonds is broken, these tetrahedral structures can form polymer-like structures by bonding with neighbouring phosphorus atoms.
With ${\text{P}}$ belonging to Group ${\text{V}}$ on the periodic table, one would expect the molecule to only be capable of making three bonds to fill its valence shell. While phosphorus forms three bonds in ${\text{P_4}}$, the molecular geometry of ${\text{P_4}}$ cannot be represented by three bonds formed by partially filled p orbitals.
${\text{P}}$ cannot form pi bonds due to its larger atomic size, so it is tetra-atomic, with each ${\text{P}}$ atom connected to three others ${\text{P}}$ atoms by three sigma bonds. As a result, the ${\text{p}}$-character plays a $75$ percent role in forming ${\text{P - P}}$ bonds in the ${\text{P_4}}$ molecule.

Note:
Ring strain and instability are caused by ${\text{P4}}$'s tetrahedral arrangement. Six single ${\text{P - P}}$ bonds are known as the molecule's structure. There are two types of crystalline forms that have been discovered. It consists of four lone pairs of electrons and ${\text{PPP}}$ angle of ${60^ \circ }$