Calculate Percentage of p Character in Phosphorus Bonds

To determine the percentage of $p$-character in the orbitals forming the $\text{P}-\text{P}$ bonds in white phosphorus ($\text{P}_4$), we analyze the hybridization of each phosphorus atom in the molecule:

1. Structure of $\text{P}_4$:
   In a tetrahedral $\text{P}_4$ molecule, each phosphorus atom has $5$ valence electrons. Each $\text{P}$ atom forms three single $\text{P}-\text{P}$ covalent bonds ($\sigma$-bonds) with three other phosphorus atoms at the corners of a tetrahedron and retains one lone pair of electrons.

2. Steric Number and Hybridization:
   The steric number ($\text{SN}$) for each phosphorus atom is given by:
   $\text{Steric Number} = (\text{Number of } \sigma\text{-bonds}) + (\text{Number of lone pairs})$
   $\text{Steric Number} = 3 + 1 = 4$

   A steric number of $4$ corresponds to $sp^3$ hybridization.

3. Percentage of $p$-character:
   An $sp^3$ hybrid orbital is formed by the combination of one $s$-orbital and three $p$-orbitals (total of $4$ hybrid orbitals).

   The fraction of $p$-character in an $sp^3$ hybrid orbital is:
   $\text{Fraction of } p\text{-character} = \frac{\text{Number of } p\text{-orbitals}}{\text{Total number of orbitals}} = \frac{3}{1 + 3} = \frac{3}{4} = 0.75$

   Converting this fraction to a percentage:
   $\text{Percentage of } p\text{-character} = 0.75 \times 100\% = 75\%$

Thus, the percentage of $p$-character in the orbitals forming $\text{P}-\text{P}$ bonds in $\text{P}_4$ is $75\%$.

Correct Option: D