Question

What is the hybridization of ${\mathrm{PF}}_5$?

Answer 1

Step 1: Formula used for calculation:

• $\mathrm{Hybridization}=\frac{1}{2}\left[\mathrm{V}+\mathrm{M}-\mathrm{C}+\mathrm{A}\right]$
• Where $\mathrm{V}$ is the number of valence electrons of the central atom, $\mathrm{M}$ is the monovalent atoms attached to the central atom, $\mathrm{C}$ is the positive charge, and $\mathrm{A}$represents the negative charge.

Step 2: Hybridization of Phosphorus pentafluoride ${\mathrm{PF}}_5$:

• The central atom is Phosphorus $\mathrm{P}$ and its atomic number is $15$, so its electronic configuration is $\left[\mathrm{Ne}\right]3{\mathrm{s}}^{2}3{\mathrm{p}}^3$.
• So, $\mathrm{P}$ has $5$ valence electrons and $5$ monovalent $\mathrm{F}$ atoms attached to it.
• $\mathrm{Hybridization}\mathrm{of}{\mathrm{PF}}_5=\frac{1}{2}\left[5+5-0+0\right]=\frac{1}{2}\times 10=5$
• The hybridization number is $5$, thus the hybridization of ${\mathrm{PF}}_5$ is ${\mathrm{sp}}^3\mathrm{d}$.