Chapter 8.4, Problem 66SSC

Interpretation Introduction

Interpretation:

The molecular shapes and hybrid orbitals of ${\text{PF}}_{\text{3 }}{\text{and PF}}_{\text{5}}$ molecules need to be compared.

Concept introduction:

The Lewis structure of an organic compound represents the bonding of atoms with lone pairs (if any). The hybridization of an atom indicates the molecular geometry of a molecule. The formula to check the hybridization can be written as:

Hybridization = Number of sigma bonds + Number of lone pair

Answer to Problem 66SSC

In ${\text{PF}}_{\text{3 }}$ molecule sp³ hybrid orbitals arrange in trigonal pyramid geometry, whereas in ${\text{PF}}_{\text{5}}$ molecule, sp³d hybrid orbitals arrange in trigonal bipyramid arrangement.

Explanation of Solution

In the ${\text{PF}}_{\text{3 }}$ molecule, P atom contains 5 valence electrons, and F has 7 valence electrons. The total number of valence electron can be determined in it as follows:

Number of valence electron = Number of atom P (valence electron in P) + Number of atom F (valence electron in F) = $\text{(1}\times \text{5) + (3}\times \text{3) = 26}$

$\begin{array}{l}\text{Hybridization = Number of sigma bonds + Number of lone pairs on bonded atoms}\\ {\text{Hybridization = 3 + 1 = sp}}^{\text{3 }}\end{array}$

With sp³ hybridization, the geometry must be tetrahedral but the presence of one lone pair on the central P atom alters the standard geometry of the molecule to pyramid shape due to repulsion between the lone pair and bond pair of the molecule.

In the ${\text{PF}}_{\text{5}}$ molecule, P atom contains 5 valence electrons, and F has 7 valence electrons. The total number of valence electron can be determined in it as follows:

Number of valence electron = Number of atom P (valence electron in P) + Number of atom F (valence electron in F) = $\text{(1}\times \text{5) + (5}\times \text{7) = 40}$

Hence the Lewis structure must be:

$\begin{array}{l}\text{Hybridization = Number of sigma bonds + Number of lone pairs on bonded atoms}\\ {\text{Hybridization = 5 + 0 = sp}}^{\text{3}}\text{d = Trigonal bipyramid}\end{array}$

Thus, in ${\text{PF}}_{\text{3 }}$ molecule sp³ hybrid orbitals arrange in trigonal pyramid geometry, whereas in ${\text{PF}}_{\text{5}}$ molecule, sp³d hybrid orbitals arrange in trigonal bipyramid arrangement.

Conclusion

The change in hybridization alters the molecular geometry of the molecule.