Q75E - Chemistry (OpenStax)

Question

Draw the Lewis structures and describe the geometry for the following:

(a) \({\rm{P}}{{\rm{F}}_4} + \) 

(b) \({\rm{P}}{{\rm{F}}_5}\)

(c) \({\rm{P}}{{\rm{F}}_6}^ - \)

(d) \({\rm{PO}}{{\rm{F}}_3}\)

Step-by-Step Solution

Answer

a) \({\rm{P}}{{\rm{F}}_4} + \):

b) \({\rm{P}}{{\rm{F}}_5}\):

c) \({\rm{P}}{{\rm{F}}_6}^-\):

d)  \({\rm{PO}}{{\rm{F}}_3}\):

Step 1: Definition of Lewis structure.

A Lewis structure is a diagram that depicts a molecule's covalent bonds and lone electron pairs.

Step 2: Write the Lewis structure .

Count all of a molecule's valent electrones to draw the Lewis structures:

\(\begin{array}{l}N\left( {{e^ - }} \right) = N\left( {F,{e^ - }} \right) + N\left( {P,{e^ - }} \right)\\N\left( {{e^ - }} \right) = 4 \cdot 7 + 1 \cdot 5\\N\left( {{e^ - }} \right) = 33\end{array}\)

To fill the octet, each atom should contain \(8\) electrons; each bond counts as two electrons, while the rest are lone electron pairs.

The \(PF_4^ + \) has a form of tetradar.

\(\begin{array}{l}N\left( {{e^ - }} \right) = N\left( {F,{e^ - }} \right) + N\left( {P,{e^ - }} \right)\\N\left( {{e^ - }} \right) = 5 \cdot 7 + 1 \cdot 5\\N\left( {{e^ - }} \right) = 40\end{array}\)

The \(P{F_5}\) has trigonal bipyramidal form.

\(\begin{array}{l}N\left( {{e^ - }} \right) = N\left( {F,{e^ - }} \right) + N\left( {P,{e^ - }} \right)\\N\left( {{e^ - }} \right) = 6 \cdot 7 + 1 \cdot 5\\N\left( {{e^ - }} \right) = 47\end{array}\)

The \(PF_6^ - \) has octahedral form.

\(\begin{array}{l}N\left( {{e^ - }} \right) = N\left( {F,{e^ - }} \right) + N\left( { + ,{e^ - }} \right) + N\left( {P,{e^ - }} \right)\\N\left( {{e^ - }} \right) = 3 \cdot 7 + 1 \cdot 6 + 1 \cdot 5\\N\left( {{e^ - }} \right) = 32\end{array}\)

The \({\rm{PO}}{{\rm{F}}_3}\) has tetrahedral form.