question_answer

The structure of \[P{{F}_{5}}\] molecule is:

A)  trigonal bipyramidal

B)  tetrahedral

C)  pentagonal bipyramidal

D)  square planar

Correct Answer: A

Solution :

\[{}_{15}P=1{{s}^{2}},\,2{{s}^{2}}2{{p}^{6}},3{{s}^{2}}3{{p}^{3}}3{{d}^{0}}\] P in excited state = \[P{{F}_{5}}=\begin{matrix}    \underset{{{F}_{{{(2px)}^{1}}}}}{\mathop{\underset{|}{\mathop{{{(s{{p}^{3}}d)}^{1}}}}\,}}\, & \underset{{{F}_{{{(2px)}^{1}}}}}{\mathop{\underset{|}{\mathop{{{(s{{p}^{3}}d)}^{1}}}}\,}}\, & \underset{{{F}_{{{(2px)}^{1}}}}}{\mathop{\underset{|}{\mathop{{{(s{{p}^{3}}d)}^{1}}}}\,}}\, & \underset{{{F}_{{{(2px)}^{1}}}}}{\mathop{\underset{|}{\mathop{{{(s{{p}^{3}}d)}^{1}}}}\,}}\, & \underset{{{F}_{{{(2px)}^{1}}}}}{\mathop{\underset{|}{\mathop{{{(s{{p}^{3}}d)}^{1}}}}\,}}\,  \\ \end{matrix}\] Thus, due to \[s{{p}^{3}}d\text{-}\]hybridization, \[P{{F}_{5}}\]molecule has trigonal bipyramidal geometry.