Draw a dot diagram for \(\mathrm{PF}_{5} .\) Which atom is an expanded octet atom?

First, we need to find the number of valence electrons for each atom in \(\mathrm{PF}_{5}\). Phosphorus belongs to Group 15 and has 5 valence electrons, while Fluorine belongs to Group 17 and has 7 valence electrons. As there are 5 fluorine atoms, the total number of valence electrons in \(\mathrm{PF}_{5}\) is: Total valence electrons = valence electrons of phosphorus + 5 (valence electrons of fluorine) Total valence electrons = 5 + 5(7) = 5 + 35 = 40

According to the octet rule, atoms tend to have eight electrons in their outer shell to achieve stability. In \(\mathrm{PF}_{5}\), we have one central phosphorous (P) atom surrounded by five fluorine (F) atoms. The central phosphorus atom forms single bonds with each of the fluorine atoms to complete their octet. However, in this case, the phosphorus atom will have more than 8 electrons in its outer shell - which is called an expanded octet.

Now we can draw the dot structure for \(\mathrm{PF}_{5}\). Place the phosphorus atom in the center and the fluorine atoms around it. Connect each fluorine atom to the phosphorus atom with a single bond, representing two shared electrons. Also, include the six remaining electrons around each fluorine atom as lone pairs. The dot structure of \(\mathrm{PF}_{5}\) is given by: F : P - F / \ F - P F \ F In this structure, each fluorine atom has an octet (8 valence electrons). However, the central phosphorus atom has 10 valence electrons, which is an expanded octet. So, the expanded octet atom in this molecule is Phosphorus (P).