Final answer:
Be can form no bonds without hybridization, P can form three, and F can form one. Hybridization would allow Be to bond by creating unoccupied orbitals, but without hybridizing, it cannot initially form bonds.
Step-by-step explanation:
The student's question pertains to the number of bonds each atom can make without hybridization. The valence electron configurations given for each atom indicate the number of unpaired electrons available for bonding.
- (a) Be2s²: Beryllium has a filled 2s subshell, indicating no unpaired electrons. Without hybridization, Be would struggle to form bonds because it does not have empty orbitals available.
- (b) P3s²3p³: Phosphorus has three unpaired electrons in its 3p orbitals, so it can form up to three bonds without hybridization.
- (c) F2s²2p⁵: Fluorine has one unpaired electron in its 2p orbitals, allowing it to form one bond without hybridization.
To summarize, Beryllium can form no bonds without hybridization, Phosphorus can form three, and Fluorine can form one. The mention of hybrid orbitals such as sp², sp³d, and sp³ d² helps illustrate the potential shapes and bond formations these atoms could undertake with hybridization. It is important to note that in reality, for Beryllium to bond with other atoms such as in BeH₂, it usually undergoes sp hybridization to accommodate bonding, despite no unpaired electrons being present in its ground state configuration.