Question

which image depicts the transfer of electrons between strontium and fluorine when they form an ionic compound?

a. \\(:\ddot{\text{f}}-\ddot{\text{sr}}-\ddot{\text{f}}:\\)

b. \\(:\ddot{\text{f}}\cdot \curvearrowleft \cdot\text{sr}\cdot \curvearrowright \cdot\ddot{\text{f}}:\\)

c. \\(\text{sr}^{+2} \\ 2\left :\ddot{\text{f}}: \
ight^{-1}\\)

Explanation:

Determine the valence electrons of strontium and fluorine

Strontium (\(\text{Sr}\)) is an alkaline earth metal in Group 2, meaning it has 2 valence electrons. Fluorine (\(\text{F}\)) is a halogen in Group 17, meaning each fluorine atom has 7 valence electrons.

Analyze the process of ionic bond formation

To achieve a stable octet, the metal strontium needs to lose its 2 valence electrons to become a \(\text{Sr}^{2+}\) cation. Each nonmetal fluorine atom needs to gain 1 electron to complete its octet and become a \(\text{F}^-\) anion. Therefore, one strontium atom transfers 1 electron to each of the two fluorine atoms.

Evaluate the given options

• Option A shows covalent bonds (represented by lines) between fluorine and strontium, which is incorrect because strontium and fluorine form an ionic compound, not a covalent one.
• Option B depicts the transfer of electrons using arrows. It shows a strontium atom with 2 valence electrons transferring one electron to each of the two fluorine atoms (each starting with 7 valence electrons). This correctly illustrates the electron transfer process.
• Option C shows the final Lewis structure of the ionic compound after the transfer has occurred, rather than the active transfer process itself.

Answer:

• (A) covalent representation with single bonds
• (B) electron transfer shown with arrows (Correct answer)
• (C) final ionic compound Lewis structure