To represent the ionic compound sodium fluoride, you should understand the following steps:
1. Identify the Ions:
- Sodium (Na) is a metal that tends to lose one electron to achieve a stable electron configuration, forming a positively charged ion (cation) denoted as [tex]\( \text{Na}^+ \)[/tex].
- Fluorine (F) is a non-metal that tends to gain one electron to achieve a stable electron configuration, forming a negatively charged ion (anion) denoted as [tex]\( \text{F}^- \)[/tex].
2. Electron Transfer:
- In forming an ionic bond, sodium (Na) will lose an electron and become [tex]\( \text{Na}^+ \)[/tex], while fluorine (F) will gain an electron to become [tex]\( \text{F}^- \)[/tex].
- When sodium loses its electron, it transfers that electron to fluorine.
3. Formation of Ionic Compound:
- The [tex]\( \text{Na}^+ \)[/tex] cation and [tex]\( \text{F}^- \)[/tex] anion are attracted to each other due to their opposite charges. This electrostatic force of attraction between oppositely charged ions forms the ionic bond.
Therefore, the correct way to represent sodium fluoride is:
[tex]\[ \text{Na}^+ \text{F}^- \][/tex]
Putting it altogether:
- Sodium (Na) loses one electron: [tex]\( \text{Na} \rightarrow \text{Na}^+ + e^- \)[/tex]
- Fluorine (F) gains one electron: [tex]\( \text{F} + e^- \rightarrow \text{F}^- \)[/tex]
Resulting in the formation of the ionic compound:
[tex]\[ \text{Na}^+ + \text{F}^- \rightarrow \text{NaF} \][/tex]
Thus, sodium fluoride is represented as [tex]\( \text{NaF} \)[/tex], comprising [tex]\( \text{Na}^+ \)[/tex] and [tex]\( \text{F}^- \)[/tex] ions.
