Question

5 drag each label to the correct location on the diagram. each label can be used more than once. what is the correct way to represent the ionic compound sodium fluoride? na + f - •na :f: → □□ → ( )

Explanation:

To determine the correct representation of sodium fluoride (\( \text{NaF} \)) as an ionic compound, we analyze the electron transfer between sodium (\( \text{Na} \)) and fluorine (\( \text{F} \)):

Step 1: Identify Valence Electrons

• Sodium (\( \text{Na} \)) is in Group 1, so it has 1 valence electron (represented as \( \boldsymbol{\cdot \text{Na}} \)).
• Fluorine (\( \text{F} \)) is in Group 17, so it has 7 valence electrons (represented as \( \boldsymbol{:\overset{\cdot \cdot}{\underset{\cdot \cdot}{{\text{F}}}}:} \) or similar Lewis dot structures).

Step 2: Electron Transfer (Ionic Bonding)

• Sodium loses its 1 valence electron to achieve a stable octet (like \( \text{Ne} \)). This forms a \( \text{Na}^+ \) ion (cation).
• Fluorine gains 1 electron to achieve a stable octet (like \( \text{Ne} \)). This forms a \( \text{F}^- \) ion (anion).

Step 3: Represent the Ionic Compound

In the diagram:

• The first empty box (after \( \text{Na} \) donates its electron) should show \( \boldsymbol{\text{Na}^+} \) (since \( \text{Na} \) loses its electron).
• The second box (around \( \text{F} \)) should show \( \boldsymbol{\text{F}^-} \) (since \( \text{F} \) gains an electron, completing its octet).
• The final ionic formula is \( \boldsymbol{\text{NaF}} \), with \( \text{Na}^+ \) and \( \text{F}^- \) ions held together by electrostatic attraction.

Final Answer (Diagram Labels)

• First empty box: \( \boldsymbol{\text{Na}^+} \)
• Second box (around F): \( \boldsymbol{\text{F}^-} \)
• Ionic formula: \( \boldsymbol{\text{NaF}} \) (with \( \text{Na}^+ \) and \( \text{F}^- \) ions in the bracket structure).

Answer:

To determine the correct representation of sodium fluoride (\( \text{NaF} \)) as an ionic compound, we analyze the electron transfer between sodium (\( \text{Na} \)) and fluorine (\( \text{F} \)):

Step 1: Identify Valence Electrons

• Sodium (\( \text{Na} \)) is in Group 1, so it has 1 valence electron (represented as \( \boldsymbol{\cdot \text{Na}} \)).
• Fluorine (\( \text{F} \)) is in Group 17, so it has 7 valence electrons (represented as \( \boldsymbol{:\overset{\cdot \cdot}{\underset{\cdot \cdot}{{\text{F}}}}:} \) or similar Lewis dot structures).

Step 2: Electron Transfer (Ionic Bonding)

• Sodium loses its 1 valence electron to achieve a stable octet (like \( \text{Ne} \)). This forms a \( \text{Na}^+ \) ion (cation).
• Fluorine gains 1 electron to achieve a stable octet (like \( \text{Ne} \)). This forms a \( \text{F}^- \) ion (anion).

Step 3: Represent the Ionic Compound

In the diagram:

• The first empty box (after \( \text{Na} \) donates its electron) should show \( \boldsymbol{\text{Na}^+} \) (since \( \text{Na} \) loses its electron).
• The second box (around \( \text{F} \)) should show \( \boldsymbol{\text{F}^-} \) (since \( \text{F} \) gains an electron, completing its octet).
• The final ionic formula is \( \boldsymbol{\text{NaF}} \), with \( \text{Na}^+ \) and \( \text{F}^- \) ions held together by electrostatic attraction.

Final Answer (Diagram Labels)

• First empty box: \( \boldsymbol{\text{Na}^+} \)
• Second box (around F): \( \boldsymbol{\text{F}^-} \)
• Ionic formula: \( \boldsymbol{\text{NaF}} \) (with \( \text{Na}^+ \) and \( \text{F}^- \) ions in the bracket structure).