To determine whether sodium (Na) and fluorine (F) form an ionic bond, we need to calculate the difference in their electronegativities and compare it to the given threshold of 1.5 to 1.7.
Here are the electronegativity values for Na and F:
- The electronegativity of sodium (Na) is 0.93.
- The electronegativity of fluorine (F) is 3.98.
The difference in electronegativities is calculated as follows:
[tex]\[
\text{Electronegativity difference} = \left| \text{Electronegativity of } F - \text{Electronegativity of } Na \right| = \left| 3.98 - 0.93 \right| = 3.05
\][/tex]
Since the difference in electronegativity (3.05) is greater than the threshold of 1.5 to 1.7, Na and F indeed form an ionic bond.
Therefore, the electronegativity difference for Na and F is 3.05.
Based on this calculation:
- The electronegativity difference of Na and F: [tex]\( 3.05 \)[/tex]
- Prediction for ionic bond formation: Yes
Thus, Na and F form an ionic bond.
