Ionic bonds generally form when the bonded elements have a difference in electronegativity greater than 1.5 to 1.7. Subtract the electronegativities for the following pairs of elements and predict whether they form ionic bonds.
Electronegativity difference of Na and F:
Answer: Yes, sodium and fluorine form ionic bonds.
Electronegativity of sodium (Na) is 0.93.
Electronegativity of fluorine (F) is 3.98.
Electronegativity difference of Na and F will be calculated as follows.
Electronegativity difference = Electronegativity of F – Electronegativity of Na
= 3.98 – 0.93
Thus, difference in electronegativity greater than 1.5 to 1.7, hence, bond between sodium and fluorine is ionic.