Answer:
AlCl₃ < NaI < HF < KBr < Li₂O
Explanation:
Electronegativity difference may help you to predict the sort of bond in a compound. We have to search for the electronegativity in the Periodic table and then, make a subtraction.
ΔEN = 0 → covalent bond, non polar
0 < ΔEN < 1.7 → covalent bond, polar
ΔEN ≥ 1, 7 → ionic bond
KBr → 3 - 0.8 ⇒ ΔEN = 2.2
AlCl₃→ 3.2 - 1.6 ⇒ ΔEN = 1.6
Li₂O → 3.4 - 1 ⇒ ΔEN = 2.4
NaI → 2.7 - 0.9 ⇒ ΔEN = 1.8
HF → 4 - 2.1 ⇒ ΔEN = 1.9
