Arrange the bonds in order of increasing ionic character in the molecules: LiF, K₂O, N₂, SO₂ and ClF₃.
N < SO < ClF < LiF < KO
Explanation
The ionic character of a bond depends on the difference in electronegativity between the bonded atoms. N consists of identical atoms sharing electron pairs (bond order 3), making it purely covalent with zero ionic character. SO and ClF involve different non-metals, resulting in polar covalent bonds with moderate ionic character. LiF and KO are formed between highly electropositive alkali metals and highly electronegative non-metals, leading to significant electron transfer and high ionic character. Among them, KO has the highest ionic character because potassium is more electropositive than lithium.
Solution Steps
Step 1: Identify purely covalent molecules. N has shared electron pairs between identical atoms (bond order 3), so it has the least ionic character.
Step 2: Identify polar covalent molecules. SO and ClF have bonds between different non-metals, giving them moderate ionic character (SO < ClF).
Step 3: Identify ionic molecules. LiF and KO are formed between highly electropositive alkali metals and highly electronegative non-metals, giving them the highest ionic character.
Step 4: Compare LiF and KO. K is more electropositive than Li, so KO has more ionic character than LiF.
Step 5: Arrange in increasing order: N < SO < ClF < LiF < KO.