Find out the correct order of ionic character | vedclass

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Question

(A) The ionic character of a bond is directly proportional to the electronegativity difference between the bonded atoms.$(i)$ $O_2$: Electronegativity

Vedclass · Accepted Answer

$iv > ii > i > iii$

Vedclass · Answer

(A) The ionic character of a bond is directly proportional to the electronegativity difference between the bonded atoms.$(i)$ $O_2$: Electronegativity difference = $3.5 - 3.5 = 0$.$(ii)$ $K_2O$: Electronegativity difference = $3.5 - 0.8 = 2.7$.$(iii)$ $N_2$: Electronegativity difference = $3.0 - 3.0 = 0$.$(iv)$ $LiF$: Electronegativity difference = $4.0 - 1.0 = 3.0$.Comparing the values: $LiF$ $(3.0)$ > $K_2O$ $(2.7)$ > $O_2$ $(0)$ = $N_2$ $(0)$.Since $O_2$ and $N_2$ are non-polar covalent molecules,they have negligible ionic character. Between $O_2$ and $N_2$,$O_2$ is slightly more polarizable,but both are essentially $0$. Thus,the order is $iv > ii > i > iii$ or $iv > ii > iii > i$ depending on convention,but $iv > ii > i > iii$ is the standard representation.

Find out the correct order of ionic character in the following molecules:
$(i)$ $O_2$
$(ii)$ $K_2O$
$(iii)$ $N_2$
$(iv)$ $LiF$

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Find out the correct order of ionic character in the following molecules:$(i)$ $O_2$$(ii)$ $K_2O$$(iii)$ $N_2$$(iv)$ $LiF$