Arrange the bonds in order of increasing ionic character in the molecules: $$LiF$$, $$K_2O$$, $$N_2$$, $$SO_2$$ and $$ClF_3$$.

To compare the ionic character of different bonds we use the Pauling concept:

• Ionic character increases with the electronegativity difference $$\Delta\chi = \chi_{\text{more}} - \chi_{\text{less}}$$ between the two bonded atoms.
• A convenient estimate of percentage ionic character is

$$\%\,\text{ionic} = \left(1 - e^{-\frac{(\Delta\chi)^2}{4}}\right)\times100$$

Although we do not need the exact percentages, the formula shows that a larger $$\Delta\chi$$ means a more ionic bond.

The Pauling electronegativity (EN) values needed are
$$\chi_{Li}=1.0,\; \chi_{K}=0.8,\; \chi_{N}=3.0,\; \chi_{S}=2.5,\; \chi_{Cl}=3.0,\; \chi_{O}=3.5,\; \chi_{F}=4.0$$

Now calculate $$\Delta\chi$$ for the representative bond in each molecule:

$$\begin{aligned} N_2: &\; \Delta\chi = 3.0-3.0 = 0 \\[2pt] SO_2\;(S{-}O): &\; \Delta\chi = 3.5-2.5 = 1.0 \\[2pt] ClF_3\;(Cl{-}F): &\; \Delta\chi = 4.0-3.0 = 1.0 \\[2pt] K_2O\;(K{-}O): &\; \Delta\chi = 3.5-0.8 = 2.7 \\[2pt] LiF\;(Li{-}F): &\; \Delta\chi = 4.0-1.0 = 3.0 \end{aligned}$$

Interpretation:

• $$N_2$$ has $$\Delta\chi = 0$$, so it is perfectly covalent.
• $$K_2O$$ and $$LiF$$ have the largest $$\Delta\chi$$ values; their bonds are highly ionic.
• $$SO_2$$ and $$ClF_3$$ tie at $$\Delta\chi = 1.0$$. However, the S-O bonds in $$SO_2$$ possess significant multiple-bond (π-bond) character because of $$p\pi\!-\!d\pi$$ back-bonding and resonance. Multiple bonding increases covalent character and therefore decreases ionic character when compared with a single S-O bond of the same $$\Delta\chi$$. The Cl-F bonds in $$ClF_3$$ are simple single bonds, so their ionic character is slightly higher than that of the S-O bonds.

Combining all the points, the bonds arranged in order of increasing ionic character are

$$N_2 \lt SO_2 \lt ClF_3 \lt K_2O \lt LiF$$

This corresponds to Option C.