Principle used

Ionic character increases with increasing electronegativity difference, i.e. larger $\Delta \chi$ generally means more ionic bond character.

Compare the given bonds

We consider the main bond in each molecule:

• $\mathrm{N_2}$ : bond is $\mathrm{N-N}$
• $\mathrm{SO_2}$ : bond is $\mathrm{S-O}$
• $\mathrm{ClF_3}$ : bond is $\mathrm{Cl-F}$
• $\mathrm{K_2O}$ : bond is $\mathrm{K-O}$
• $\mathrm{LiF}$ : bond is $\mathrm{Li-F}$

Use electronegativity differences

Approximate Pauling electronegativities:

• $\chi(\mathrm{N}) \approx 3.0$
• $\chi(\mathrm{S}) \approx 2.5$
• $\chi(\mathrm{O}) \approx 3.5$
• $\chi(\mathrm{Cl}) \approx 3.0$
• $\chi(\mathrm{F}) \approx 4.0$
• $\chi(\mathrm{K}) \approx 0.8$
• $\chi(\mathrm{Li}) \approx 1.0$

Therefore,

• For $\mathrm{N_2}$: $\Delta \chi = 3.0 - 3.0 = 0$
• For $\mathrm{SO_2}$: $\Delta \chi = 3.5 - 2.5 = 1.0$
• For $\mathrm{ClF_3}$: $\Delta \chi = 4.0 - 3.0 = 1.0$
• For $\mathrm{K_2O}$: $\Delta \chi = 3.5 - 0.8 = 2.7$
• For $\mathrm{LiF}$: $\Delta \chi = 4.0 - 1.0 = 3.0$

Refine comparison between $\mathrm{SO_2}$ and $\mathrm{ClF_3}$

Both have similar $\Delta \chi$, but $\mathrm{S-O}$ bond is generally more polar than $\mathrm{Cl-F}$ based on actual electronegativity values:

$\chi(\mathrm{S}) \approx 2.58,\quad \chi(\mathrm{O}) \approx 3.44 \Rightarrow \Delta \chi \approx 0.86$ $\chi(\mathrm{Cl}) \approx 3.16,\quad \chi(\mathrm{F}) \approx 3.98 \Rightarrow \Delta \chi \approx 0.82$

So, $\mathrm{ClF_3} < \mathrm{SO_2}$ in ionic character.

Overall increasing order

Thus the order is:

$\mathrm{N_2} < \mathrm{ClF_3} < \mathrm{SO_2} < \mathrm{K_2O} < \mathrm{LiF}$

Match with options

This corresponds to Option D.

Comparison with stored answer

Stored answer is A: $\mathrm{N_2} < \mathrm{SO_2} < \mathrm{ClF_3} < \mathrm{K_2O} < \mathrm{LiF}$

But since $\Delta \chi(\mathrm{S-O}) > \Delta \chi(\mathrm{Cl-F})$, the correct order should place $\mathrm{ClF_3}$ before $\mathrm{SO_2}$.

Hence, I disagree with the stored answer.