The order of increasing polarity in HCl, $\mathrm{CO}_{2}$ , $\mathrm{H}_{2}\mathrm{O}$ and HF molecules is
A$\mathrm{CO}_{2}$ , HCl, $\mathrm{H}_{2}\mathrm{O}$ , HF
BHF, $\mathrm{H}_2\mathrm{O}$ , HCl, $\mathrm{CO}_{2}$
C$\mathrm{CO}_{2}$ , HCl, HF, $\mathrm{H}_{2} \mathrm{O}$
D$\mathrm{CC}_2$ , HF, $\mathrm{H}_2\mathrm{O}$ , HC
Answer & Solution
Correct answer: C. $\mathrm{CO}_{2}$ , HCl, HF, $\mathrm{H}_{2} \mathrm{O}$
Polarity depends on the net dipole moment of the molecule.
$\mathrm{CO}_{2}$ is linear, so the two bond dipoles cancel and its polarity is zero.
For the others, the approximate dipole moments increase as:
$$\mathrm{HCl} < \mathrm{HF} < \mathrm{H_2O}$$
This is because $\mathrm{H_2O}$ has a bent shape, so its bond dipoles add to give a larger net dipole moment than $\mathrm{HF}$.
Therefore, the increasing order is:
$$\mathrm{CO}_{2} < \mathrm{HCl} < \mathrm{HF} < \mathrm{H_2O}$$
Now compare with the given options. This matches option $\mathrm{C}$.
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