bond - Why is H2S more polar than HCl?

From EN alone we might expect $\ce{HCl}$ to be more polar but a book says that the bent structure of $\ce{H2S}$ gives rise to its polarity - which is also apparently more than $\ce{HCl}$. Why? Does this have to do with the lone pairs? Lone pairs supply electron density, right?

Answer

I have some additional comments. As mentioned previously, by the most straightforward measure of polarity for small molecules (net molecular dipole moment), $\ce{HCl}$ is undoubtedly more polar than $\ce{H2S}$. If some other measure of polarity is being considered, then it needs to be stated for clarity.

I suspect the authors of the book and some of its readers are taking the higher boiling point of $\ce{H2S}$ compared to $\ce{HCl}$ to mean the former is a more polar molecule. In fact, this behaviour holds for all of the group 15/16/17 molecular hydrides. However, this is only an indirect measure of polarity, and there are other confounding factors. In this particular case, I believe the reason $\ce{H2S}$ has a higher boiling point does not have to do with the strength of each intermolecular interaction, but the average number of intermolecular interactions which can be held at once, compared to its neighbouring hydrides ($\ce{PH3}$ and $\ce{HCl}$).

A hand-waving explanation is that the group 16 hydrides have a better "balance" of partially charged atoms compared to the group 15 and 17 hydrides in the same period. For example, looking at the hydrogen bonded molecules of the second period and considering an optimal hydrogen bonded network, $\ce{NH3}$ is capable of being the donor atom for three H-bonds and being the acceptor atom for only one H-bond. In $\ce{HF}$, the situation is inverted; now there are three H-bond acceptor sites, and only one H-bond donor. $\ce{H2O}$, however, is a convenient match of being a H-bond double donor and double acceptor. Thus, water performs more hydrogen bonds per molecule on average than either ammonia or hydrogen fluoride, and has a significantly higher boiling point, even though the actual hydrogen bonds are arguably strongest for $\ce{HF}$. The logic would be similar for the group 15/16/17 hydrides of the third period.

Lastly, I'd like to clarify the role of non-bonding pairs in the dipole moment of a molecule. Electrons in pure atomic orbitals are symmetric with respect to the nucleus and produce no net dipole. In a hybridization picture, one might say that non-hybridized $s$ and $p$ orbitals possess no dipole moment, while hybrid orbitals are not symmetric and do possess a dipole moment. In the case of the second period element hydrides, the lone pairs in hybridized $sp^3$ orbitals would provide a net dipole moment. However, for the hydrides in the third period and below, the molecular geometries are consistent with no hybridization (bond angles close to 90°), and therefore the lone pairs would be present in effectively $s$ and $p$ atomic orbitals, with no effect on a molecule's polarity.

In a more accurate picture without invoking hybridization, the molecular orbitals which contain the lone pairs are almost the same as the central atom's atomic orbitals in the heavier group 15/16/17 hydrides (and hence produce no net dipole), while for the second period hydrides the molecular orbitals are not as symmetric and hence actually produce a dipole.