entropy of vaporization vs entropy of fusion
Here I'm assuming you are interested in comparing the magnitude of $\Delta S^{\circ}_{fus}$ at $T_{fus}$ with that of $\Delta S^{\circ}_{vap}$ at $T_{vap}$, for the elements in their standard states (so, for instance, hydrogen would be $\ce{H_2}$ rather than $\ce{H}$).
Like you, I was unable to find a comparative tabulation of these values on the internet.  Fortunately, one can readily generate such a table using Wolfram Mathematica's chemical database.
For each element:
$$\Delta S^{\circ}_{fus} \text{ at } T_{fus} = \frac{\Delta H^{\circ}_{fus}}{T_{fus}}$$
$$\Delta S^{\circ}_{vap} \text{ at } T_{vap} = \frac{\Delta H^{\circ}_{vap}}{T_{vas}}$$
Wolfram has the above data for all of elements 1–93 (hydrogen through neptunium), except for helium (which can't be solidified at standard pressure, which is 1 bar), astatine, and francium.*
Here I've plotted $|\Delta S^{\circ}_{vap}|$ vs. $|\Delta S^{\circ}_{fus}|$* for these 90 elements, and added a y = x line. From the placement of the points relative to this line, you can see that all except one of the elements have $|\Delta S^{\circ}_{vap}| &gt; |\Delta S^{\circ}_{fus}|$.
That single exception is hydrogen, for which:
$$|\Delta S^{\circ}_{fus}| \text{ at } T_{fus} = 39.8 \frac{J}{mol K}$$
$$|\Delta S^{\circ}_{vap}| \text{ at } T_{vap} = 22.3 \frac{J}{mol K}$$

*Note, however, this complication:  Most, but not all, of these measurements were done at standard pressure (1 bar). For instance:
&quot;When heated at standard atmospheric pressure, arsenic changes directly from a solid to a gas, or sublimates, at a temperature of 887 K. In order to form liquid arsenic, the atmospheric pressure must be increased. At 28 times standard atmospheric pressure, arsenic melts at a temperature of 1090 K. If it were also measured at a pressure of 28 atmospheres, arsenic's boiling point would be higher than its melting point, as you would expect.&quot;
https://education.jlab.org/itselemental/ele033.html