Answer
$H_{2}O, H_{2}S, H_{2}O_{2}$
$H_{2}O_{2}$ is a more complex molecule than $H_{2}O$ and $H_{2}S$.
Therefore, entropy of $H_{2}O_{2}$ is greater than that of $H_{2}O$ and $H_{2}S$.
Among $H_{2}O$ and $H_{2}S$, molar mass of $H_{2}S$ is higher. Therefore, entropy of $H_{2}S$ should be greater than that of $H_{2}O$.
Work Step by Step
$H_{2}O_{2}$ is a more complex molecule than $H_{2}O$ and $H_{2}S$.
Therefore, entropy of $H_{2}O_{2}$ is greater than that of $H_{2}O$ and $H_{2}S$.
Among $H_{2}O$ and $H_{2}S$, molar mass of $H_{2}S$ is higher. Therefore, entropy of $H_{2}S$ should be greater than that of $H_{2}O$.
The ranking in order of increasing $S^{\circ}$ is
$H_{2}O, H_{2}S, H_{2}O_{2}$
