Statement-I: ${∆}_{\mathrm{f}}^0\mathrm{H} \mathrm{of} {\mathrm{H}}_2\mathrm{O} \mathrm{and} {\mathrm{H}}_2\mathrm{S}$ is negative

Statement-II: ${\mathrm{TeO}}_2$ is an oxidizing agent

I) Thermal stability of ${\mathrm{H}}_2\mathrm{O}>{\mathrm{H}}_2\mathrm{S}>{\mathrm{H}}_2\mathrm{Se}>{\mathrm{H}}_2\mathrm{Te}>{\mathrm{H}}_2\mathrm{Po}.$ ${∆}_{\mathrm{f}}^0\mathrm{H} \mathrm{of} {\mathrm{H}}_2\mathrm{O} \mathrm{and} {\mathrm{H}}_2\mathrm{S}$ is negative indicating their high thermal stability.

II) ${\mathrm{TeO}}_2\left({\mathrm{Te}}^{+4}\right)$ is an oxidizing agent because ${\mathrm{Te}}^{+4}$ has a great tendency to reduce to ${\mathrm{Te}}^{+2}$. Down the group lower oxidation states become stable.