A diol of formula $\mathrm{C}{}_5\mathrm{H}{}_{12}\mathrm{O}{}_2$ consumes one mole of $\mathrm{HIO}{}_4$ and gives only methanal and butan-2-one. The diol is

Cleavage must separate a $\mathrm{CH}{}_2\mathrm{OH}$ carbon (giving $\mathrm{HCHO}$) from a carbon bearing OH, $\mathrm{CH}{}_3$ and $\mathrm{C}{}_2\mathrm{H}{}_5$ (giving butan-2-one). That is $\mathrm{HOCH}{}_2-\mathrm{C}(\mathrm{OH})(\mathrm{CH}{}_3)-\mathrm{CH}{}_2\mathrm{CH}{}_3$, 2-methylbutane-1,2-diol. Pentane-1,2-diol gives butanal, the 3-methyl isomer gives 2-methylpropanal, and 2-methylbutane-2,3-diol gives propanone and ethanal.