One mole of open-chain D-glucose, $\mathrm{HOCH}{}_2(\mathrm{CHOH}){}_4\mathrm{CHO}$, is treated with excess $\mathrm{HIO}{}_4$. Which statements are correct?

All six carbons are oxygenated and adjacent, so there are 5 C-C bonds to cleave and 5 mol of $\mathrm{HIO}{}_4$ are used. The terminal $\mathrm{CH}{}_2\mathrm{OH}$ gives $\mathrm{HCHO}$, and the other five carbons (C1 to C5) each end up as $\mathrm{HCOOH}$. No carbon becomes $\mathrm{CO}{}_2$ because there is no ketone carbon. Statements 1, 2 and 3 are correct.