Why glucose and other similar carbohydrates are oxidised fully by HIO4? $\ce{HIO4_{(aq)}}$ requires the vicinal diols to have syn configuration, as it forms a cyclic periodate ester.



But glucose does not have syn configuration at C3, and if you consider cyclic forms of glucose it shouldn't oxidise fully at all because alternate hydroxy groups with anti configuration will be unreactive. But it still gets oxidised by $\ce{HIO4}$ to give 5 molecules of methanoic acid, $\ce{HCOOH}$, and one molecule of formaldehyde, $\ce{HCHO}$.
Glucose can react with $\ce{HIO4}$, which can be understood easily by looking at chair conformation of glucose.

Here is example of a diol, for more information looks at this answer https://chemistry.stackexchange.com/a/114836/67254