Let's analyze each statement:
A) It is an aldohexose:
This statement is accurate. Glucose is a hexose (six-carbon sugar) featuring an aldehyde functional group, classifying it as an aldohexose.
B) On heating with HI it forms n-hexane:
This statement is also accurate. HI, a potent reducing agent, eliminates all oxygen atoms from glucose, reducing it to n-hexane.
C) It exists in furanose form:
This statement is partially true but can be misleading. Although glucose can exist in both pyranose (six-membered ring) and furanose (five-membered ring) forms, the pyranose form is significantly more prevalent in solution. The furanose form is present only in minor quantities.
D) It does not give Schiff's test:
This statement is false. Glucose yields a positive Schiff's test due to its open-chain form, which exists in equilibrium with its cyclic forms and contains an aldehyde group. Schiff's reagent reacts with aldehyde groups to produce a pink-magenta coloration.
Conclusion:
Statements A, B, and C contain elements of truth; however, statement D is entirely incorrect. The statement that is not true is (D).
Final Answer:
The incorrect statement is (D).