Let's analyze each statement:
A) It is an aldohexose:
This is true. Glucose is a six-carbon sugar (hexose) with an aldehyde functional group (aldo-).
B) On heating with HI it forms n-hexane:
This is also true. HI is a strong reducing agent that removes all oxygen atoms from glucose, reducing it to n-hexane.
C) It exists in furanose form:
This is partially true but misleading. While glucose can exist in both pyranose (six-membered ring) and furanose (five-membered ring) forms, the pyranose form is overwhelmingly predominant in solution. The furanose form exists only in minor amounts.
D) It does not give Schiff's test:
This is false. Glucose does give Schiff's test because its open-chain form contains an aldehyde group (in equilibrium with cyclic forms). Schiff's reagent reacts with aldehydes to produce a pink-magenta color.
Conclusion:
While statements A, B, and C have elements of truth, statement D is completely incorrect. The statement that is not true is (D).
Final Answer:
The incorrect statement is (D).