Identify the functional behavior of glucose

Glucose is an aldohexose, so in its open-chain form it contains an aldehyde group. However, in solution and in crystalline state, glucose exists predominantly in cyclic hemiacetal forms ($\alpha$ and $\beta$ forms).

Because of this, many aldehyde tests are affected by the very small concentration of free aldehyde form.

Check each option

Option A: Glucose gives Schiff's test for aldehyde

Schiff's reagent is a test for free aldehydes. Although glucose is a reducing sugar and can reduce Tollens' reagent and Fehling's solution, it does not give Schiff's test readily because the free aldehyde form is present in very small amount due to cyclic hemiacetal formation.

Therefore, this statement is not true.

Option B: Glucose reacts with hydroxylamine to form oxime

Hydroxylamine reacts with the carbonyl group of the open-chain form of glucose to form an oxime.

So this statement is true.

Option C: The pentaacetate of glucose does not react with hydroxylamine to give oxime

In glucose pentaacetate, the cyclic form is locked and the free aldehyde form is not available. Hence it does not react with hydroxylamine to form an oxime.

So this statement is true.

Option D: Glucose exists in two crystalline forms $\alpha$ and $\beta$

This is correct. Glucose exists as $\alpha$-D-glucose and $\beta$-D-glucose.

So this statement is true.

Conclusion

The only statement that is not true is: $\boxed{\text{A}}$