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(A) The cyclic structure of glucose is supported by the following facts:$(i)$ Glucose does not give Schiff's test because the $-CHO$ group is involved

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$(i)$,$(ii)$ and $(iv)$

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(A) The cyclic structure of glucose is supported by the following facts:$(i)$ Glucose does not give Schiff's test because the $-CHO$ group is involved in hemiacetal formation and is not free.$(ii)$ Glucose exists in two different crystalline forms ($\alpha$-$D$-glucose and $\beta$-$D$-glucose) due to the formation of an anomeric carbon.$(iv)$ Pentaacetate of glucose does not react with hydroxylamine because the aldehyde group is not free (it is in the cyclic hemiacetal form).Statement $(iii)$ describes the oxidation of glucose to saccharic acid,which is a property of the open-chain structure,not the cyclic structure.Therefore,statements $(i)$,$(ii)$,and $(iv)$ support the cyclic structure of glucose.

Which of the following statements support the cyclic form for glucose?
$(i)$ It does not give Schiff's test.
$(ii)$ It is found to exist in two different crystalline forms.
$(iii)$ It oxidises with nitric acid to give saccharic acid.
$(iv)$ Pentaacetate of glucose does not react with hydroxylamine.

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Which of the following statements support the cyclic form for glucose?$(i)$ It does not give Schiff's test.$(ii)$ It is found to exist in two different crystalline forms.$(iii)$ It oxidises with nitric acid to give saccharic acid.$(iv)$ Pentaacetate of glucose does not react with hydroxylamine.