All naturally occuring polysaccharide have one terminal residue a free anomeric carbon why do these polysaccharide not give a positive chemical test for reducing sugar

 The concentration of reducing ends is too small to detect.

Explanation:

The reducing sugar can be defined as the sugar that is capable of acting as a reducing agent because it possesses an aldehyde group and a ketone group. Some polysaccharides are reducing sugars but all the monosaccharides are reducing sugars. The polysaccharide with reducing sugars do not give a positive test for reducing sugars because the concentration or amount of reducing ends is too small to detect.

you did not give any options thus this is the best answer i can give to you with this little information you have given

> they are both natural and not man made

> they can be tracked and predicted by the right tools

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c is having 4 valence electrons and is bonded to three hydrogens and one sulphur atom, therefore, the number of bond pair is equal to four. as all the valence electrons are used in forming the bonds so, there is no lone pair left on carbon.

s is having 6 valence electrons and is bonded to one other sulfur atom and one carbon atom, therefore, the number of bond pair is equal to 2. as only two valence electrons are used in forming the bonds so, four electrons are left. therefore, two lone pairs are present on s.