3. As we discussed, the Oxygen atom of each water molecule can form two hydrogen

Question

3. As we discussed, the Oxygen atom of each water molecule can form two hydrogen bonds with its neighbor water molecules. As each water molecule is sitting at the center of a tetrahedral network made up of four other water molecules, (a) What's the number of different configurations for the water molecule in the center? (show your thinking or intermediate steps). (b) If one of the water molecules of the tetrahedron is replaced by a hydrophobic molecule and can no long form hydrogen bonds, what's the number of different configurations for the water molecule in the center now? (c) What's the change in entropy?

Explanation / Answer

a) the water molecule at the centre of the tetrahedron can change its orientation pointing to the four corner of tetrahedran. However they are all identical structurally. Water molecules can rearrange themselves into a cubic structure, more specifically diamond cubic. In diamond cubic structure, the carbon atoms in diamond (sp3 hybridized) are located at the centre of tetrahedron with four corners occupied by other C atoms. Therefore, number of configurations of water molecules is 2.

b) If a non-hydrogen bonding molecule is placed in the tetrahedral corner, then the hydrogen bonding network is broken around that molecule. That is, four tetrahedral sites are broken and hence a 'defective' tetrahedral or diamond cubic structure is formed. Therefore, there can only be one configuration for the central water molecule as the tetralhedral structure become planar trianguar structure.

c) The structure around the non-bonding molecule is 'frozen' - meaning that its movement is restricted - compared to the more dynamic H-bonding network, the entropy of system decreases.

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