Decreasing order of the hydrogen bonding in following forms of water is correctly represented by
A. Liquid water
B. Ice
C. Impure water

Hydrogen bonding depends on the structural arrangement and purity of water. In ice, every water molecule is hydrogen-bonded to exactly four neighbouring water molecules in a rigid, three-dimensional tetrahedral lattice. This maximises the number of hydrogen bonds per molecule.

In liquid water, thermal energy partially disrupts this ordered network. While extensive hydrogen bonding still exists, some bonds are continuously broken and reformed, so the average number of hydrogen bonds per molecule is fewer than in ice.

In impure water, dissolved solutes (ions, molecules, or other impurities) interfere with the hydrogen-bonding network. The impurity particles occupy space and interact with water molecules differently, further reducing the extent of hydrogen bonding compared to pure liquid water.

Therefore, the decreasing order of hydrogen bonding is $$\text{Ice (B)} > \text{Liquid Water (A)} > \text{Impure Water (C)}$$, i.e., $$B > A > C$$.

Hence, the correct answer is Option 2.