Understand the structure of $\mathrm{CuSO_4\cdot 5H_2O}$

The compound is better represented as: $[\mathrm{Cu(H_2O)_4}]\mathrm{SO_4}\cdot \mathrm{H_2O}$

This means:

• $4$ water molecules are directly coordinated to $\mathrm{Cu^{2+}}$.
• $1$ water molecule is present as crystal water.
• $\mathrm{SO_4^{2-}}$ is outside the primary coordination sphere.

Examine Statement I

Statement I says: In $\mathrm{CuSO_4\cdot 5H_2O}$, Cu–O bonds are present.

Since $\mathrm{Cu^{2+}}$ is coordinated by water molecules through oxygen atoms, there are definitely $\mathrm{Cu-O}$ coordinate bonds.

So, Statement I is correct.

Examine Statement II

Statement II says: In $\mathrm{CuSO_4\cdot 5H_2O}$, ligands coordinating with $\mathrm{Cu(II)}$ ion are O- and S-based ligands.

This is incorrect because:

• The coordinated ligands are water molecules, which are O-donor ligands.
• Sulfate is not acting as an S-donor ligand.
• Even if sulfate interacts weakly in the crystal structure, it would coordinate through oxygen, not sulfur.

Therefore, Statement II is incorrect.

Choose the correct option

• Statement I: Correct
• Statement II: Incorrect

Hence, the correct option is: $\boxed{\text{C}}$

Comparison with stored answer

Stored correct answer: $\mathrm{C}$

Our derived answer matches the stored answer.