Given below are two statements: Statement I: In CuSO₄·5H₂O, Cu–O bonds are present. Statement II: In CuSO₄·5H₂O,…

🧠 Read the Crystal Structure of Blue Vitriol

$\mathrm{CuSO_4 \cdot 5H_2O}$ is best written as $[\mathrm{Cu(H_2O)_4}]\mathrm{SO_4 \cdot H_2O}$.

• 4 water molecules sit in the coordination sphere bonded to $\mathrm{Cu^{2+}}$ via Cu–O bonds (water donates through O lone pairs).
• The 5th water is outside the coordination sphere, hydrogen-bonded to $\mathrm{SO_4^{2-}}$ and to the coordinated waters.
• Sulfate provides 2 weakly-axial Cu–O contacts (or none, depending on phase) — also Cu–O.

So Statement I is true: every Cu-ligand bond involves oxygen.

Statement II says the ligands are O- and S-based. Sulfur in sulfate is the central S; it sits buried with four oxygens around it and never approaches Cu directly. Cu coordinates only to O, not S. Statement II is false.

🗺️ Verdict

I correct ✓, II incorrect ✗ → option (3).

⚡ The "Sulfate Donor = O" Rule

Whenever sulfate (or phosphate, nitrate, carbonate) coordinates to a metal, it does so through oxygen — never the central S/P/N/C atom. Sulfate is an O-donor, full stop.

⚠️ The "It Has S, So It Must Donate via S" Trap

Just because $\mathrm{SO_4^{2-}}$ contains sulfur doesn't make it an S-donor. The donor atom is whichever atom carries the lone pair facing the metal. For sulfate, that's always O.

$\boxed{\text{Answer: (3) Statement I correct, Statement II incorrect}}$