The number of unpaired d-electrons in [Co(H2O)6]3+ is:

🧠 The "$\mathrm{Co^{3+}}$ Exception"

Water is normally a moderate-field ligand, but with the high-charge $\mathrm{Co^{3+}}$ centre something special happens: the strong electrostatic pull of the $+3$ ion contracts the metal $\mathrm{d}$-orbitals enough to make $\Delta_o$ unusually large. So $[\mathrm{Co(H_2O)_6}]^{3+}$ is low-spin, against the usual "$\mathrm{H_2O}$ → high-spin" intuition. This is one of the most-tested exceptions in CFT.

🗺️ Walk the Configuration

Cobalt: $Z = 27$, $[\mathrm{Ar}]\,3\mathrm{d}^7\,4\mathrm{s}^2$. Ionise to $\mathrm{Co^{3+}}$ → $[\mathrm{Ar}]\,3\mathrm{d}^6$.

Ligand $\mathrm{H_2O}$, oxidation state +3 → low-spin $\mathrm{d}^6$ on $\mathrm{Co}$:

• $\mathrm{t_{2g}^6\,e_g^0}$
• All six electrons paired in the lower set.

Number of unpaired electrons = 0.

⚡ The Two-Tier $\mathrm{H_2O}$ Rule

Memorise this exception:

• $\mathrm{H_2O}$ + $\mathrm{M^{2+}}$ (or weaker pull) → high-spin (e.g. $[\mathrm{Fe(H_2O)_6}]^{2+}$, $[\mathrm{Mn(H_2O)_6}]^{2+}$).
• $\mathrm{H_2O}$ + $\mathrm{Co^{3+}}$ → low-spin.

So whenever you see $\mathrm{Co^{3+}}$ with water, expect 0 unpaired electrons.

⚠️ The Sloppy "H₂O is Always Weak" Reflex

Many students learn the spectrochemical series as a strict ladder and apply "H₂O = weak field" reflexively. That gives high-spin $\mathrm{t_{2g}^4\,e_g^2}$ → 4 unpaired, which is option (4) — the JEE distractor. Always check the metal's oxidation state alongside the ligand: $+3$ centres often push borderline ligands into the strong-field regime.

$\boxed{\text{Answer: (3) 0}}$