Question

Identify the number of unpaired electrons present and geometry respectively of \([\text{Co(NH}_3)_6]^{3+}\) complex.

• A. 0, square planar
• B. 2, square planar
• C. 4, octahedral
• D. 0, octahedral

Hint:

Co3+ is d6; ammonia is a strong field ligand giving a low spin octahedral complex.

Answer 1

The Correct Option is D

Step 1: Coordination number:
Six ligands means octahedral geometry.

Step 2: Electron count:
$\text{Co}^{3+}$ has six d electrons. A strong field ligand like $\text{NH}_3$ makes the splitting larger than the pairing energy.

Step 3: Fill orbitals:
All six electrons fill the three $t_{2g}$ orbitals in pairs, leaving none unpaired. This is option (D).

Final Answer:
The complex has 0 unpaired electrons and octahedral shape. \[ \boxed{\text{(D) }0,\ \text{octahedral}} \]