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: Understanding the Concept:
Count the d electrons of the metal ion, then decide spin state from the ligand strength. Six ligands around the metal give an octahedral geometry.

Step 2: Detailed Explanation:
Co has atomic number 27 and \([\text{Ar}]3d^7 4s^2\). So \(\text{Co}^{3+}\) is \(3d^6\).
\(\text{NH}_3\) is a strong field ligand, so the crystal field splitting is large and electrons pair up: \(t_{2g}^6 e_g^0\).
Unpaired electrons = 0, so the complex is diamagnetic.
Six \(\text{NH}_3\) ligands give an octahedral shape (\(d^2sp^3\) hybridisation).
Options (A) and (B) call it square planar, which needs 4 ligands. Option (C) gives 4 unpaired electrons, which is the high spin case and does not occur with ammonia.

Step 3: Final Answer:
0 unpaired electrons and octahedral, option (D).

Final Answer:
Low spin d6 gives 0 unpaired electrons, octahedral. \[ \boxed{\text{(D) }0,\ \text{octahedral}} \]