Question

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

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

Hint:

$\text{Co}^{3+}$ is a classic $d^6$ system. Whenever a $d^6$ central metal ion encounters a strong-field ligand like $\text{NH}_3$ or $\text{CN}^-$, it pairs up perfectly to form a low-spin, completely diamagnetic (0 unpaired electrons) complex. Since the coordination number is 6, the shape must be octahedral.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
The question asks us to find the total number of unpaired electrons and determine the corresponding coordination geometry for the hexaamminecobalt(III) coordination complex ion, $[\text{Co}(\text{NH}_3)_6]^{3+}$.

Step 2: Key Formula or Approach:
1. Determine the oxidation state of the central metal ion (Co).
2. Write down its $d$-electron configuration.
3. Analyze the ligand field strength ($\text{NH}_3$) according to Crystal Field Theory (CFT) or Valence Bond Theory (VBT) to check for electron pairing.
4. Use the coordination number ($\text{CN} = 6$) to identify the geometric shape.

Step 3: Detailed Explanation:
5.

Oxidation State: Let the oxidation state of Cobalt be $x$. Since ammonia ($\text{NH}_3$) is a neutral molecule, its charge is 0: $$x + 6(0) = +3 \implies x = +3$$ The central metal ion is $\text{Co}^{3+}$. 6.

Electron Configuration: Neutral cobalt (Co, $Z=27$) has the configuration $[\text{Ar}] 3d^7 4s^2$. Stripping 3 electrons to form the trivalent cation yields: $$\text{Co}^{3+} = [\text{Ar}] 3d^6 4s^0 4p^0$$ 7.

Ligand Field and Pairing: In the presence of $\text{Co}^{3+}$, ammonia ($\text{NH}_3$) behaves as a strong-field ligand. This strong crystal field splitting energy ($\Delta_0 > P$) forces the 6 electrons in the $3d$ subshell to fully pair up in the lower-energy $t_{2g}$ orbitals ($t_{2g}^6 e_g^0$).
8. Because all 6 electrons are completely paired up, the total number of unpaired electrons is exactly

0.
9.

Geometry: The coordination number is 6, and the ion undergoes inner-orbital $d^2sp^3$ hybridization. A coordination number of 6 universally establishes an

octhedral geometry.

Step 4: Final Answer:
The complex has 0 unpaired electrons and an octahedral geometry, which perfectly maps to option (D).