Question

For the metal complex \([\text{Co}(\text{NH}_3)_5\text{SO}_4]\text{Br}\), coordination number, oxidation number, number of \(d\)-electrons and number of unpaired \(d\)-electrons are respectively

• A. $6, 3, 6, 0$
• B. $7, 2, 6, 2$
• C. $6, 2, 6, 0$
• D. $6, 2, 7, 0$

Hint:

Cobalt(III) ($d^6$) octahedral complexes are almost always low-spin (except with exceptionally weak-field ligands like $\text{F}^-$ or $[\text{CoF}_6]^{3-}$). In low-spin $d^6$ systems, all electrons are paired up in the $t_{2g}$ level, meaning the number of unpaired electrons is always $0$.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We need to find the coordination number, oxidation state, total number of $d$-electrons, and the number of unpaired $d$-electrons for the central metal ion in the coordination complex $[\text{Co}(\text{NH}_3)_5\text{SO}_4]\text{Br}$.


Step 2: Detailed Explanation:
Let us analyze each property systematically:
1. Coordination Number:
- Inside the coordination sphere, there are five ammine ($\text{NH}_3$) ligands and one sulfato ($\text{SO}_4^{2-}$) ligand.
- Both are monodentate ligands, so they form a total of $5 + 1 = 6$ coordinate bonds with the central cobalt ion.
- Therefore, the coordination number is $6$.
2. Oxidation Number:
- Let $x$ be the oxidation state of Cobalt ($\text{Co}$).
- Ammine ($\text{NH}_3$) is a neutral ligand (charge = 0).
- Sulfato ($\text{SO}_4^{2-}$) has a charge of $-2$.
- Bromide ($\text{Br}^-$) as a counterion has a charge of $-1$.
\[ x + 5(0) + (-2) + (-1) = 0 \implies x = +3 \]
- Therefore, Cobalt is in the $+3$ oxidation state ($\text{Co}^{3+}$).
3. Number of $d$-electrons:
- Neutral Cobalt ($\text{Co}$, atomic number 27) has the configuration: $[\text{Ar}] 3\text{d}^7 4\text{s}^2$.
- For $\text{Co}^{3+}$, we remove 3 electrons (2 from $4\text{s}$ and 1 from $3\text{d}$), giving the configuration: $[\text{Ar}] 3\text{d}^6$.
- Thus, the number of $d$-electrons is $6$.
4. Number of unpaired $d$-electrons:
- $\text{Co}^{3+}$ ($d^6$) in an octahedral complex with strong-field ammine ($\text{NH}_3$) ligands forms a low-spin complex.
- All 6 $d$-electrons pair up in the lower energy $t_{2g}$ orbitals:
\[ t_{2g}^6 e_g^0 \]
- Hence, the number of unpaired $d$-electrons is $0$.
Comparing these values with the options: $6$, $3$, $6$, $0$ corresponds to option (A).


Step 3: Final Answer:
The correct option is (A).