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 Concept:
This question tests fundamental concepts of coordination chemistry. We need to analyze the complex ion $[Co(NH_3)_5SO_4]^+$.
- Coordination Number: The total number of coordinate bonds formed between the central metal and ligands.
- Oxidation Number: The residual charge on the central metal when all ligands are removed.
- d-electrons: The remaining electrons in the d-subshell of the metal ion.
- Unpaired electrons: Determined by crystal field theory, considering the ligand field strength.
Step 2: Key Formula or Approach:
1. Sum of (number of ligands $\times$ denticity) = Coordination Number.
2. (Metal charge) + (Ligand charges) = (Net complex charge).
3. $Co$ is atomic number 27. Electronic configuration $[Ar] 3d^7 4s^2$.
4. $Co^{3+}$ configuration: $[Ar] 3d^6$.
5. Octahedral field splitting: $t_{2g}$ and $e_g$. Ammine ($NH_3$) is a strong field ligand for $Co(III)$.
Step 3: Detailed Explanation:
1. Coordination Number: There are 5 $NH_3$ (monodentate) and 1 $SO_4^{2-}$ (monodentate in this complex). Total coordinate bonds = $5 + 1 = 6$.
2. Oxidation Number: Let $x$ be the charge on $Co$. Ammine is neutral ($0$), Sulfate is $-2$, Bromide is $-1$.
\[ x + 5(0) + (-2) + (-1) = 0 \rightarrow x = +3 \]
So, the oxidation state is +3.
3. Number of d-electrons: Neutral $Co$ has 27 electrons. $Co^{3+}$ has $27 - 3 = 24$ electrons. Relative to Argon (18), it has $24 - 18 = 6$ electrons in the $3d$ shell.
4. Unpaired d-electrons: $Co(III)$ in an octahedral field with strong ligands like $NH_3$ results in a low-spin configuration. The 6 d-electrons fill the lower energy $t_{2g}$ level and pair up completely: $t_{2g}^6 e_g^0$.
Thus, there are $0$ unpaired electrons.
The sequence is 6, 3, 6, 0.
Step 4: Final Answer:
The values are 6, 3, 6, and 0. The correct option is (A).