Consider the following: `[Co(CO_(3))(NH_(3))_(5)]ClO_(4)`
mark the correct option:

Options

A. `{:("Coordination no.","Oxidation no.","No. of d-electrons","Unpaired d-electron"),(6,3,6,0):}`
B. `{:("Coordination no.","Oxidation no.","No. of d-electrons","Unpaired d-electron"),(7,2,7,1):}`
C. `{:("Coordination no.","Oxidation no.","No. of d-electrons","Unpaired d-electron"),(7,1,6,4):}`
D. `{:("Coordination no.","Oxidation no.","No. of d-electrons","Unpaired d-electron"),(6,2,7,3):}`

Answer

To solve the problem regarding the coordination compound \([Co(CO_3)(NH_3)_5]ClO_4\), we will go through the following steps:

### Step 1: Identify the Coordination Number
The coordination number is determined by the number of ligands attached to the central metal atom. In this complex, we have:
- 5 ammonia ligands (NH₃), which are monodentate ligands.
- 1 carbonate ligand (CO₃²⁻), which is also a monodentate ligand.

Thus, the coordination number is:
\[ \text{Coordination Number} = 5 \, (\text{from } NH_3) + 1 \, (\text{from } CO_3) = 6 \]

### Step 2: Determine the Oxidation State of Cobalt
To find the oxidation state of cobalt (Co) in the complex, we can set up the following equation based on the charges:
Let the oxidation state of Co be \(x\).
- The charge from the carbonate ion (CO₃²⁻) is -2.
- The ammonia ligands are neutral.
- The overall charge of the complex is +1 (since it is balanced by the ClO₄⁻ ion).

Thus, we have:
\[ x + (-2) + 0 = +1 \]
Solving for \(x\):
\[ x - 2 = 1 \implies x = 3 \]
So, the oxidation state of cobalt is +3.

### Step 3: Calculate the Number of d-Electrons
Cobalt (Co) has an atomic number of 27, and its electronic configuration is:
\[ [Ar] \, 4s^2 \, 3d^7 \]
In the +3 oxidation state, cobalt loses 2 electrons from the 4s orbital and 1 electron from the 3d orbital:
\[ 3d^6 \]
Thus, the number of d-electrons in Co³⁺ is 6.

### Step 4: Determine the Number of Unpaired d-Electrons
Since CO₃²⁻ is a strong field ligand, it causes pairing of electrons in the d-orbitals. The distribution of the 6 d-electrons in the presence of strong field ligands would be:
\[ \text{d-orbital filling: } \uparrow\downarrow \, \uparrow\downarrow \, \uparrow\downarrow \]
All six electrons are paired, so the number of unpaired d-electrons is 0.

### Step 5: Determine the Hybridization
With a coordination number of 6, the hybridization can be determined as follows:
- For a coordination number of 6, the hybridization is \(d^2sp^3\).

### Summary of Results
- Coordination Number: 6
- Oxidation State of Co: +3
- Number of d-Electrons: 6
- Number of Unpaired d-Electrons: 0
- Hybridization: \(d^2sp^3\)

### Final Answer
The correct option is **A** (Coordination number = 6, Oxidation state = +3, d-electrons = 6, unpaired d-electrons = 0).