In the complex `[Co(NH_3)_(6)][CdCl_x]` the oxidation number of cobalt is +3 . The value of x is

Options

A. 3
B. 4
C. 2
D. 5

Answer

To find the value of \( x \) in the complex \([Co(NH_3)_6][CdCl_x]\) given that the oxidation number of cobalt (Co) is +3, we can follow these steps:

### Step 1: Identify the oxidation states of the components
1. The oxidation state of cobalt (Co) is given as +3.
2. The ammonia (NH₃) is a neutral ligand, so its oxidation state is 0.
3. The cadmium (Cd) in the complex is in the +2 oxidation state because it is in group 12 of the periodic table.

### Step 2: Write the equation for the overall charge of the complex
The overall charge of the complex must be neutral (0). Therefore, we can set up the equation based on the oxidation states:

\[ \text{Charge of } [Co(NH_3)_6] + \text{Charge of } [CdCl_x] = 0 \]

### Step 3: Calculate the charge of the cobalt complex
The charge of the cobalt complex can be calculated as follows:
- The charge contributed by cobalt is +3.
- The charge contributed by the six ammonia ligands is \( 6 \times 0 = 0 \).

Thus, the total charge from the cobalt complex is:

\[ +3 + 0 = +3 \]

### Step 4: Set up the equation for the cadmium complex
Now, we need to consider the cadmium complex \([CdCl_x]\):
- The charge of cadmium is +2.
- The charge contributed by \( x \) chloride ions (Cl) is \( -1 \times x \).

So, the total charge from the cadmium complex is:

\[ +2 - x \]

### Step 5: Set the total charge to zero
Now we can set up the equation:

\[ 3 + (2 - x) = 0 \]

### Step 6: Solve for \( x \)
Now, simplify and solve for \( x \):

\[ 3 + 2 - x = 0 \]
\[ 5 - x = 0 \]
\[ x = 5 \]

### Conclusion
The value of \( x \) is 5.

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