Oxidation number of 'Co' in `Hg[Co(SCN)_(4)]`

Options

A. `+2`
B. `+1`
C. `+3`
D. `+5`

Explanation

To determine the oxidation number of cobalt (Co) in the coordination compound \( \text{Hg[Co(SCN)_4]} \), we will follow a systematic approach.

### Step-by-Step Solution:

1. **Identify the Components of the Compound:**
The compound consists of mercury (Hg), cobalt (Co), and four thiocyanate ligands (SCN).

2. **Assign Known Oxidation States:**
- Mercury (Hg) typically has an oxidation state of +2 in coordination compounds.
- The thiocyanate ion (SCN) has a charge of -1. Since there are four SCN ligands, the total contribution from SCN is \( 4 \times (-1) = -4 \).

3. **Set Up the Equation:**
The overall charge of the coordination complex is neutral (0). Therefore, we can set up the following equation to find the oxidation state of cobalt (let's denote it as \( x \)):
\[ \text{Oxidation state of Hg} + \text{Oxidation state of Co} + \text{Total charge from SCN} = 0 \]
Substituting the known values:
\[ (+2) + x + (-4) = 0 \]

4. **Solve for \( x \):**
Rearranging the equation gives:
\[ x - 2 = 0 \]
Therefore:
\[ x = +2 \]

5. **Conclusion:**
The oxidation number of cobalt (Co) in \( \text{Hg[Co(SCN)_4]} \) is +2.

### Final Answer:
The oxidation number of Co in \( \text{Hg[Co(SCN)_4]} \) is +2.