Spin only magnetic moment of the compound `Hg[Co(SCN)_(4)]` is `sqrt(x)` B.M. The value of 'x' is

Answer

To find the value of 'x' in the spin-only magnetic moment of the compound \( \text{Hg[Co(SCN)}_4] \), we will follow these steps:

### Step 1: Identify the oxidation state of cobalt
The compound can be dissociated as follows:
\[ \text{Hg}^{2+} + \text{[Co(SCN)}_4]^{2-} \]
Here, we need to find the oxidation state of cobalt (Co) in the complex \( \text{[Co(SCN)}_4]^{2-} \).

### Step 2: Calculate the oxidation state of cobalt
Let the oxidation state of cobalt be \( x \). The thiocyanate ion (SCN) has a charge of -1, and there are 4 SCN ions:
\[ x + 4(-1) = -2 \]
This simplifies to:
\[ x - 4 = -2 \implies x = +2 \]

### Step 3: Determine the electronic configuration of cobalt
Cobalt has an atomic number of 27. Its ground state electronic configuration is:
\[ \text{Co}: [\text{Ar}] 3d^7 4s^2 \]
In the +2 oxidation state, cobalt loses two electrons (typically from the 4s orbital first), resulting in:
\[ \text{Co}^{2+}: 3d^7 \]

### Step 4: Determine the number of unpaired electrons
The \( 3d^7 \) configuration can be represented as:
\[ \text{3d: } \uparrow \downarrow \, \uparrow \, \uparrow \, \uparrow \]
This shows that there are 3 unpaired electrons.

### Step 5: Use the formula for magnetic moment
The spin-only magnetic moment (\( \mu \)) is given by the formula:
\[ \mu = \sqrt{n(n + 2)} \, \text{B.M.} \]
where \( n \) is the number of unpaired electrons. Here, \( n = 3 \):
\[ \mu = \sqrt{3(3 + 2)} = \sqrt{3 \times 5} = \sqrt{15} \, \text{B.M.} \]

### Step 6: Relate to the given expression
The problem states that the magnetic moment is \( \sqrt{x} \, \text{B.M.} \). Therefore:
\[ \sqrt{x} = \sqrt{15} \]
Squaring both sides gives:
\[ x = 15 \]

### Final Answer
The value of \( x \) is \( \boxed{15} \).

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