The spin magnetic moment of cobalt in the compound `Hg[Co(SCN)_(4)]` is

Options

A. `sqrt(3)`
B. `sqrt(8)`
C. `sqrt(15)`
D. `sqrt(24)`

Answer

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

### Step 1: Identify the oxidation state of cobalt
The given compound is \( \text{Hg[Co(SCN)}_4] \). We know that mercury (Hg) in this compound is in the +1 oxidation state. The thiocyanate ion (SCN) has a charge of -1. Since there are four SCN ions, the total negative charge contributed by them is \( 4 \times (-1) = -4 \).

Let the oxidation state of cobalt be \( x \). The overall charge of the coordination complex must be neutral, so we can set up the equation:

\[ x + (-4) + (+1) = 0 \]

Solving for \( x \):

\[ x - 4 + 1 = 0 \\ x - 3 = 0 \\ x = +3 \]

Thus, cobalt is in the +3 oxidation state.

### Step 2: Determine the electronic configuration of cobalt in the +3 state
Cobalt (Co) has an atomic number of 27, and its ground state electronic configuration is:

\[ \text{Co: } [\text{Ar}] 4s^2 3d^7 \]

When cobalt loses three electrons to form \( \text{Co}^{3+} \), the configuration becomes:

\[ \text{Co}^{3+}: [\text{Ar}] 3d^6 \]

### Step 3: Determine the number of unpaired electrons
The \( 3d^6 \) configuration can be represented in terms of the distribution of electrons in the d-orbitals. The filling of the d-orbitals follows Hund's rule, which states that electrons will fill degenerate orbitals singly before pairing up.

The distribution of 6 electrons in the d-orbitals is as follows:

- 1st orbital: 2 electrons (paired)
- 2nd orbital: 2 electrons (paired)
- 3rd orbital: 2 electrons (paired)

Thus, in the \( 3d^6 \) configuration, there are **4 unpaired electrons**.

### Step 4: Calculate the spin magnetic moment
The formula for calculating the spin magnetic moment (\( \mu \)) is given by:

\[ \mu = \sqrt{n(n + 2)} \]

where \( n \) is the number of unpaired electrons. In this case, \( n = 4 \):

\[ \mu = \sqrt{4(4 + 2)} \\ \mu = \sqrt{4 \times 6} \\ \mu = \sqrt{24} \\ \mu = 2\sqrt{6} \]

### Final Answer
The spin magnetic moment of cobalt in the compound \( \text{Hg[Co(SCN)}_4] \) is \( 2\sqrt{6} \) BM (Bohr Magneton).

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