In compound `FeCl_(2)` the orbital angular momentum of last electron in its cation `&` magnetic moments (in Bohr Megneton) of this compound are :-

Options

A. `(sqrt(6))h,sqrt(35)`
B. `(sqrt(6))h,sqrt(24)`
C. `0,sqrt(35)`
D. none of these

Explanation

To solve the problem regarding the orbital angular momentum and magnetic moment of the last electron in the cation of `FeCl2`, we will follow these steps:

### Step 1: Determine the oxidation state of Iron in `FeCl2`
In `FeCl2`, chlorine has a charge of -1. Therefore, to balance the two chlorine atoms, iron must have a charge of +2. Thus, the oxidation state of iron (Fe) in `FeCl2` is +2.

### Step 2: Write the electronic configuration of `Fe^2+`
Iron (Fe) has an atomic number of 26. The electronic configuration of neutral Fe is:
\[ \text{Fe: } [Ar] 3d^6 4s^2 \]
When iron loses two electrons to form `Fe^2+`, it loses the two 4s electrons first:
\[ \text{Fe}^{2+}: [Ar] 3d^6 \]

### Step 3: Identify the last electron in `Fe^2+`
In the `Fe^2+` ion, the last electron is in the 3d subshell. The configuration is `3d^6`, meaning there are 6 electrons in the d-orbitals.

### Step 4: Calculate the orbital angular momentum (L)
The orbital angular momentum (L) for a d-orbital can be calculated using the formula:
\[ L = \sqrt{l(l + 1)} \hbar \]
For d-orbitals, \( l = 2 \):
\[ L = \sqrt{2(2 + 1)} \hbar = \sqrt{6} \hbar \]
Thus, the orbital angular momentum of the last electron is:
\[ L = \sqrt{6} \hbar \]

### Step 5: Calculate the magnetic moment
The magnetic moment (μ) can be calculated using the formula:
\[ \mu = \sqrt{n(n + 2)} \, \text{Bohr magneton} \]
where \( n \) is the number of unpaired electrons. In `Fe^2+`, the 3d subshell has 6 electrons, and the distribution of these electrons in the d-orbitals can be represented as follows:
- The 3d subshell can hold a maximum of 10 electrons, and with 6 electrons, we can have 4 unpaired electrons (following Hund's rule).

Thus, \( n = 4 \):
\[ \mu = \sqrt{4(4 + 2)} \]
\[ \mu = \sqrt{4 \times 6} = \sqrt{24} \, \text{Bohr magneton} \]

### Final Answers
- Orbital angular momentum \( L = \sqrt{6} \hbar \)
- Magnetic moment \( \mu = \sqrt{24} \, \text{Bohr magneton} \)