`[RhF_(6)]^(3-)` complex ion is :

Options

A. Outer orbital complex
B. Inner orbital complex
C. Neither outer nor inner orbital complex
D. Heteroleptic complex

Explanation

To determine the nature of the complex ion \([RhF_6]^{3-}\), we will follow these steps:

### Step 1: Determine the Oxidation State of Rhodium (Rh)
Let the oxidation state of Rhodium be \(x\). The charge on each fluorine atom (F) is \(-1\), and since there are 6 fluorine atoms, the total contribution from fluorine is \(-6\). The overall charge of the complex ion is \(-3\).

Setting up the equation:
\[ x + 6(-1) = -3 \]
\[ x - 6 = -3 \]
\[ x = +3 \]

### Step 2: Determine the Electronic Configuration of Rhodium
The atomic number of Rhodium (Rh) is 45. The ground state electronic configuration is:
\[ [Kr] 4d^8 5s^1 \]
When Rhodium is in the +3 oxidation state, it loses 3 electrons. The electrons are lost from the 5s and 4d orbitals:
\[ Rh^{3+} : [Kr] 4d^6 \]

### Step 3: Identify the Type of Complex
The coordination number of the complex \([RhF_6]^{3-}\) is 6, indicating an octahedral geometry. In octahedral complexes, there are two possible hybridizations:
- **Inner Orbital Complex**: \(d^2sp^3\) (using \(n-1 d\) orbitals)
- **Outer Orbital Complex**: \(sp^3d^2\) (using \(nd\) orbitals)

### Step 4: Analyze the Ligand Field Strength
Fluoride (F\(^-\)) is a weak field ligand, which means it does not cause significant pairing of electrons in the d-orbitals. Therefore, we expect that the electrons in the \(4d\) orbitals will remain unpaired.

### Step 5: Determine the Hybridization
Since Rhodium in the +3 state has the electronic configuration \(4d^6\) and the ligand is weak field, the electrons will occupy the higher energy \(4d\) orbitals without pairing. Therefore, the hybridization will be \(d^2sp^3\), which corresponds to an inner orbital complex.

### Conclusion
The complex ion \([RhF_6]^{3-}\) is an **inner orbital complex**.

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