`[Fe(CN)_(6)]^(4-)` is

Options

A. inner complex
B. outer complex
C. square planar
D. trigonal bipyramidal

Answer

To determine the geometry of the complex ion \([Fe(CN)_6]^{4-}\), we can follow these steps:

### Step 1: Determine the oxidation state of Iron (Fe)
The overall charge of the complex ion is \(-4\). The cyanide ion (CN) has a charge of \(-1\). Since there are 6 cyanide ions, the total negative charge contributed by CN is:
\[ 6 \times (-1) = -6 \]
Let the oxidation state of Fe be \(x\). The equation for the overall charge can be set up as follows:
\[ x + (-6) = -4 \]
Solving for \(x\):
\[ x - 6 = -4 \implies x = +2 \]
Thus, the oxidation state of Fe in \([Fe(CN)_6]^{4-}\) is \(+2\).

**Hint:** Remember to account for the charges of the ligands when calculating the oxidation state.

### Step 2: Identify the electron configuration of Fe
Iron (Fe) has an atomic number of 26. The electron configuration of neutral Fe is:
\[ [Ar] 3d^6 4s^2 \]
Since we have determined that the oxidation state of Fe is \(+2\), we need to remove two electrons from the outermost shell (4s). Therefore, the electron configuration for \(Fe^{2+}\) becomes:
\[ [Ar] 3d^6 \]

**Hint:** The oxidation state helps you determine how many electrons to remove from the electron configuration.

### Step 3: Determine the nature of the ligand
Cyanide (CN) is a strong field ligand. Strong field ligands cause pairing of electrons in the d-orbitals. In the case of \(Fe^{2+}\) with a \(d^6\) configuration, the electrons will pair up in the lower energy \(d\) orbitals.

**Hint:** Strong field ligands lead to electron pairing, which affects the geometry of the complex.

### Step 4: Determine the hybridization
With \(d^6\) configuration and pairing of electrons, the hybridization for the complex can be determined. Since we have 6 ligands (CN) surrounding the central metal ion (Fe), the hybridization will be:
\[ d^2sp^3 \]
This hybridization is consistent with octahedral geometry.

**Hint:** Count the number of ligands to determine the hybridization type.

### Step 5: Determine the geometry
The hybridization \(d^2sp^3\) corresponds to an octahedral geometry. Therefore, the geometry of the complex ion \([Fe(CN)_6]^{4-}\) is octahedral.

**Hint:** The hybridization can often give a direct indication of the geometry of the complex.

### Final Answer
The geometry of \([Fe(CN)_6]^{4-}\) is octahedral.