To solve the question, we will analyze both the assertion and the reason step by step.
### Step 1: Determine the oxidation state of iron in `[Fe(CN)_(6)]^(3-)`
- The formula for the complex ion is `[Fe(CN)_(6)]^(3-)`.
- Each cyanide ion (CN) has a charge of -1. Therefore, for six cyanide ions, the total charge contributed is -6.
- Let the oxidation state of iron be \( x \).
- The overall charge of the complex is -3.
Setting up the equation:
\[ x + (-6) = -3 \]
\[ x - 6 = -3 \]
\[ x = +3 \]
**Hint:** To find the oxidation state, set up an equation based on the total charge of the complex and the charges of the ligands.
### Step 2: Determine the electron configuration of iron in the +3 oxidation state
- The atomic number of iron (Fe) is 26. Its ground state electron configuration is \( [Ar] 4s^2 3d^6 \).
- For \( Fe^{3+} \), we remove 3 electrons:
- The two 4s electrons and one from the 3d subshell.
Thus, the electron configuration for \( Fe^{3+} \) is:
\[ 3d^5 \]
**Hint:** Remember that when determining the electron configuration for a cation, you remove electrons starting from the outermost shell.
### Step 3: Analyze the magnetic properties of `[Fe(CN)_(6)]^(3-)`
- The presence of cyanide (CN) as a ligand is important. CN is a strong field ligand, which means it causes pairing of electrons in the d-orbitals.
- In a strong field ligand environment, the 3d electrons will pair up, leading to a low-spin configuration.
Filling the d-orbitals:
- The 3d orbitals will fill as follows for \( Fe^{3+} \):
- T2g: ↑↓ ↑↓ ↑ (pairing occurs)
- Eg: ↑ (one unpaired electron remains)
Thus, in the case of \( [Fe(CN)_{6}]^{3-} \), there is only **one unpaired electron**.
**Hint:** Strong field ligands can lead to pairing of electrons, affecting the magnetic properties of the complex.
### Step 4: Determine the hybridization of `[Fe(CN)_(6)]^(3-)`
- The coordination number of iron in this complex is 6, which suggests octahedral geometry.
- The hybridization for an octahedral complex involving d-orbitals is typically \( d^2sp^3 \).
In this case, the hybridization is indeed \( d^2sp^3 \).
**Hint:** For octahedral complexes, remember that the hybridization involves two d-orbitals, one s-orbital, and three p-orbitals.
### Conclusion
- **Assertion (A)**: `[Fe(CN)_(6)]^(3-)` shows magnetic moment corresponding to **two unpaired electrons**. This is **false**; it actually has **one unpaired electron**.
- **Reason (R)**: It has `d^(2)sp^(3)` type hybridization. This is **true**.
Thus, the correct answer is that the assertion is false and the reason is true.
### Final Answer
The assertion is false and the reason is true.
