Which of the following is/are inner orbital complex ?

Options

A. `[Fe(CN)_(6)]^(4-)`
B. `[Cr(NH_(3))_(6)]^(3+)`
C. `[Mn(CN)_(6)]^(3-)`
D. All of these

Answer

To determine which of the given complexes are inner orbital complexes, we need to analyze each complex based on the oxidation state of the central metal ion, its electron configuration, and the hybridization involved. Inner orbital complexes involve the use of inner d orbitals in hybridization.

### Step-by-Step Solution:

1. **Identify the Complexes**: The complexes given are:
- A) [Fe(CN)6]^(4-)
- B) [Cr(NH3)6]^(3+)
- C) [Mn(CN)6]^(3-)

2. **Determine the Oxidation State and Electron Configuration**:
- **Complex A: [Fe(CN)6]^(4-)**
- Let the oxidation state of Fe be \( x \).
- The charge of CN is -1, and there are 6 CN ligands:
\[ x + 6(-1) = -4 \implies x - 6 = -4 \implies x = +2 \]
- Iron (Fe) has an atomic number of 26, so its electron configuration is [Ar] 3d^6 4s^2.
- For Fe^2+, the configuration becomes [Ar] 3d^6 4s^0.
- In the presence of strong field ligands (CN^-), electron pairing occurs in the 3d orbitals. The hybridization is d^2sp^3, which involves the 3d orbitals. Thus, this is an inner orbital complex.

- **Complex B: [Cr(NH3)6]^(3+)**
- Let the oxidation state of Cr be \( y \).
- NH3 is a neutral ligand, so:
\[ y = +3 \]
- Chromium (Cr) has an atomic number of 24, so its electron configuration is [Ar] 3d^5 4s^1.
- For Cr^3+, the configuration becomes [Ar] 3d^3 4s^0.
- NH3 is a weak field ligand, so no pairing occurs. The hybridization is also d^2sp^3, which involves the 3d orbitals. Thus, this is also an inner orbital complex.

- **Complex C: [Mn(CN)6]^(3-)**
- Let the oxidation state of Mn be \( z \).
- The charge of CN is -1, and there are 6 CN ligands:
\[ z + 6(-1) = -3 \implies z - 6 = -3 \implies z = +3 \]
- Manganese (Mn) has an atomic number of 25, so its electron configuration is [Ar] 3d^5 4s^2.
- For Mn^3+, the configuration becomes [Ar] 3d^4 4s^0.
- CN is a strong field ligand, which leads to electron pairing. The hybridization is d^2sp^3, which involves the 3d orbitals. Thus, this is also an inner orbital complex.

3. **Conclusion**: All three complexes involve the inner d orbitals in their hybridization. Therefore, the answer is:
- **Option D: All of these are inner orbital complexes.**