Which of the following complex is an outer orbital complex?

Options

A. `[Ni(NH_(3))_(6)]^(2+)`
B. `[Mn(CN)_(6)]^(4-)`
C. `[Co(NH_(3))_(6)]^(3+)`
D. `[Fe(CN)_(6)]^(4-)`

Answer

To determine which of the following complexes is an outer orbital complex, we need to analyze the oxidation states and electronic configurations of the central metal ions in each complex. An outer orbital complex is formed when the hybridization involves the outer d orbitals (i.e., the d orbitals from the penultimate shell).

### Step-by-Step Solution:

1. **Identify the Complexes**: We need to analyze the given complexes. Let's assume we have the following complexes:
- A) [Ni(NH3)6]²⁺
- B) [Co(NH3)6]²⁺
- C) [Fe(NH3)6]²⁺
- D) [Cu(NH3)6]²⁺

2. **Determine the Oxidation State**:
- For each complex, we will denote the oxidation state of the metal as \( X \).
- The amine ligands (NH3) are neutral, contributing 0 to the charge.
- The overall charge of the complex is +2.

For example, for [Ni(NH3)6]²⁺:
\[ X + 6(0) = +2 \implies X = +2 \]

3. **Write the Electronic Configuration**:
- For Nickel (Ni), the atomic number is 28. The electron configuration is:
\[ \text{Ni: } [Ar] 4s^2 3d^8 \quad \text{(for Ni in the 0 oxidation state)} \]
- For Ni²⁺, we remove the 4s electrons:
\[ \text{Ni}^{2+}: [Ar] 3d^8 \]

4. **Hybridization**:
- In the case of Ni²⁺, the 3d orbitals are involved in bonding with the ligands. Since NH3 is a strong field ligand, it will cause pairing of electrons in the 3d orbitals.
- The hybridization will involve the outer orbitals (4s and 4p) along with the 3d orbitals, leading to:
\[ \text{Hybridization: } d^2sp^3 \quad \text{(outer orbital complex)} \]

5. **Analyze Other Complexes**:
- Repeat the steps for Co²⁺, Fe²⁺, and Cu²⁺:
- **Cobalt (Co)**:
\[ \text{Co: } [Ar] 4s^2 3d^7 \implies \text{Co}^{2+}: [Ar] 3d^7 \quad \text{(inner orbital complex)} \]
- **Iron (Fe)**:
\[ \text{Fe: } [Ar] 4s^2 3d^6 \implies \text{Fe}^{2+}: [Ar] 3d^6 \quad \text{(inner orbital complex)} \]
- **Copper (Cu)**:
\[ \text{Cu: } [Ar] 4s^2 3d^{10} \implies \text{Cu}^{2+}: [Ar] 3d^9 \quad \text{(inner orbital complex)} \]

6. **Conclusion**:
- After analyzing all the complexes, we find that only [Ni(NH3)6]²⁺ is an outer orbital complex, while the others are inner orbital complexes.

### Final Answer:
The complex that is an outer orbital complex is **[Ni(NH3)6]²⁺**.