Which of the following is an outer orbital complex according to valence bond theory ?

Explanation

To determine which of the given complexes is an outer orbital complex according to valence bond theory, we will analyze each option step by step.

### Step 1: Analyze Option A - [CoO4]3-
1. **Identify the oxidation state of cobalt (Co)**:
- Let the oxidation state of Co be \( X \).
- The complex has 4 oxygen atoms, each with a charge of -2.
- The overall charge of the complex is -3.
- Therefore, the equation becomes:
\[ X + 4(-2) = -3 \implies X - 8 = -3 \implies X = +5 \]
2. **Determine the hybridization**:
- Co in +5 oxidation state will have the electronic configuration of \( 3d^4 \).
- Since Co is in a high oxidation state and typically forms inner orbital complexes, this complex is likely to be an inner orbital complex.

### Step 2: Analyze Option B - [PtCl6]2-
1. **Identify the oxidation state of platinum (Pt)**:
- Let the oxidation state of Pt be \( X \).
- The complex has 6 chloride ions, each with a charge of -1.
- The overall charge of the complex is -2.
- Therefore, the equation becomes:
\[ X + 6(-1) = -2 \implies X - 6 = -2 \implies X = +4 \]
2. **Determine the hybridization**:
- Pt in +4 oxidation state belongs to the 4d series.
- Members of the 4d and 5d series typically form inner orbital complexes. Thus, this complex is also an inner orbital complex.

### Step 3: Analyze Option C - [Mn(CN)6]3-
1. **Identify the oxidation state of manganese (Mn)**:
- Let the oxidation state of Mn be \( X \).
- The complex has 6 cyanide ions, each with a charge of -1.
- The overall charge of the complex is -3.
- Therefore, the equation becomes:
\[ X + 6(-1) = -3 \implies X - 6 = -3 \implies X = +3 \]
2. **Determine the hybridization**:
- Mn in +3 oxidation state will have the electronic configuration of \( 3d^4 \).
- In the presence of strong field ligands like CN, pairing occurs, leading to the formation of an inner orbital complex.

### Step 4: Analyze Option D - [Fe(H2O)5NO]2+
1. **Identify the oxidation state of iron (Fe)**:
- Let the oxidation state of Fe be \( X \).
- The complex has 5 water molecules (neutral) and one NO molecule with a charge of +1.
- The overall charge of the complex is +2.
- Therefore, the equation becomes:
\[ X + 5(0) + 1 = +2 \implies X + 1 = +2 \implies X = +1 \]
2. **Determine the hybridization**:
- Fe in +1 oxidation state will have the electronic configuration of \( 3d^6 4s^2 \).
- The electronic configuration for Fe+1 is \( 3d^7 \).
- Since there are no strong field ligands, pairing does not occur, leading to the formation of an outer orbital complex with \( sp^3d^2 \) hybridization.

### Conclusion:
- **Option A**: Inner orbital complex (false).
- **Option B**: Inner orbital complex (false).
- **Option C**: Inner orbital complex (false).
- **Option D**: Outer orbital complex (true).

Thus, the correct answer is **Option D**.