Which amongst the following metal carbonyls are inner orbital complexes with diamagnetic property ?
(I)`Ni(CO)_4` , (II)`Fe(CO)_5` , (III)`V(CO)_6` , (IV)`Cr(CO)_6`
Select the correct answer from the codes given below :

Options

A. I and II only
B. II, III and IV only
C. II and IV only
D. I,II and IV only

Explanation

To determine which metal carbonyls are inner orbital complexes with diamagnetic properties, we need to analyze each option step by step.

### Step 1: Analyze `Ni(CO)₄`
1. **Oxidation State**: Nickel (Ni) is in the zero oxidation state.
2. **Electronic Configuration**: Ni has an atomic number of 28, so its configuration is \( [Ar] 4s^2 3d^8 \).
3. **Hybridization**:
- CO is a strong field ligand, which causes pairing of electrons.
- The 4s electrons will pair with the 3d electrons, leading to the configuration \( 3d^{10} \).
- The hybridization is \( sp^3 \) (outer orbital complex).
4. **Magnetic Property**: Since all electrons are paired, `Ni(CO)₄` is diamagnetic.

### Step 2: Analyze `Fe(CO)₅`
1. **Oxidation State**: Iron (Fe) is in the zero oxidation state.
2. **Electronic Configuration**: Fe has an atomic number of 26, so its configuration is \( [Ar] 4s^2 3d^6 \).
3. **Hybridization**:
- Again, CO is a strong field ligand, causing pairing.
- The configuration becomes \( 3d^6 \) with paired electrons.
- The hybridization is \( dsp^3 \) (inner orbital complex).
4. **Magnetic Property**: All electrons are paired, making `Fe(CO)₅` diamagnetic.

### Step 3: Analyze `V(CO)₆`
1. **Oxidation State**: Vanadium (V) is in the zero oxidation state.
2. **Electronic Configuration**: V has an atomic number of 23, so its configuration is \( [Ar] 4s^2 3d^3 \).
3. **Hybridization**:
- With CO as a strong field ligand, pairing occurs.
- The configuration becomes \( 3d^3 \) with one unpaired electron remaining.
- The hybridization is \( d^2sp^3 \) (inner orbital complex).
4. **Magnetic Property**: Since there is one unpaired electron, `V(CO)₆` is paramagnetic.

### Step 4: Analyze `Cr(CO)₆`
1. **Oxidation State**: Chromium (Cr) is in the zero oxidation state.
2. **Electronic Configuration**: Cr has an atomic number of 24, so its configuration is \( [Ar] 4s^1 3d^5 \) (an exceptional case).
3. **Hybridization**:
- CO is a strong field ligand, leading to pairing.
- The configuration becomes \( 3d^5 \) with all paired electrons.
- The hybridization is \( d^2sp^3 \) (inner orbital complex).
4. **Magnetic Property**: All electrons are paired, making `Cr(CO)₆` diamagnetic.

### Conclusion
From the analysis:
- `Ni(CO)₄` is an outer orbital complex but is diamagnetic.
- `Fe(CO)₅` is an inner orbital complex and is diamagnetic.
- `V(CO)₆` is an inner orbital complex but is paramagnetic.
- `Cr(CO)₆` is an inner orbital complex and is diamagnetic.

Thus, the correct answers are `Fe(CO)₅` and `Cr(CO)₆`.

### Final Answer
The correct options are (II) `Fe(CO)₅` and (IV) `Cr(CO)₆`.