Ligands are broadly classified into two classes classical and non-classical ligands, depending on their donor annd acceptor ability. Classical ligands form classical complexes while non-classical ligands form non-classical complex. Bonding mechanism in non-classical is called synergic bonding.
Q. Synergic bonding is absent in:

Options

A. `[Mo(CO)_(6)]`
B. `[Cr(CO)_(3)(B_(3)N_(3)H_(6)]`
C. `[Sc(CO)_(6)]^(3+)`
D. `[Ni(CN)_(4)]^(4-)`

Explanation

To determine where synergic bonding is absent, we need to analyze the ligands and the oxidation states of the metal ions involved. Here's a step-by-step solution:

### Step 1: Understand Synergic Bonding
Synergic bonding involves the back donation of electrons from the metal ion to the anti-bonding orbitals of the ligands. This typically occurs when the metal has available d-electrons that can be donated.

**Hint:** Remember that synergic bonding requires the presence of d-electrons in the metal ion.

### Step 2: Identify the Metal Ions and Their Oxidation States
We need to look at the metal ions in the given options and their oxidation states:
- **Molybdenum (Mo)**: Neutral ligand, oxidation state = 0.
- **Chromium (Cr)**: Neutral ligand, oxidation state = 0.
- **Scandium (Sc)**: This is a +3 oxidation state, which means it has lost its 3d and 4s electrons.
- **Nickel (Ni)**: Neutral ligand, oxidation state = 0.

**Hint:** Check the oxidation states of each metal ion to see if they have available d-electrons.

### Step 3: Determine Electron Configurations
Next, we write the electron configurations for the relevant metal ions:
- **Molybdenum (Mo)**: [Kr] 4d5 5s1
- **Chromium (Cr)**: [Ar] 4s1 3d5
- **Scandium (Sc)**: [Ar] 4s2 3d1 (but in +3 state, it becomes empty)
- **Nickel (Ni)**: [Ar] 4s2 3d8

**Hint:** The key here is to identify which metal ion has no d-electrons available for back donation.

### Step 4: Analyze Each Case for Synergic Bonding
- **Molybdenum**: Has d-electrons available for synergic bonding.
- **Chromium**: Also has d-electrons available for synergic bonding.
- **Scandium**: In the +3 oxidation state, it has no d-electrons available (3d0), hence no synergic bonding.
- **Nickel**: Has d-electrons available for synergic bonding.

**Hint:** Focus on the metal ion that has no d-electrons left after ionization.

### Step 5: Conclusion
From the analysis, we conclude that synergic bonding is absent in **Scandium (Sc)** in its +3 oxidation state.

**Final Answer:** Synergic bonding is absent in Scandium (option C).