Select the correct statement:

Options

A. a.`[Co(H_(2)O)_(6)]^(3+)` is Co(III), low spin, 0 unpaired electrons diamagnetic .
B. b.`[CoF_(6)]^(3-)` is Co(III), high spin `d^(6),4` unpaired electron paramagnetic.
C. c.`[RhF_(6)]^(3-)` is `Rh(III)` low spin `d^(6),0` unpaired eletrons diamagnetic .
D. d.`[Fe(CN)_(6)]^(4-)` high spin `d^(6),0` unpaired electron dimagentic .

Explanation

To solve the question of selecting the correct statement regarding coordination compounds, we will analyze each option based on the information provided in the video transcript.

### Step-by-Step Solution:

1. **Understanding the Complex [Co(H2O)6]³⁺:**
- Cobalt (Co) has an atomic number of 27, with an electron configuration of [Ar] 3d⁷ 4s².
- In the +3 oxidation state, cobalt loses 3 electrons, resulting in a configuration of 3d⁶ 4s⁰.
- Since H2O is a weak field ligand, it does not cause pairing of electrons in the d-orbitals.
- The 3d orbitals will have 6 electrons, and there will be no unpaired electrons, making it **diamagnetic** and **low spin**.

**Hint:** For weak field ligands, electrons tend to remain unpaired.

2. **Analyzing CuFe-3:**
- The complex CuFe-3 involves cobalt in the +3 oxidation state, which again has a configuration of 3d⁶ 4s⁰.
- However, in the presence of a strong field ligand (like F⁻), pairing occurs.
- The 3d orbitals will have all 6 electrons paired, resulting in **4 unpaired electrons** and making it **paramagnetic** and **high spin**.

**Hint:** Strong field ligands promote electron pairing, leading to paramagnetism.

3. **Evaluating [Rh(Fe(CN)6)]³⁻:**
- Rhodium (Rh) has an atomic number of 45, with a configuration of [Kr] 4d⁸ 5s².
- In the +3 oxidation state, its configuration becomes 4d⁶ 5s⁰.
- The presence of CN⁻, a strong field ligand, causes pairing of electrons.
- Therefore, all 6 electrons in the 4d orbitals are paired, resulting in **no unpaired electrons**, making it **diamagnetic** and **low spin**.

**Hint:** Strong field ligands lead to low spin configurations with paired electrons.

4. **Conclusion:**
- Based on the analysis:
- Option 1: Correct (diamagnetic, low spin, no unpaired electrons)
- Option 2: Correct (paramagnetic, high spin, 4 unpaired electrons)
- Option 3: Correct (diamagnetic, low spin, no unpaired electrons)
- Option 4: Incorrect (not high spin, low spin instead)

Thus, the correct statements are options 1, 2, and 3.

### Final Answer:
Options 1, 2, and 3 are correct. Option 4 is incorrect.