Which is a low spin complex?

Options

A. `[Fe(CN_(6))]^(3-)`
B. `[Co(NO_(2))_(6)]^(3-)`
C. `[Mn(CN)_(6)]^(3-)`
D. all of these

Answer

To determine which complex is a low spin complex, we need to analyze the given coordination compounds and their magnetic properties based on the ligands and the oxidation states of the central metal ions.

### Step-by-Step Solution:

1. **Identify the Complexes:**
We are given three complexes:
- \( \text{[Fe(CN)_6]}^{3-} \)
- \( \text{[Co(NO_2)_6]}^{3-} \)
- \( \text{[Mn(CN)_6]}^{3-} \)

2. **Calculate the Oxidation State of the Metal Ions:**
- For \( \text{[Fe(CN)_6]}^{3-} \):
- Let the oxidation state of Fe be \( x \).
- The charge of CN is -1, and there are 6 CN ligands.
- The equation becomes: \( x + 6(-1) = -3 \) → \( x - 6 = -3 \) → \( x = +3 \).

- For \( \text{[Co(NO_2)_6]}^{3-} \):
- Let the oxidation state of Co be \( y \).
- The charge of NO2 is -1, and there are 6 NO2 ligands.
- The equation becomes: \( y + 6(-1) = -3 \) → \( y - 6 = -3 \) → \( y = +3 \).

- For \( \text{[Mn(CN)_6]}^{3-} \):
- Let the oxidation state of Mn be \( z \).
- The charge of CN is -1, and there are 6 CN ligands.
- The equation becomes: \( z + 6(-1) = -3 \) → \( z - 6 = -3 \) → \( z = +3 \).

3. **Determine the Electronic Configuration:**
- For \( \text{Fe}^{3+} \) (Atomic number 26):
- Ground state: \( \text{[Ar]} 4s^2 3d^6 \)
- Configuration for \( \text{Fe}^{3+} \): \( \text{[Ar]} 3d^5 \)

- For \( \text{Co}^{3+} \) (Atomic number 27):
- Ground state: \( \text{[Ar]} 4s^2 3d^7 \)
- Configuration for \( \text{Co}^{3+} \): \( \text{[Ar]} 3d^6 \)

- For \( \text{Mn}^{3+} \) (Atomic number 25):
- Ground state: \( \text{[Ar]} 4s^2 3d^5 \)
- Configuration for \( \text{Mn}^{3+} \): \( \text{[Ar]} 3d^4 \)

4. **Analyze the Ligands:**
- CN (cyanide) is a strong field ligand, which causes pairing of electrons.
- NO2 (nitrite) is also a strong field ligand, which causes pairing of electrons.

5. **Determine the Spin State:**
- For \( \text{[Fe(CN)_6]}^{3-} \):
- \( 3d^5 \) configuration, strong field ligand → pairs up electrons.
- Result: 1 unpaired electron (low spin).

- For \( \text{[Co(NO_2)_6]}^{3-} \):
- \( 3d^6 \) configuration, strong field ligand → all electrons pair up.
- Result: 0 unpaired electrons (diamagnetic).

- For \( \text{[Mn(CN)_6]}^{3-} \):
- \( 3d^4 \) configuration, strong field ligand → pairs up electrons.
- Result: 0 unpaired electrons (diamagnetic).

6. **Conclusion:**
- The complexes \( \text{[Co(NO_2)_6]}^{3-} \) and \( \text{[Mn(CN)_6]}^{3-} \) are low spin complexes because they have no unpaired electrons.
- Therefore, the answer is that all the complexes listed are low spin complexes.

### Final Answer:
All of the given complexes are low spin complexes.