Choose the incorrect regarding stability
(a) `[Fe("phen")_(3)]^(+2) gt [Fe("phen")_(3)]^(+3)`
(b) `[Fe(CN)_(6)]^(-4) gt [Fe(CN)_(6)]^(-3)`
(c) `Fe("dipy")_(3)^(+3) gt Fe("dipy")_(3)^(+2)`

Options

A. a,b
B. b,c
C. a,c
D. a,b,c

Answer

To determine the incorrect statement regarding the stability of the given coordination complexes, we will analyze each option based on the oxidation states of iron and the nature of the ligands involved.

### Step-by-Step Solution:

1. **Analyze Option (a):**
- The complex `[Fe(phen)₃]²⁺` has iron in the +2 oxidation state, while `[Fe(phen)₃]³⁺` has iron in the +3 oxidation state.
- Since phen (phenanthroline) is a neutral ligand, it does not contribute to the charge.
- Higher oxidation state (in this case +3) generally leads to greater stability due to increased positive charge attracting more electron density from the ligands.
- Therefore, the statement `[Fe(phen)₃]²⁺ < [Fe(phen)₃]³⁺` is correct, meaning the order given in option (a) is incorrect.

2. **Analyze Option (b):**
- For the complex `[Fe(CN)₆]⁴⁻`, we need to determine the oxidation state of iron.
- The cyanide ion (CN⁻) has a charge of -1, and there are 6 cyanide ligands, contributing a total charge of -6.
- Setting up the equation: \( x - 6 = -4 \) (where x is the oxidation state of Fe), we find \( x = +2 \).
- For the complex `[Fe(CN)₆]³⁻`, we set up the equation: \( x - 6 = -3 \), giving \( x = +3 \).
- Again, the +3 oxidation state indicates greater stability than +2, so the order `[Fe(CN)₆]⁴⁻ < [Fe(CN)₆]³⁻` is also incorrect.

3. **Analyze Option (c):**
- The complex `Fe(dipy)₃³⁺` has iron in the +3 oxidation state, while `Fe(dipy)₃²⁺` has iron in the +2 oxidation state.
- Dipyridyl (dipy) is also a neutral ligand, contributing no charge.
- As previously established, a higher oxidation state results in greater stability. Thus, the order `Fe(dipy)₃³⁺ > Fe(dipy)₃²⁺` is correct.

### Conclusion:
The incorrect statements regarding stability are found in options (a) and (b). Therefore, the answer is:

**Incorrect options: (a) and (b)**