Out of `[Fe(CN)_6]^(4-)`,`[Ni(CN)_4]^(2-)`, and `[Ni(CO)_4]`: select the incorrect Statement(s):

Options

A. All have identical geometry
B. All are paramagnetic
C. all are diamagnetic
D. `[Fe(CN)_6]^(4-)` is dimagnetic but `[Ni(CN)_4]^(2-)` and `[Ni(CO)_4]` are paramagnetic

Answer

To solve the question regarding the incorrect statements about the complexes \([Fe(CN)_6]^{4-}\), \([Ni(CN)_4]^{2-}\), and \([Ni(CO)_4]\), we will analyze each complex step by step.

### Step 1: Analyze \([Fe(CN)_6]^{4-}\)

1. **Determine the oxidation state of Iron (Fe)**:
- The charge of the complex is \(-4\).
- Each cyanide ion (CN) has a charge of \(-1\).
- Therefore, for 6 cyanide ions, the total charge is \(-6\).
- Let the oxidation state of Fe be \(x\).
- The equation is:
\[ x + 6(-1) = -4 \implies x - 6 = -4 \implies x = +2 \]

2. **Electronic configuration of Fe in +2 state**:
- The electronic configuration of Fe is \([Ar] 3d^6\).
- Since CN is a strong field ligand, it causes pairing of electrons.
- The configuration becomes: \(3d^{6}\) (all electrons paired).

3. **Geometry and magnetic property**:
- The hybridization is \(d^2sp^3\), leading to an octahedral geometry.
- Since all electrons are paired, \([Fe(CN)_6]^{4-}\) is **diamagnetic**.

### Step 2: Analyze \([Ni(CN)_4]^{2-}\)

1. **Determine the oxidation state of Nickel (Ni)**:
- The charge of the complex is \(-2\).
- For 4 cyanide ions, the total charge is \(-4\).
- Let the oxidation state of Ni be \(y\).
- The equation is:
\[ y + 4(-1) = -2 \implies y - 4 = -2 \implies y = +2 \]

2. **Electronic configuration of Ni in +2 state**:
- The electronic configuration of Ni is \([Ar] 3d^8\).
- CN being a strong field ligand causes pairing of electrons.
- The configuration becomes: \(3d^{8}\) (all electrons paired).

3. **Geometry and magnetic property**:
- The hybridization is \(dsp^2\), leading to a square planar geometry.
- Since all electrons are paired, \([Ni(CN)_4]^{2-}\) is **diamagnetic**.

### Step 3: Analyze \([Ni(CO)_4]\)

1. **Determine the oxidation state of Nickel (Ni)**:
- The charge of the complex is \(0\) (neutral).
- CO is a neutral ligand, so the oxidation state of Ni is \(0\).

2. **Electronic configuration of Ni in 0 state**:
- The electronic configuration of Ni is \([Ar] 3d^8 4s^2\).
- CO being a strong field ligand causes pairing of electrons.
- The configuration becomes: \(3d^{10}\) (all electrons paired).

3. **Geometry and magnetic property**:
- The hybridization is \(sp^3\), leading to a tetrahedral geometry.
- Since all electrons are paired, \([Ni(CO)_4]\) is **diamagnetic**.

### Conclusion

Now, we summarize the findings for each complex:
- \([Fe(CN)_6]^{4-}\): Octahedral, Diamagnetic
- \([Ni(CN)_4]^{2-}\): Square planar, Diamagnetic
- \([Ni(CO)_4]\): Tetrahedral, Diamagnetic

### Incorrect Statements:
1. All have identical geometry. **(Incorrect)**
2. All are paramagnetic. **(Incorrect)**
3. All are diamagnetic. **(Correct)**
4. \([Fe(CN)_6]^{4-}\) is diamagnetic but \([Ni(CN)_4]^{2-}\) and \([Ni(CO)_4]\) are paramagnetic. **(Incorrect)**

### Final Answer:
The incorrect statements are: 1, 2, and 4.

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