Which of the following statements is (are) true for `[Cu(NO_2)_6]^(4-)` and `[Fe(H_2O)_6]^(3+)` ?

Options

A. Both are outer orbital complexes
B. Both are inner orbital complexes with magnetic moment equal to 1.73 B.M.
C. The former complex is paramagnetic with one unpaired electron while later one is paramagnetic with five unpaired electrons.
D. The former is outer orbital complex and later is inner orbital complexes respectively

Explanation

To determine which statements are true for the complexes \([Cu(NO_2)_6]^{4-}\) and \([Fe(H_2O)_6]^{3+}\), we will analyze the oxidation states, hybridization, geometry, and magnetic properties of both complexes.

### Step 1: Determine the oxidation state of Copper in \([Cu(NO_2)_6]^{4-}\)

1. The overall charge of the complex is \(-4\).
2. Each \((NO_2)^-\) ligand contributes a \(-1\) charge, and there are 6 of them:
\[ \text{Charge from ligands} = 6 \times (-1) = -6 \]
3. Let \(X\) be the oxidation state of Copper (Cu):
\[ X + (-6) = -4 \implies X = -4 + 6 = +2 \]
4. Therefore, the oxidation state of Copper is \(+2\).

### Step 2: Determine the electronic configuration of Copper

1. The atomic number of Copper (Cu) is 29, so its ground state electronic configuration is:
\[ [Ar] 3d^{10} 4s^1 \]
2. In the \(+2\) oxidation state, Copper loses two electrons, typically from the \(4s\) and \(3d\) orbitals:
\[ [Ar] 3d^9 \]

### Step 3: Determine the hybridization of \([Cu(NO_2)_6]^{4-}\)

1. With 6 ligands, we expect the hybridization to be \(sp^3d^2\).
2. The geometry of the complex is octahedral due to the arrangement of 6 ligands around the central metal ion.

### Step 4: Determine the number of unpaired electrons in \([Cu(NO_2)_6]^{4-}\)

1. The \(3d^9\) configuration means there is one unpaired electron.
2. Thus, \([Cu(NO_2)_6]^{4-}\) is paramagnetic.

### Step 5: Determine the oxidation state of Iron in \([Fe(H_2O)_6]^{3+}\)

1. The overall charge of the complex is \(+3\).
2. Water (\(H_2O\)) is a neutral ligand, contributing \(0\) to the charge.
3. Let \(Y\) be the oxidation state of Iron (Fe):
\[ Y + 0 = +3 \implies Y = +3 \]
4. Therefore, the oxidation state of Iron is \(+3\).

### Step 6: Determine the electronic configuration of Iron

1. The atomic number of Iron (Fe) is 26, so its ground state electronic configuration is:
\[ [Ar] 3d^6 4s^2 \]
2. In the \(+3\) oxidation state, Iron loses three electrons, typically from the \(4s\) and \(3d\) orbitals:
\[ [Ar] 3d^5 \]

### Step 7: Determine the hybridization of \([Fe(H_2O)_6]^{3+}\)

1. With 6 ligands, we expect the hybridization to be \(sp^3d^2\).
2. The geometry of the complex is octahedral due to the arrangement of 6 ligands around the central metal ion.

### Step 8: Determine the number of unpaired electrons in \([Fe(H_2O)_6]^{3+}\)

1. The \(3d^5\) configuration means there are 5 unpaired electrons.
2. Thus, \([Fe(H_2O)_6]^{3+}\) is also paramagnetic.

### Conclusion

1. Both complexes are outer orbital complexes.
2. The first complex is paramagnetic with one unpaired electron, while the second complex is paramagnetic with five unpaired electrons.

### Final Statements

- **True Statements**:
- Both are outer orbital complexes.
- The former complex is paramagnetic with one unpaired electron, while the latter one is paramagnetic with five unpaired electrons.