The magnetic moments of complexes given below are in the order: [Atomic no. Ti = 22 , Cr = 24 , Fe = 26 , Ni = 28 ]
`[FeF_(6)]^(3-)`
`[Ti(H_(2)O)_(6)]^(3+)`
`[Cr(NH_(3))_(6)]^(3+)`
`[Ni(H_(2)O)_(6)]^(2+)`

Options

A. `I gt III gt IV gt II`
B. `I lt III lt II lt IV`
C. `IV lt II lt I lt III`
D. `I gt III gt II gt IV`

Explanation

To determine the order of magnetic moments of the given complexes, we will analyze each complex step by step, calculating the oxidation state, electronic configuration, number of unpaired electrons, and finally the magnetic moment using the formula.

### Step 1: Calculate the magnetic moment of \([FeF_6]^{3-}\)

1. **Determine the oxidation state of Fe**:
\[ x + 6(-1) = -3 \implies x - 6 = -3 \implies x = +3 \]
So, Fe is in the +3 oxidation state.

2. **Electronic configuration of Fe in +3 state**:
\[ \text{Fe: } [Ar] 3d^5 \]
Since there are no paired electrons in the 3d subshell, all 5 electrons are unpaired.

3. **Calculate the magnetic moment**:
\[ \mu = \sqrt{n(n+2)} = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \text{ Bohr magneton} \]

### Step 2: Calculate the magnetic moment of \([Ti(H_2O)_6]^{3+}\)

1. **Determine the oxidation state of Ti**:
\[ x + 6(0) = +3 \implies x = +3 \]
So, Ti is in the +3 oxidation state.

2. **Electronic configuration of Ti in +3 state**:
\[ \text{Ti: } [Ar] 3d^1 \]
There is 1 unpaired electron.

3. **Calculate the magnetic moment**:
\[ \mu = \sqrt{n(n+2)} = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \text{ Bohr magneton} \]

### Step 3: Calculate the magnetic moment of \([Cr(NH_3)_6]^{3+}\)

1. **Determine the oxidation state of Cr**:
\[ x + 6(0) = +3 \implies x = +3 \]
So, Cr is in the +3 oxidation state.

2. **Electronic configuration of Cr in +3 state**:
\[ \text{Cr: } [Ar] 3d^3 \]
There are 3 unpaired electrons.

3. **Calculate the magnetic moment**:
\[ \mu = \sqrt{n(n+2)} = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \text{ Bohr magneton} \]

### Step 4: Calculate the magnetic moment of \([Ni(H_2O)_6]^{2+}\)

1. **Determine the oxidation state of Ni**:
\[ x + 6(0) = +2 \implies x = +2 \]
So, Ni is in the +2 oxidation state.

2. **Electronic configuration of Ni in +2 state**:
\[ \text{Ni: } [Ar] 3d^8 \]
There are 2 unpaired electrons.

3. **Calculate the magnetic moment**:
\[ \mu = \sqrt{n(n+2)} = \sqrt{2(2+2)} = \sqrt{8} \approx 2.83 \text{ Bohr magneton} \]

### Summary of Magnetic Moments
- \([FeF_6]^{3-}\): 5.92 Bohr magneton
- \([Cr(NH_3)_6]^{3+}\): 3.87 Bohr magneton
- \([Ni(H_2O)_6]^{2+}\): 2.83 Bohr magneton
- \([Ti(H_2O)_6]^{3+}\): 1.73 Bohr magneton

### Final Order of Magnetic Moments
The order of magnetic moments from highest to lowest is:
1. \([FeF_6]^{3-}\) (5.92)
2. \([Cr(NH_3)_6]^{3+}\) (3.87)
3. \([Ni(H_2O)_6]^{2+}\) (2.83)
4. \([Ti(H_2O)_6]^{3+}\) (1.73)

### Conclusion
Thus, the correct order of magnetic moments is:
\[ [FeF_6]^{3-} > [Cr(NH_3)_6]^{3+} > [Ni(H_2O)_6]^{2+} > [Ti(H_2O)_6]^{3+} \]