Arrange the following cyano complexes in decreasing order of their magnetic moment.

Options

A. `[Cr(CN)_(6)]^(3-)gt [Mn(CN)_(6)]^(3-)gt [Fe(CN)_(6)]^(3-) gt [Co(CN)_(6)]^(3-)`
B. `[Mn(CN)_(6)]^(3-) gt [Cr(CN)_(6)]^(3-) gt [Fe(CN)_(6)]^(3-) gt [Co(CN)_(6)]^(3-)`
C. `[Fe(CN)_(6)]^(3-) gt [Cr(CN)_(6)]^(3-) gt [Mn(CN)_(6)]^(3-) gt [Co(CN)_(6)]^(3-)`
D. `[Co(CN)_(6)]^(3-) gt [Cr(CN)_(6)]^(3-) gt [Mn(CN)_(6)^(3-) gt [Fe(CN)_(6)]^(3-)`

Answer

To arrange the given cyano complexes in decreasing order of their magnetic moment, we need to determine the number of unpaired electrons (n) in each complex. The magnetic moment (μ) can be calculated using the formula:

\[ \mu = \sqrt{n(n + 2)} \]

where n is the number of unpaired electrons. The more unpaired electrons there are, the higher the magnetic moment.

### Step-by-Step Solution:

1. **Identify the complexes and their oxidation states:**
- **Chromium complex:** \([Cr(CN)_6]^{3-}\)
- **Manganese complex:** \([Mn(CN)_6]^{3-}\)
- **Iron complex:** \([Fe(CN)_6]^{3-}\)
- **Cobalt complex:** \([Co(CN)_6]^{3-}\)

2. **Determine the oxidation states:**
- For \([Cr(CN)_6]^{3-}\):
- Let the oxidation state of Cr be \(x\).
- \(x + 6(-1) = -3\) → \(x = +3\)
- For \([Mn(CN)_6]^{3-}\):
- Let the oxidation state of Mn be \(x\).
- \(x + 6(-1) = -3\) → \(x = +3\)
- For \([Fe(CN)_6]^{3-}\):
- Let the oxidation state of Fe be \(x\).
- \(x + 6(-1) = -3\) → \(x = +3\)
- For \([Co(CN)_6]^{3-}\):
- Let the oxidation state of Co be \(x\).
- \(x + 6(-1) = -3\) → \(x = +3\)

3. **Determine the electronic configurations:**
- **Chromium (Cr):** Atomic number = 24 → Configuration = \( [Ar] 4s^1 3d^5 \)
- In \(Cr^{3+}\): Loses 3 electrons → \(3d^3\) (3 unpaired electrons)
- **Manganese (Mn):** Atomic number = 25 → Configuration = \( [Ar] 4s^2 3d^5 \)
- In \(Mn^{3+}\): Loses 3 electrons → \(3d^4\) (2 unpaired electrons)
- **Iron (Fe):** Atomic number = 26 → Configuration = \( [Ar] 4s^2 3d^6 \)
- In \(Fe^{3+}\): Loses 3 electrons → \(3d^5\) (5 unpaired electrons)
- **Cobalt (Co):** Atomic number = 27 → Configuration = \( [Ar] 4s^2 3d^7 \)
- In \(Co^{3+}\): Loses 3 electrons → \(3d^6\) (4 unpaired electrons)

4. **Count the number of unpaired electrons (n):**
- Chromium: \(n = 3\)
- Manganese: \(n = 2\)
- Iron: \(n = 5\)
- Cobalt: \(n = 4\)

5. **Calculate the magnetic moments:**
- For Chromium:
\[ \mu = \sqrt{3(3 + 2)} = \sqrt{15} \approx 3.87 \, \mu_B \]
- For Manganese:
\[ \mu = \sqrt{2(2 + 2)} = \sqrt{8} \approx 2.83 \, \mu_B \]
- For Iron:
\[ \mu = \sqrt{5(5 + 2)} = \sqrt{35} \approx 5.92 \, \mu_B \]
- For Cobalt:
\[ \mu = \sqrt{4(4 + 2)} = \sqrt{24} \approx 4.89 \, \mu_B \]

6. **Arrange in decreasing order of magnetic moment:**
- Iron: \(5.92 \, \mu_B\)
- Cobalt: \(4.89 \, \mu_B\)
- Chromium: \(3.87 \, \mu_B\)
- Manganese: \(2.83 \, \mu_B\)

### Final Order:
\[ [Fe(CN)_6]^{3-} > [Co(CN)_6]^{3-} > [Cr(CN)_6]^{3-} > [Mn(CN)_6]^{3-} \]