The ratio of magnetic of `Fe` (III) and `Co` (II) is :

Options

A. `sqrt(5) : sqrt(7)`
B. `sqrt(35) : sqrt(15)`
C. `7 : 3`
D. `sqrt(24) : sqrt(15)`

Answer

To find the ratio of the magnetic moments of Fe (III) and Co (II), we will follow these steps:

### Step 1: Determine the electronic configuration of Fe (III)
- The atomic number of iron (Fe) is 26. Its ground state electronic configuration is:
\[ \text{Fe: } [\text{Ar}] 3d^6 4s^2 \]
- For Fe (III), it loses 3 electrons, which means:
\[ \text{Fe (III): } 3d^5 \]

### Step 2: Count the number of unpaired electrons in Fe (III)
- In the 3d subshell of Fe (III), there are 5 electrons. According to Hund's rule, these electrons will occupy separate orbitals before pairing. Therefore, all 5 electrons in the 3d subshell are unpaired:
\[ \text{Unpaired electrons in Fe (III) = 5} \]

### Step 3: Calculate the magnetic moment for Fe (III)
- The formula for calculating the magnetic moment (\(\mu\)) is:
\[ \mu = \sqrt{n(n + 2)} \text{ Bohr Magneton} \]
- Substituting \(n = 5\):
\[ \mu_{\text{Fe (III)}} = \sqrt{5(5 + 2)} = \sqrt{5 \times 7} = \sqrt{35} \text{ Bohr Magneton} \]

### Step 4: Determine the electronic configuration of Co (II)
- The atomic number of cobalt (Co) is 27. Its ground state electronic configuration is:
\[ \text{Co: } [\text{Ar}] 3d^7 4s^2 \]
- For Co (II), it loses 2 electrons, which means:
\[ \text{Co (II): } 3d^7 \]

### Step 5: Count the number of unpaired electrons in Co (II)
- In the 3d subshell of Co (II), there are 7 electrons. According to Hund's rule, the distribution will be:
- 5 electrons will occupy separate orbitals (all unpaired).
- The next two electrons will pair up in the first two orbitals.
- Therefore, there are 3 unpaired electrons:
\[ \text{Unpaired electrons in Co (II) = 3} \]

### Step 6: Calculate the magnetic moment for Co (II)
- Using the same formula for magnetic moment:
\[ \mu_{\text{Co (II)}} = \sqrt{3(3 + 2)} = \sqrt{3 \times 5} = \sqrt{15} \text{ Bohr Magneton} \]

### Step 7: Calculate the ratio of the magnetic moments
- The ratio of the magnetic moments of Fe (III) to Co (II) is:
\[ \text{Ratio} = \frac{\mu_{\text{Fe (III)}}}{\mu_{\text{Co (II)}}} = \frac{\sqrt{35}}{\sqrt{15}} = \sqrt{\frac{35}{15}} = \sqrt{\frac{7}{3}} \]

### Final Answer
The ratio of the magnetic moments of Fe (III) to Co (II) is:
\[ \sqrt{35} : \sqrt{15} \]

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