Transition metals and many of their compounds show paramagnetic behaviour where there are unpaired electron or electrons. The magnetic moment arises from the spin and orbital motions in ions or molecule. Magnetic moment of n unpaired electrons is given as
`mu=sqrt(n(n+2))` Bohr magneton
Magnetic moment increases as the number of unpaired electrons increases.
Q. Magnetic moments of `Cr(Z=24),Mn^(o+)(Z=35)` and `Fe^(2+)(Z=36)` are x,y and z respectively. Which of the following order is correct?

Options

A. `xltyltz`
B. `x=yltz`
C. `zltx=y`
D. `x=y=z`

Explanation

To solve the problem of determining the magnetic moments of Chromium (Cr), Manganese (Mn²⁺), and Iron (Fe²⁺), we will follow these steps:

### Step 1: Determine the electronic configuration of Chromium (Cr)
- Atomic number of Cr = 24
- The electronic configuration is:
\[ \text{Cr: } [\text{Ar}] 3d^5 4s^1 \]
- Count the unpaired electrons:
- 5 unpaired electrons in 3d and 1 unpaired electron in 4s = **6 unpaired electrons**.

### Step 2: Calculate the magnetic moment for Chromium (Cr)
- Using the formula for magnetic moment:
\[ \mu = \sqrt{n(n+2)} \]
where \( n = 6 \) (the number of unpaired electrons).
- Calculation:
\[ \mu = \sqrt{6(6+2)} = \sqrt{6 \times 8} = \sqrt{48} \approx 6.92 \text{ Bohr magneton} \]
- Thus, \( x = 6.92 \).

### Step 3: Determine the electronic configuration of Manganese (Mn²⁺)
- Atomic number of Mn = 25
- The electronic configuration is:
\[ \text{Mn: } [\text{Ar}] 3d^5 4s^2 \]
- For Mn²⁺, we remove 2 electrons (1 from 4s and 1 from 3d):
\[ \text{Mn}^{2+}: [\text{Ar}] 3d^5 \]
- Count the unpaired electrons:
- 5 unpaired electrons in 3d = **5 unpaired electrons**.

### Step 4: Calculate the magnetic moment for Manganese (Mn²⁺)
- Using the same formula:
\[ n = 5 \]
- Calculation:
\[ \mu = \sqrt{5(5+2)} = \sqrt{5 \times 7} = \sqrt{35} \approx 5.92 \text{ Bohr magneton} \]
- Thus, \( y \approx 5.92 \).

### Step 5: Determine the electronic configuration of Iron (Fe²⁺)
- Atomic number of Fe = 26
- The electronic configuration is:
\[ \text{Fe: } [\text{Ar}] 3d^6 4s^2 \]
- For Fe²⁺, we remove 2 electrons (both from 4s):
\[ \text{Fe}^{2+}: [\text{Ar}] 3d^6 \]
- Count the unpaired electrons:
- In 3d, we can have 4 unpaired electrons (due to Hund's rule).
- Thus, **4 unpaired electrons**.

### Step 6: Calculate the magnetic moment for Iron (Fe²⁺)
- Using the formula:
\[ n = 4 \]
- Calculation:
\[ \mu = \sqrt{4(4+2)} = \sqrt{4 \times 6} = \sqrt{24} \approx 4.89 \text{ Bohr magneton} \]
- Thus, \( z \approx 4.89 \).

### Step 7: Compare the magnetic moments
- We have:
- \( x \approx 6.92 \) (for Cr)
- \( y \approx 5.92 \) (for Mn²⁺)
- \( z \approx 4.89 \) (for Fe²⁺)

### Conclusion
The order of magnetic moments is:
\[ x > y > z \]
Thus, the correct order is \( x = y > z \).