Magnetic moment of `Cr^(+2)(Z=24),Mn^(+2)(Z=25)` and `Fe^(2+) (Z=26)` are x,y,z. they are in order

Options

A. `xltyltz`
B. `xgtygtz`
C. `zltx=y`
D. `x=zlty`

Answer

To determine the magnetic moments of the ions Cr²⁺, Mn²⁺, and Fe²⁺, we will follow these steps:

### Step 1: Determine the electronic configuration of Cr²⁺
- Chromium (Cr) has an atomic number of 24.
- The electronic configuration of Cr is:
\[ \text{Cr: } [Ar] 3d^5 4s^1 \]
- For Cr²⁺, we remove two electrons (one from 4s and one from 3d):
\[ \text{Cr}^{2+}: [Ar] 3d^4 \]

### Step 2: Count the number of unpaired electrons in Cr²⁺
- The 3d subshell has 4 electrons. The distribution of electrons in the 3d subshell will be:
\[ \uparrow \downarrow \uparrow \uparrow \]
- There are 4 unpaired electrons (n = 4).

### Step 3: Calculate the magnetic moment for Cr²⁺
- The formula for magnetic moment (μ) is:
\[ \mu = \sqrt{n(n + 2)} \]
- Substituting n = 4:
\[ \mu = \sqrt{4(4 + 2)} = \sqrt{24} \approx 4.89 \, \text{BM} \]

### Step 4: Determine the electronic configuration of Mn²⁺
- Manganese (Mn) has an atomic number of 25.
- The electronic configuration of Mn is:
\[ \text{Mn: } [Ar] 3d^5 4s^2 \]
- For Mn²⁺, we remove two electrons (both from 4s):
\[ \text{Mn}^{2+}: [Ar] 3d^5 \]

### Step 5: Count the number of unpaired electrons in Mn²⁺
- The 3d subshell has 5 electrons, which will be distributed as:
\[ \uparrow \uparrow \uparrow \uparrow \uparrow \]
- All 5 electrons are unpaired (n = 5).

### Step 6: Calculate the magnetic moment for Mn²⁺
- Using the formula:
\[ \mu = \sqrt{n(n + 2)} \]
- Substituting n = 5:
\[ \mu = \sqrt{5(5 + 2)} = \sqrt{35} \approx 5.92 \, \text{BM} \]

### Step 7: Determine the electronic configuration of Fe²⁺
- Iron (Fe) has an atomic number of 26.
- The electronic configuration of Fe is:
\[ \text{Fe: } [Ar] 3d^6 4s^2 \]
- For Fe²⁺, we remove two electrons (both from 4s):
\[ \text{Fe}^{2+}: [Ar] 3d^6 \]

### Step 8: Count the number of unpaired electrons in Fe²⁺
- The 3d subshell has 6 electrons, which will be distributed as:
\[ \uparrow \uparrow \uparrow \uparrow \uparrow \downarrow \]
- There are 4 unpaired electrons (n = 4).

### Step 9: Calculate the magnetic moment for Fe²⁺
- Using the formula:
\[ \mu = \sqrt{n(n + 2)} \]
- Substituting n = 4:
\[ \mu = \sqrt{4(4 + 2)} = \sqrt{24} \approx 4.89 \, \text{BM} \]

### Step 10: Compare the magnetic moments
- We have:
- \( \mu_{\text{Cr}^{2+}} = 4.89 \, \text{BM} \) (X)
- \( \mu_{\text{Mn}^{2+}} = 5.92 \, \text{BM} \) (Y)
- \( \mu_{\text{Fe}^{2+}} = 4.89 \, \text{BM} \) (Z)

### Conclusion
- The order of magnetic moments is:
\[ X = Z < Y \]