`M^(3+)` ion of the first transition series metal 'M' has a magnetic moment 1.73 BM. The atomic number of the metal 'M' is

Options

A. 21
B. 24
C. 29
D. 22

Answer

To find the atomic number of the metal 'M' that corresponds to the \( M^{3+} \) ion with a magnetic moment of 1.73 BM, we can follow these steps:

### Step 1: Understand the magnetic moment formula
The magnetic moment (\( \mu \)) is given by the formula:
\[ \mu = \sqrt{n(n + 2)} \]
where \( n \) is the number of unpaired electrons.

### Step 2: Set the magnetic moment equal to 1.73 BM
Given that \( \mu = 1.73 \) BM, we can set up the equation:
\[ 1.73 = \sqrt{n(n + 2)} \]

### Step 3: Square both sides to eliminate the square root
Squaring both sides gives:
\[ (1.73)^2 = n(n + 2) \]
Calculating \( (1.73)^2 \):
\[ 1.73^2 = 2.9929 \approx 3 \]
So, we have:
\[ 3 = n(n + 2) \]

### Step 4: Solve for \( n \)
This simplifies to the quadratic equation:
\[ n^2 + 2n - 3 = 0 \]
Factoring the quadratic:
\[ (n + 3)(n - 1) = 0 \]
Thus, \( n = 1 \) (since \( n \) cannot be negative).

### Step 5: Identify the metal with \( n = 1 \)
Now, we know that the \( M^{3+} \) ion has 1 unpaired electron. We need to find a metal in the first transition series (Sc to Zn) that has 1 unpaired electron in its \( 3d \) subshell when it is in the \( +3 \) oxidation state.

### Step 6: Check the elements in the first transition series
1. **Scandium (Sc, Z=21)**: \( [Ar] 3d^1 4s^2 \) → \( Sc^{3+} \): \( [Ar] \) (0 unpaired electrons)
2. **Titanium (Ti, Z=22)**: \( [Ar] 3d^2 4s^2 \) → \( Ti^{3+} \): \( [Ar] 3d^1 \) (1 unpaired electron)
3. **Vanadium (V, Z=23)**: \( [Ar] 3d^3 4s^2 \) → \( V^{3+} \): \( [Ar] 3d^3 \) (3 unpaired electrons)
4. **Chromium (Cr, Z=24)**: \( [Ar] 3d^5 4s^1 \) → \( Cr^{3+} \): \( [Ar] 3d^3 \) (3 unpaired electrons)
5. **Manganese (Mn, Z=25)**: \( [Ar] 3d^5 4s^2 \) → \( Mn^{3+} \): \( [Ar] 3d^4 \) (4 unpaired electrons)
6. **Iron (Fe, Z=26)**: \( [Ar] 3d^6 4s^2 \) → \( Fe^{3+} \): \( [Ar] 3d^5 \) (5 unpaired electrons)
7. **Cobalt (Co, Z=27)**: \( [Ar] 3d^7 4s^2 \) → \( Co^{3+} \): \( [Ar] 3d^6 \) (6 unpaired electrons)
8. **Nickel (Ni, Z=28)**: \( [Ar] 3d^8 4s^2 \) → \( Ni^{3+} \): \( [Ar] 3d^7 \) (7 unpaired electrons)
9. **Copper (Cu, Z=29)**: \( [Ar] 3d^{10} 4s^1 \) → \( Cu^{3+} \): \( [Ar] 3d^9 \) (9 unpaired electrons)
10. **Zinc (Zn, Z=30)**: \( [Ar] 3d^{10} 4s^2 \) → \( Zn^{3+} \): \( [Ar] 3d^{10} \) (0 unpaired electrons)

### Conclusion
The only metal in the first transition series that has 1 unpaired electron in the \( M^{3+} \) state is Titanium (Ti), which has an atomic number of 22.

Thus, the atomic number of the metal 'M' is:
\[ \text{Atomic number of M} = 22 \]