Magnetic moment of `X^(+3)` ion of 3d series is `sqrt(35)B.M` What is the atomic number of `X^(+3)`?

Options

A. (a)24
B. (b)25
C. (c)26
D. (d)27

Answer

To find the atomic number of the ion \( X^{+3} \) with a magnetic moment of \( \sqrt{35} \) Bohr Magneton, we can follow these steps:

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

### Step 2: Set up the equation using the given magnetic moment
We know from the problem that:
\[ \mu = \sqrt{35} \text{ B.M.} \]
Thus, we can set up the equation:
\[ \sqrt{n(n + 2)} = \sqrt{35} \]

### Step 3: Square both sides to eliminate the square root
Squaring both sides gives us:
\[ n(n + 2) = 35 \]

### Step 4: Rearrange the equation into standard quadratic form
Rearranging the equation, we have:
\[ n^2 + 2n - 35 = 0 \]

### Step 5: Factor the quadratic equation
To solve for \( n \), we can factor the quadratic:
\[ (n + 7)(n - 5) = 0 \]
This gives us two potential solutions for \( n \):
\[ n + 7 = 0 \quad \Rightarrow \quad n = -7 \quad (\text{not valid since } n \text{ cannot be negative}) \]
\[ n - 5 = 0 \quad \Rightarrow \quad n = 5 \]

### Step 6: Determine the electronic configuration of the neutral atom
Since \( n = 5 \), this means the ion \( X^{+3} \) has 5 unpaired electrons. The ion has lost 3 electrons, so we need to find the electronic configuration of the neutral atom \( X \).

### Step 7: Identify the electronic configuration
For transition metals in the 3d series, the typical electronic configuration is:
\[ [X] = 4s^2 3d^n \]
Since \( X^{+3} \) has lost 3 electrons (2 from \( 4s \) and 1 from \( 3d \)), the configuration of the neutral atom \( X \) must be:
\[ 4s^2 3d^6 \]
This means that the neutral atom \( X \) has a total of \( 2 + 6 = 8 \) electrons in the \( 3d \) subshell.

### Step 8: Determine the atomic number
The complete electronic configuration of \( X \) can be written as:
\[ 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^6 \]
Counting the electrons, we find:
\[ 2 (1s) + 2 (2s) + 6 (2p) + 2 (3s) + 6 (3p) + 2 (4s) + 6 (3d) = 26 \]
Thus, the atomic number of \( X \) is:
\[ \text{Atomic Number} = 26 \]

### Final Answer
The atomic number of \( X^{+3} \) is **26**.

---