There are three elements A, B and C. their atomic number are `Z_(1), Z_(2)` and `Z_(3)` respectively. If `Z_(1)-Z_(2)=2 and (Z_(1)+Z_(2))/(2)=Z_(3)-2` and the electronic configuration of element A is `[Ar]3d^(6)4s^(2)`, then correct order of magnetic momentum is/are:

Options

A. `B^(+) gt A^(2+) gt C^(2+)`
B. `A^(3+) gt B^(2+) gt C`
C. `B gt A gt C^(2+)`
D. `B=A^(3+) gt C^(3+)`

Answer

To solve the problem, we need to follow these steps:

### Step 1: Identify Atomic Numbers
We are given the equations:
1. \( Z_1 - Z_2 = 2 \)
2. \( \frac{Z_1 + Z_2}{2} = Z_3 - 2 \)

We also know that the electronic configuration of element A is \([Ar] 3d^6 4s^2\). This configuration corresponds to Iron (Fe), which has an atomic number \( Z_1 = 26 \).

### Step 2: Calculate \( Z_2 \) and \( Z_3 \)
Using the first equation:
\[ Z_1 - Z_2 = 2 \implies 26 - Z_2 = 2 \implies Z_2 = 24 \]
Element B with atomic number \( Z_2 = 24 \) corresponds to Chromium (Cr).

Now, we can use the second equation to find \( Z_3 \):
\[ \frac{Z_1 + Z_2}{2} = Z_3 - 2 \]
Substituting the values:
\[ \frac{26 + 24}{2} = Z_3 - 2 \implies \frac{50}{2} = Z_3 - 2 \implies 25 = Z_3 - 2 \implies Z_3 = 27 \]
Element C with atomic number \( Z_3 = 27 \) corresponds to Cobalt (Co).

### Step 3: Determine Electronic Configurations
Now we can write the electronic configurations for all three elements:
- Element A (Fe): \([Ar] 3d^6 4s^2\)
- Element B (Cr): \([Ar] 3d^5 4s^1\)
- Element C (Co): \([Ar] 3d^7 4s^2\)

### Step 4: Determine Unpaired Electrons
Next, we need to find the number of unpaired electrons in each element:
- For Fe (A): The configuration \([Ar] 3d^6 4s^2\) has 4 unpaired electrons.
- For Cr (B): The configuration \([Ar] 3d^5 4s^1\) has 6 unpaired electrons.
- For Co (C): The configuration \([Ar] 3d^7 4s^2\) has 3 unpaired electrons.

### Step 5: Calculate Magnetic Moments
The magnetic moment (μ) can be calculated using the formula:
\[ \mu = \sqrt{n(n + 2)} \]
where \( n \) is the number of unpaired electrons.

- For Fe (A): \( n = 4 \)
\[ \mu_A = \sqrt{4(4 + 2)} = \sqrt{24} \]

- For Cr (B): \( n = 6 \)
\[ \mu_B = \sqrt{6(6 + 2)} = \sqrt{48} \]

- For Co (C): \( n = 3 \)
\[ \mu_C = \sqrt{3(3 + 2)} = \sqrt{15} \]

### Step 6: Order of Magnetic Moments
Now we can order the magnetic moments based on the number of unpaired electrons:
- B (Cr) has the highest magnetic moment due to 6 unpaired electrons.
- A (Fe) comes next with 4 unpaired electrons.
- C (Co) has the lowest magnetic moment with 3 unpaired electrons.

Thus, the correct order of magnetic moments is:
\[ \text{B} > \text{A} > \text{C} \]

### Final Answer
The correct order of magnetic moments is:
- B (Cr) > A (Fe) > C (Co)