A metallic ion `M^(2+)` has an electronic configuration of 2, 8, 14 and the ionic weight is 56 amu. The number of neutrons in its nucleus is

Options

A. 30
B. 32
C. 34
D. 42

Explanation

To find the number of neutrons in the nucleus of the metallic ion \( M^{2+} \) with an electronic configuration of 2, 8, 14 and an ionic weight of 56 amu, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Ionic State**: The ion is given as \( M^{2+} \). This means it has lost 2 electrons compared to its neutral state.

2. **Determine the Electronic Configuration of the Neutral Atom**: The electronic configuration of \( M^{2+} \) is 2, 8, 14. To find the configuration of the neutral atom \( M \), we need to add 2 electrons back:
\[ \text{Neutral atom configuration} = 2, 8, 14 + 2 = 2, 8, 16 \]

3. **Calculate the Atomic Number (Z)**: The atomic number is the total number of electrons in the neutral atom, which corresponds to the number of protons. From the configuration 2, 8, 16:
\[ Z = 2 + 8 + 16 = 26 \]

4. **Identify the Atomic Weight (A)**: The atomic weight (mass number) is given as 56 amu.

5. **Calculate the Number of Neutrons (N)**: The number of neutrons can be calculated using the formula:
\[ N = A - Z \]
Substituting the values we have:
\[ N = 56 - 26 = 30 \]

6. **Conclusion**: The number of neutrons in the nucleus of the metallic ion \( M^{2+} \) is 30.

### Final Answer:
The number of neutrons in the nucleus is **30**.