A metallic ion `m^(2+)` has an electronic configuration of 2,8,8,6 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 \( 2, 8, 8, 6 \) and an ionic weight of \( 56 \) amu, we can follow these steps:

### Step 1: Determine the Mass Number
The mass number of an ion is typically given as its ionic weight. In this case, the ionic weight is \( 56 \) amu, which means the mass number (A) is \( 56 \).

### Step 2: Determine the Atomic Number
The atomic number (Z) can be determined from the electronic configuration of the ion. The electronic configuration \( 2, 8, 8, 6 \) indicates the distribution of electrons in the shells.

- The total number of electrons in the configuration is:
\[ 2 + 8 + 8 + 6 = 24 \]

Since the ion is \( m^{2+} \), it has lost 2 electrons compared to its neutral state. Therefore, the number of protons (which is equal to the atomic number) in the neutral atom is:
\[ 24 + 2 = 26 \]
Thus, the atomic number \( Z \) is \( 26 \).

### Step 3: Calculate the Number of Neutrons
The number of neutrons (N) can be calculated using the formula:
\[ N = A - Z \]
Substituting the values we found:
\[ N = 56 - 26 = 30 \]

### Conclusion
The number of neutrons in the nucleus of the metallic ion \( m^{2+} \) is \( 30 \).

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