An element M forms an ion `M^(-)` . The electronic configuration of this ion is (2,8). Which of the following statements is correct?

Options

A. M is in group 16 and period 2
B. M is in group 17 and period 2
C. M is in group 17 and period 3
D. M is in group 16 and period 3

Explanation

To solve the question regarding the element M that forms the ion \( M^{-} \) with an electronic configuration of (2,8), we can follow these steps:

### Step 1: Understand the Ion Configuration
The electronic configuration of the ion \( M^{-} \) is given as (2,8). This means that the ion has a total of 10 electrons.

### Step 2: Determine the Number of Electrons in Element M
Since \( M^{-} \) indicates that the element M has gained one electron to form the anion, we can deduce the number of electrons in the neutral atom M:
- Number of electrons in \( M^{-} \) = 10
- Therefore, the number of electrons in M = 10 - 1 = 9

### Step 3: Identify the Element
The element with 9 electrons corresponds to the atomic number 9. Referring to the periodic table, the element with atomic number 9 is Fluorine (F).

### Step 4: Determine the Group and Period of Element M
- **Group**: Fluorine belongs to Group 17 (the halogens).
- **Period**: Fluorine is located in Period 2 of the periodic table.

### Conclusion
Based on the above deductions, we can conclude that:
- Element M is in Group 17 and Period 2.

### Final Answer
The correct statement is that M is in Group 17 and Period 2.

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