An element A belongs to IIA group and another element B belongs to VIA group. The compound formed between A and B contains

Options

A. `A^(2+), B^(-)` ions,
B. `A^(2+),B^(-2)` ions
C. `A^(-2), B^(+2)` ion
D. `A^(+), B^(-2)` ions

Answer

To solve the problem, we need to determine the ions formed when element A from group IIA (alkaline earth metals) combines with element B from group VIA (chalcogens).

### Step-by-Step Solution:

1. **Identify the Elements and Their Groups**:
- Element A belongs to group IIA (alkaline earth metals).
- Element B belongs to group VIA (chalcogens).

2. **Determine the Electron Configuration**:
- Group IIA elements have the outermost electron configuration of ns². This means they have 2 electrons in their outer shell.
- Group VIA elements have the outermost electron configuration of ns²np⁴. This means they have 6 electrons in their outer shell.

3. **Determine the Ionization of Element A**:
- Group IIA elements tend to lose their outermost electrons easily.
- When element A loses 2 electrons, it becomes a divalent cation:
\[ A \rightarrow A^{2+} \]

4. **Determine the Electron Acceptance of Element B**:
- Group VIA elements need 2 more electrons to complete their octet (8 electrons in the outer shell).
- Therefore, element B will accept 2 electrons and become a divalent anion:
\[ B + 2 \rightarrow B^{2-} \]

5. **Combine the Ions to Form the Compound**:
- The compound formed between A and B will consist of the cation \(A^{2+}\) and the anion \(B^{2-}\).
- The formula for the compound will be \(AB\), where A is the cation and B is the anion.

6. **Final Conclusion**:
- The compound formed between element A and element B contains \(A^{2+}\) and \(B^{2-}\).

### Final Answer:
The compound formed between A and B contains \(A^{2+}\) and \(B^{2-}\).

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