Two elements A and B have the following electronic configurations. The formula of the compound formed between them can be
`A=1s^(2) 2s^(2) 2p^(6) 3s^(2) 3p^(1), B=1s^(2) 2s^(2) 2p^(4)`

Options

A. AB
B. `AB_(2)`
C. `A_(2)B_(3)`
D. `A_(3)B_(2)`

Answer

To determine the formula of the compound formed between elements A and B based on their electronic configurations, we can follow these steps:

### Step 1: Identify the Elements
- **Element A** has the electronic configuration: \(1s^2 2s^2 2p^6 3s^2 3p^1\). This indicates that A has 13 electrons, which corresponds to Aluminum (Al) in the periodic table.
- **Element B** has the electronic configuration: \(1s^2 2s^2 2p^4\). This indicates that B has 16 electrons, which corresponds to Sulfur (S) in the periodic table.

### Step 2: Determine the Valence Electrons
- **Element A (Al)** has 3 valence electrons (2 from \(3s^2\) and 1 from \(3p^1\)).
- **Element B (S)** has 6 valence electrons (2 from \(2s^2\) and 4 from \(2p^4\)).

### Step 3: Determine the Electron Transfer
- To achieve a stable octet configuration:
- **Element A (Al)** can lose 3 electrons to achieve a stable configuration (becoming \(Al^{3+}\)).
- **Element B (S)** needs 2 additional electrons to complete its octet (becoming \(S^{2-}\)).

### Step 4: Determine the Ratio of A and B in the Compound
- Since \(Al\) loses 3 electrons and \(S\) gains 2 electrons, we need to find the lowest common multiple of the charges to balance the compound.
- The least common multiple of 3 (from \(Al^{3+}\)) and 2 (from \(S^{2-}\)) is 6.
- Therefore, we need:
- 2 atoms of \(Al\) to provide a total of \(2 \times 3 = 6\) positive charges.
- 3 atoms of \(S\) to provide a total of \(3 \times 2 = 6\) negative charges.

### Step 5: Write the Formula of the Compound
- The formula of the compound formed between A and B is \(A_2B_3\), which translates to \(Al_2S_3\).

### Final Answer
The formula of the compound formed between elements A and B is \(Al_2S_3\).

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