The ions present in `Al_(4)C_(3), CaC_(2)` and `Mg_(2)C_(3)` are respectively

Options

A. `C^(4-), C_(2)^(2-), C_(3)^(4-)`
B. `C_(2)^(2-), C^(4-), C_(3)^(4-)`
C. `C_(2)^(2-), C_(3)^(2-), C^(4-)`
D. `C_(3)^(4-), C^(4-), C_(2)^(2-)`

Answer

To determine the ions present in the compounds \( \text{Al}_4\text{C}_3 \), \( \text{CaC}_2 \), and \( \text{Mg}_2\text{C}_3 \), we will analyze each compound step by step.

### Step 1: Analyze \( \text{Al}_4\text{C}_3 \)

1. **Identify the oxidation state of Aluminum (Al)**:
- Aluminum typically has an oxidation state of +3.
- In \( \text{Al}_4\text{C}_3 \), there are 4 Aluminum atoms, contributing a total charge of \( 4 \times (+3) = +12 \).

2. **Determine the charge contributed by Carbon (C)**:
- The compound must be electrically neutral, so the total negative charge from Carbon must balance the +12 charge from Aluminum.
- Let the charge on Carbon be \( x \). Since there are 3 Carbon atoms, the total charge from Carbon is \( 3x \).
- Therefore, we set up the equation:
\[ +12 + 3x = 0 \]
- Solving for \( x \):
\[ 3x = -12 \quad \Rightarrow \quad x = -4 \]
- Thus, the ion present is \( \text{C}^{4-} \).

### Step 2: Analyze \( \text{CaC}_2 \)

1. **Identify the oxidation state of Calcium (Ca)**:
- Calcium typically has an oxidation state of +2.
- In \( \text{CaC}_2 \), there is 1 Calcium atom contributing a total charge of +2.

2. **Determine the charge contributed by Carbon (C)**:
- Let the charge on Carbon be \( y \). Since there are 2 Carbon atoms, the total charge from Carbon is \( 2y \).
- The compound must be electrically neutral, so we set up the equation:
\[ +2 + 2y = 0 \]
- Solving for \( y \):
\[ 2y = -2 \quad \Rightarrow \quad y = -1 \]
- Thus, the ion present is \( \text{C}^{2-} \).

### Step 3: Analyze \( \text{Mg}_2\text{C}_3 \)

1. **Identify the oxidation state of Magnesium (Mg)**:
- Magnesium typically has an oxidation state of +2.
- In \( \text{Mg}_2\text{C}_3 \), there are 2 Magnesium atoms contributing a total charge of \( 2 \times (+2) = +4 \).

2. **Determine the charge contributed by Carbon (C)**:
- Let the charge on Carbon be \( z \). Since there are 3 Carbon atoms, the total charge from Carbon is \( 3z \).
- The compound must be electrically neutral, so we set up the equation:
\[ +4 + 3z = 0 \]
- Solving for \( z \):
\[ 3z = -4 \quad \Rightarrow \quad z = -\frac{4}{3} \]
- Thus, the ion present is \( \text{C}^{4-} \).

### Summary of Ions Present
- For \( \text{Al}_4\text{C}_3 \): \( \text{C}^{4-} \)
- For \( \text{CaC}_2 \): \( \text{C}^{2-} \)
- For \( \text{Mg}_2\text{C}_3 \): \( \text{C}^{4-} \)

### Final Answer
The ions present in \( \text{Al}_4\text{C}_3 \), \( \text{CaC}_2 \), and \( \text{Mg}_2\text{C}_3 \) are \( \text{C}^{4-} \), \( \text{C}^{2-} \), and \( \text{C}^{4-} \) respectively.

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