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 Al₄C₃, CaC₂, and Mg₂C₃, we need to analyze the oxidation states of carbon in each compound.

### Step-by-Step Solution:

1. **Analyzing Al₄C₃:**
- The formula indicates there are 4 aluminum (Al) atoms and 3 carbon (C) atoms.
- The oxidation state of aluminum is +3.
- Let the oxidation state of carbon be \( x \).
- The overall charge of the compound is neutral (0).
- Therefore, we can set up the equation:
\[ 4(+3) + 3x = 0 \]
\[ 12 + 3x = 0 \]
\[ 3x = -12 \]
\[ x = -4 \]
- This means each carbon has an oxidation state of -4, leading to the presence of the ion \( C^{4-} \).

2. **Analyzing CaC₂:**
- The formula indicates there is 1 calcium (Ca) atom and 2 carbon (C) atoms.
- The oxidation state of calcium is +2.
- Let the oxidation state of carbon be \( y \).
- The overall charge of the compound is neutral (0).
- Therefore, we can set up the equation:
\[ +2 + 2y = 0 \]
\[ 2y = -2 \]
\[ y = -1 \]
- This means each carbon has an oxidation state of -1, leading to the presence of the ion \( C^{2-} \).

3. **Analyzing Mg₂C₃:**
- The formula indicates there are 2 magnesium (Mg) atoms and 3 carbon (C) atoms.
- The oxidation state of magnesium is +2.
- Let the oxidation state of carbon be \( z \).
- The overall charge of the compound is neutral (0).
- Therefore, we can set up the equation:
\[ 2(+2) + 3z = 0 \]
\[ 4 + 3z = 0 \]
\[ 3z = -4 \]
\[ z = -\frac{4}{3} \]
- This means the average oxidation state of carbon in this compound is -\(\frac{4}{3}\), leading to the presence of the ion \( C^{\frac{4}{3}-} \) for the three carbon atoms.

### Summary of Ions Present:
- For Al₄C₃: \( C^{4-} \)
- For CaC₂: \( C^{2-} \)
- For Mg₂C₃: \( C^{\frac{4}{3}-} \)

### Final Answer:
The ions present in Al₄C₃, CaC₂, and Mg₂C₃ are respectively \( C^{4-}, C^{2-}, \) and \( C^{\frac{4}{3}-} \).