`Mg_(2)C_(3)` reacts with water forming propyne. `C_(3)^(4-)` has

Options

A. two sigma and two pi bonds
B. three sigma and one pi bond
C. two sigma and one pi bond
D. two sigma and three pi bonds

Answer

To solve the problem, we need to analyze the reaction of \( \text{Mg}_2\text{C}_3 \) with water and determine the structure and bonding of the \( \text{C}_3^{4-} \) ion.

### Step 1: Understand the Reaction
When \( \text{Mg}_2\text{C}_3 \) reacts with water, it hydrolyzes to form magnesium hydroxide and propyne (not propane as mentioned in the transcript). The reaction can be represented as:
\[ \text{Mg}_2\text{C}_3 + 6 \text{H}_2\text{O} \rightarrow 2 \text{Mg(OH)}_2 + 2 \text{C}_3\text{H}_4 \]

### Step 2: Identify the Anion
The \( \text{C}_3^{4-} \) ion is derived from the carbide \( \text{C}_3^{4-} \) which is a carbon anion with a charge of -4. This indicates that it has gained four electrons.

### Step 3: Draw the Structure of \( \text{C}_3^{4-} \)
The structure of \( \text{C}_3^{4-} \) can be drawn as follows:
- It consists of three carbon atoms.
- The first and second carbon atoms are bonded by a double bond, and the third carbon has a lone pair of electrons.

The structure can be represented as:
\[ \text{C} \equiv \text{C} - \text{C}^{2-} \]
This shows that there are two double bonds between the carbon atoms.

### Step 4: Count the Sigma and Pi Bonds
In the structure of \( \text{C}_3^{4-} \):
- Each double bond consists of one sigma bond and one pi bond.
- Therefore, for two double bonds, we have:
- **Sigma Bonds**: 2 (one from each double bond)
- **Pi Bonds**: 2 (one from each double bond)

### Conclusion
Thus, the \( \text{C}_3^{4-} \) ion has 2 sigma bonds and 2 pi bonds.

### Final Answer
The correct statement regarding \( \text{C}_3^{4-} \) is that it has **2 sigma bonds and 2 pi bonds**.

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