Question:
Aluminium carbide on reaction with heavy water gives a carbon compound X. The hybridization in X is
Aluminium carbide on reaction with heavy water gives a carbon compound X. The hybridization in X is
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$\text{Al}_4\text{C}_3$ and $\text{Be}_2\text{C}$ are methanides because they produce methane ($\text{CH}_4$) on hydrolysis.
Since deuterium is an isotope of hydrogen, the hybridization of $\text{CD}_4$ remains identical to that of $\text{CH}_4$ ($\text{sp}^3$).
Since deuterium is an isotope of hydrogen, the hybridization of $\text{CD}_4$ remains identical to that of $\text{CH}_4$ ($\text{sp}^3$).
Updated On: Jul 22, 2026
- sp
- $\text{sp}^2$
- $\text{sp}^3$
- $\text{dsp}^2$
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Question:
The question is about the chemical reaction of aluminium carbide ($\text{Al}_4\text{C}_3$) with heavy water ($\text{D}_2\text{O}$) and identifying the hybridization of the carbon atom in the product.
Step 2: Key Formula or Approach:
Aluminium carbide is an ionic carbide containing methanide ($\text{C}^{4-}$) ions.
The reaction of such carbides with water or heavy water yields methane or its deuterated analogue.
The hybridization is determined by the steric number of the central atom (number of $\sigma$ bonds + number of lone pairs).
Step 3: Detailed Explanation:
• The reaction between aluminium carbide and heavy water is:
\[ \text{Al}_4\text{C}_3 + 12\text{D}_2\text{O} \rightarrow 4\text{Al(OD)}_3 + 3\text{CD}_4 \]
• The carbon compound 'X' formed is deuterated methane ($\text{CD}_4$).
• In a molecule of $\text{CD}_4$, the central carbon atom forms 4 single covalent $\sigma$ bonds with 4 deuterium atoms.
• The carbon atom has 4 valence electrons and forms 4 bonds, leaving 0 lone pairs.
• Steric Number = 4 ($\sigma$-bonds) + 0 (lone pairs) = 4.
• A steric number of 4 corresponds to $\text{sp}^3$ hybridization with a tetrahedral geometry.
Step 4: Final Answer:
The hybridization in compound X ($\text{CD}_4$) is $\text{sp}^3$.
The question is about the chemical reaction of aluminium carbide ($\text{Al}_4\text{C}_3$) with heavy water ($\text{D}_2\text{O}$) and identifying the hybridization of the carbon atom in the product.
Step 2: Key Formula or Approach:
Aluminium carbide is an ionic carbide containing methanide ($\text{C}^{4-}$) ions.
The reaction of such carbides with water or heavy water yields methane or its deuterated analogue.
The hybridization is determined by the steric number of the central atom (number of $\sigma$ bonds + number of lone pairs).
Step 3: Detailed Explanation:
• The reaction between aluminium carbide and heavy water is:
\[ \text{Al}_4\text{C}_3 + 12\text{D}_2\text{O} \rightarrow 4\text{Al(OD)}_3 + 3\text{CD}_4 \]
• The carbon compound 'X' formed is deuterated methane ($\text{CD}_4$).
• In a molecule of $\text{CD}_4$, the central carbon atom forms 4 single covalent $\sigma$ bonds with 4 deuterium atoms.
• The carbon atom has 4 valence electrons and forms 4 bonds, leaving 0 lone pairs.
• Steric Number = 4 ($\sigma$-bonds) + 0 (lone pairs) = 4.
• A steric number of 4 corresponds to $\text{sp}^3$ hybridization with a tetrahedral geometry.
Step 4: Final Answer:
The hybridization in compound X ($\text{CD}_4$) is $\text{sp}^3$.
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