Question:
The hybridizations of carbon in graphite, diamond and \(C_{60}\) are respectively
The hybridizations of carbon in graphite, diamond and \(C_{60}\) are respectively
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Carbon with three sigma bonds generally shows \(sp^2\) hybridization, while carbon with four sigma bonds shows \(sp^3\) hybridization.
Updated On: Jun 22, 2026
- \(sp^2,\; sp^3,\; sp\)
- \(sp^2,\; sp^3,\; sp^2\)
- \(sp,\; sp^2,\; sp^3\)
- \(sp,\; sp^3,\; sp\)
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The Correct Option is B
Solution and Explanation
Step 1: Determine the hybridization in graphite.
Graphite consists of layers of hexagonal carbon atoms.
Each carbon atom in graphite is bonded to three other carbon atoms in the same plane.
Thus, each carbon atom forms:
\[ 3\;\sigma\text{-bonds} \] and has one unhybridized \(p\)-orbital involved in \(\pi\)-bonding.
Therefore, the hybridization of carbon in graphite is:
\[ sp^2 \] The structure of graphite is planar and trigonal in geometry.
Step 2: Determine the hybridization in diamond.
In diamond, each carbon atom is tetrahedrally bonded to four other carbon atoms.
Thus, each carbon atom forms:
\[ 4\;\sigma\text{-bonds} \] There are no free electrons or \(\pi\)-bonds present.
Hence, the hybridization of carbon in diamond is:
\[ sp^3 \] The geometry around each carbon atom is tetrahedral.
Step 3: Determine the hybridization in \(C_{60}\).
\(C_{60}\) is buckminsterfullerene or fullerene, which contains carbon atoms arranged in a spherical cage-like structure.
Each carbon atom is bonded to three neighboring carbon atoms.
Therefore, each carbon atom undergoes:
\[ sp^2\;\text{hybridization} \] One unhybridized \(p\)-orbital remains available for delocalized \(\pi\)-bonding.
Step 4: Write the sequence of hybridizations.
For:
\[ \text{Graphite} \rightarrow sp^2 \] \[ \text{Diamond} \rightarrow sp^3 \] \[ C_{60} \rightarrow sp^2 \] Thus, the required sequence is:
\[ sp^2,\; sp^3,\; sp^2 \]
Step 5: Match with the given options.
The correct option is:
\[ (2)\; sp^2,\; sp^3,\; sp^2 \]
Step 6: Final conclusion.
Hence, the correct answer is:
\[ \boxed{(2)\; sp^2,\; sp^3,\; sp^2} \]
Graphite consists of layers of hexagonal carbon atoms.
Each carbon atom in graphite is bonded to three other carbon atoms in the same plane.
Thus, each carbon atom forms:
\[ 3\;\sigma\text{-bonds} \] and has one unhybridized \(p\)-orbital involved in \(\pi\)-bonding.
Therefore, the hybridization of carbon in graphite is:
\[ sp^2 \] The structure of graphite is planar and trigonal in geometry.
Step 2: Determine the hybridization in diamond.
In diamond, each carbon atom is tetrahedrally bonded to four other carbon atoms.
Thus, each carbon atom forms:
\[ 4\;\sigma\text{-bonds} \] There are no free electrons or \(\pi\)-bonds present.
Hence, the hybridization of carbon in diamond is:
\[ sp^3 \] The geometry around each carbon atom is tetrahedral.
Step 3: Determine the hybridization in \(C_{60}\).
\(C_{60}\) is buckminsterfullerene or fullerene, which contains carbon atoms arranged in a spherical cage-like structure.
Each carbon atom is bonded to three neighboring carbon atoms.
Therefore, each carbon atom undergoes:
\[ sp^2\;\text{hybridization} \] One unhybridized \(p\)-orbital remains available for delocalized \(\pi\)-bonding.
Step 4: Write the sequence of hybridizations.
For:
\[ \text{Graphite} \rightarrow sp^2 \] \[ \text{Diamond} \rightarrow sp^3 \] \[ C_{60} \rightarrow sp^2 \] Thus, the required sequence is:
\[ sp^2,\; sp^3,\; sp^2 \]
Step 5: Match with the given options.
The correct option is:
\[ (2)\; sp^2,\; sp^3,\; sp^2 \]
Step 6: Final conclusion.
Hence, the correct answer is:
\[ \boxed{(2)\; sp^2,\; sp^3,\; sp^2} \]
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