Question

The hybridizations of carbon in graphite, diamond and \(C_{60}\) are respectively

• A. \(sp^2,\; sp^3,\; sp\)
• B. \(sp^2,\; sp^3,\; sp^2\)
• C. \(sp,\; sp^2,\; sp^3\)
• D. \(sp,\; sp^3,\; sp\)

Hint:

Carbon with three sigma bonds generally shows \(sp^2\) hybridization, while carbon with four sigma bonds shows \(sp^3\) hybridization.

Answer 1

The Correct Option is B

Step 1: Understand the question.
We must find the hybridization of carbon in three allotropes, graphite, diamond and $C_{60}$ (buckminsterfullerene), in that order.
Step 2: Hybridization in graphite.
In graphite each carbon is bonded to three neighbouring carbons in a flat hexagonal sheet, forming three sigma bonds and leaving one unhybridized $p$ orbital for delocalized pi bonding. This is \[ sp^2 \] hybridization.
Step 3: Hybridization in diamond.
In diamond each carbon is bonded to four other carbons in a rigid three dimensional tetrahedral network, forming four sigma bonds. This is \[ sp^3 \] hybridization.
Step 4: Hybridization in $C_{60}$.
In $C_{60}$ each carbon is bonded to three others, forming a closed cage of fused five and six membered rings with delocalized pi electrons, just like a curved graphite sheet. This is \[ sp^2 \] hybridization.
Step 5: Arrange the answers in order.
So for graphite, diamond and $C_{60}$ respectively the hybridizations are $sp^2$, $sp^3$, $sp^2$.
Step 6: State the answer.
The correct set is $sp^2, sp^3, sp^2$, matching the key.
\[ \boxed{sp^2,\; sp^3,\; sp^2} \]