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Q8.1:
What are hybridisation states of each carbon atom in the following compounds ? $CH_2=C=O$, $CH_3CH=CH_2$, $(CH_3)_2CO$, $CH_2=CHCN$, $C_6H_6$

Solution :

Theoretical Basis for Hybridization

The hybridization state of a carbon atom is determined by the number of sigma ($\sigma$) bonds and lone pairs attached to it, as defined by the Steric Number ($SN$):

$SN = \text{Number of } \sigma \text{ bonds} + \text{Number of lone pairs}$

  • $SN = 4 \implies sp^3$ hybridization (Tetrahedral geometry)
  • $SN = 3 \implies sp^2$ hybridization (Trigonal planar geometry)
  • $SN = 2 \implies sp$ hybridization (Linear geometry)

Step 1: Analysis of $CH_2=C=O$ (Ketene)

Structure: $H_2C=C=O$

  • Terminal $C$ (in $CH_2$): Attached to 2 $H$ atoms ($\sigma$) and 1 double bond ($1 \sigma, 1 \pi$). Total $\sigma = 3$. Hybridization: $sp^2$.
  • Central $C$: Attached to 1 double bond on left ($1 \sigma$) and 1 double bond on right ($1 \sigma$). Total $\sigma = 2$. Hybridization: $sp$.

Step 2: Analysis of $CH_3CH=CH_2$ (Propene)

Structure: $H_3C^1-C^2H=C^3H_2$

  • $C^1$ ($CH_3$): Attached to 3 $H$ atoms and 1 $C$. Total $\sigma = 4$. Hybridization: $sp^3$.
  • $C^2$: Attached to 1 $C$, 1 $H$, and 1 double bond ($1 \sigma$). Total $\sigma = 3$. Hybridization: $sp^2$.
  • $C^3$ ($CH_2$): Attached to 2 $H$ atoms and 1 double bond ($1 \sigma$). Total $\sigma = 3$. Hybridization: $sp^2$.

Step 3: Analysis of $(CH_3)_2CO$ (Acetone)

Structure: $CH_3-C(=O)-CH_3$

  • Methyl carbons ($CH_3$): Each attached to 3 $H$ and 1 $C$. Total $\sigma = 4$. Hybridization: $sp^3$.
  • Carbonyl carbon ($C=O$): Attached to 2 methyl groups and 1 oxygen ($1 \sigma$). Total $\sigma = 3$. Hybridization: $sp^2$.

Step 4: Analysis of $CH_2=CHCN$ (Acrylonitrile)

Structure: $H_2C^1=C^2H-C^3 \equiv N$

  • $C^1$: 2 $H$ and 1 $\sigma$ from double bond. Total $\sigma = 3$. Hybridization: $sp^2$.
  • $C^2$: 1 $H$, 1 $C^1$ ($\sigma$), 1 $C^3$ ($\sigma$). Total $\sigma = 3$. Hybridization: $sp^2$.
  • $C^3$: 1 $C^2$ ($\sigma$) and 1 triple bond to $N$ ($1 \sigma$). Total $\sigma = 2$. Hybridization: $sp$.

Step 5: Analysis of $C_6H_6$ (Benzene)

Benzene is a planar hexagonal ring where each carbon is bonded to two adjacent carbons and one hydrogen. Each carbon participates in one double bond ($1 \sigma + 1 \pi$).

  • Each $C$: 2 $\sigma$ (C-C) + 1 $\sigma$ (C-H) = 3 $\sigma$. Hybridization: $sp^2$ for all six carbons.

Summary Table

CompoundCarbon AtomsHybridization
$CH_2=C=O$$C_1, C_2$$sp^2, sp$
$CH_3CH=CH_2$$C_1, C_2, C_3$$sp^3, sp^2, sp^2$
$(CH_3)_2CO$$CH_3, C=O, CH_3$$sp^3, sp^2, sp^3$
$CH_2=CHCN$$C_1, C_2, C_3$$sp^2, sp^2, sp$
$C_6H_6$All 6 $C$$sp^2$

Final Solution: The hybridization states are: $CH_2=C=O$ ($sp^2, sp$); $CH_3CH=CH_2$ ($sp^3, sp^2, sp^2$); $(CH_3)_2CO$ ($sp^3, sp^2, sp^3$); $CH_2=CHCN$ ($sp^2, sp^2, sp$); $C_6H_6$ (all $sp^2$).


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