CBSE Class 11 Chemistry Chapter 12: Organic Chemistry – Some Basic Principles and Techniques NCERT Solutions

The hybridization state of each carbon atom is determined by the number of sigma bonds and lone pairs it forms around itself. A carbon atom with four single bonds is sp³ hybridised. A carbon atom with one double bond and two single bonds is sp² hybridised. A carbon atom involved in a triple bond or two double bonds is sp hybridised.

1. $CH\_2=C=O$ (Propadienone)

   • The first carbon atom ($CH\_2=$) is bonded to two hydrogen atoms and one carbon atom via a double bond. It has 3 sigma bonds and 0 lone pairs, so it is $sp^2$ hybridised.
   • The second carbon atom ($=C=$) is bonded to two other carbon atoms via double bonds. It has 2 sigma bonds and 0 lone pairs, so it is sp hybridised.
   • The third carbon atom ($=O$) is double bonded to the second carbon and single bonded to oxygen. It has 3 sigma bonds and 0 lone pairs, so it is $sp^2$ hybridised.

2. $CH\_3-CH=CH\_2$ (Propene)

   • The first carbon atom ($CH\_3-$) is bonded to three hydrogen atoms and one carbon atom via single bonds. It has 4 sigma bonds and 0 lone pairs, so it is $sp^3$ hybridised.
   • The second carbon atom ($-CH=$) is bonded to one hydrogen atom, one carbon atom via a double bond, and one carbon atom via a single bond. It has 3 sigma bonds and 0 lone pairs, so it is $sp^2$ hybridised.
   • The third carbon atom ($=CH\_2$) is bonded to two hydrogen atoms and one carbon atom via a double bond. It has 3 sigma bonds and 0 lone pairs, so it is $sp^2$ hybridised.

3. $(CH\_3)\_2CO$ (Acetone)

   The structure is $CH\_3\; -\; \backslash overset\{O\}\{||\}C\; -\; CH\_3$.

   • The two methyl carbon atoms ($CH\_3-$) are each bonded to three hydrogen atoms and the central carbon atom via single bonds. Each has 4 sigma bonds and 0 lone pairs, making them $sp^3$ hybridised.
   • The central carbon atom ($C=O$) is double bonded to oxygen and single bonded to two methyl carbons. It has 3 sigma bonds and 0 lone pairs, making it $sp^2$ hybridised.

4. $CH\_2=CH-C\; \backslash equiv\; N$ (Propenenitrile)

   • The first carbon atom ($CH\_2=$) is bonded to two hydrogen atoms and one carbon atom via a double bond. It has 3 sigma bonds and 0 lone pairs, so it is $sp^2$ hybridised.
   • The second carbon atom ($=CH-$) is bonded to one hydrogen atom, one carbon atom via a double bond, and one carbon atom via a single bond. It has 3 sigma bonds and 0 lone pairs, so it is $sp^2$ hybridised.
   • The third carbon atom ($-C\; \backslash equiv\; N$) is triple bonded to nitrogen and single bonded to the second carbon. It has 2 sigma bonds and 0 lone pairs, so it is sp hybridised.

5. $C\_6H\_6$ (Benzene)

   In benzene, each of the six carbon atoms is bonded to two other carbon atoms and one hydrogen atom. Each carbon atom forms one double bond and two single bonds (considering resonance). Each carbon atom has 3 sigma bonds and 0 lone pairs. Therefore, all 6 carbon atoms in benzene are $sp^2$ hybridised.

Sigma ($\backslash sigma$) bonds are formed by the head-on overlap of atomic orbitals, while pi ($\backslash pi$) bonds are formed by the sideways overlap of p-orbitals. Single bonds are $\backslash sigma$ bonds. Double bonds consist of one $\backslash sigma$ and one $\backslash pi$ bond. Triple bonds consist of one $\backslash sigma$ and two $\backslash pi$ bonds.

1. $C\_6H\_6$ (Benzene)

   Benzene has a ring structure. Each carbon atom is $sp^2$ hybridised. There are 6 C–C sigma bonds forming the ring, 6 C–H sigma bonds, and 3 pi ($\backslash pi$) bonds delocalized across the ring due to the overlap of unhybridized p-orbitals on each carbon atom. These pi bonds are often represented as alternating double bonds, but in reality, they are delocalized.

2. $C\_6H\_\{12\}$ (Cyclohexane or other isomers)

   Assuming the most common isomer, cyclohexane, which is a saturated cyclic hydrocarbon. Each carbon atom is $sp^3$ hybridised. There are 6 C–C sigma bonds forming the ring and 12 C–H sigma bonds. There are no pi bonds in cyclohexane.

3. $CH\_2CI\_2$ (Dichloromethane)

   The central carbon atom is $sp^3$ hybridised. It forms two C–H sigma bonds and two C–Cl sigma bonds. There are a total of 4 sigma bonds and no pi bonds in this molecule.

4. $CH\_2\; =\; C\; =\; CH\_2$ (Propadiene or Allene)

   The terminal carbon atoms ($CH\_2=$) are $sp^2$ hybridised, and the central carbon atom ($=C=$) is sp hybridised. There are three sigma bonds: two C=C sigma bonds and two C-H sigma bonds. There are two pi bonds: one between the first and second carbon, and another between the second and third carbon. So, there are 5 sigma bonds and 2 pi bonds in total.

5. $CH\_3NO\_2$ (Nitromethane)

   The carbon atom is $sp^3$ hybridised, forming 3 C–H sigma bonds and one C–N sigma bond. The nitrogen atom is bonded to the carbon atom and also forms a double bond with one oxygen atom and a single bond with another oxygen atom (often represented with coordinate bonds or resonance). This structure contains sigma bonds (C-H, C-N, N-O) and one pi bond (N=O).

6. $HCONHCH\_3$ (N-methylformamide)

   This molecule contains several sigma bonds: C-H (in CHO), N-H, C-N, C-O, and C-H (in CH₃). It also contains one pi bond between the carbon and oxygen atoms (C=O). The carbon atom of the carbonyl group is $sp^2$ hybridised.