Q.Which of the following compounds contain all the carbon atoms in the same hybridisation state? (Note: more than one of the given options may be correct.)

The key idea to solve this problem is Sigma Pi Bond Counting to determine the hybridisation state of each carbon atom.

Here's the essential reasoning:

1. Count the number of sigma bonds around each carbon atom. (Lone pairs are not present on carbon in these neutral organic compounds).
2. A carbon atom forming 4 sigma bonds is $sp^3$ hybridised.
3. A carbon atom forming 3 sigma bonds is $sp^2$ hybridised.
4. A carbon atom forming 2 sigma bonds is $sp$ hybridised.
5. Apply these rules to each carbon in the given compounds:
   • (i) H—C≡C—C≡C—H: All four carbon atoms form 2 sigma bonds each (one C-H or C-C single bond, and one sigma bond within the triple bond), making them all $sp$ hybridised.
   • (ii) CH3—C≡C—CH3: The two terminal CH3 carbons form 4 sigma bonds each ($sp^3$), while the two internal carbons in the triple bond form 2 sigma bonds each ($sp$). …

Understanding the hybridization state of carbon atoms is fundamental in organic chemistry, as it dictates the geometry and reactivity of molecules. For carbon, hybridization is primarily determined by the number of sigma ($\sigma$) bonds it forms. Each single bond is a sigma bond, a double bond consists of one sigma and one pi ($\pi$) bond, and a triple bond consists of one sigma and two pi bonds.

The general rule for determining hybridization for a central atom (like carbon) is to calculate its steric number, which is the sum of the number of sigma bonds and the number of lone pairs around it. For carbon in stable organic compounds, there are typically no lone pairs. Therefore, the hybridization of a carbon atom simplifies to:

• Steric Number = 4 (4 sigma bonds): The carbon atom is $sp^3$ hybridized. This leads to a tetrahedral geometry.
• Steric Number = 3 (3 sigma bonds): The carbon atom is $sp^2$ hybridized. This leads to a trigonal planar geometry.
• Steric Number = 2 (2 sigma bonds): The carbon atom is $sp$ hybridized. This leads to a linear geometry.

Let's apply this concept to each given compound.

1. Analyze Option (i): H—C≡C—C≡C—H

   This compound is butadiyne. Let's label the carbon atoms from left to right as C1, C2, C3, and C4.

   • C1: Forms one C—H single bond ($\sigma$) and one C≡C triple bond (which contains one $\sigma$ and two $\pi$ bonds). So, C1 forms a total of $1 + 1 = 2$ sigma bonds. Therefore, C1 is $sp$ hybridized.
   • C2: Forms one C≡C triple bond to C1 (one $\sigma$) and one C≡C triple bond to C3 (one $\sigma$). So, C2 forms a total of $1 + 1 = 2$ sigma bonds. Therefore, C2 is $sp$ hybridized.
   • C3: Forms one C≡C triple bond to C2 (one $\sigma$) and one C≡C triple bond to C4 (one $\sigma$). So, C3 forms a total of $1 + 1 = 2$ sigma bonds. Therefore, C3 is $sp$ hybridized.
   • C4: Forms one C≡C triple bond to C3 (one $\sigma$) and one C—H single bond ($\sigma$). So, C4 forms a total of $1 + 1 = 2$ sigma bonds. Therefore, C4 is $sp$ hybridized. All carbon atoms in H—C≡C—C≡C—H are $sp$ hybridized.

2. Analyze Option (ii): CH3—C≡C—CH3

   This compound is 2-butyne. Let's label the carbon atoms from left to right as C1, C2, C3, and C4.

   • C1 (methyl group): Forms three C—H single bonds (3 $\sigma$) and one C—C single bond to C2 (1 $\sigma$). So, C1 forms a total of $3 + 1 = 4$ sigma bonds. Therefore, C1 is $sp^3$ hybridized.
   • C2: Forms one C—C single bond to C1 (1 $\sigma$) and one C≡C triple bond to C3 (which contains one $\sigma$ and two $\pi$ bonds). So, C2 forms a total of $1 + 1 = 2$ sigma bonds. Therefore, C2 is $sp$ hybridized.
   • C3: Forms one C≡C triple bond to C2 (one $\sigma$) and one C—C single bond to C4 (1 $\sigma$). So, C3 forms a total of $1 + 1 = 2$ sigma bonds. Therefore, C3 is $sp$ hybridized.
   • C4 (methyl group): Forms one C—C single bond to C3 (1 $\sigma$) and three C—H single bonds (3 $\sigma$). So, C4 forms a total of $1 + 3 = 4$ sigma bonds. Therefore, C4 is $sp^3$ hybridized. The carbon atoms in CH3—C≡C—CH3 are $sp^3$ and $sp$ hybridized. They are not all in the same hybridization state.

3. Analyze Option (iii): CH2=C=CH2

   This compound is allene. Let's label the carbon atoms from left to right as C1, C2, and C3.

   • C1: Forms two C—H single bonds (2 $\sigma$) and one C=C double bond to C2 (which contains one $\sigma$ and one $\pi$ bond). So, C1 forms a total of $2 + 1 = 3$ sigma bonds. Therefore, C1 is $sp^2$ hybridized. …