How is the electronegativity of carbon atoms related to their state of hybridisationin an organic compound?

The electronegativity of carbon atoms in an organic compound is related to their state of hybridization in the following ways:

1. sp Hybridization: In sp hybridization, the carbon atom is bonded to two other atoms and has two unshared electrons. This type of hybridization results in a linear molecular geometry. The carbon atom in this state is more electronegative because the s-character is 50%. The more the s-character, the closer the electrons are to the nucleus, and hence, the greater the electronegativity.

2. sp2 Hybridization: In sp2 hybridization, the carbon atom is bonded to three other atoms and has one unshared electron. This type of hybridization results in a trigonal planar molecular geometry. The carbon atom in this state is less electronegative than in the sp state because the s-character is 33.3%.

3. sp3 Hybridization: In sp3 hybridization, the carbon atom is bonded to four other atoms and has no unshared electrons. This type of hybridization results in a tetrahedral molecular geometry. The carbon atom in this state is the least electronegative because the s-character is 25%.

In summary, the electronegativity of carbon atoms decreases as the state of hybridization changes from sp to sp2 to sp3. This is because the percentage of s-character decreases, which means the electrons are less close to the nucleus, and hence, the electronegativity decreases.