Question

The correct order regarding the electronegativity of hybrid orbitals of carbon is

• A. $sp > sp^2 < sp^3$
• B. $sp > sp^2 > sp^3$
• C. $sp < sp^2 > sp^3$
• D. $sp < sp^2 < sp^3$

Answer 1

The Correct Option is B

To determine the correct order of electronegativity for the hybrid orbitals of carbon, it's important to understand the relationship between the type of hybridization and electronegativity. The degree of s-character in a carbon hybrid orbital affects its electronegativity:

1. sp Hybridization: In sp hybridization, there is a 50% s-character and 50% p-character. As s-electrons are closer to the nucleus, sp hybrid orbitals have higher electronegativity.
2. sp² Hybridization: This has approximately 33% s-character and 67% p-character. Therefore, sp² orbitals are less electronegative than sp orbitals.
3. sp³ Hybridization: These have 25% s-character and 75% p-character. Consequently, sp³ orbitals are the least electronegative among the three types.

The higher the s-character, the closer the electrons are held to the nucleus, increasing the electronegativity. Thus, the order of electronegativity based on the hybridization character is:

sp > sp^2 > sp^3

Therefore, considering the factors outlined above, the correct answer is sp > sp^2 > sp^3, and this corresponds with our understanding of hybrid orbitals' electronegativity based on their s-character content.