Question

The order of relative strengths of bonds formed by sp, \(\mathrm{sp}^{2}\) and \(\mathrm{sp}^{3}\) hybrid orbitals is (1) \(\mathrm{sp}>\mathrm{sp}^{2}>\mathrm{sp}^{3}\) (2) \(\mathrm{sp}<\mathrm{sp}^{2}<\mathrm{sp}^{3}\) (3) \(\mathrm{sp}=\mathrm{sp}^{2}=\mathrm{sp}^{3}\) (4) \(\mathrm{sp}<\mathrm{sp}^{2}>\mathrm{sp}^{3}\)

Short answer

Option (1): sp > sp^{2} > sp^{3}.

Step-by-step solution

- Understand Hybridization

Hybridization refers to the process where atomic orbitals mix to form new hybrid orbitals. The type of hybrid orbitals (sp, sp^{2}, sp^{3}) indicates the number of orbitals involved and their bonding strength.

- Analyze Bond Strength with Respect to Hybridization

The bond strength is influenced by the s-character of the hybrid orbitals. Higher s-character results in stronger and shorter bonds. The s-character in hybrid orbitals are: sp (50% s, 50% p), sp^{2} (33% s, 67% p), sp^{3} (25% s, 75% p).

- Compare the Hybrid Orbitals

Given the s-character for each type of hybrid orbital, we can compare their bond strengths. The greater the s-character, the stronger the bond. Therefore, sp has the highest bond strength, followed by sp^{2}, and then sp^{3}.

- Select the Correct Option

Based on the comparison, the order of relative strengths of bonds formed by hybrid orbitals is sp > sp^{2} > sp^{3}. Therefore, the correct option is (1).

Key concepts

sp hybrid orbitals

When an atom undergoes sp hybridization, it involves the mixing of one s orbital and one p orbital. This combination results in two sp hybrid orbitals. Each sp hybrid orbital has 50% s-character and 50% p-character. The high percentage of s-character leads to stronger bonds because s orbitals are closer to the nucleus than p orbitals and thus, they can better overlap with another atomic orbital.
Sp hybridization typically occurs in molecules with a linear shape, such as acetylene (C₂H₂). The atoms involved in sp hybridization form bonds that are both strong and short. The linear geometry makes the bond angles 180°, providing the most effective bonding from an orbital overlap perspective. Therefore, sp hybridized bonds are the strongest among sp, sp², and sp³ due to the highest s-character.

sp2 hybrid orbitals

In sp² hybridization, one s orbital mixes with two p orbitals to form three sp² hybrid orbitals. Each of these hybrid orbitals holds 33% s-character and 67% p-character. The decrease in s-character from sp to sp² means that sp² bonds are not as strong as sp bonds, but they are still relatively strong compared to sp³ hybrid orbitals.
Sp² hybrid orbitals are found in trigonal planar molecules, such as ethene (C₂H₄). In these molecules, the bond angles are 120°, creating a planar structure. The presence of 33% s-character allows for moderate overlap, forming bonds that are fairly strong but not as strong as sp hybridized bonds. The specific geometry of sp² hybridization contributes to unique properties such as the ability to create double bonds.

sp3 hybrid orbitals

Sp³ hybridization involves the mixing of one s orbital with three p orbitals, resulting in four sp³ hybrid orbitals. These have 25% s-character and 75% p-character. Due to the lowest s-character among sp, sp², and sp³, sp³ hybridized bonds are the weakest in comparison.
Sp³ hybrid orbitals are typically present in tetrahedral molecules like methane (CH₄). In these molecules, the bond angles are around 109.5°, providing a three-dimensional shape.
This geometry and low s-character contribute to relatively longer and weaker bonds, compared to sp and sp² hybridized bonds. However, the tetrahedral arrangement allows for effective spacing of electrons, which helps minimize repulsion between them, resulting in a stable compound.