Find the predicted bond angle of hydrogen sulfide (H2S) using the valence bond theory. Also describe the bonding in H2S based on the theory.
The Correct Answer and Explanation is:
Answer:
The predicted bond angle of hydrogen sulfide (H₂S) using valence bond theory is approximately 92°.
Explanation
According to valence bond theory, bonding in molecules is explained through the overlap of atomic orbitals from adjacent atoms. In the case of hydrogen sulfide (H₂S), the central atom is sulfur (S), which is in group 16 of the periodic table. It has the electron configuration:
[Ne] 3s² 3p⁴, meaning it has six valence electrons.
In H₂S, sulfur forms two single bonds with two hydrogen atoms. This uses two of sulfur’s valence electrons, one for each S–H bond. The remaining four valence electrons exist as two lone pairs on the sulfur atom. Therefore, the electron domain geometry around sulfur consists of four regions of electron density (two bonding pairs and two lone pairs), which suggests a tetrahedral electron geometry.
However, due to the repulsion caused by lone pairs, the molecular geometry is not tetrahedral but rather bent (angular). Lone pairs exert greater repulsion than bonding pairs, so they push the bonding pairs closer together, reducing the bond angle.
While an ideal tetrahedral angle is 109.5°, the actual H–S–H bond angle in H₂S is much smaller. Due to the large size of sulfur and its relatively diffuse 3p orbitals, the overlap with hydrogen’s 1s orbitals is weak, and no significant hybridization (like sp³) occurs. As a result, the S–H bonds are considered to arise mainly from unhybridized 3p orbitals overlapping with the hydrogen 1s orbitals.
This absence of hybridization and the presence of two lone pairs result in a bond angle close to the p-orbital angle of 90°. Experimentally, the angle is about 92°, consistent with this explanation. Therefore, valence bond theory predicts a bent structure with a bond angle close to 92°, based on the geometry and bonding orbitals involved.
