The bond angle in H2 O is 104.5 ^∘; the bond angles in H2 S H2 Se, and H2 Te are very close to 9

step 1 Water with a bond angle of 104 .5 degrees, which is very close to a predicted bond angle of 109

The bond angle in H2 O is 104.5 ^∘; the bond angles in H2 S...

The bond angle in $\mathrm{H}_{2} \mathrm{O}$ is 104.5 $^\circ$; the bond angles in $\mathrm{H}_{2} \mathrm{S}$ $\mathrm{H}_{2} \mathrm{Se},$ and $\mathrm{H}_{2} \mathrm{Te}$ are very close to $90^{\circ} .$ Which theory would you apply to describe the geometry in $\mathrm{H}_{2} \mathrm{S}, \mathrm{H}_{2} \mathrm{Se},$ and $\mathrm{H}_{2} \mathrm{Te}:$ VSEPR? Valence bond without invoking hybrid orbitals? Valence bond theory, using hybrid orbitals? Why?

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Key Concepts

VSEPR Theory

VSEPR (Valence Shell Electron Pair Repulsion) theory describes molecular geometry based on the repulsions between electron pairs on a central atom. It predicts shapes by arranging electron pairs to minimize repulsion. While it works well for many molecules, its predictions are most accurate when the central atom has low-lying hybrid orbitals, which may not be the case for heavier atoms like S, Se, and Te.

Valence Bond Theory

Valence bond theory explains molecular structure by considering the overlap of atomic orbitals to form bonds. For heavier chalcogen hydrides, the theory suggests that bonds are formed using largely unhybridized atomic orbitals. This accounts for bond angles close to 90° since pure p-orbital bonding results in perpendicular orientations between bonds.

Hybridization

Hybridization is a concept where atomic orbitals mix to form new orbitals that can explain the geometry of molecules such as water, where sp3 hybridization leads to a bond angle of about 104.5°. However, for heavier atoms like S, Se, and Te, the energy difference between available orbitals leads to minimal hybridization, which results in bond angles that remain near 90°.

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