The bond angle in $H_{2} O$ is approximately $105^{\circ}$ while the bond angle in $H_{2} S$ is approximately $90^{\circ}$ . Which explanation best accounts for this difference?

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$H - S$ bonds are longer than $H - S$ bonds,

S has d orbitals available for bonding, O does not

$H - S$ bonds are less polar than $H - O$ bonds.

O uses ${sp}^{3}$ hybrid orbitals for bonding, S uses its $3 p$ orbitals.

answer is D.

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Detailed Solution

According to Drago's rule, in the following circumstances, it is not necessary to pay attention to an element's hybridization: Case 1: The centre atom of the molecule has at least one lone pair of electrons. Case 2: The centre atom can be made up of any of the elements from groups 13, 14, 15, or 16 or from the third to seventh period.

Hence, the correct answer is D.

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