Question

what is the closest bond angle around the central atom in the h₂s molecule? 90° 120° 109.5° 180°

Explanation:

Step1: Determine the electron - pair geometry

In \(H_2S\), the central atom \(S\) has 6 valence electrons. It forms 2 single bonds with \(H\) atoms (\(2\) bonding pairs) and has \(2\) lone pairs (\(6 - 2=4\) non - bonding electrons, \(4\div2 = 2\) lone pairs). According to VSEPR theory, the electron - pair geometry is tetrahedral.

Step2: Consider the effect of lone pairs on bond angle

For a tetrahedral electron - pair geometry, the ideal bond angle is \(109.5^{\circ}\). But due to the repulsion between lone pairs and bonding pairs (lone - pair - bond - pair repulsion is stronger than bond - pair - bond - pair repulsion), the bond angle in \(H_2S\) is slightly less than \(109.5^{\circ}\), but the closest value among the given options is \(90^{\circ}\) is too small (for octahedral with lone pairs in some cases), \(120^{\circ}\) is for trigonal planar, \(180^{\circ}\) is for linear. The bond angle in \(H_2S\) is closest to \(90^{\circ}\) compared to the other options when considering the deviation from the ideal tetrahedral angle due to the relatively large size of \(S\) and the nature of the bonds.

Answer:

\(90^{\circ}\)