Question

use vsepr theory to predict the shape of the hydrogen sulfide molecule, h₂s. *
tetrahedral
linear
octahedral
angular

Explanation:

Step1: Determine the central atom

The central atom in \(H_2S\) is sulfur (\(S\)).

Step2: Calculate the number of valence electrons

Sulfur has \(6\) valence electrons, and each hydrogen has \(1\) valence electron. So total valence electrons \(=6 + 2\times1=8\).

Step3: Find the electron - pair geometry

Using the formula \(VSEPR\) formula \(AX_nE_m\), where \(A\) is the central atom, \(X\) is the number of bonded atoms, and \(E\) is the number of lone pairs. For \(H_2S\), \(n = 2\) (two \(H\) atoms bonded to \(S\)), and the number of lone pairs \(m=\frac{6 - 2\times1}{2}=2\). The electron - pair geometry is tetrahedral (\(AX_2E_2\) has a tetrahedral electron - pair geometry as per \(VSEPR\) theory: the sum \(n + m=2 + 2 = 4\), and for \(4\) electron - pairs, the electron - pair geometry is tetrahedral).

Step4: Determine the molecular geometry

The molecular geometry (shape) ignores the lone pairs. For \(AX_2E_2\) (tetrahedral electron - pair geometry with two lone pairs), the molecular geometry is angular (bent). The two lone pairs on \(S\) push the \(H - S\) bonds closer together.

Answer:

angular