Question

part b
describe the correct structure for h₂se.
the structure should have
◯ trigonal planar electron geometry and bent molecular geometry
◯ tetrahedral electron geometry and bent molecular geometry.
◯ tetrahedral electron geometry and tetrahedral molecular geometry
◯ linear electron geometry and linear molecular geometry
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Explanation:

Step1: Calculate the number of valence electrons

Selenium (\(Se\)) has \(6\) valence electrons and each hydrogen (\(H\)) has \(1\) valence electron. So for \(H_{2}Se\), the total number of valence electrons is \(6+(1\times2)=8\).

Step2: Determine the electron - group geometry

The central atom is \(Se\). Using the VSEPR (Valence - Shell Electron - Pair Repulsion) theory, the formula for the number of electron groups around the central atom is \(n=\frac{V + M - C + A}{2}\), where \(V\) is the valence electrons of the central atom, \(M\) is the number of monovalent atoms (\(H\) is monovalent), \(C\) is the charge of cation and \(A\) is the charge of anion. For \(H_{2}Se\) (neutral molecule, \(C = A=0\)), \(n=\frac{6 + 2-0 + 0}{2}=4\). When \(n = 4\), the electron - group geometry is tetrahedral.

Step3: Determine the molecular geometry

In \(H_{2}Se\), there are \(2\) bonding pairs (\(Se - H\) bonds) and \(2\) lone pairs on the \(Se\) atom. According to VSEPR theory, when there are \(2\) bonding pairs and \(2\) lone pairs around the central atom, the molecular geometry is bent.

Answer:

tetrahedral electron geometry and bent molecular geometry.