Question

which of the following compounds exhibits a linear molecular geometry? select two correct answers.
h₂o
co₂
h₂s
bef₂
ch₄

Explanation:

• For \(CO_2\): The central \(C\) atom has \(4\) valence electrons. It forms double - bonds with two \(O\) atoms (\(O::C::O\)). Using VSEPR (Valence - Shell Electron - Pair Repulsion) theory, the electron - pair geometry and molecular geometry are linear because there are two bonding pairs and no lone pairs around the central \(C\) atom.
• For \(BeF_2\): The central \(Be\) atom has \(2\) valence electrons. It forms single bonds with two \(F\) atoms (\(F - Be - F\)). According to VSEPR theory, with two bonding pairs and no lone pairs around the central \(Be\) atom, the molecular geometry is linear.
• For \(H_2O\): The central \(O\) atom has \(6\) valence electrons. It forms single bonds with two \(H\) atoms and has two lone pairs (\(H - O - H\) with two lone pairs on \(O\)). Using VSEPR theory, the electron - pair geometry is tetrahedral, but the molecular geometry is bent (or V - shaped).
• For \(H_2S\): The central \(S\) atom has \(6\) valence electrons. It forms single bonds with two \(H\) atoms and has two lone pairs (\(H - S - H\) with two lone pairs on \(S\)). According to VSEPR theory, the electron - pair geometry is tetrahedral, and the molecular geometry is bent (or V - shaped).
• For \(CH_4\): The central \(C\) atom has \(4\) valence electrons. It forms single bonds with four \(H\) atoms (\(H - C - H\) in a tetrahedral arrangement). Using VSEPR theory, with four bonding pairs and no lone pairs around the central \(C\) atom, the molecular geometry is tetrahedral.

Answer:

\(CO_2\), \(BeF_2\)