Question

name: date: class period: u6d3 - 4 test review: chemical bonding part i: application (10 points) 1. please draw the structural formulas for the following and write the correct name for the molecular geometry. a) sbfcl₂ b) h₂s c) cs₂ d) sihcl₃ e) siof₂

Explanation:

Step1: Determine the central atom

For \(SbFCl_{2}\), the central atom is \(Sb\). For \(H_{2}S\), the central atom is \(S\). For \(CS_{2}\), the central atom is \(C\). For \(SiHCl_{3}\), the central atom is \(Si\). For \(SiOF_{2}\), the central atom is \(Si\).

Step2: Calculate the number of valence electrons

• \(Sb\) has \(5\) valence electrons. \(F\) has \(7\) and \(Cl\) has \(7\). So for \(SbFCl_{2}\), total valence electrons \(=5 + 7+2\times7=26\).
• \(S\) has \(6\) valence electrons and \(H\) has \(1\). For \(H_{2}S\), total valence electrons \(=6 + 2\times1=8\).
• \(C\) has \(4\) valence electrons and \(S\) has \(6\). For \(CS_{2}\), total valence electrons \(=4+2\times6 = 16\).
• \(Si\) has \(4\) valence electrons, \(H\) has \(1\) and \(Cl\) has \(7\). For \(SiHCl_{3}\), total valence electrons \(=4 + 1+3\times7=26\).
• \(Si\) has \(4\) valence electrons, \(O\) has \(6\) and \(F\) has \(7\). For \(SiOF_{2}\), total valence electrons \(=4+6 + 2\times7=24\).

Step3: Draw the Lewis structure

• For \(SbFCl_{2}\): \(Sb\) is bonded to \(F\) and two \(Cl\) atoms. There are lone pairs on \(Sb\).
• For \(H_{2}S\): \(S\) is bonded to two \(H\) atoms with two lone pairs on \(S\).
• For \(CS_{2}\): \(C\) is double - bonded to two \(S\) atoms.
• For \(SiHCl_{3}\): \(Si\) is bonded to \(H\) and three \(Cl\) atoms.
• For \(SiOF_{2}\): \(Si\) is bonded to \(O\) (double bond) and two \(F\) atoms.

Step4: Determine the molecular geometry

• \(SbFCl_{2}\): The central atom \(Sb\) has \(3\) bonding pairs and \(2\) lone pairs. Using VSEPR theory, the molecular geometry is trigonal bipyramidal (if we consider electron - pair geometry) but since there are lone pairs, the molecular geometry is T - shaped (the student wrote "Point" which is incorrect).
• \(H_{2}S\): The central atom \(S\) has \(2\) bonding pairs and \(2\) lone pairs. Using VSEPR theory, the molecular geometry is bent.
• \(CS_{2}\): The central atom \(C\) has \(2\) bonding pairs (double bonds count as one region of electron density). Using VSEPR theory, the molecular geometry is linear.
• \(SiHCl_{3}\): The central atom \(Si\) has \(4\) bonding pairs. Using VSEPR theory, the molecular geometry is tetrahedral.
• \(SiOF_{2}\): The central atom \(Si\) has \(3\) bonding pairs (double bond to \(O\) counts as one region and two single bonds to \(F\)). Using VSEPR theory, the molecular geometry is trigonal planar (the student wrote "Tetrahedral" which is incorrect as there are \(3\) regions of electron density around \(Si\)).

Answer:

a) Structural formula: \(Cl - Sb - Cl\) with \(F\) attached to \(Sb\) (and lone pairs on \(Sb\)), Geometry name: T - shaped
b) Structural formula: \(H - S - H\) (with lone pairs on \(S\)), Geometry name: Bent
c) Structural formula: \(S = C = S\), Geometry name: Linear
d) Structural formula: \(Cl - Si - Cl\) with \(H\) and another \(Cl\) attached to \(Si\), Geometry name: Tetrahedral
e) Structural formula: \(F - Si - F\) with a double bond \(Si = O\), Geometry name: Trigonal planar