Question

determine the electron geometry for each molecule. drag the items into the appropriate bins. ch₃f cs₂ pcl₃ scl₂ linear trigonal planar tetrahedral trigonal bipyramidal

Explanation:

Step1: Analyze \( CS_2 \)

The central atom in \( CS_2 \) is \( C \). Carbon has 4 valence electrons, and each \( S \) contributes 6. The Lewis structure shows \( C \) is double - bonded to two \( S \) atoms, with no lone pairs on \( C \). The number of electron groups (bonding groups) is 2. For a central atom with 2 electron groups, the electron geometry is linear.

Step2: Analyze \( CH_3F \)

The central atom is \( C \). Carbon has 4 valence electrons. It is bonded to 3 \( H \) atoms and 1 \( F \) atom. So, the number of electron groups (all bonding groups here, since there are no lone pairs on \( C \)) is 4. A central atom with 4 electron groups has a tetrahedral electron geometry.

Step3: Analyze \( PCl_3 \)

The central atom is \( P \). Phosphorus has 5 valence electrons. It is bonded to 3 \( Cl \) atoms, and there is 1 lone pair on \( P \). The total number of electron groups (3 bonding + 1 lone pair) is 4. A central atom with 4 electron groups has a tetrahedral electron geometry.

Step4: Analyze \( SCl_2 \)

The central atom is \( S \). Sulfur has 6 valence electrons. It is bonded to 2 \( Cl \) atoms, and there are 2 lone pairs on \( S \). The total number of electron groups (2 bonding + 2 lone pairs) is 4. A central atom with 4 electron groups has a tetrahedral electron geometry.

Answer:

Linear: \( CS_2 \)
Trigonal planar: None
Tetrahedral: \( CH_3F \), \( PCl_3 \), \( SCl_2 \)
Trigonal bipyramidal: None