Question

name:
q3 chemistry: u5l3 vsepr theory day 2 practice set 2

1. **cs₂

lewis dot structure
compound name

of electron groups

shape of molecule around central atom
degree between terminal atoms
s - c - s

1. **scl₂

lewis dot structure
compound name

of electron groups

shape of molecule around central atom
degree between terminal atoms
i - sc - i

1. *ccl₄

lewis dot structure
compound name

of electron groups

shape of molecule around central atom
degree between terminal atoms
i - cc - i (with two more i attached to cc vertically)

Explanation:

Question 4: $\boldsymbol{\ce{CS2}}$

Step 1: Lewis Dot Structure

Carbon (C) is central, with 4 valence electrons. Sulfur (S) has 6 valence electrons. In $\ce{CS2}$, C forms double bonds with two S atoms (each double bond is 2 shared pairs). The Lewis structure is $\ce{S::C::S}$ (or with lone pairs: each S has 2 lone pairs, C has no lone pairs).

Step 2: Compound Name

Binary molecular compound: name first element (carbon), second element (sulfur) with -ide suffix, and prefix for 2 S: carbon disulfide.

Step 3: # of Electron Groups

Electron groups = bonding groups (no lone pairs on C). C has 2 double bonds (each counts as 1 electron group) → $\boldsymbol{2}$ electron groups.

Step 4: Molecular Shape

2 electron groups, no lone pairs → linear (VSEPR: $AX_2$).

Step 5: Bond Angle

Linear molecules have bond angle $\boldsymbol{180^\circ}$.

Question 5: $\boldsymbol{\ce{SCl2}}$

Step 1: Lewis Dot Structure

Sulfur (S) is central (6 valence e⁻), Cl has 7 each. S forms single bonds with 2 Cl, and has 2 lone pairs. Structure: $\ce{Cl - S - Cl}$ with 2 lone pairs on S.

Step 2: Compound Name

Binary molecular: sulfur dichloride (sulfur + chlorine → chloride, prefix di- for 2 Cl).

Step 3: # of Electron Groups

Bonding groups (2 Cl) + lone pairs (2) → $2 + 2 = \boldsymbol{4}$ electron groups.

Step 4: Molecular Shape

4 electron groups, 2 lone pairs → bent (VSEPR: $AX_2E_2$).

Step 5: Bond Angle

Lone pairs repel, so angle is less than 109.5° (tetrahedral parent) → ~$\boldsymbol{103^\circ}$ (approximate, due to lone pair repulsion).

Question 6: $\boldsymbol{\ce{CCl4}}$

Step 1: Lewis Dot Structure

Carbon (C) central (4 valence e⁻), forms single bonds with 4 Cl (each Cl has 3 lone pairs). Structure: $\ce{Cl - C - Cl}$ (tetrahedral, 4 single bonds, no lone pairs on C).

Step 2: Compound Name

Binary molecular: carbon tetrachloride (carbon + chlorine → chloride, prefix tetra- for 4 Cl).

Step 3: # of Electron Groups

4 single bonds (each is 1 electron group) → $\boldsymbol{4}$ electron groups.

Step 4: Molecular Shape

4 electron groups, no lone pairs → tetrahedral (VSEPR: $AX_4$).

Step 5: Bond Angle

Tetrahedral bond angle: $\boldsymbol{109.5^\circ}$.

Summary Tables

Compound	Lewis Dot Structure	Name	# Electron Groups	Shape	Bond Angle
$\ce{SCl2}$	$\ce{Cl - S - Cl}$ (2 lone pairs on S)	Sulfur Dichloride	4	Bent	~$103^\circ$
$\ce{CCl4}$	$\ce{Cl - C - Cl}$ (tetrahedral, 4 bonds)	Carbon Tetrachloride	4	Tetrahedral	$109.5^\circ$

Final Answers (Key Fields)

• $\ce{CS2}$: Name = Carbon Disulfide; Electron Groups = 2; Shape = Linear; Angle = $180^\circ$
• $\ce{SCl2}$: Name = Sulfur Dichloride; Electron Groups = 4; Shape = Bent; Angle = ~$103^\circ$
• $\ce{CCl4}$: Name = Carbon Tetrachloride; Electron Groups = 4; Shape = Tetrahedral; Angle = $109.5^\circ$

Answer:

Question 4: $\boldsymbol{\ce{CS2}}$

Step 1: Lewis Dot Structure

Carbon (C) is central, with 4 valence electrons. Sulfur (S) has 6 valence electrons. In $\ce{CS2}$, C forms double bonds with two S atoms (each double bond is 2 shared pairs). The Lewis structure is $\ce{S::C::S}$ (or with lone pairs: each S has 2 lone pairs, C has no lone pairs).

Step 2: Compound Name

Binary molecular compound: name first element (carbon), second element (sulfur) with -ide suffix, and prefix for 2 S: carbon disulfide.

Step 3: # of Electron Groups

Electron groups = bonding groups (no lone pairs on C). C has 2 double bonds (each counts as 1 electron group) → $\boldsymbol{2}$ electron groups.

Step 4: Molecular Shape

2 electron groups, no lone pairs → linear (VSEPR: $AX_2$).

Step 5: Bond Angle

Linear molecules have bond angle $\boldsymbol{180^\circ}$.

Question 5: $\boldsymbol{\ce{SCl2}}$

Step 1: Lewis Dot Structure

Sulfur (S) is central (6 valence e⁻), Cl has 7 each. S forms single bonds with 2 Cl, and has 2 lone pairs. Structure: $\ce{Cl - S - Cl}$ with 2 lone pairs on S.

Step 2: Compound Name

Binary molecular: sulfur dichloride (sulfur + chlorine → chloride, prefix di- for 2 Cl).

Step 3: # of Electron Groups

Bonding groups (2 Cl) + lone pairs (2) → $2 + 2 = \boldsymbol{4}$ electron groups.

Step 4: Molecular Shape

4 electron groups, 2 lone pairs → bent (VSEPR: $AX_2E_2$).

Step 5: Bond Angle

Lone pairs repel, so angle is less than 109.5° (tetrahedral parent) → ~$\boldsymbol{103^\circ}$ (approximate, due to lone pair repulsion).

Question 6: $\boldsymbol{\ce{CCl4}}$

Step 1: Lewis Dot Structure

Carbon (C) central (4 valence e⁻), forms single bonds with 4 Cl (each Cl has 3 lone pairs). Structure: $\ce{Cl - C - Cl}$ (tetrahedral, 4 single bonds, no lone pairs on C).

Step 2: Compound Name

Binary molecular: carbon tetrachloride (carbon + chlorine → chloride, prefix tetra- for 4 Cl).

Step 3: # of Electron Groups

4 single bonds (each is 1 electron group) → $\boldsymbol{4}$ electron groups.

Step 4: Molecular Shape

4 electron groups, no lone pairs → tetrahedral (VSEPR: $AX_4$).

Step 5: Bond Angle

Tetrahedral bond angle: $\boldsymbol{109.5^\circ}$.

Summary Tables

Compound	Lewis Dot Structure	Name	# Electron Groups	Shape	Bond Angle
$\ce{SCl2}$	$\ce{Cl - S - Cl}$ (2 lone pairs on S)	Sulfur Dichloride	4	Bent	~$103^\circ$
$\ce{CCl4}$	$\ce{Cl - C - Cl}$ (tetrahedral, 4 bonds)	Carbon Tetrachloride	4	Tetrahedral	$109.5^\circ$

Final Answers (Key Fields)

• $\ce{CS2}$: Name = Carbon Disulfide; Electron Groups = 2; Shape = Linear; Angle = $180^\circ$
• $\ce{SCl2}$: Name = Sulfur Dichloride; Electron Groups = 4; Shape = Bent; Angle = ~$103^\circ$
• $\ce{CCl4}$: Name = Carbon Tetrachloride; Electron Groups = 4; Shape = Tetrahedral; Angle = $109.5^\circ$