for each of the following covalent compounds, provide the missing formula or name and draw the lewis dot diagram.
1. *formula: br₂o
name:
lewis dot diagram:
2. **formula:
name: sulfur dioxide
lewis dot diagram:

classify each compound below as ionic or covalent. use this information to help you write either the name or formula and draw the lewis dot diagram.
3. **cacl₂
a. ionic or covalent?
b. name:
c. lewis dot diagram:
4. cbr₄
a. ionic or covalent?
b. name:
c. lewis dot diagram:

Question

for each of the following covalent compounds, provide the missing formula or name and draw the lewis dot diagram.
1. *formula: br₂o
   name:
   lewis dot diagram:
2. **formula:
   name: sulfur dioxide
   lewis dot diagram:

classify each compound below as ionic or covalent. use this information to help you write either the name or formula and draw the lewis dot diagram.
3. **cacl₂
   a. ionic or covalent?
   b. name:
   c. lewis dot diagram:
4. cbr₄
   a. ionic or covalent?
   b. name:
   c. lewis dot diagram:

Accepted Answer

### Problem 1:
# Explanation:
## Step1: Identify elements and prefixes
The formula is $ \text{Br}_2\text{O} $. The element $ \text{Br} $ is bromine, and there are 2 bromine atoms (so prefix "di-"), and $ \text{O} $ is oxygen (prefix "mono-" but often omitted for the second element in covalent naming).
## Step2: Name the compound
Using covalent compound naming rules, the name is dibromine monoxide.
## Step3: Draw Lewis Dot diagram
- Bromine (Br) has 7 valence electrons, oxygen (O) has 6.
- The structure: O is the central atom (less electronegative? Wait, O is more electronegative than Br? Wait, actually, in $ \text{Br}_2\text{O} $, the structure is Br - O - Br. Each Br has 3 lone pairs, O has 2 lone pairs. So the Lewis dot diagram: Br (with 3 lone pairs) - O (with 2 lone pairs) - Br (with 3 lone pairs). The bonding pairs: each Br - O is a single bond (2 electrons).

# Answer:
Name: dibromine monoxide
Lewis Dot diagram: Br (:::Br) - O (::) - Br (:::Br) (where : represent lone pairs, and - is a single bond; each Br has 3 lone pairs (6 electrons) and O has 2 lone pairs (4 electrons), and 2 bonding pairs (1 between each Br and O))

### Problem 2:
# Explanation:
## Step1: Identify elements and subscripts
Name is sulfur dioxide. Sulfur is S, dioxide means 2 oxygen atoms (O).
## Step2: Write the formula
So the formula is $ \text{SO}_2 $.
## Step3: Draw Lewis Dot diagram
- Sulfur (S) has 6 valence electrons, oxygen (O) has 6.
- S is central, bonded to two O atoms. S has a double bond with one O and a single bond with the other? Wait, formal charge: S has 6 valence, in $ \text{SO}_2 $, the structure is O=S=O (double bonds), with S having a lone pair? Wait, no: total valence electrons: 6 (S) + 2*6 (O) = 18.
- Arrange: S in center, two O atoms. Each O - S bond: let's calculate. If we do two double bonds: S=O=O? No, O=S=O. Each double bond is 4 electrons, so two double bonds (8 electrons) + lone pairs: S has 1 lone pair (2 electrons), each O has 2 lone pairs (4 electrons each). Wait, total: 2 (S lone) + 2*4 (O lone) + 8 (bonding) = 2 + 8 + 8 = 18, which matches. So Lewis dot diagram: O (::)=S( : )=O (::), where S has a lone pair (2 electrons), each O has 2 lone pairs (4 electrons), and double bonds (4 electrons each) between S and O.

# Answer:
Formula: $ \text{SO}_2 $
Lewis Dot diagram: O (::)=S( : )=O (::) (S has a lone pair, each O has 2 lone pairs, double bonds between S and each O)

### Problem 3:
#### Part a:
# Explanation:
## Step1: Identify bond type
$ \text{CaCl}_2 $: Ca is a metal (calcium), Cl is a nonmetal (chlorine). Ionic compounds form between metals and nonmetals (transfer of electrons).
## Step2: Classify
So it's ionic.

#### Part b:
# Explanation:
## Step1: Ionic compound naming
For ionic compounds, name the cation (Ca²⁺, calcium) and anion (Cl⁻, chloride). No prefixes (since it's ionic, based on charges). Ca has +2, Cl has -1, so 2 Cl⁻ to balance Ca²⁺. Name is calcium chloride.

#### Part c:
# Explanation:
## Step1: Ionic Lewis dot (or electron dot)
Ionic compounds: show transfer of electrons. Ca loses 2 electrons to form Ca²⁺, each Cl gains 1 electron to form Cl⁻.
- Lewis dot: Ca (with 2 valence electrons) → Ca²⁺ + 2e⁻; each Cl (with 7 valence electrons) + e⁻ → Cl⁻ (with 8 valence electrons, octet). So the diagram: [Ca]²⁺ + 2[Cl]⁻, where Cl⁻ has 8 electrons (4 lone pairs) and Ca²⁺ has no valence electrons (lost them).

# Answer:
a. Ionic
b. Name: calcium chloride
c. Lewis dot diagram: $ \text{Ca}^{2+} + 2[:\!\!\overset{..}{Cl}\!\!\!:]^- $ (each Cl⁻ has 8 electrons (4 lone pairs), Ca²⁺ has no valence electrons)

### Problem 4:
#### Part a:
# Explanation:
## Step1: Identify bond type
$ \text{CBr}_4 $: C (carbon) and Br (bromine) are both nonmetals. Covalent compounds form between nonmetals (sharing electrons).
## Step2: Classify
So it's covalent.

#### Part b:
# Explanation:
## Step1: Covalent compound naming
Carbon is C, bromine is Br (prefix "tetra-" for 4 Br atoms). So name is carbon tetrabromide.

#### Part c:
# Explanation:
## Step1: Draw Lewis Dot diagram
- Carbon (C) has 4 valence electrons, bromine (Br) has 7.
- C is central, bonded to 4 Br atoms. Each C - Br bond is a single bond (2 electrons).
- Total valence electrons: 4 (C) + 4*7 (Br) = 4 + 28 = 32.
- Each single bond: 4 bonds × 2 = 8 electrons. Lone pairs: 32 - 8 = 24. Each Br has 3 lone pairs (6 electrons), so 4 Br × 6 = 24, which matches.
- Lewis dot diagram: Br (:::Br) - C - Br (:::Br) (with Br atoms at four corners, each Br has 3 lone pairs, C in center with 4 single bonds (no lone pairs on C, since 4 bonds use all 4 valence electrons)).

# Answer:
a. Covalent
b. Name: carbon tetrabromide
c. Lewis dot diagram: $ \begin{array}{c} \!\!\!\!\!\!\!\!\text{Br (:::Br)}\ \text{Br (:::Br)} - \text{C} - \text{Br (:::Br)}\ \!\!\!\!\!\!\!\!\text{Br (:::Br)} \end{array} $ (each Br has 3 lone pairs, C is bonded to 4 Br with single bonds)

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QUESTION IMAGE

Question

Explanation:

Problem 1:

Step1: Identify elements and prefixes

Step2: Name the compound

Step3: Draw Lewis Dot diagram

Step1: Identify elements and subscripts

Step2: Write the formula

Step3: Draw Lewis Dot diagram

Step1: Identify bond type

Step2: Classify

Part b:

Answer:

Problem 2: