lewis dot structure worksheet
draw the lewis dot structure of the given molecules.
ni₃ h₂se i₃⁻
hcn brf₅ co
sif₄ of₂ ibr
nh₄cl bf₃ ch₂f₂
ch₃br po₃³⁻ nh₄⁺
chemistrylearner.com

Question

lewis dot structure worksheet
draw the lewis dot structure of the given molecules.
ni₃  h₂se  i₃⁻
hcn  brf₅  co
sif₄  of₂  ibr
nh₄cl  bf₃  ch₂f₂
ch₃br  po₃³⁻  nh₄⁺
chemistrylearner.com

Accepted Answer

### For $ \boldsymbol{NI_3} $
# Explanation:
## Step1: Calculate valence electrons
Nitrogen (N) has 5 valence electrons, each Iodine (I) has 7, and there are 3 I atoms. So total valence electrons: $ 5 + 3\times7 = 5 + 21 = 26 $.
## Step2: Arrange atoms and form bonds
N is the central atom. Connect N to each I with a single bond (3 bonds, using $ 3\times2 = 6 $ electrons).
## Step3: Distribute remaining electrons
Remaining electrons: $ 26 - 6 = 20 $. Each I gets 3 lone pairs ($ 3\times6 = 18 $ electrons), and N gets 1 lone pair (2 electrons). So the Lewis structure has N in the center, single - bonded to 3 I atoms. N has 1 lone pair, each I has 3 lone pairs.

### For $ \boldsymbol{H_2Se} $
# Explanation:
## Step1: Calculate valence electrons
Hydrogen (H) has 1 valence electron (2 H atoms: $ 2\times1 = 2 $), Selenium (Se) has 6. Total: $ 2 + 6 = 8 $.
## Step2: Arrange atoms and form bonds
Se is the central atom. Connect Se to each H with a single bond (2 bonds, using $ 2\times2 = 4 $ electrons).
## Step3: Distribute remaining electrons
Remaining electrons: $ 8 - 4 = 4 $, which form 2 lone pairs on Se. So Se is in the center, single - bonded to 2 H atoms, with 2 lone pairs on Se.

### For $ \boldsymbol{I_3^-} $
# Explanation:
## Step1: Calculate valence electrons
Each Iodine (I) has 7, 3 I atoms: $ 3\times7 = 21 $, plus 1 for the negative charge. Total: $ 21+ 1=22 $.
## Step2: Arrange atoms and form bonds
The central I is bonded to two outer I atoms (2 single bonds, using $ 2\times2 = 4 $ electrons).
## Step3: Distribute remaining electrons
Remaining electrons: $ 22 - 4 = 18 $. The central I gets 3 lone pairs (6 electrons), each outer I gets 3 lone pairs (6 electrons each). So the structure is $ [\text{I}-\text{I}-\text{I}]^- $ with 3 lone pairs on each I (central I has 3, each outer I has 3).

### For $ \boldsymbol{HCN} $
# Explanation:
## Step1: Calculate valence electrons
Hydrogen (H): 1, Carbon (C): 4, Nitrogen (N): 5. Total: $ 1 + 4+5 = 10 $.
## Step2: Arrange atoms and form bonds
H - C - N. Form a single bond between H and C (2 electrons), and a triple bond between C and N (6 electrons).
## Step3: Distribute remaining electrons
Electrons used: $ 2 + 6 = 8 $, remaining: $ 10 - 8 = 2 $, which is a lone pair on N. So the structure is $ \text{H}-\text{C}\equiv\text{N} $ with a lone pair on N.

### For $ \boldsymbol{BrF_5} $
# Explanation:
## Step1: Calculate valence electrons
Bromine (Br): 7, Fluorine (F): 7 (5 F atoms). Total: $ 7+5\times7 = 7 + 35=42 $.
## Step2: Arrange atoms and form bonds
Br is the central atom, bonded to 5 F atoms (5 single bonds, using $ 5\times2 = 10 $ electrons).
## Step3: Distribute remaining electrons
Remaining electrons: $ 42 - 10 = 32 $. Each F gets 3 lone pairs (6 electrons), so $ 5\times6 = 30 $ electrons, and Br gets 1 lone pair (2 electrons). The structure has Br in the center, single - bonded to 5 F atoms, with 1 lone pair on Br.

### For $ \boldsymbol{CO} $
# Explanation:
## Step1: Calculate valence electrons
Carbon (C): 4, Oxygen (O): 6. Total: $ 4 + 6 = 10 $.
## Step2: Arrange atoms and form bonds
C and O are bonded. Form a triple bond (6 electrons) and a lone pair on C and a lone pair on O? Wait, correct way: A triple bond (6 electrons) between C and O, and a lone pair on C (2 electrons) and a lone pair on O (2 electrons)? Wait, no. Let's recalculate. Valence electrons: C (4) + O (6)=10. A triple bond uses 6 electrons, leaving 4 electrons. But C has 4 valence, O has 6. The correct Lewis structure is $ \text{C}\equiv\text{O} $ with a lone pair on C (2 electrons) and a lone pair on O (2 electrons)? Wait, no, actually, the formal charge: For C: $ 4-(2 + 3)= - 1 $, for O: $ 6-(2 + 3)=+1 $, but the common Lewis structure is $ \text{C}\equiv\text{O} $ with a lone pair on C and a lone pair on O (total 10 electrons: 6 in triple bond, 2 on C, 2 on O).

### For $ \boldsymbol{SiF_4} $
# Explanation:
## Step1: Calculate valence electrons
Silicon (Si): 4, Fluorine (F): 7 (4 F atoms). Total: $ 4+4\times7 = 4 + 28 = 32 $.
## Step2: Arrange atoms and form bonds
Si is the central atom, bonded to 4 F atoms (4 single bonds, using $ 4\times2 = 8 $ electrons).
## Step3: Distribute remaining electrons
Remaining electrons: $ 32 - 8 = 24 $. Each F gets 3 lone pairs (6 electrons), so $ 4\times6 = 24 $ electrons. The structure has Si in the center, single - bonded to 4 F atoms, each F with 3 lone pairs.

### For $ \boldsymbol{OF_2} $
# Explanation:
## Step1: Calculate valence electrons
Oxygen (O): 6, Fluorine (F): 7 (2 F atoms). Total: $ 6+2\times7 = 6 + 14 = 20 $.
## Step2: Arrange atoms and form bonds
O is the central atom, bonded to 2 F atoms (2 single bonds, using $ 2\times2 = 4 $ electrons).
## Step3: Distribute remaining electrons
Remaining electrons: $ 20 - 4 = 16 $. O gets 2 lone pairs (4 electrons), each F gets 3 lone pairs (6 electrons each). The structure has O in the center, single - bonded to 2 F atoms, with 2 lone pairs on O and 3 lone pairs on each F.

### For $ \boldsymbol{IBr} $
# Explanation:
## Step1: Calculate valence electrons
Iodine (I): 7, Bromine (Br): 7. Total: $ 7 + 7=14 $.
## Step2: Arrange atoms and form bonds
Form a single bond between I and Br (2 electrons).
## Step3: Distribute remaining electrons
Remaining electrons: $ 14 - 2 = 12 $. Each atom (I and Br) gets 3 lone pairs (6 electrons each). So the structure is $ \text{I}-\text{Br} $ with 3 lone pairs on I and 3 lone pairs on Br.

### For $ \boldsymbol{NH_4Cl} $
# Explanation:
## Step1: Analyze ions
$ NH_4Cl $ consists of $ NH_4^+ $ and $ Cl^- $. For $ NH_4^+ $: N (5) + 4H (4×1) - 1 (for positive charge) = 8 valence electrons. For $ Cl^- $: Cl (7)+1 (for negative charge)=8 valence electrons.
## Step2: Lewis structure of $ NH_4^+ $
N is central, bonded to 4 H atoms (4 single bonds, 8 electrons, no lone pairs on N). $ Cl^- $ has 4 lone pairs (8 electrons). So the structure shows $ NH_4^+ $ (N bonded to 4 H) and $ Cl^- $ as separate ions.

### For $ \boldsymbol{BF_3} $
# Explanation:
## Step1: Calculate valence electrons
Boron (B): 3, Fluorine (F): 7 (3 F atoms). Total: $ 3+3\times7 = 3 + 21 = 24 $.
## Step2: Arrange atoms and form bonds
B is central, bonded to 3 F atoms (3 single bonds, using $ 3\times2 = 6 $ electrons).
## Step3: Distribute remaining electrons
Remaining electrons: $ 24 - 6 = 18 $. Each F gets 3 lone pairs (6 electrons each). The structure has B in the center, single - bonded to 3 F atoms, each F with 3 lone pairs. (B has an incomplete octet, which is acceptable for B).

### For $ \boldsymbol{CH_2F_2} $
# Explanation:
## Step1: Calculate valence electrons
C (4) + 2H (2×1)+2F (2×7)=4 + 2+14 = 20.
## Step2: Arrange atoms and form bonds
C is central, bonded to 2 H and 2 F atoms (4 single bonds, using $ 4\times2 = 8 $ electrons).
## Step3: Distribute remaining electrons
Remaining electrons: $ 20 - 8 = 12 $. Each F gets 3 lone pairs (6 electrons each), H has no lone pairs. So C is in the center, bonded to 2 H and 2 F, with 3 lone pairs on each F.

### For $ \boldsymbol{CH_3Br} $
# Explanation:
## Step1: Calculate valence electrons
C (4)+3H (3×1)+Br (7)=4 + 3+7 = 14.
## Step2: Arrange atoms and form bonds
C is central, bonded to 3 H and 1 Br (4 single bonds, using $ 4\times2 = 8 $ electrons).
## Step3: Distribute remaining electrons
Remaining electrons: $ 14 - 8 = 6 $, which form 3 lone pairs on Br. So C is in the center, bonded to 3 H and 1 Br, with 3 lone pairs on Br.

### For $ \boldsymbol{PO_3^{3 - }} $
# Explanation:
## Step1: Calculate valence electrons
P (5)+3O (3×6)+3 (for negative charge)=5 + 18+3 = 26.
## Step2: Arrange atoms and form bonds
P is central, bonded to 3 O atoms. Let's form single bonds first (3 bonds, 6 electrons).
## Step3: Distribute remaining electrons
Remaining electrons: $ 26 - 6 = 20 $. Each O gets 3 lone pairs (6 electrons each), but we have 3 O atoms: $ 3\times6 = 18 $ electrons, and P gets 1 lone pair (2 electrons). But we can also have resonance. The more stable structure has one double bond and two single bonds. Let's recalculate with one double bond: P (5)+3O (3×6)+3 - 2 (for double bond)=5 + 18+3 - 2 = 24? Wait, no. The correct way: Valence electrons: P (5) + 3O (18) + 3 (charge) = 26. The Lewis structure has P in the center, with one double - bonded O and two single - bonded O atoms. The double - bonded O has 2 lone pairs, each single - bonded O has 3 lone pairs, and P has no lone pairs. The negative charge is on the single - bonded O atoms.

### For $ \boldsymbol{NH_4^+} $
# Explanation:
## Step1: Calculate valence electrons
Nitrogen (N): 5, Hydrogen (H): 4×1, minus 1 (for positive charge). So $ 5 + 4 - 1=8 $.
## Step2: Arrange atoms and form bonds
N is central, bonded to 4 H atoms (4 single bonds, using 8 electrons). There are no lone pairs on N. So the structure is a tetrahedral arrangement with N in the center and 4 H atoms bonded to it.

(Note: Since the problem is to draw Lewis dot structures, the above explanations provide the step - by - step process to determine the number of valence electrons, bond formation, and lone pair distribution for each molecule/ion. The actual Lewis dot structures can be drawn based on these steps, showing the atoms, bonds, and lone pairs as described.)

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QUESTION IMAGE

Question

Explanation:

For \( \boldsymbol{NI_3} \)

Step1: Calculate valence electrons

Step2: Arrange atoms and form bonds

Step3: Distribute remaining electrons

For \( \boldsymbol{H_2Se} \)

Step1: Calculate valence electrons

Step2: Arrange atoms and form bonds

Step3: Distribute remaining electrons

For \( \boldsymbol{I_3^-} \)

Step1: Calculate valence electrons

Step2: Arrange atoms and form bonds

Step3: Distribute remaining electrons

For \( \boldsymbol{HCN} \)

Answer:

Step1: Calculate valence electrons

Step2: Arrange atoms and form bonds