Formula
Step 1
Step 2
Step 3
Step 4
Step 5
Step 6
N2H4
H N N H H H
14
7
H N N
H H
H
H
N H
2
N H
H
Unnecessary (octets complete)
SiO2
O Si O
O Si
O
Name Date Class
TEACHING TRANSPARENCY WORKSHEET
Lewis Structures
29
Use with Chapter 8,
Section 8.3
1. Step 1 in drawing the Lewis structure for a molecule is to decide which atoms of the molecule are most likely the terminal ones. In the transparency, why are the hydrogen
(H) atoms in hydrazine (N2H4) shown as the terminal atoms?
2. Step 2 in drawing a Lewis structure involves determining the total number of valence electrons in the atoms in the molecule. Explain why the total number of valence electrons in N2H4 is 14.
3. Step 3 in drawing a Lewis structure requires finding the number of bonding pairs. What must be done to the result of step 2 to find the number of bonding pairs? Verify that this is so in the case of N2H4 in the transparency.
4. In step 4 in the transparency, one bonding pair has been placed between each pair of bonded atoms in N2H4. How many such bonding pairs are shown in step 4, and what symbol is used to represent them?
5. Step 5 requires subtraction of the number of bonding pairs used in step 4 from the number of bonding pairs determined in step 3. Verify that the result is 2 for N2H4. Lone pairs are then placed around each terminal atom to achieve a full outer level, and any remaining pairs are assigned to the central atom(s). Explain the drawing that has resulted for N2H4.
6. In step 6, if any central atom drawn in step 5 does not have an octet, lone pairs from the terminal atoms must be converted to double or triple bonds involving the central atom. Why was this extra step unnecessary in the case of N2H4?
7. What number should be placed in the blank for step 2 for the silicon dioxide (SiO2) molecule?
8. What number should be placed in the blank for step 3 for SiO2?
58 Chemistry: Matter and Change • Chapter 9 Teaching Transparency Worksheets
Name Date Class
Name Date Class
TEACHING TRANSPARENCY WORKSHEET
Lewis Structures
29
Use with Chapter 8,
Section 8.3
TEACHING TRANSPARENCY WORKSHEET
VSEPR Model and Molecular Shape
30
Use with Chapter 8,
Section 8.4
1. Step 1 in drawing the Lewis structure for a molecule is to decide which atoms of the molecule are most likely the terminal ones. In the transparency, why are the hydrogen
(H) atoms in hydrazine (N2H4) shown as the terminal atoms?
A hydrogen atom can form only one bond, so it cannot be a central atom.
2. Step 2 in drawing a Lewis structure involves determining the total number of valence electrons in the atoms in the molecule. Explain why the total number of valence electrons in N2H4 is 14.
Each nitrogen (N) atom has five valence electrons and each hydrogen (H) atom has one valence electron, resulting in a total of (2 x 5) + (4 - 1) = 14.
3. Step 3 in drawing a Lewis structure requires finding the number of bonding pairs. What must be done to the result of step 2 to find the number of bonding pairs? Verify that this is so in the case of N2H4 in the transparency.
The total number of valence electrons from step 2 must be divided by 2; in the case of N H , 14/2 = 7.
1. The shapes of the molecules shown have been determined by means of the VSEPR model. What is the basic assumption of this model?
Pairs of electrons, either shared or unshared, repel each other as much as possible around a central atom.
2. How many lone pairs and how many shared pairs of electrons surround the boron (B) atom in the borane (BH3) molecule shown? no lone pairs; three shared pairs
3. What is the shape of the BH3 molecule? Explain why.
The BH3 molecule is trigonal planar. There are three electron pairs that repel so that they are as far as possible from one another. The shape that maximizes the distance is trigonal planar.
2 4
4. How many lone pairs and how many shared pairs of electrons surround the carbon (C)
4. In step 4 in the transparency, one bonding pair has been placed between each pair of bonded atoms in N H . How many such bonding pairs are shown in step 4, and what atom in the methane (CH4) molecule shown? What is the shape of the molecule?
2 4 symbol is used to them? no lone pairs; four shared pairs; tetrahedral
five; a line represent 5. How many lone pairs and how many shared pairs of electrons surround the nitrogen (N)
5. Step 5 requires subtraction of the number of bonding pairs used in step 4 from the number of bonding pairs determined in step 3. Verify that the result is 2 for N2H4. Lone pairs are then placed around each terminal atom to achieve a full outer level, and any remaining pairs are assigned to the central atom(s). Explain the drawing that has resulted for N2H4.
7 - 5 = 2. Because the H atoms already had a complete outer level of electrons, a lone pair of electrons was positioned next to each N atom.
6. In step 6, if any central atom drawn in step 5 does not have an octet, lone pairs from the terminal atoms must be converted to double or triple bonds involving the central atom. Why was this extra step unnecessary in the case of N2H4?
The central N atoms already had complete octets, and the Lewis structure was already correct.
7. What number should be placed in the blank for step 2 for the silicon dioxide (SiO2) molecule?
4 + 6 + 6 = 16, the total number of valence electrons
8. What number should be placed in the blank for step 3 for SiO2?
16/2 = 8, the number of bonding pairs atom in the ammonia (NH3) molecule shown? What is the shape of the molecule? one lone pair; three shared pairs; trigonal pyramidal
6. How many lone pairs and how many shared pairs of electrons surround the oxygen (O) atom in the water (H2O) molecule shown? What is the shape of the molecule? two lone pairs; two shared pairs; bent
7. How many lone pairs and how many shared pairs of electrons surround the fluorine (F) atom in the hydrogen fluoride (HF) molecule shown? What is the shape of the molecule? three lone pairs; one shared pair; linear
58 Chemistry: Matter and Change • Chapter 8 Teaching Transparency Worksheets
60 Chemistry: Matter and Change • Chapter 8 Teaching Transparency Worksheets
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