Chapter 8 Covalent Bonding

Section 8.3 Molecular Structures

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Lewis Structures1. Step 1 in drawing the Lewis structure for a molecule is to decide which atoms of the molecule are

most likely the terminal ones. 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.

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______________________________________________________________________________________________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.

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______________________________________________________________________________________________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.

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______________________________________________________________________________________________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?

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______________________________________________________________________________________________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?

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TEACHER GUIDE AND ANSWERS

Teaching Transparency 26 – Lewis Structures1. A hydrogen atom can form only one bond, so it

cannot be a central atom.2. Each nitrogen (N) atom has five valence electrons and

each hydrogen (H) atom has one valence electron, resulting in a total of (2 5) (4 1) 14.

3. The total number of valence electrons from step 2 must be divided by 2; in the case of N2H4, 14/2 7.

4. five; a line

5. 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. The central N atoms already had complete octets, and the Lewis structure was already correct.

7. 4 6 6 16, the total number of valence electrons8. 16/2 8, the number of bonding pairs

Chemistry: Matter and Change Teacher Guide and Answers3