1A_exercise_key_8_9 - Chapters 8 and 9 Exercise Key 1...

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Chapters 8 and 9 Exercise Key 1 Chapters 8 & 9 Exercise Key Exercise 8,9.1 – Explanation of Bonding Patterns: Explain the following bonding patterns. a. C - 4 bonds and no lone pairs, and no formal charge. Only the highest energy electrons participate in bonding. Carbon is in group 4A, so it has four valence electrons per atom. It is as if one electron is promoted from the 2 s orbital to the 2 p orbital. The 2 s and the three 2 p orbitals blend to form four equivalent sp 3 hybrid orbitals. Covalent bonds form to in order to pair unpaired electrons. The four unpaired electrons lead to four covalent bonds. Because there are no pairs of electrons, carbon atoms have no lone pairs when they form four bonds. b. N - 4 bonds, no lone pairs, and a +1 formal charge Nitrogen is in group 5A, so it has five valence electrons per atom. If a nitrogen atom loses one electron from the pair, it will have four unpaired electrons. The 2 s and the three 2 p orbitals blend to form four equivalent sp 3 hybrid orbitals. Covalent bonds form to in order to pair unpaired electrons. The four unpaired electrons lead to four bonds. There are no pairs of electrons remaining, so there are no lone pairs. The lost electron leads to the +1 formal charge. Copyright 2004 Mark Bishop
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Chapters 8 and 9 Exercise Key 2 c. O - 1 bond, 3 lone pairs, and a –1 formal charge. Only the highest energy electrons participate in bonding. Oxygen is in group 6A, so it has six valence electrons per atom. If it gains one electron, it will have a total of seven. The 2 s and the three 2 p orbitals blend to form four equivalent sp 3 hybrid orbitals. Covalent bonds form to in order to pair unpaired electrons. The one unpaired electron leads to one bond, and the three pairs of electrons give oxygen atoms with an extra electron three lone pairs. The gained electron leads to the 1 formal charge. d.
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1A_exercise_key_8_9 - Chapters 8 and 9 Exercise Key 1...

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