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Get Full Access to Chemistry: The Central Science - 14 Edition - Chapter 7 - Problem 7.53
Get Full Access to Chemistry: The Central Science - 14 Edition - Chapter 7 - Problem 7.53

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# Consider the first ionization energy of neon and the electron affinity of fluorine. (a) Write equations, including electron configurations, for each

ISBN: 9780134414232 1274

## Solution for problem 7.53 Chapter 7

Chemistry: The Central Science | 14th Edition

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Problem 7.53 Consider the first ionization energy of neon and the electron affinity of fluorine. (a) Write equations, including electron configurations, for each process. (b) These two quantities have opposite signs. Which will be positive, and which will be negative? (c) Would you expect the magnitudes of these two quantities to be equal? If not, which one would you expect to be larger?
Step-by-Step Solution:

Step 1 of 5) In almost all cases, single bonds are s bonds. A double bond consists of one s bond and one p bond, and a triple bond consists of one s bond and two p bonds: Consider ethylene 1C2H42, which has a C“C double bond. As illustrated by the balland-stick model of Figure 9.21, the three bond angles about each carbon are all approximately 120°, suggesting that each carbon atom uses sp2 hybrid orbitals (Figure 9.16) to form s bonds with the other carbon and with two hydrogens. Because carbon has four valence electrons, after sp2 hybridization one electron in each carbon remains in the unhybridized 2p orbital. Note that this unhybridized 2p orbital is directed perpendicular to the plane that contains the three sp2 hybrid orbitals. Let’s go through the steps of building the bonds in the ethylene molecule. Each sp2 hybrid orbital on a carbon atom contains one electron. Figure 9.22 shows how we can first envision forming the C¬C s bond by the overlap of two sp2 hybrid orbitals, one on each carbon atom. Two electrons are used in forming the C¬C s bond. Next, the C¬H s bonds are formed by overlap of the remaining sp2 hybrid orbitals on the C atoms with the 1s orbitals on each H atom. We use eight more electrons to form these four C¬H bonds. Thus, 10 of the 12 valence electrons in the C2H4 molecule are used to form five s bonds.

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