The xs represent the electrons from the h atoms and

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Unformatted text preview: veral compounds involving H 1s1 and C 1s2 2s2 2p2 . Since an H atom requires only one additional electron to achieve a full valence shell, each H atom bonds to one additional H atom to form diatomic hydrogen. Since both atoms in the H2 structure have filled valence shells, this diatomic molecule is stable and will not form any additional primary bonds. | v v 46 pgm 1-19-98 plm 3-21-98 MP | e-Text Main Menu | Textbook Table of Contents pg047 [R] G1 7-27060 / IRWIN / Schaffer pgm 1-19-98 plm 3-21-98 MP Chapter 2 H H C H or H H C C H H C H C C C or C C H C H C C H C C (a) Atomic Scale Structures H H (b) C H Double bond H C or H (c) FIGURE 2.6–2 Schematic illustrations of the covalent bond structure in a series of compounds: (a) CH4 , (b) pure carbon in the diamond structure, and (c) C2 H4 . The x’s represent the electrons from the H atoms and the •’s and ’s represent those from the C atoms. Note the double bond in the compound C2 H4 . The bond structure for carbon is more complex. Since C has only four electrons in its outer shell, it must share four pairs of electrons in order to achieve a full complement of eight valence electrons. In the compound CH4 , the C atom shares one electron with each H atom (see Figure 2.6–2a). The resulting methane molecule is stable. Next, consider the bonding arrangement for pure carbon. Each C atom may form a single covalent bond with four other C atoms, giving rise to the highly stable three-dimensional diamond structure shown in Figure 2.6–2b. What is the bond structure in the molecule C2 H4? Each H can satisfy its bonding requirements by forming a single covalent bond with one of the C atoms. In turn, each C atom will be covalently bonded to two H atoms. Each C atom, however, must still form two additional covalent bonds in order to achieve a filled valence shell. This can be accomplished if the C atoms share two pairs of electrons. The C2H4 molecule is shown schematically in Figure 2.6–2c. This molecule is fundamentally different from eith...
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