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8 Chemical Bonding Solutions to Exercises for C-C (1.34 Å, 1.54 Å), N-N (1.24 Å, 1.47 Å) and 0-0 (1.21 Å, 1.48 Å) bonds from Table 8.5, we see that, on average, a double bond is approximately 0.23 Å shorter than a single bond. Applying this difference to the S-N single bond length, we estimate the S-N double bond length as 1.54 Å. Finally, the intermediate S-N bond length in should be between these two values, approximately 1.60-165 Å. (The measured bond length is 1.62 Å.) (d) 4S(g) + 4N(g) = = 4(222.8 kJ) + 4(472.7 480 kJ = 2302 kJ This energy, 2302 kJ, represents the dissociation of 8 S-N bonds in the molecule; the average dissociation energy of one S-N bond in is then 2302 kJ/8 bonds = 287.8 kJ. 8.110 (a) Yes. In the structure shown in the exercise, each P atom needs 1 unshared pair to complete its octet. This is confirmed by noting that only 6 of the 10 valence pairs are bonding pairs. (b) There are six P-P bonds in (c) 20 10 pr In this Lewis structure, the octet rule is satisfied for all atoms. However, it requires P=P, which is uncommon because P has a covalent radius that is too large to accommodate the side-to-side π overlap of parallel P orbitals required for double bond formation. (d) From left to right, the formal charges are on the P atoms in the linear structure are -1, +1, +1, -1. In the tetrahedral structure, all formal charges are zero. Clearly the linear structure does not minimize formal charge and is probably less stable than the tetrahedral structure, owing to the difficulty of P=P bond formation (see above). 8.111 (a) = H(g) + C(g) = 6(217.94) kJ + 6(718.4) kJ 82.9 = 5535 (b) 6CH(g) (c) C₆H₆(g) 6H(g)+6C(g) 5535 kJ 6H(g)+6C(g) + 6CH(g) -6D(C-H) -6(413)kJ C₆H₆(g) 6CH(g) 3057 kJ 3057 kJ is the energy required to break the six C-C bonds in C₆H₆(g). The average bond dissociation energy for one carbon-carbon bond in is 3057 kJ = 509.5 bonds 228