Question 9.SIE.1: Elemental sulfur is a yellow solid that consists of S8 molec...
Elemental sulfur is a yellow solid that consists of S_8 molecules. The structure of the S_8 molecule is a puckered, eight-membered ring (see Figure 7.26). Heating elemental sulfur to high temperatures produces gaseous S_2 molecules:
S_8(s)→ 4 S_2(g)
(a) The electron configuration of which Period 2 element is most similar to that of sulfur? (b) Use the VSEPR model to predict the S—S—S bond angles in S_8 and the hybridization at S in S_8.(c) Use MO theory to predict the sulfur–sulfur bond order in S_2. Do you expect this molecule to be diamagnetic or paramagnetic? (d) Use average bond enthalpies (Table 8.3) to estimate the enthalpy change for this reaction. Is the reaction exothermic or endothermic?
| TABLE 8.3 Average Bond Enthalpies (kJ/mol) | |||||||
| Single Bonds | |||||||
| C—H | 413 | N—H | 391 | O—H | 463 | F—F | 155 |
| C—C | 348 | N—N | 163 | O—O | 146 | ||
| C—N | 293 | N—O | 201 | O—F | 190 | Cl—F | 253 |
| C—O | 358 | N—F | 272 | O—Cl | 203 | Cl—Cl | 242 |
| C—F | 485 | N—Cl | 200 | O—I | 234 | ||
| C—Cl | 328 | N—Br | 243 | Br—F | 237 | ||
| C—Br | 276 | S—H | 339 | Br—Cl | 218 | ||
| C—I | 240 | H—H | 436 | S—F | 327 | Br—Br | 193 |
| C—S | 259 | H—F | 567 | S—Cl | 253 | ||
| H—Cl | 431 | S—Br | 218 | I—Cl | 208 | ||
| Si—H | 323 | H—Br | 366 | S—S | 266 | I—Br | 175 |
| Si—Si | 226 | H—I | 299 | I—I | 151 | ||
| Si—C | 301 | ||||||
| Si—O | 368 | ||||||
| Si—Cl | 464 | ||||||
| Multiple Bonds | |||||||
| C=C | 614 | N=N | 418 | O=O | 495 | ||
| C ≡ C | 839 | N≡N | 941 | ||||
| C=N | 615 | N=O | 607 | S=O | 523 | ||
| C ≡ N | 891 | S=S | 418 | ||||
| C=O | 799 | ||||||
| C ≡ O | 1072 | ||||||
(a) Sulfur is a Group 16 element with an [Ne]3s²3p^4 electron configuration.
It is expected to be most similar electronically to oxygen (electron configuration,[He]2s²2p^4),which is immediately above it in the periodic table.
(b) The Lewis structure of S_8 is
There is a single bond between each pair of S atoms and two nonbonding electron pairs on each S atom. Thus, we see four electron domains around each S atom and expect a tetrahedral electron-domain geometry corresponding to sp³ hybridization. Because of the nonbonding pairs, we expect the S—S—S angles to be somewhat less than 109.5°, the tetrahedral angle. Experimentally, the S—S—S angle in S_8 is 108°, in good agreement with this prediction. Interestingly, if S_8 were a planar ring, it would have S—S—S angles of 135°. Instead, the S_8 ring puckers to accommodate the smaller angles dictated by sp³ hybridization.
(c) The MOs of S_2 are analogous to those of O_2, although the MOs for S_2 are constructed from the 3s and 3p atomic orbitals of sulfur. Further, S_2 has the same number of valence electrons as O_2. Thus, by analogy with O_2, we expect S_2 to have a bond order of 2 (a double bond) and to be paramagnetic with two unpaired electrons in the \pi_{3 p}^* molecular orbitals of S_2.
(d) We are considering the reaction in which an S_8 molecule falls apart into four S_2 molecules. From parts (b) and (c), we see that S_8 has S—S single bonds and S_2 has S=S double bonds. During the reaction, therefore, we are breaking eight S—S single bonds and forming four S=S double bonds. We can estimate the enthalpy of the reaction by using Equation 5.33 and the average bond enthalpies in Table 8.3:
H – CH _3(g)+ Cl – Cl (g) \longrightarrow Cl – CH _3(g)+ H – Cl (g) \quad \Delta H_{rxn}=? (5.33)
\begin{aligned}\Delta H_{rxn}&=8 D( S – S )-4 D( S=S ) \\&=8(266~kJ )-4(418~kJ )=+456~ kJ\end{aligned}
Recall that D(X—Y) represents the X—Y bond enthalpy. Because \Delta H_{rxn}>0, the reaction is endothermic (Section 5.3). The very positive value of \Delta H_{rxn} suggests that high temperatures are required to cause the reaction to occur.
