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Ch.8 - Covalent Compounds: Bonding Theories and Molecular Structure

Chapter 8, Problem 52

Oceanographers study the mixing of water masses by releasing tracer molecules at a site and then detecting their presence at other places. The molecule trifluoromethylsulfur pentafluoride is one such tracer. Draw an electron-dot structure for CF3SF5, and predict the bond angles around both carbon and sulfur.

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hey everyone, we're told that per floor in ated compounds. PFCs are used in tracer applications wherein they are released in one location and then their concentration is measured at a different place. One application is to observe air movement and trace the flow pollutants. Hexafluoride propylene is one example of PFCs which is a very potent greenhouse gas because of its long lifetime in the atmosphere, draw the complete lewis structure of hexafluoride prop line and determine the bond angles around each carbon center first, let's go ahead and determine the total number of electrons will have in our hexafluoride propylene. Starting with our carbon, we know that carbon is in our group for a So this will give us four valence electrons and looking at our formula, we can see that we have three carbon. So multiplying four times 3 we get 12. Now looking at our flooring, we know that flooring is in our group seven a. So this gives us seven veils electrons. We're going to multiply this by six since we have six of flooring, this gets us to a total of 42. Now, when we add these two values up we get a total of 54 electrons. Now let's go ahead and draw out our lewis structure. So we have our three carbons bonded to one another and carbon will be our central atom since it is least electro negative. Our carbon # one will be bonded to three florins. Our carbon number two will be double bonded to our third carbon and single bonded to our flooring while our carbon number three will be bonded to two florins. Now to complete our octet and fill up those 54 electrons, we will need to add three lone pairs onto each of our florins. Now let's go ahead and determine the bond angles around each carbon center. Starting with carbon number one, we can see that we have four electron groups which are four atoms and zero loan pairs. So this will be a tetrahedron Geometry with a bond angle of 109.5°. Now looking at carbon too, We can see that we have three electron groups. So we have three atoms and zero lone pairs. So that's going to be tribunal planer And our bond angle is going to be 120°. And lastly looking at carbon number three, we can see that we have three electron groups as well, which is going to be three atoms and zero lone pairs. So again, that's going to be tribunal planer With a bond angle of 120°. So this is going to be the final answer for each of our carbons. Now, I hope that made sense. And let us know if you have any questions