Sulfur tetrafluoride 1SF42 reacts slowly with O2 to form sulfur
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Question

Sulfur tetrafluoride 1SF42 reacts slowly with O2 to form sulfur
tetrafluoride monoxide 1OSF42 according to the following
unbalanced reaction:
SF41g2 + O21g2¡OSF41g2
The O atom and the four F atoms in OSF4 are bonded to a
central S atom. (b) Write a Lewis
structure of OSF4 in which the formal charges of all atoms
are zero.

Accepted Answer

hey everyone, we're asked to predict the lewis structure of the following compound. And we need to make sure that all of our atoms have zero formal charges first. Let's go ahead and determine the total number of valence electrons will have for our compound. Looking at our oxygen, we have one of oxygen and we're going to multiply this by six valence electrons since oxygen is in our group six a. This will get us the sticks. Next looking at our selenium, we have one of selenium and we're going to multiply this by six as well since it's also in our group six A. This will get assistance valence electrons And lastly looking at our chlorine we have four of chlorine and we're going to multiply this by seven since it's in our group seven A. This will get us to 28 valence electrons. When we add these three values up, we get a total of 40 valence electrons. Now to draw our lewis structure, we were told that selenium was our central atom and we have our four chlorine and our oxygen attached to it as well To complete our chlorine octet we will add three lone pairs onto each chlorine And to get 40 valence electrons, we can go ahead and create a double bond between selenium and oxygen and add two lone pairs to calculate the formal charge of each atom. We can go ahead and take our group number and subtract the sum of our bonds plus our non bonding electrons. Starting off with the formal charge of oxygen. We're going to take our group number of six and subtract the sum of two plus four. Since we have two bonds and four non bonding electrons, this will get us so formal charge of zero, calculating the former charge of selenium. We're going to take our group number of six and subtract the sum of six plus zero, which will also get us to zero. And lastly calculating the formal charge of chlorine. We're going to take our group number of seven and subtract The sum of 1-plus 6, which will also get us to a formal charge of zero. So each atom has a formal charge of zero and this will be our final lewis structure for our compound. Now, I hope that made sense. And let us know if you have any questions.

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