Give the geometry and approximate bond angles around the
central ...

Question

Give the geometry and approximate bond angles around the
central atom in CCl3-. (LO 8.1)
(a) Trigonal planar, 120°
(b) Trigonal pyramidal, 109.5°
(c) Trigonal pyramidal, 120°
(d) Bent, 109.5°

Accepted Answer

hey everyone were asked which of the following is the correct geometry and estimated bond angles around roaming in br cl four minus first. We need to determine the total number of electrons will have in our lewis structure. So starting with bro mean we know that bromine is in our group seven a. So this will give us seven valence electrons. Since we only have one of bromine, we're going to multiply this by one and in total will get seven. Looking at our chlorine, we know that chlorine is also in our group seven a. And so this will give us seven valence electrons and we're going to multiply this by four since we have four of chlorine, this will get us to a total of 28. Now since we have that -1 charge, this means that we will have to add one electron into our total. So adding all these values up, we need to draw 36 electrons in our Lewis structure. Now to draw our Lewis structure, we know that grooming is going to be our central atom since it is less electro negative than chlorine and we're going to draw our four chlorine surrounding it. Now to complete our chlorine soaked, it's we will draw three lone pairs onto each one of them. But since we have a remainder of four electrons we will add those two lone pairs onto our booming but this will give our booming a formal charge of -1. Now to determine our geometry, we can see that we have six electron groups which means that our electron geometry tells us that we have Octa hydro, which gives us 90 degrees and 180 degrees now surrounding our bruning. We can see that we have two lone pairs and four, atoms. which tells us that our molecular geometry is square plainer. So this means that we have a 90° bond angle around our bro mean. So our final answer here is going to be our molecular geometry where we have square planner and 90°. So answer choice c. Now, I hope that made sense and let us know if you have any questions.

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