{"id":3300,"date":"2024-05-17T18:08:09","date_gmt":"2024-05-17T18:08:09","guid":{"rendered":"https:\/\/www.aceorganicchem.com\/blog\/?p=3300"},"modified":"2024-05-17T18:08:11","modified_gmt":"2024-05-17T18:08:11","slug":"molecular-geometry-of-brf5-with-video-and-free-study-guide","status":"publish","type":"post","link":"https:\/\/www.aceorganicchem.com\/blog\/molecular-geometry-of-brf5-with-video-and-free-study-guide\/","title":{"rendered":"Molecular Geometry of BrF5 [with video and free study guide]"},"content":{"rendered":"\n

What is the molecular geometry of BrF5<\/sub>?<\/strong><\/h1>\n\n\n\n

The molecular shape of BrF5<\/sub> is square pyramidal, or AX5<\/sub>E using Valence Shell Electron Pair Repulsion (VSEPR) theory. Hence, the molecular geometry of BrF5<\/sub> has only 90 degree bond angles in the molecule. BrF5<\/sub> looks like this:<\/p>\n\n\n\n

\"Molecular<\/a><\/figure>\n\n\n\n

How do you find the molecular geometry of BrF5<\/sub><\/strong>? <\/strong><\/h2>\n\n\n\n

There is an easy three-step process for determining the geometry of molecules with one central atom.<\/p>\n\n\n\n

Step 1<\/strong>: Determine the Lewis structure of the molecule.<\/em>
For BrF5<\/sub>, it is as shown below: For a full-explanation of how to figure out the Lewis structure, please go to Lewis Structure of BrF5<\/sub>. However, here is what it looks like. It is different because bromine is hypervalent (bromine usually prefers one bond and three lone pairs. Here it has five bonds and one lone pair)<\/p>\n\n\n\n

\"Lewis<\/a><\/figure>\n\n\n\n

Step 2<\/strong>: Apply the VSEPR notation to the molecule.<\/em>
Apply VSEPR notation, A X E
A=Number of central atoms
X=Number of surrounding atoms
E= Number of lone pairs on central atom
For this one, we can see that it has one central atom (Br), five surrounding atoms (F), and one lone pair of electrons around the central atom, making it AX5<\/sub>E.<\/p>\n\n\n\n

Step 3<\/strong>: Use the VSEPR table to determine the geometry of BrF5<\/sub>.<\/em><\/p>\n\n\n\n

\"VSEPR<\/a><\/figure>\n\n\n\n

As you can see from the chart, the AX5<\/sub>E molecule is square pyramidal. <\/strong><\/h2>\n\n\n\n

Bond angles help show molecular geometry of BrF5<\/sub><\/strong><\/h3>\n\n\n\n

The only bond angles in this molecule are the F-Br-F angles. There are two different types of F-Br-F angles: equatorial-axial and equatorial-equatorial. Below is a diagram which will explain this more.<\/p>\n\n\n\n

\"Bond<\/a><\/figure>\n\n\n\n

As you can see from the diagram above, the equatorial-equatorial bond angle in BrF5<\/sub> is 90 degrees, and the axial-equatorial is 90 degrees. This geometry allows for substituents (atoms and lone pairs) to be as far apart from each other as possible. <\/p>\n\n\n\n

What about that lone pair?<\/strong><\/h3>\n\n\n\n

The lone pair of the molecule resides in one of the axial positions, below the square as shown below. This means that the ELECTRONIC geometry is octahedral, even though the MOLECULAR geometry of BrF5 is square pyramidal. See more about about concept in the FAQs below.<\/p>\n\n\n\n

\"Electronic<\/a><\/figure>\n\n\n\n

More about VSEPR and the molecular geometry of BrF5<\/sub>:<\/strong><\/h3>\n\n\n\n

Let’s not forget, the whole purpose of VSEPR is to minimize interactions between the substituents (atoms and lone pairs) of a molecule. We also know that electrons repel each other. Hence, simple molecules (like the ones we are looking) at will tend to place substituent atoms as far from each other as possible. We know this because of the bond angles associated with each of the four types of shapes.<\/p>\n\n\n\n

\"VSEPR<\/a><\/figure>\n\n\n\n

Here is one way to remember this chart: Think about each lone pair as just replacing an atom. In the chart above we have tried to show how this works by just blurring out an atom for a lone pair. <\/p>\n\n\n\n

For the 3 and 4 substituent molecules (AX3<\/sub> group and AX4 <\/sub>group, respectively) it is easy to do this because each one of the substituent atoms is the same. So for AX2<\/sub>E, it is simple to see that we get trigonal pyramidal as the answer because we can replace any of the atoms with a lone pair because they are all geometrically equivalent. Same for AX3<\/sub>E because all of the atoms are geometrically equivalent. <\/p>\n\n\n\n

It gets a little trickier when we get to the 5 and 6 substituent molecules (AX5<\/sub> group and AX6 <\/sub>group, respectively). Here, there is a geometric difference between the atoms on the axis (called axial substituents) and the ones around the middle, called the equatorial substituents. Thus, we can’t just substitute a lone pair for any old atom. So…..what we need to remember is that for the AX5<\/sub> group, you need to replace equatorial atoms with lone pairs AND for the AX6<\/sub> group, you need to replace the atoms on the axis with lone pairs, as we have shown above.<\/p>\n\n\n\n

Some video to make it a little simpler:<\/strong><\/h3>\n\n\n\n
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