Write the Lewis structure for each of these molecules or ions. (a) BrF5 (b) IF 5 (c) IBr2

Name: Problem 67, Chapter 8: Chemistry: The Molecular Science
Uploaded: 2022-03-21T05:52:08-08:00
Duration: 6 min 10 s
Channel: Sandra Lundell
Description: Write the Lewis structure for each of these molecules or ions. (a) BrF5 (b) IF 5 (c) IBr2

step 1 First, we're drawing the Lewis structure of BRF5. The molecules in this problem are going to be

Write the Lewis structure for each of these molecules or...

Key Concepts

Molecular Geometry

The arrangement of bonds and lone pairs around a central atom determines its molecular geometry. This shape, which can be predicted using concepts like VSEPR theory, is essential for understanding the chemical reactivity and physical properties of the molecule.

Valence Electrons

The total number of valence electrons from all atoms in a molecule is the foundation of constructing a Lewis structure. This count helps determine how electrons will be allocated between bonds and lone pairs to satisfy stability requirements.

Electron Counting and Distribution

After determining the total valence electrons, these electrons must be distributed among the atoms. This involves creating bonds to share electrons and assigning remaining electrons as lone pairs, ensuring each atom (as applicable) meets required stability criteria.

Octet Rule and Exceptions

The octet rule posits that atoms tend to form structures in which they have eight electrons in their valence shell, leading to stability. However, elements in the third period or below or beyond can have expanded octets due to available d-orbitals, which is crucial for understanding the bonding patterns in certain molecules.

Formal Charge

Calculating formal charges helps assess the stability of a Lewis structure. It is important to assign electrons in such a way that formal charges are minimized across the structure, confirming that the distribution of electrons is appropriate.

Transcript

00:01 First, we're drawing the lewis structure of brf5.

00:09 The molecules in this problem are going to be breaking octets, meaning they're going to be central atoms that are capable of having more than eight electrons.

00:18 So to start, let's add up our balance electrons.

00:22 Bromene has seven, as does fluorine, and there's five of those.

00:30 So that gives us a total of 42 electrons.

00:36 We're going to put bromine in the middle, and then we're going to put five fluorines around it.

00:43 And it doesn't have to be done in some specific way.

00:47 You just need to fit five around that.

00:49 The reason that it's possible to have more than four bonds is because anything that comes in the third row of the periodic table or lower, technically can have more.

01:03 More than eight electrons, valence electrons.

01:06 It can break the octets due to its availability of the d orbitals.

01:12 There's empty orbitals that are very near the same energy that their valence electrons are already in.

01:18 That allows them that if you give them more than eight electrons, it allows them to pop them up into a higher orbital without too much trouble.

01:25 Now, if you're in an earlier column like carbon, nitrogen, oxygen, they cannot break octets.

01:32 They don't have an available d orbital because there is no 2d orbital.

01:38 It starts at 3d.

01:40 All right.

01:41 So these can break octets.

01:42 So bromine already has more than eight electrons, but that's okay.

01:46 All right.

01:47 So what we've drawn so far has five bonds.

01:49 That's 10 electrons, two in each bond, which drops us down to 32 to go.

01:56 So let's fill in the octets of our outer electrons.

02:00 Each fluorine wants six more to get it up to eight.

02:11 6 times 5 subtracts off another 30.

02:15 We've got two electrons left over.

02:18 Those are going to go on the central atom.

02:21 It doesn't matter where you put them on the bromine as long as you keep them together, so they look like a pair.

02:27 That's important.

02:29 All righty? so that is actually your final structure.

02:33 Okay.

02:34 And that is the max amount that we're going to have you put on a compound.

02:38 We're going to say that these atoms can go between having 8 to 12 electrons, but they can't break that 12 mark.

02:46 And this one currently, bromine has 10 coming from the bonds, plus the lone pair gets it to 12.

02:52 If you've done shapes, this makes it an octahedral.

02:58 Okay, going to the next one.

03:02 Now we're doing if5.

03:05 Same exact setup.

03:07 Iodine has seven valens.

03:11 Each fluorine has seven as well, which means we're working with 42 electrons...

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