Draw a Lewis structure for the following molecules:
e. BeCl₂ (Note: This molecule does not follow the octet rule.)

Question

Draw a Lewis structure for the following molecules:

e. BeCl₂ (Note: This molecule does not follow the octet rule.)

Accepted Answer

Welcome back. Everyone show the Lewis dot structure for beryllium. I died a compound that is not compliant with the octet rule. Let's begin by recalling that we need to calculate total valence electrons for beryllium. I died. Beginning with our single atom of beryllium. We have one atom of beryllium which we recall is in group two a which will correspond to two valence electrons. So we would multiply Rather instead we would say two valence electrons for beryllium multiplied by one atom of beryllium. This is going to contribute to valence electrons from beryllium. Next we recognize we have iodine recall that iodine is in group seven a Corresponding to seven valence electrons Multiplied by our two atoms of iodine. This will correspond to a total of 14 valence electrons contributed by iodine. So just to note, we have brilliant in group two a Again corresponding to two valence electrons and we multiply it by one atom of beryllium. So sorry for that confusion there. Now, with this outlined, we can add up our total contribution of valence electrons to get our total valence electrons equal to 16 valence electrons total for our compound of beryllium. I died recall that we want the least electro negative atom in the center. So we can see that based on our electro negativity trend, which increases towards the top right of our periodic table, we would say that the electro negativity of beryllium is less than the electro negativity of our compound iding or Adam Aydin. And so therefore, beryllium is our central atom. So in our Lewis structure, we have beryllium surrounded by two atoms of iodine, we're going to begin by making our base connections. So we're going to connect both two atoms of iodine with single bonds to beryllium. Now, in doing so, notice that we have directly attached to valence electrons which are shared in a Covalin bond with the two iodine atoms. So these two valence electrons are attached directly to beryllium. And this means that our beryllium is stable and happy because as we stated, beryllium is a group to a atom and does have two valence electrons to be stable and have a formal charge of zero. So how we've connected it makes sense and now for the rest of our electrons, we're going to say 16 minus two, which leaves us with a balance of 14 electrons to use to make our Aydin Adam stable. And as we stated, iodine is a group seven A Adam and corresponds to seven valence electrons to be stable. Currently, each atom has one valence electron directly attached to it, meaning that it's not stable. So we're gonna need to add six more. So we would add lone pairs around our iodine atoms. We have 246, 8, 10, 12. And going back to our balance of valence electrons, I'd like to correct myself and recall that each of our bonds represent two electrons each. So we have a difference of four electrons. So subtracting four electrons leaves us with a balance of 12 electrons. And because we added a total of three lone pairs around each iodine atom. Each idea atom now has directly attached 1234567 valence electrons, which is what they want being in group 78 to be stable and have a formal charge of zero. And so therefore, that uses up the rest of our 12 valence electrons leaving us with a balance of zero and completing our lewis structure recall that we can check our formal charge for each atom by taking the group number subtracted from bonding electrons divided by two plus non bonding electrons. So in this case, just to show the work for our formal charge of beryllium, we would take its group number being two Subtracted from its number of bonding electrons in which it has four total bonding electrons in the two bonds that it has divided by two added to the non bonding electrons, which is zero. This is going to simplify to two minus two, giving us a formal charge equal to zero for beryllium. Now, for the formal charge of Aydin will label each Aydin as I D N A and then I dine B on the left. So for idea a calculating its formal charge, we take its group number being in group seven on the periodic table subtracted from its bonding electron which would be two bonding electrons since it just has one bond to our central atom, beryllium divided by two plus, it's non bonding electrons in which we added a total of three lone pairs being a total of six non bonding electrons. So this would be equal to seven minus one plus six, which would be equal to zero. So formal charge of iodine A has a formal charge of zero, which is what we want. So it's stable. And now our formal charge of I D and B on the left will say is equal to its group number seven on the periodic table subtracted from two divided by two plus 6, again equal to seven minus one plus six also equal to a formal charge of zero because we see that idea A and B are bonded in the same way. So their formal charges will both equal the same calculation zero. Notice that beryllium indeed does break our octet rule because it only has two valence electrons directly attached to itself and a total of four electrons surrounding beryllium in Covalin bonds. And this would make sense because we should recall that group two a Adams do not follow the octet rule. So they're typically going to have four valence electrons around themselves in a structure. So ultimately, after everything we reviewed, we can confirm that this Lewis structure box in yellow would be our final answer to complete this example, which corresponds to choice D in the multiple choice. I hope this made sense. And let us know if you have any question.

Draw a Lewis structure for the following molecules:
e. BeCl₂ (Note: This molecule does not follow the octet rule.)

Verified video answer for a similar problem:

Key Concepts

Lewis Structure

Octet Rule

Molecular Geometry

Draw a Lewis structure for the following molecules: e. BeCl2 (Note: This molecule does not follow the octet rule.)

Verified video answer for a similar problem:

Key Concepts

Lewis Structure

Octet Rule

Molecular Geometry

Draw a Lewis structure for the following molecules:
e. BeCl₂ (Note: This molecule does not follow the octet rule.)