Match each of the formulas (a to c) with the correct diagram (1 t...

Question

Match each of the formulas (a to c) with the correct diagram (1 to 3) of its shape, and name the shape; indicate if each molecule is polar or nonpolar. (6.6, 6.8, 6.9)1.

2.

3.

a. PBr₃

Accepted Answer

Welcome back everyone. The formula for nitrogen tri ID is N I three identify its molecular shape representation from diagrams one through four and give the name of the shape determine whether the molecule is polar or nonpolar. We're going to need to analyze electron groups bonding groups as well as nonbinding electrons. A K a lone pairs on our central atom beginning with electron groups, we can observe that our central atom nitrogen is bonded to three or rather four Adams of iodine. And that would correspond to a total of four electron groups as far as bonding groups. Our four electron groups all happen to be bonds to bonds from our central atom to the four ID atoms. So we have four bonding groups as well. And as far as lone pairs and nonbinding electrons, we observe a total of zero non-binding electrons on our central atom nitrogen. So we have zero lone pairs recall that when we have four electron groups, four bonding groups and zero lone pairs on our central atom that corresponds to a molecular geometry, which is going to be in this case, tetrahedral. Let's move on to diagram two in diagram two, we observe that our central atom nitrogen is bonded to three iodine atoms. So that counts for a total of three electron groups. And our three electron groups all happen to be bonding groups. We observe zero lone pairs on our central atom nitrogen. And so thus, our electron geometry we should recall is going to be trigonal planer. Now, moving on to diagram three, we observe that our central atom nitrogen is bonded to three atoms. However, we no longer have a planar arrangement to these three atoms. So what we we have occurring is an influence of another factor that is pushing these bonds in this downward direction. And we should recall that bond, that nitrogen has a bonding preference of being bonded to three atoms. So it's bonding preference is to be bonded to three atoms and have one lone pair on itself. So we can actually draw in a lone pair to our central atom nitrogen here. And due to the presence of this lone pair, it is interrupting the symmetry of our structure and pushing the bonds to the three iodine atoms in a downward position. And so thus, we can count a total of four electron groups where three are our bonds to iodine. And one is our lone pair on nitrogen. As far as bonding groups, our three bonds to iodine would represent our three bonding groups. And we have one lone pair on our central atom nitrogen that we drew in. And thus, this corresponds to a molecular geometry which is trigonal pyramidal. And last, we have diagram four in which we can observe that we have two atoms bonded to our central atom nitrogen here. And we also observe that some factor is pushing these bonds to iodine in a downward position. So we're going to want to also recognize that our octet for nitrogen here is incomplete. It has two bonds to iodine counting for a total of four electrons. We're going to need to add in two more or four more electrons to complete its octet, which we can do. So by adding a set of two lone pairs to our central atom nitrogen. And via these two lone pairs, they would be influencing the shape of our structure by pushing our bonds from our central atom to our iodine atoms in a downward position. So we can count a total of four electron groups again where two are our lone pairs and two are the bonds to iodine that would count for two bonds, two bonding groups. And then as far as lone pairs, we added a total of two, two sets. So we have two lone pairs on our central atom nitrogen. And this will all correspond to a molecular geometry which is bent, also known as V shaped and also known as angular. So now we need to consider the shape of nitrogen tri iodide. Now that we understand our three given diagrams, so we're going to need to calculate total advance electrons for nitrogen tri iodide beginning with nitrogen, which we should recall on our periodic table is located in group five A and corresponds to five valence electrons. We're then going to move on to iodine, which we recall on the periodic table is located in group seven A corresponding to seven valence electrons. Now notice that we have that subscript of three that tells us that we're going to need to multiply by a total of three atoms of iodine, which will yield a contribution of 21 val electrons. So taking our sum between our five and 21 valence electrons, we would have a total of 26 valence electrons to draw our structure of nitrogen tri ide. So making more room to draw our structure, we'll place nitrogen in the center since it's our singular atom and we're going to surround it by our three iodine atoms. As we stated earlier, nitrogen has a bonding preference and sorry, this should be a I here. So as we stated earlier, nitrogen has a bonding preference of having three bonds and one lone pair in itself. So we're going to go ahead and create our three bonds from our central atom nitrogen to our three surrounding iodine atoms. And then we can fill in a total of one lone pair on nitrogen. So counting what we've used so far out of our total valence electrons, we have 246 for our three bonds and then electrons used so far. So subtracting the eight electrons used from our total, we would have a balance remaining of 18 valence electrons to include in our structure. As we stated earlier, iodine is a group 78 atom and should have seven valence electrons for a formal charge of zero. And currently each of our iodine atoms have one directly attached valence electron that it shares in its covalent bond to nitrogen. We're going to need to place a total of three lon pairs on each of our iodine atoms so that they can all have directly attached their preference of seven valence electrons. However, in doing so, we would have used up a balance of 246, 8, 10, 12, 14, 16, 18 of our remaining electrons leaving us with a balance of zero and rendering our structure complete for nitrogen tri iodide with a net formal charge of zero. Now, we're going to consider electron groups bonding groups and any lone pairs on nitrogen, our central atom. So beginning with electron groups, we observe that we have three surrounding iodine atoms bonded to nitrogen and alone pairs on nitrogen that would count for a total of four electron groups. And out of our four electron groups, we have three bonds to iodine from nitrogen. So that is a total of three bonding groups. And we stated that we observe one lone pair on our central atom nitrogen that we drew in And so we have just one set of non-binding electrons or one lone pair. Now, based on this presence of one lone pair, we therefore have an influence of how our iodine atoms are bonded to nitrogen. So we therefore have an asymmetrical, so we have an asymmetrical structure. So really how we want to draw our structure in is so that our bonds to iodine are now in a downward facing position. So we would have bonds to iodine all in a downward position. And our lone pair of nitrogen at the top. Now, based on this asymmetrical structure, we want to recall the electron negativity trend on our periodic table which increases in the direction of fluorine, our most electron negative atom. And thus, we would observe that the electron negativity of nitrogen is greater than that of our iodine atoms. So we have dipoles that face the direction of nitrogen where we also have a net dipole in the upper direction because our electrons are even more negative than our nitrogen itself. And so we have a net dipole due to the presence of those electrons, those lone electrons on nitrogen specifically. So that creates our asymmetrical structure, which therefore creates our net dipole. And therefore, we can say we have a polar molecule and as far as our electron geometry or molecular geometry rather labeling that because we have that one loan pair are four electron groups and three bonding groups that corresponds to trigonal pure middle electron or molecular geometry. And we confirm that we have a polar molecule. So going back to our diagrams, we would agree that the only trigonal pyramidal diagram was diagram three, which corresponds to the shape of nitrogen tri iodide as to the name. And we did determine that our molecule of nitrogen tri iddy is polar. So our final answers our diagram three trigonal pyramidal as the name of our molecular geometry for nitrogen tri ide. And our third answer is that it is a polar molecule since it has a net dipole. So I hope that everything we went through made sense. If you have any questions, let us know and I'll see everyone in the next video.

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Match each of the formulas (a to c) with the correct diagram (1 to 3) of its shape, and name the shape; indicate if each molecule is polar or nonpolar. (6.6, 6.8, 6.9)1.
2.
3.
a. PBr₃

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