Compare the Lewis structures of CH₄ and H₂O Why do these molecule...

Question

Compare the Lewis structures of CH₄ and H₂O Why do these molecules have similar bond angles but different molecular shapes?

Accepted Answer

Welcome back everyone. Why do ammonia and sulfur chloride? Despite having comparable bond angles have different molecular geometries. Choice. A states that ammonia and sulfur di chloride have different molecular geometries because they have different electron arrangements. Choice B states that ammonia su dilor have different molecular geometries because they have different numbers of lone pairs of electrons. Choice C states that ammonia and sulfur dilor have different molecular geometries because they have different number of electron groups and choice D states all of the above. So we're going to need to compare our blueish structures of each of our atoms. Beginning with ammonia, we need to calculate total valence electrons. First recall that nitrogen is located in group five A on our periodic table and group number corresponds to valence electrons. So we have five valence electrons. Next, we have hydrogen which we recall is in group one A on our periodic table corresponding to one valance electron. And given by our subscript of three in our chemical formula, we would multiply by three atoms of hydrogen. This is going to yield a contribution of three valence electrons for hydrogen. So taking our sum of valence electrons, we would have three plus five, which would give us eight valence electrons total. For our structure being that we have a single nitrogen atom, we would make that our central atom and it's going to be surrounded by our three hydrogen atoms. So just making more room, We have our three hydrogen atoms. And we're going to want to recall our bonding preference of nitrogen, which is to be bonded to three atoms. So we'll make three covalent bonds from our hydrogens on the outside to our central atom nitrogen. And we also have a bonding preference of having a lone pair of nitrogen. So we'll drop in two lone electrons paired on nitrogen at the top, counting up our electrons used, we would have used 246 for the three bonds, 78 for the lone pair. So that uses up our balance of eight electrons leaving us with zero. And because we have fulfilled our octet for nitrogen, as well as our valence electrons for hydrogen, which are directly attached and covalent shared in a bond to nitrogen. We have the best le structure for ammonia And thus a net formal charge of zero. We're now going to need to consider electron groups and bonding groups in order to determine our molecular geometry of our molecule. Now, beginning with electron groups, we're going to recall that electron groups include bonds plus any loan pairs on our central atom. So looking at our central atom nitrogen, we have three atoms of hydrogen bonded to our central atom nitrogen. And we have one lone pair on nitrogen that will correspond to a total of, We'll say three H is bonded Plus one Lon pair will correspond to four electron groups in ammonia total. Next, as far as bonding groups, we said we have three hydrogen atoms bonded to our central atom nitrogen. So we have three bonding groups. Now we're going to recall that when we have three bonding groups with Our 4th bonding group being a lone pair, this will correspond to a molecular geometry which will be trigonal th pure middle. So we have trigonal by pyramidal molecular geometry as the shape of our molecule, which again is affected by that presence of the lone pair, pushing our hydrogen atoms towards the bottom of our central atom nitrogen in the trigonal Biram shape. And we're going to recall that our bond angle for a trigonal biya molecular geometry Will equal about 109.5°. Now, as far as our electron geometry of our ammonia, because we have four electron groups total, this will correspond to an electron geometry which is tetrahedral. We recall from our textbooks. I'm sorry. So we have a tetrahedral electron geometry just based off of the fact that we have four electron groups total, including our three hydrogen atoms bonded to nitrogen and the one lone pair where our molecular geometry differed because we consider the loan pair which affects our the way our three bonding groups are arranged in our structure giving us the tri trigonal pyramidal shape of our molecule with bond angles of one oh 9.5. Now let's move on to compare sulfur dilor, we have S C L two, we need to calculate total valence electrons. Beginning with sulfur recall that it's located in group six A like oxygen corresponding to six val electrons. And chlorine B recall is in our halogen group group seven A on the periodic table corresponding to seven valence electrons given by the subscript of two in our chemical formula, we multiply by two atoms of chlorine yielding a contribution of 14 valence electrons and taking the sum we have 14 plus six, which is going to give us a total of 20 valence electrons for our lower structure. Placing our least electron negative atom in the center being our single sulfur and surrounding it by our two chlorine atoms. We're next going to want to recall our bonding preference of oxygen which like sulfur as we stated is in group six A and recall that oxygen prefers to be bonded to two atoms with two lone pairs on itself. So like oxygen, we're going to form two covalent bonds from our chlorine to our central atom sulfur and place two lone pairs on sulfur. This is going to use up a total of Eight of our 20 vals electrons total leaving us with a balance of 12 valence electrons remaining next, as we stated, chlorine prefers to have seven vals electrons total. But in our current structure has one directly attached bound electron shared in a covalent bond of sulfur. So we're going to need to add a set of three lone pairs to each of our chlorine atoms. And in doing so, we would have used up a total of our 12 remaining balance electrons leaving us with a balance of zero. And this way, we know we have formal charges of zero for all of our atoms as the best and most stable form of our structure of sulfur chloride, giving us a net formal charge of zero for our entire molecule. So now with this Lewis structure, we're going to again consider total electron groups, total bonding groups. And we would find our molecular geometry based on our bonding groups and loan pairs. So beginning with total electron groups, we're going to recognize that we have two chlorine atoms bonded to our central atom sulfur and two lone pairs on chlorine our central atom. This will yield a total of four electron groups. Next, considering bonding groups. As we stated, we have two chlorine atoms bonded to our central atom sulfur. So we have two bonding groups. And as far as lone pairs, we stated, we have two lone pairs. So based on our two bonding groups and our two lone pairs, this will correlate to a molecular geometry which is bent. Recall that for bent bent molecular geometry, we have bond angles which equal Also 109.5° like in our trigonal pyramidal shape. Now we're going to define our electron geometry, which as we stated comes from our total electron groups being that we have four electron groups recall that this corresponds to an electron geometry which is going to be tetrahedral. So again, distinguishing our molecular geometry from electron geometry, we were able to determine our molecular geometry via our bonding groups being two and our two long pairs on our central atom sulfur, which correlated to a bent molecular geometry. But for electron geometry based on our total four electron groups, which again included our bonds and lone pairs, we have four electron groups corresponding to a tetrahedral electron geometry. And going back to our molecule of ammonia based on our lone pair on nitrogen and three bonding groups, we were able to determine our molecular geometry as trigonal pyramidal. And then based on our four total electron groups being our three atoms bonded and lone pair, we have our electron geometry as tetrahedral. So going to our answer choices, we would rule out choice c because we observed that ammonia and silicon actually have the same number of electron groups being a total of four. We can also rule out choice d all of the above. And we would also be able to rule out choice a since we've observed that they have the same number or the same type of electron arrangement, which is tetrahedral from the four electron groups. So this leaves us with choice B which states that ammonia and sulfur chloride have different molecular geometries because they have different numbers of lone pairs of electrons. For ammonia, we saw that we had one lone pair of electrons and for sulfur dilor, we saw that we have two lone pairs of electrons in our lower structure. So choice B is our final answer as explaining why despite having comparable bond angles, why ammonia and sulfur chloride have different molecular geometries? I hope that this made sense. And let us know if you have any questions.

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Compare the Lewis structures of CH₄ and H₂O Why do these molecules have similar bond angles but different molecular shapes?

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