The following models represent the structures of binary hydrides of second-row elements:

b. Draw an electron-dot structure for each hydride. For which hydride is there a problem in drawing the structure? Explain.

Video transcript

Hi, everybody. Welcome back. Let's take a look at our next problem. It says, consider the molecular models shown below. And we have five different molecular models that we'll talk through. And then it says, which are possible structures for binary hydride of second period elements provide an electron dot structure for each hydride and explain why the others are not possible. So first, we need to think about which are the elements in the second period. And then let's work our way through these structures to see which are possible structures for some of these hyd rides. So our second period elements are lithium beryllium boron, carbon, nitrogen, oxygen and fluorine course is also neon, but it's a noble gas not going to make a hydride. So let's think about which what their formulas are as hydride. Lithium is going to be lih lithium hydride, but lithium will form an ionic hydride and these are all covalent structures. So we're going to x out lithium, it's not going to be on our structures here. We will also eliminate boron because boron with just its three valence electrons, no ability to have a lone pair and its general electron deficiency does not form the hydride, bh three. Instead, it exists as B two H six with these hydrogen bridges between the boons and this sort of funny um three center two electron bonding there. And there's no molecule with this many atoms in this arrangement. So we'll eliminate boron, it doesn't match any of our structures. So we're looking at brilli carbon nitrogen, oxygen and fluorine. Now, you might be held up at beryllium because it's a group two A element and the group two a elements usually form ionic hydride. But beryllium is the exception. It forms a covalent hydride. Uh due to its very small size and high electronegativity, its electron negativity is actually pretty close to that of hydrogen and it does form a covalent bond. So our possible formulas here would be beh two since Brilli has two valence electrons, carbon has four valence electrons. So ch four nitrogen has five valence electrons. But as we can recall, forms the hydride, NH three with just three valence electrons participating in bonding and a lone pair oxygen of course, has two lone pairs and forms water h2o and fluorine with its seven valence electrons shares a bond with a single hydrogen to be HF So we've got our possible molecules that could be represented by these structures. Let's turn our attention to the structures. So structure one, we have four atoms in a trigonal parameter arrangement. So we've got a sort of top atom with three extending down like a pyramid structural two is octahedral with four atoms in a plane and a square plane and then two atoms extending above and below that plane. However, this will not be possible because our second period elements cannot have an expanded octet. They have no de orbitals available to make this possible in this arrangement. Number two, the central atom has six bonds surrounding it and that requires an expanded octet. So structure two is not possible. So we're right expanded octet as a reason why this is not possible. Structure three is just a linear two atoms in a line structure. Four is trigonal bi parameter. Our central atom with three atoms surrounding it in in a trigonal plane and then atoms above and below. But this also involves an expanded octet because the central atom has five bonds. So structure four also not possible for our second period elements. And then finally, structure five is tetrahedral, a central atom with four, excuse me, not three with four atoms surrounding it in a tetrahedral arrangement. So let's we've now got down to three possible structures. Let's see if they match with our second period hydride. So structure one which is trigonal parameter has our central atom with three hydrogens. And we see we have that in NH three. And indeed, that will be a trigonal parameter arrangement because our nitrogen has to accommodate one lone pair which will take up that space. It has four sp three hybridized orbitals which would be a tetrahedral arrangement. But one of those orbitals contains the lone pair giving it a trigonal parameter arrangement or geometry. So structure number one will correspond to NH three and there's our Lewis dot structure that we drew a circle around that and label that as structure number one. Now let's look at structure number three, it's linear, just two atoms and we see that HF is our one hydride here that has just two atoms. And it would be a linear arrangement. We have our hydrogen bonded to our fluorine. Fluorine is group seven A. So then we have three lone pairs surrounding the fluorine and it corresponds to structure number three. And there's our Lewis dot Structure for hydrogen fluoride mark that there. And then finally, we have structure number five, our tetrahedral arrangement. So we've got our central atom with four hydrogen surrounding it. And that of course corresponds to ch four with just four single bonds. Those are sp three hybridized orbitals. We have our central carbon and it's got four hydrogens surrounding it, four single bonds and a tetrahedral arrangement, no lone pairs. So very straightforward Lewis dot Structure corresponding to structure number five. So once again, we look back to what our problem asks us. It says, which are the possible structures for binary hydrates of the second period elements that would be 13 and 52 and four are not possible because they require an expanded octet that's not available for second period structure, second period elements excuse me and we provided electron dot structures for NH three HF and CH four corresponding to structures 13 and five. See you in the next video.