Consider the elements C, Se, B, Sn, and Cl. Identify which of these elements:

e. Forms a hydride with the empirical formula 

Video transcript

All right. Hi everyone. So this question is asking us which of the following elements, silicone boron oxygen or nitrogen will form a binary hydride with an empirical formula of H two X. So to start us off, let's recall the fact that hydrogen can only form one bond with a different atom. So in order to find which of these elements will form a hydride with the formula H two X, we have to consider how many covalent bonds each respective element listed can actually form starting off with silicone. Now because silicon is in group four a of the periodic table, it's going to have four valence electrons. So if it has four valence electrons, then four more are going to be needed in order for silicone to complete its octet, right, which means that silicone can form four covalent bonds with four different hydrogens in order to complete that octet, right. So the empirical formula which would be equal to the molecular formula would be xh four or simply SI H four because we're referring to silicon specifically. So this doesn't quite fit the criteria. So let's go ahead and proceed to the next element which is boron and boron boron resides in group three a of the periodic table, which means that it has three valence electrons. So boron can go ahead and share its three unpaired valence electrons to three different hydrogens, therefore creating three covalent bonds. However, recall that bh three is actually relatively unstable, right. So instead of forming bh three, the molecular formula that forms or rather the molecular formula of the compound that actually results is going to be die boring, right? Or B two H six. No, this is the molecular formula of di boring. But recall that the empirical formula is the simplest ratio between all atoms present in the compound, right. So I can simplify B two H six to get an empirical formula of bh three four xh three. So again, this is not quite what we're looking for because we're looking for an empirical formula or a simplest mole ratio of H two X. Now up next is hydrogen or excuse me, oxygen. So here I'm going to scroll down and give myself some space. An oxygen being from group six A is going to have six valence electrons. Now because two more electrons are missing to complete the octet of oxygen, then oxygen can form two covalent bonds to two different hydrogen atoms in order to complete that octet. So the molecular formula of our hydride would be H2O which is also equal to the empirical formula because I cannot simplify this ratio. Further, I saw this can translate to H two X which is actually what we're looking for. However, I do want to talk about nitrogen regardless. So I'm going to scroll down one more time to do so, right? Nitrogen is from group five a of the periodic table, which means that it has five valence electrons. So if it has five valent electrons, then that means nitrogen can form three covalent bonds with hydrogen in order to complete its octet right, which gives me a molecular formula of NH three, which is also equal to the empirical formula. So when with all this said and done right, the only element in this case that can create a hydride with the formula H two X is going to be oxygen, it's going to be oxygen because oxygen having six valence electrons already can form two covalent bonds with hydrogen to complete its octet. And with that being said, thank you so very much for watching. And I hope you found this helpful.