Hydrogen Compounds - Video Tutorials & Practice Problems
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1
concept
Ionic Hydrides
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Elemental hydrogen reacts with different elements to produce hydrides. Now hydrides are just binary compounds containing hydrogen and a metal or non metal. There are 3 types of hydrides that exist, and they are ionic, covalent, and metallic hydrides. Here in this first one we're going to take a look at ionic hydrides. Now, they represent white crystalline solids, formed when diatomic hydrogen reacts with group 1a or 2a metals. The exception here is Beryllium, Beryllium is in group 2a, but when it combines with a diatomic hydrogen it doesn't classify as an ionic hydride. Now, here with these ionic hydrides, the Hydrogen within our solid possesses an oxidation number of -one. And this is key to understanding how the reactants combine to give me my solid product. If we take a look here for group 1 a Metals. Group 1 a Metals have a charge of plus 1. Hydrogen which has an oxidation of minus 1, its ion would also be minus 1. Remember when the numbers are the same within the charges, they just cancel out, so we'll be left with m h solid. We still need to balance this, I'd have to put a 2 here to have 2 hydrogens on both sides, and then I'd put a 2 here to have 2 metals on both sides. For group 2a metals, their charge is plus 2, and then we're going to say here Hydrogen is still minus 1. The numbers and the charges are different. When they're different, they don't cancel out, they criss cross. So what we get at the end is m h2, this equation is already balanced because we have 1 metal on both sides, and 2 hydrogens on both sides. These represent the ionic hydrates that exist when diatomic hydrogen combines with a group 1A or a 2A metal.
2
example
Hydrogen Compounds Example
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43s
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Complete and balance the following reaction. Here we have solid strontium reacting with diatomic hydrogen gas. Strontium is in group 2a, so it's charge is plus 2 or 2+. Hydrogen here will have a charge and oxidation number of -one. The numbers in the charges are different and when that happens they don't cancel out, they criss cross. 2 comes here, 1 comes here. At the end that gives us SRH2, and this will be a solid. Here our equation is already balanced because we have 1 strontium on each side and 2 hydrogens on each side. So this would be our final balanced chemical reaction.
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Problem
Problem
Write and balance a reaction for formation of an ionic hydride with sodium.
A
2 Na(s) + 2 H2(g) → 2 NaH2(s)
B
Na2(s) + H2(g) → 2 NaH(s)
C
2 Na(s) + H2(g) → 2 NaH(s)
D
Na2(s) + 2 H(g) → Na2H2(s)
4
concept
Covalent (Molecular) Hydrides
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In this video we now take a look at covalent or molecular hydrides. Now they're formed when Diatonic Hydrogen reacts with nonmetals, and metalloids. Now, if we take a look here at this periodic table, we've outlined in red the nonmetals and metalloids that are involved in creating covalent or molecular hydrides. Notice that Boron and Astatine are not involved here. Okay? So just remember, it's just the ones that are within the red border. So we have carbon, we have iodine, so everything within here, as well as these 5 metalloids. Now when it comes to covalent or molecular hydrides, hydrogens will have an oxidation number or charge of plus 1. And here when we're talking about common covalent hydrides, we have Methane, Ammonia, Water, and Hydrogen Fluoride. Now, here with their groups, we have types of oxidation numbers that are pretty common, oxidation numbers slash charges. So for group 1 A we'd say it's minus 4, for group 3 A, m 5a, it's minus 3, for group 6a, it's minus 2, and for group 7a, it's minus 1. A good example of the creation of a covalent or molecular hydride, here we have an example is the formation of Ammonia through the following reaction. Here we have natural Diatomic Nitrogen reacting with natural Diatomic Hydrogen to produce ammonia, which is NH 3. This happens with the use of heat and a catalyst. It doesn't happen very easily, so we need help in terms of heat and a catalyst. Here we'd have to balance this equation, we'd have to put a 2 here so that both sides have 2 Nitrogens, but then we have 2 times 3, 6 Hydrogens on the product side, so we put a 3 here. So this would be the formation of ammonia, the typical reaction. Ammonia itself represents a covalent or molecular hydride.
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example
Hydrogen Compounds Example
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Complete and balance the following reaction. So here we have selenium reacting with diatomic hydrogen gas. Selenium is in group 6A, so its charge is minus 2. Hydrogen here will be plus 1. When the numbers and the charges are different, they don't cancel out, they criss cross, 2 comes here, 1 comes here. So here we could write this as s e h two, but it's in the same group as oxygen in terms of selenium. Oxygen reacting with hydrogen gives us water which is written as h two o, and to keep with this theme we're gonna write this as h two s e instead. K? This is still a good way to write it as well, it's just that you normally see it written in this way. When we look both sides are balanced, we have 2 Hydrogens on both sides, 1 Selenium on both sides, so this is our balanced overall reaction. Here, this is Hydrogen Solenide, it's a coal gas that's also flammable. This would be our final answer.
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Problem
Problem
Provide a balanced equation for the reaction of hydrogen gas with bromine gas.
A
H2(g) + Br2(g) → 2 HBr(g)
B
2 H(g) + 2 Br(g) → 2 HBr(g)
C
H2(g) + Br2(g) → H2Br2(g)
D
H2(g) + 2 Br(g) → 2 HBr(g)
7
concept
Metallic Hydrides
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In this video, we now take a look at metallic hydrides. Now, these are formed when Diatonic Hydrogen reacts with transition metals. Here we have an example of tantalum hydride. Now, you might look at this and say, hey, this isn't a whole number, this is 0.9. Well, what we need to understand here is that sometimes from metallic hydrides, many do not follow a stoichiometric ratio of metals to hydrogen atoms. H fills in between the gaps of the metal lattice structure. So here we have an example of this, our larger red spheres represent our transition metal, Hydrogen which is a very small element within itself, just kind of fills in the gaps between each of these, Transition Metal Atoms, so this lattice structure. Right? So just remember when it comes to metallic hydride, sometimes you won't get whole numbers for the transition metal and the hydrogen.
8
example
Hydrogen Compounds Example
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1m
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Here it says to identify a metallic hydride out of the following. So for the first one we have hydrogen connected to tellurium. Now, this would not represent a metallic hydride, this would be a covalent or molecular hydride. And that's because we have hydrogen connected to a non metal, so this would be covalent. B. Here we have titanium connected to hydrogen. This would represent a metallic hydride because titanium is a transition metal. Remember a metallic hydride is formed when diatomic hydrogen reacts with a transition metal. Next one we have Rubidium connected to Hydrogen. This here would not represent a metallic hydride. This is an ionic one. Because it's connected to a group 1 A metal. Then finally here we have Aluminum and Hydrogen here. By the definition, Aluminum is not a transition metal, so it couldn't be a metallic hydride, so this is out. Right? So here the only option that is correct is option b.