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 hydride. Now, hydride are just binary compounds containing hydrogen and a metal or non metal, there are three types of hydride that exist and they are ionic covalent and metallic hydride. Here in this first one, we're gonna take a look at ionic hydride. Now, they represent white crystalline solids formed when diatomic hydrogen reacts with group one A or two A metals. The exception here is beryllium. Beryllium is in group two A. But when it combines with a diatomic hydrogen, it doesn't classify as an ionic hydride. Now, here with these ionic hydride, the hydrogen within our solid possesses an oxidation number of minus 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 18 medals, group one, a medals have a charge of plus one hydrogen which has an oxidation of minus one, it would also be minus one. Remember when the numbers are the same within the charges they just cancel out. So we'll be left with MH solid. We still need to balance this. I'd have to put a two here to have two hydrogens on both sides. And then I put it to here to have two medals on both sides for group two A medals. Their charge is plus two. And then we're gonna say here hydrogen is still minus one. The numbers and the charges are different when they're different, they don't cancel out, they crisscross. So it would get at the end is mh two, this equation is already balanced because we have one metal on both sides and two hydrogens on both sides. These represent the ionic hydrate that exist when diatomic hydrogen combines with a group, one A or a two, a metal.
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example
Hydrogen Compounds Example
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43s
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Complete and balance the following reaction here, we have solid strong team reacting with diatomic hydrogen gas. Stra team is in group two A. So it's charges plus two or two plus hydrogen. He will have a charged and oxidation number of minus one. The numbers and the charges are different. And when that happens, they don't cancel out. They crisscross two comes here, one comes here at the end that gives us srh two and this will be a solid here. Our equation is already balanced because we have one ST strong team on each side and two 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)
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concept
Covalent (Molecular) Hydrides
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2m
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In this video, we now take a look at covalent or molecular hydride. Now, they're formed when diatomic hydrogen reacts with nonmetals and metalloid. Now, if we take a look here at this periodic table, we've outlined in red, the non metals and metalloid that are involved in creating covalent or molecular hydride. Notice that boron and astatine are not involved here. OK. 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 five metalloid. Now, when it comes to covalent or molecular hydride, hydrogens will have an oxidation number or charge of plus one. And here when we're talking about common covalent hydride, we have ammo and we have methane, ammonia, water and hydrogen fluoride. Now, here with their groups, we have types of oxidation numbers that are pretty common oxidation number slash charges. So for group one A would say it's minus four, for group three A, um five A, it's minus three for group six A, it's minus two. And for group seven A, it's minus one, 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 three. 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 two here so that both sides have two nitrogens. But then we have two times 36 hydrogens on the product side. So we put a three 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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1m
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Complete and balance of following reaction. So here we have selenium reacting with diatomic hydrogen gas. Selenium is in group six A. So it's charges minus two hydrogen here will be plus one when the numbers and the charges are different, they don't cancel out. They crisscross two comes here. One comes here. So here we could write this as seh two, but it's in the same group as oxygen in terms of selenium, oxygen reacting with W with hydrogen gives us water, which is written as H2O. And to keep with this theme, we're gonna write this as H two se instead. OK. 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 two hydrogens on both sides, one selenium on both sides. So this is our balanced overall reaction here. This is hydrogen cide. 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)
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concept
Metallic Hydrides
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1m
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In this video, we now take a look at metallic hydride. Now, these are formed when diatomic 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 hydride, many do not follow a stoichiometry 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 uh 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.
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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 delirium. 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 nonmetal. 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, one, a metal. And 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