Hydrogen Compounds - Online Tutor, Practice Problems & Exam Prep

• Ionic hydrides are formed when diatomic hydrogen reacts with Group 1A or 2A metals (excluding Beryllium), resulting in compounds like MH or MH2, where hydrogen has an oxidation number of -1.
• Covalent or molecular hydrides are created when diatomic hydrogen reacts with nonmetals and metalloids (excluding Boron and Astatine), leading to compounds such as methane, ammonia, water, and hydrogen fluoride, with hydrogen typically having an oxidation number of +1.
• Metallic hydrides form when diatomic hydrogen reacts with transition metals, often resulting in non-stoichiometric ratios due to hydrogen atoms filling gaps in the metal lattice, as seen in compounds like tantalum hydride (TaH0.9).

Ionic Hydrides

Elemental hydrogen reacts with different elements to produce hydrides. Now, hydrides are just binary compounds containing hydrogen and a metal or nonmetal. There are three 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 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 -1 and this is key to understanding how the reactants combine to give me my solid product.

If we take a look here for group 1A metals, group 1A metals have a charge of plus one hydrogen which has an oxidation of -1. Its ion would also be -1. 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 2 here to have two hydrogens on both sides, and then I put a 2 here to have two metals on both sides.

For group 2A metals, their charge is +2 and then we're going to say here hydrogen is still -1. The numbers in the charges are different. When they're different, they don't cancel out, they crisscross, so it get at the end is MH2. This equation is already balanced because we have one metal on both sides and two hydrogens on both sides. These represent the ionic hydride that exist when diatomic hydrogen combines with a group 1A or 2A metal.

Hydrogen Compounds Example

Complete and balance the following reaction. Here we have solid strontium reacting with diatomic hydrogen gas. Strontium is in Group 2A so it's charges +2 or two. Plus, hydrogen here will have a charge and oxidation number of -1.

The numbers in the charges are different and when that happens they don't cancel out. The crisscross 2 comes here, one comes here at the end. That gives us SrH2 and this will be a solid.

Here our equation is already balanced because we have one strontium on each side and two hydrogens on each side. So this would be our final balanced chemical reaction.

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)

Covalent (Molecular) Hydrides

In this video, we now take a look at covalent or molecular hydrides. Now they're formed when diatomic 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 acetene are not involved here. OK. So just remember, it's just the ones that are within the red border.

So we have carbon and we have iodine, so everything within here as well as these five metalloids. Now when it comes to covalent or molecular hydrides, hydrogens will have an oxidation number or charge of plus one. And here when we're talking about common covalent hydrides, we have 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 1A we'd say it's -4, for group 3A and 5A it's -3, for group 6A it's -2 and for group 7A it's -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 NH₃. 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 two nitrogens, but then we have 2 × 3 six 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.

Hydrogen Compounds Example

Complete and balance of following reaction. So here we have selenium reacting with diatomic hydrogen gas. Selenium is in Group 6A so its charges -2. Hydrogen here will be plus one. When the numbers in the charges are different, they don't cancel out. They crisscross 2 comes here, one comes here.

So here we could write this as SE2H2, but it's in the same group as oxygen. In terms of selenium, Oxygen reacting with hydrogen gives us water, which is written as H2O. And to keep with this theme, we're going to write this as H2SE 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 balance overall reaction here. This is hydrogen solenoid. It's a coalesced gas that's also flammable. This would be our final answer.

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)

Metallic Hydrides

In this video we now take a look at metallic hydrides. 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 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 are 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.

Hydrogen Compounds Example

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 1A 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.

Classify each hydride according to its type.

i) SbH₃          _____________________

ii) NaH          _____________________

iii) PdH_0.5      _____________________

iv) CaH₂        _____________________

• A. i) covalent hydride
  ii) metallic hydride
  iii) ionic hydride
  iv) metallic hydride
• B. i) ionic hydride
  ii) covalent hydride
  iii) metallic hydride
  iv) covalent hydride
• C. i) covalent hydride
  ii) ionic hydride
  iii) metallic hydride
  iv) ionic hydride
• D. i) ionic hydride
  ii) metallic hydride
  iii) covalent hydride
  iv) metallic hydride
• E. i) metallic hydride
  ii) ionic hydride
  iii) covalent hydride
  iv) ionic hydride