Hydrogen Compounds: Videos & Practice Problems

Hydrides are binary compounds of hydrogen with metals or nonmetals, categorized into ionic, covalent, and metallic types. Ionic hydrides form with group 1A or 2A metals, where hydrogen has an oxidation state of -1. Covalent hydrides, like ammonia (NH₃), arise from reactions with nonmetals, with hydrogen at +1 oxidation. Metallic hydrides result from diatomic hydrogen reacting with transition metals, often lacking stoichiometric ratios. Understanding these classifications aids in grasping chemical bonding and reactions involving hydrogen.

Ionic Hydrides

Hydrides are binary compounds formed when elemental hydrogen reacts with metals or nonmetals. They can be categorized into three types: ionic, covalent, and metallic hydrides. This summary focuses on ionic hydrides, which are characterized as white crystalline solids produced when diatomic hydrogen reacts with metals from Group 1A or 2A of the periodic table, with the notable exception of beryllium.

In ionic hydrides, hydrogen has an oxidation number of -1. Understanding this oxidation state is crucial for predicting how these compounds form. For Group 1A metals, which have a charge of +1, the reaction with hydrogen can be represented as follows:

For a Group 1A metal (M):

M⁺¹ + H₂ → MH₂

To balance the equation, we need two hydrogen atoms, leading to:

2M + H₂ → 2MH

In contrast, Group 2A metals have a charge of +2. When these metals react with hydrogen, the charges do not cancel out but instead crisscross. The reaction can be represented as:

For a Group 2A metal (M):

M⁺² + H₂ → MH₂

Since the charges are different, they crisscross, resulting in the formula MH₂, which is already balanced with one metal and two hydrogen atoms on both sides.

In summary, ionic hydrides form when diatomic hydrogen reacts with Group 1A or 2A metals, leading to compounds such as MH and MH₂, depending on the metal's group. Understanding the oxidation states and charge interactions is essential for predicting the formation of these ionic compounds.

Hydrogen Compounds Example

In the reaction between solid strontium (Sr) and diatomic hydrogen gas (H₂), we start by identifying the charges of the reactants. Strontium, being an alkaline earth metal in group 2A, has a charge of +2 (Sr²⁺), while hydrogen, in this context, has a charge of -1 (H^-). To balance the charges, we apply the crisscross method, which involves swapping the charges to determine the subscripts in the formula.

Since strontium has a +2 charge, we need two hydrogen atoms to balance it, leading to the formula SrH₂. This compound is a solid. The balanced chemical equation for the reaction can be written as:

Sr (s) + H₂ (g) → SrH₂ (s)

In this equation, we have one strontium atom and two hydrogen atoms on both sides, confirming that the reaction is balanced. Therefore, the final balanced chemical reaction is:

Sr (s) + H₂ (g) → SrH₂ (s)

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)

Answer: (C)

Covalent (Molecular) Hydrides

Covalent or molecular hydrides are formed when diatomic hydrogen (H₂) reacts with nonmetals and metalloids. In the periodic table, specific nonmetals and metalloids are highlighted, excluding boron and astatine. The relevant elements include carbon, iodine, and five metalloids. In these compounds, hydrogen typically has an oxidation number of +1.

Common examples of covalent hydrides include methane (CH₄), ammonia (NH₃), water (H₂O), and hydrogen fluoride (HF). The oxidation states for various groups are as follows: group 1A elements typically have an oxidation state of -4, group 3A and 5A elements have -3, group 6A elements have -2, and group 7A elements have -1.

To illustrate the formation of a covalent hydride, consider the synthesis of ammonia. The reaction involves diatomic nitrogen (N₂) reacting with diatomic hydrogen (H₂) to produce ammonia (NH₃). This reaction requires heat and a catalyst due to its relatively low spontaneity. The balanced chemical equation for this reaction is:

2 N₂ + 6 H₂ → 4 NH₃

In this equation, a coefficient of 2 is placed before nitrogen to balance the number of nitrogen atoms, while a coefficient of 3 is used for hydrogen to ensure that the total number of hydrogen atoms is also balanced. Ammonia exemplifies a covalent or molecular hydride, showcasing the interaction between hydrogen and nitrogen.

Hydrogen Compounds Example

In the reaction between selenium (Se) and diatomic hydrogen gas (H₂), we start by identifying the charges of the elements involved. Selenium, being in group 6A, typically has a charge of -2, while hydrogen has a charge of +1. To balance the charges, we apply the crisscross method, which leads us to the empirical formula H₂Se. This indicates that two hydrogen atoms combine with one selenium atom.

In a similar manner to how oxygen reacts with hydrogen to form water (H₂O), selenium reacts with hydrogen to form hydrogen selenide, which can also be represented as H₂Se. This compound is known as a flammable gas, often referred to as coal gas.

When we examine the balanced equation, we find that there are two hydrogen atoms and one selenium atom on both sides of the reaction, confirming that the equation is balanced. Thus, the final balanced reaction can be expressed as:

Se + H₂ → H₂Se

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)

Answer: (A)

Metallic Hydrides

Metallic hydrides are compounds formed when diatomic hydrogen reacts with transition metals. A notable example is tantalum hydride, which illustrates an important concept in the study of these compounds: the stoichiometric ratio of metals to hydrogen atoms is not always a whole number. This is due to the unique way hydrogen occupies the spaces within the metal lattice structure.

In a metallic hydride, the transition metal atoms, represented by larger spheres, create a lattice framework. The much smaller hydrogen atoms fit into the gaps between these metal atoms. This arrangement can lead to non-integer ratios, such as 0.9, indicating that hydrogen does not always fill the lattice in a simple, whole-number manner. Understanding this behavior is crucial for grasping the properties and formation of metallic hydrides, as it highlights the complexity of their structure and composition.

Hydrogen Compounds Example

In the context of identifying metallic hydrides, it is essential to understand the distinctions between different types of hydrides based on the elements involved. A metallic hydride is formed when diatomic hydrogen (H₂) reacts with a transition metal. Transition metals are typically found in the d-block of the periodic table and exhibit metallic properties.

For example, when examining the compounds:

• Hydrogen connected to Tellurium: This represents a covalent or molecular hydride, as tellurium is a non-metal. Therefore, this compound does not qualify as a metallic hydride.
• Titanium connected to Hydrogen: This is a metallic hydride. Titanium, being a transition metal, can form a hydride when it reacts with diatomic hydrogen.
• Rubidium connected to Hydrogen: This compound is classified as an ionic hydride. Rubidium is a Group 1A metal, and the bond formed here is ionic rather than metallic.
• Aluminum connected to Hydrogen: Although aluminum can form hydrides, it is not classified as a transition metal. Thus, this does not meet the criteria for a metallic hydride.

In summary, the only correct example of a metallic hydride from the options provided is the compound formed with titanium and hydrogen. Understanding these classifications is crucial for distinguishing between different types of hydrides in chemistry.

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

Answer: (C)

Here’s what students ask on this topic:

What are the different types of hydrides and how are they formed?

Hydrides are binary compounds containing hydrogen and another element. There are three types of hydrides: ionic, covalent (molecular), and metallic. Ionic hydrides form 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 (CH₄), ammonia (NH₃), water (H₂O), and hydrogen fluoride (HF), 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 (TaH_0.9).

How do ionic hydrides differ from covalent hydrides in terms of formation and properties?

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. These hydrides are typically white crystalline solids. In contrast, covalent or molecular hydrides are formed when diatomic hydrogen reacts with nonmetals and metalloids (excluding Boron and Astatine), leading to compounds such as methane (CH₄), ammonia (NH₃), water (H₂O), and hydrogen fluoride (HF), with hydrogen having an oxidation number of +1. Covalent hydrides involve sharing of electrons between hydrogen and the other element, whereas ionic hydrides involve the transfer of electrons, resulting in ionic bonds.

What is the role of hydrogen's oxidation number in the formation of hydrides?

Hydrogen's oxidation number plays a crucial role in the formation of hydrides. In ionic hydrides, hydrogen has an oxidation number of -1, which allows it to form compounds with Group 1A or 2A metals, resulting in MH or MH₂. In covalent or molecular hydrides, hydrogen typically has an oxidation number of +1, enabling it to form compounds with nonmetals and metalloids, such as methane (CH₄), ammonia (NH₃), water (H₂O), and hydrogen fluoride (HF). The oxidation number determines the type of bond (ionic or covalent) and the resulting chemical properties of the hydride.

Why do metallic hydrides often have non-stoichiometric ratios?

Metallic hydrides often have non-stoichiometric ratios because hydrogen atoms fill the gaps in the metal lattice structure rather than forming fixed ratios with the metal atoms. This results in compounds where the ratio of metal to hydrogen is not a whole number, as seen in tantalum hydride (TaH_0.9). The small size of hydrogen atoms allows them to occupy interstitial sites within the metal lattice, leading to variable compositions depending on the specific metal and conditions of formation.

Can you provide an example of a reaction that forms a covalent hydride?

An example of a reaction that forms a covalent hydride is the formation of ammonia (NH₃). Diatomic nitrogen (N₂) reacts with diatomic hydrogen (H₂) in the presence of heat and a catalyst to produce ammonia. The balanced chemical equation for this reaction is:

$1 N_2 + 3 H_2 \to 2 {\mathrm{NH}}_3$

This reaction requires heat and a catalyst because it does not occur easily under standard conditions. Ammonia is a covalent hydride where hydrogen has an oxidation number of +1.