What is the oxidation number of Li in LiAlH4 ?

To determine the oxidation number of lithium (Li) in lithium aluminum hydride (LiAlH4), we can break down the compound and analyze the oxidation states of its constituent elements. This will help us arrive at the correct oxidation number for lithium.

Understanding the Components

LiAlH4 consists of three elements: lithium (Li), aluminum (Al), and hydrogen (H). Each of these elements has a typical oxidation state that we can use as a reference:

• Lithium (Li): Typically has an oxidation state of +1.
• Aluminum (Al): Generally has an oxidation state of +3.
• Hydrogen (H): Usually has an oxidation state of -1 when bonded to metals.

Calculating the Oxidation Number

In LiAlH4, we can set up an equation based on the overall charge of the compound, which is neutral (0). The oxidation states of the elements must sum to zero:

Let’s denote the oxidation state of lithium as x. The oxidation states of aluminum and hydrogen are known, so we can express the equation as follows:

x (for Li) + 3 (for Al) + 4 * (-1) (for H) = 0

Now, substituting the known values:

x + 3 - 4 = 0

Simplifying this gives:

x - 1 = 0

Thus, we find:

x = +1

Conclusion on Oxidation State

From this calculation, we can conclude that the oxidation number of lithium in LiAlH4 is +1. This aligns with the typical behavior of lithium in compounds, where it usually exhibits a +1 oxidation state. Understanding these oxidation states is crucial for predicting the behavior of compounds in chemical reactions.

Real-World Application

LiAlH4 is widely used as a reducing agent in organic chemistry. Knowing the oxidation states helps chemists understand how the compound will react with other substances, particularly in reduction reactions where it donates electrons. This knowledge is essential for designing synthetic pathways in chemical synthesis.