Problem 34 - Chemistry and Chemical Reactivity

Question

Diborane can be prepared by the reaction of \(\mathrm{NaBH}_{4}\) and I \(_{2} .\) Which substance is oxidized and which is reduced?

Step-by-Step Solution

Answer

Iodine is reduced; sodium borohydride acts as the reducing agent.

Step 1: Identify Oxidation Numbers Before Reaction

Determine the oxidation states of elements in \(\mathrm{NaBH}_{4}\) and \(\mathrm{I}_{2}\) before the reaction. In \(\mathrm{NaBH}_{4}\), \(\mathrm{B}\) is at -3, and \(\mathrm{H}\) is at +1. In \(\mathrm{I}_{2}\), \(\mathrm{I}\) is at 0.

Step 2: Identify Oxidation Numbers After Reaction

After the reaction, \(\mathrm{I}_{2}\) is converted to \(\mathrm{I}^{-}\), with \(\mathrm{I}\) having an oxidation state of -1. The \(\mathrm{B}\) converts to part of \(\mathrm{B}_{2}\mathrm{H}_{6}\) still at -3.

Step 3: Determine Which Element is Oxidized

The oxidation number of iodine decreases from 0 in \(\mathrm{I}_{2}\) to -1 in \(\mathrm{I}^{-}\). Since oxidation refers to an increase in oxidation number, iodine is the element that is reduced, not oxidized.

Step 4: Determine Which Element is Reduced

The oxidation number of boron remains at -3; however, we analyze the reaction stoichiometrically to consider any hydrogen from \(\mathrm{NaBH}_{4}\). Since hydrogen and other components are balanced in this context, reduction involves \(\mathrm{I}_{2}\), signaling iodine's reduction.

Step 5: Conclusion on Oxidation and Reduction

Given the reaction, no specific numerical oxidation occurs on boron; rather, the transformation focuses on iodine. Thus, iodine is reduced, and sodium borohydride (\(\mathrm{NaBH}_{4}\)), indirectly acts as the reducing agent, not being oxidized itself but facilitating iodine's reduction.

Key Concepts

Oxidation States

Understanding oxidation states is crucial for identifying oxidation-reduction reactions. Oxidation states, or oxidation numbers, indicate the degree of oxidation of an atom in a compound. It reflects how many electrons are gained or lost compared to the neutral atom.
To determine the oxidation state of each element, recall these simple rules:

• In elements, the oxidation state is always zero. For example, in \(\mathrm{I}_{2}\), iodine has an oxidation state of 0.
• For compounds, such as \(\mathrm{NaBH}_{4}\), we determine states based on known values: hydrogen is typically +1, and boron here is -3 because the sum must equal the overall compound charge.

Oxidation involves an increase in oxidation number, while reduction involves a decrease. In the exercise, iodine goes from an oxidation state of 0 to -1, indicating reduction.

Reducing Agent

A reducing agent is a substance that donates electrons to another species in a redox reaction. This causes the other species to be reduced, while the reducing agent itself is oxidized. In simpler terms, reducing agents lose electrons and help others gain them.
In the reaction forming diborane (\(\mathrm{B}_{2}\mathrm{H}_{6}\)), sodium borohydride (\(\mathrm{NaBH}_{4}\)) acts as the reducing agent. It provides electrons that lead to the reduction of iodine from 0 in \(\mathrm{I}_{2}\) to -1 in \(\mathrm{I}^{-}\).
Despite \(\mathrm{NaBH}_{4}\) not experiencing an increase in boron's oxidation state, it facilitates the transfer of electrons needed for the reduction of iodine. This action emphasizes its role as a reducing agent.

Diborane Preparation

Diborane (\(\mathrm{B}_{2}\mathrm{H}_{6}\)) is a fascinating compound in boron chemistry. It can be synthesized through the chemical reaction between sodium borohydride (\(\mathrm{NaBH}_{4}\)) and iodine (\(\mathrm{I}_{2}\)).
The preparation hinges on a redox reaction, where \(\mathrm{NaBH}_{4}\) provides hydrogen and electrons, leading to iodine's reduction and simultaneous formation of diborane. An essential part of this process is balancing the stoichiometry, ensuring all reactants transform correctly to products.
Synthesizing diborane is significant because it highlights unique characteristics of boron-hydrogen chemistry and showcases how redox reactions can be used to create complex chemical species.

Boron Chemistry

Boron chemistry explores the behavior and compounds of the element boron, well-regarded for its diverse and sometimes surprising characteristics. Boron is unique because it's a metalloid, meaning it shares properties typical of both metals and non-metals.
Diborane, (\(\mathrm{B}_{2}\mathrm{H}_{6}\)), is a prime exhibit of boron's versatility. Its structure consists of boron atoms forming bonds with hydrogen pairs, creating an intriguing "banana bond" or multi-center bond, not classically observed in organic compounds.
Diborane itself serves as an example of boron's potential in forming varied structures and compounds, showcasing the element's versatility in producing both industrially significant chemicals and compounds with interesting geometry and reactivity. Understanding diborane's preparation is a stepping stone into the broader field of boron chemistry.