Question

Compounds such as \(\mathrm{NaBH}_{4}, \mathrm{Al}\left(\mathrm{BH}_{4}\right)_{3},\) and \(\mathrm{LiAlH}_{4}\) are com- plex hydrides used as reducing agents in many syntheses. (a) Give the oxidation state of each element in these compounds. (b) Write a Lewis structure for the polyatomic anion in \(\mathrm{NaBH}_{4}\), and predict its shape.

Short answer

Oxidation states: Na: +1, B: +3, H: -1 in NaBH_4. Al: +3, B: +3, H: -1 in Al(BH_4)_3. Li: +1, Al: +3, H: -1 in LiAlH_4. Shape of BH_4^- is tetrahedral.

Step-by-step solution

Oxidation States in \(\text{NaBH}_4\)

Determine the oxidation states of each element in \(\text{NaBH}_4\). Sodium \(\) (Na) is in Group 1, so it has an oxidation state of \(\pm 1\). Hydrogen (H) typically has an oxidation state of \(-1\) in metal hydrides. Therefore, Boron (B)'s oxidation state can be calculated by realizing the compound is neutral. Total oxidation sum for \(\text{NaBH}_4\) must be zero: \(+1 + x + 4(-1) = 0\), where \(x\) is B's oxidation state. This simplifies to: \[ x = +3 \]. Hence, Sodium: +1, Boron: +3, Hydrogen: -1.

Oxidation States in \(\text{Al(BH}_4)_3\)

Determine the oxidation states of each element in \(\text{Al(BH}_4)_3\). Aluminum (Al) typically has an oxidation state of +3. For the \(\) \( \text{BH}_4^{-} \) \(\) anion, from the previous step, we already know B: +3 and H: -1. The compound needs to be neutral. Total oxidation sum for \(\text{Al(BH}_4)_3\) must be zero: \(+3 + 3(-1) = 0\). Hence, Aluminum: +3, Boron: +3 (in each \(\) \( \text{BH}_4^{-} \) \(\)), Hydrogen: -1 (in each \(\) \( \text{BH}_4^{-} \)).

Oxidation States in \(\text{LiAlH}_4\)

Determine the oxidation states of each element in \(\text{LiAlH}_4\). Lithium (Li) is in Group 1, so it has an oxidation state of +1. Again, we know Hydrogen (H) typically has an oxidation state of \(-1\) in metal hydrides. Hence, Aluminum (Al)'s oxidation state can be calculated by: \(+1 + x + 4(-1) = 0\), where \(x\) is Al's oxidation state. This simplifies to: \[ x = +3 \]. Hence, Lithium: +1, Aluminum: +3, Hydrogen: -1.

Lewis Structure of \(BH_4^{-}\)

Draw the Lewis structure for the polyatomic anion \(BH_4^{-}\). Start by calculating the total number of valence electrons. B has 3 valence electrons, and H has 1 each (total 4 H gives 4 electrons) and an additional 1 for the charge: \((3 + 4 + 1 = 8 \) valence electrons). Place B at the center and surround it with 4 H atoms. Distribute the electrons around B to complete each atom with a valence shell of 8 (Lewis Octet) electrons. The structure will have 4 single bonds connecting B to each H and an additional lone pair on B.

Shape Prediction of \(BH_4^{-}\)

Predict the shape of the \(\) \( \text{BH}_4^- \) \(\) anion. Based on the Lewis structure, there are 4 bonding pairs around B and no lone pairs, making the geometry tetrahedral using VSEPR (Valence Shell Electron Pair Repulsion) theory.

Key concepts

Oxidation State Determination

Understanding oxidation states helps us determine how electrons are distributed in a molecule. First, let's look at \(\text{NaBH}_4\). Sodium (\text{Na}) is in Group 1 and has an oxidation state of \(+1\). Hydrogen (\text{H}) has an unusual oxidation state of \(-1\) in metal hydrides. Since the compound is neutral, the sum of oxidation states must be zero. We calculate Boron's (\text{B}) state as follows: \(1 + x + 4(-1) = 0\), which simplifies to \(x = +3\).
In \(\text{Al(BH}_4)_3\), Aluminum (\text{Al}) typically has an oxidation state of \(+3\). We already know the \(\text{BH}_4^-\) anion’s states from previous calculations: Boron (\text{B}) \(+3\) and Hydrogen (\text{H}) \(-1\). Thus, for neutrality, the sum is \(+3 + 3(-1) = 0\).
Lastly, for \(\text{LiAlH}_4\). Lithium (\text{Li}) in Group 1 has an oxidation state of \(+1\). Using similar logic, we get \(1 + x + 4(-1) = 0\) for Aluminum (\text{Al}) which results in \(+3\). Understanding this is fundamental to exploring the electronic structure of these compounds.

Lewis Structure

The Lewis structure is a visual representation of a molecule's valence electrons. For \(\text{BH}_4^-\), we first calculate the total valence electrons: Boron (B) has 3, each Hydrogen (H) has 1, which gives 4 from H atoms, and we add 1 for the extra negative charge. This total is \(3 + 4 + 1 = 8\) valence electrons. We place B at the center, surrounded by four H atoms. Each H forms a single bond with B, utilizing 8 electrons. This creates a stable structure by applying the octet rule, showing how electrons are shared within the molecule. Such visualizations can solidify your understanding of molecular shapes and interactions.

VSEPR Theory

Valence Shell Electron Pair Repulsion (VSEPR) theory predicts the shape of a molecule based on repulsion between electron pairs. In \(\text{BH}_4^-\), we have four bonding pairs around the central Boron (B) atom, with no lone pairs. VSEPR theory tells us these pairs will position themselves as far apart as possible to minimize repulsion. This concept is crucial as it explains many molecular geometries and their associated chemical properties. By understanding VSEPR, you get insights into molecular angles and shapes, which can be applied broadly across diverse chemical compounds. Breaking down complex shapes into simple electron pair interactions makes them less daunting and more intuitive.

Tetrahedral Geometry

A tetrahedral geometry occurs when a central atom is surrounded by four substituents positioned as the corners of a tetrahedron. In \(\text{BH}_4^-\), Boron (B) forms such a structure with its four Hydrogen (H) atoms, resulting in bond angles of approximately 109.5°. This shape minimizes electron pair repulsion as per VSEPR theory. Recognizing tetrahedral structures is quite vital, as they're common in organic and inorganic chemistry. Such geometries impact the molecule's reactivity, stability, and overall behavior. Visualizing molecules in three dimensions enhances comprehension and can be pivotal when predicting or rationalizing molecular interactions.

Reduction Agents in Synthesis

Reduction agents like \(\text{NaBH}_4\), \(\text{Al(BH}_4)_3\), and \(\text{LiAlH}_4\) donate electrons to other substances, reducing them. These complex hydrides are potent reducing agents due to their ability to donate hydride ions (\text{H}-). These agents play crucial roles in synthetic chemistry, enabling selective reductions that are otherwise difficult or impossible. For example, \(\text{NaBH}_4\) can reduce ketones to secondary alcohols selectively. Such applications are invaluable in fields ranging from pharmaceuticals to materials science. Understanding the role and mechanisms of these reducing agents provides a fundamental toolset for any budding chemist aiming to master synthesis techniques.