Problem 3 - Chemistry: Introducing Inorganic, Organic and Physical Chemistry

Question

Assign oxidation states to the elements in the following hydride compounds \(\mathrm{LiH}, \quad \mathrm{HI}, \quad \mathrm{NH}_{3}, \quad \mathrm{SiH}_{4}, \quad \mathrm{B}_{2} \mathrm{H}_{6}(\text { Section } 25.2)\)

Step-by-Step Solution

Answer

Li: +1 in LiH, I: -1 in HI, N: -3 in NH₃, Si: +4 in SiH₄, B: +3 in B₂H₆.

Step 1: Introduction to Oxidation States

Oxidation states help us understand the electron distribution in molecules. The sum of the oxidation states of all atoms in a neutral compound must be zero.

Step 2: Determine Known Oxidation States

Begin by determining known oxidation states in the molecules. Hydrogen is typically +1, but when bonded with metals like in hydrides, it is -1. Halogens like iodine (in HI) have an oxidation state of usually -1.

Step 3: Assign Oxidation State for Li in LiH

Lithium (Li) is in Group 1 of the periodic table and typically has an oxidation state of +1. Since hydrogen in LiH is a hydride, it adopts an oxidation state of -1.

Step 4: Assign Oxidation State for I in HI

Iodine (I) is a halogen and typically has an oxidation state of -1. Given that HI is neutral, hydrogen will assume an oxidation state of +1.

Step 5: Assign Oxidation State for N in NH₃

In ammonia (NH₃), nitrogen (N) typically has an oxidation state of -3, as each hydrogen is +1, and the sum must equal zero.

Step 6: Assign Oxidation State for Si in SiH₄

In silicon tetrhydride (SiH₄), each hydrogen is -1 as hydrides, so silicon must be +4 to balance the four -1 from the four hydrogens.

Step 7: Assign Oxidation State for B in B₂H₆

In diborane (B₂H₆), hydrogens are -1 as hydrides. Boron needs oxidation state +3 (with 2 Borons giving a total of +6) to balance six hydrogens (-6).

Step 8: Conclusion

Summarize the oxidation states: Li in LiH is +1, H is -1; H in HI is +1, I is -1; N in NH₃ is -3, each H is +1; Si in SiH₄ is +4, each H is -1; and B in B₂H₆ is +3, each H is -1.

Key Concepts

Hydrides

Hydrides are compounds formed when hydrogen bonds with another element, usually a metal or a non-metal. The special thing about hydrides is the oxidation state of hydrogen in these compounds. Normally, hydrogen carries an oxidation state of +1, but in hydrides, it takes on an oxidation state of -1. This occurs because, in these bonds, hydrogen receives an electron from the other element, contrary to its usual status of losing an electron. Some common hydride examples include lithium hydride (LiH) and silane (SiH₄). It's important to understand that by recognizing the oxidation state of hydrogen in these hydrides, we can easily assign the oxidation numbers to the other element in the compound. This shifts our perspective on how electrons are distributed within these molecules.

Electron Distribution

Electron distribution in molecules plays a crucial role in understanding chemical bonding and reactivity. At the core of this concept is the idea that electrons are shared between atoms to form bonds. In simple terms, electrons are the currency of chemistry, and how they are distributed within a molecule tells us a lot about the chemical behavior of that molecule. In hydrides, where hydrogen is present as a hydride ion (H⁻), electrons from the other element are shifted toward hydrogen. This shift modifies the oxidation states, dictating how electrons are counted in these molecules. For example, in SiH₄, silicon shares its electrons with four hydrogen atoms, maintaining the electron distribution that results in silicon having an oxidation state of +4 and each hydrogen having -1. Understanding electron distribution is key to assigning oxidation states correctly.

Periodic Table

The periodic table is a valuable tool in determining the properties of elements, including their potential oxidation states. It categorizes elements based on their atomic structure and properties, revealing patterns such as ionic charges and bonding abilities. For example, the Group 1 elements, like lithium in LiH, usually have an oxidation state of +1 since they tend to lose one electron. Transitioning between different groupings, we find that elements like iodine, a halogen in Group 17, often have an oxidation state of -1 as they gain an electron to complete their outer shell. By using the periodic table, we can predict these tendencies and help assign correct oxidation states. This understanding assists in explaining why, for instance, nitrogen typically has an oxidation state of -3 in NH₃, balancing the +1 state of the hydrogen atoms.

Oxidation State Assignment

Assigning oxidation states is a systematic process crucial for understanding chemical reactions and bonding. Each element in a compound is assigned an oxidation state based on some general rules.

• For instance, the oxidation state of any elemental substance is 0.
• Hydrogen typically has an oxidation state of +1 but is -1 in hydrides.
• Oxygen is typically -2, except in peroxides.
• The sum of oxidation states in a neutral compound must equal zero.

Given these rules, we analyze compounds step by step. Starting with known states, like hydrogen or halogens, we deduce the unknowns by applying simple arithmetic to the oxidation states. For example, in B₂H₆, once we know hydrogen is -1, we can deduce boron must be +3 to balance the total charge in the compound. This systematic approach ensures consistency and understanding in how electrons are distributed in chemical compounds.