Question

In most compounds, H has an oxidation state of +1 and 0 has an oxidation state of \(-2 .\) The following compounds are exceptions to this rule. Assume each metal has the oxidation state of its most common ion and that \(F\) has an oxidation state of \(-1 .\) Find the oxidation state of \(\mathrm{H}\) or \(\mathrm{O}\) in each compound: (a) \(\mathrm{KO}_{2}\) (b) \(\mathrm{Na}_{2} \mathrm{O}_{2} ;(\mathrm{c}) \mathrm{MgH}_{2}\) (d) \(\mathrm{LiAIH}_{4}\) (e) \(\left.\mathrm{OF}_{2} \text { . (Section } 1.4\right)\)

Short answer

(a) O: -1/2, (b) O: -1, (c) H: -1, (d) H: -1, (e) O: +2.

Step-by-step solution

Determine the Oxidation State in KO2

First, identify the oxidation state of potassium (K), which is typically +1. Since KO2 is a compound, the sum of all oxidation states must equal 0. Let the oxidation state of oxygen in KO2 be represented as \(x\). In superoxides like KO2, the typical oxidation state of oxygen is -1/2. With one K and two O atoms, we write the equation: \(+1 + 2(-1/2) = 0\). Solving this, we find that each O in KO2 has an oxidation state of -1/2.

Determine the Oxidation State in Na2O2

Sodium (Na) has a typical oxidation state of +1. In peroxides like Na2O2, oxygen has a characteristic oxidation state of -1. This compound is made up of two Na atoms and two O atoms, so we write the equation: \(2(+1) + 2(-1) = 0\). Solving this equation, we see that each O in Na2O2 has an oxidation state of -1.

Determine the Oxidation State in MgH2

Magnesium (Mg) usually has an oxidation state of +2. In metal hydrides like MgH2, hydrogen takes an unusual oxidation state of -1. This compound has one Mg atom and two H atoms. We set up the equation: \(+2 + 2(-1) = 0\). Solving this, we confirm that the oxidation state of H in MgH2 is -1.

Determine the Oxidation State in LiAlH4

Lithium (Li) generally has an oxidation state of +1 and aluminum (Al) typically has an oxidation state of +3. In metal hydrides, hydrogen has an oxidation state of -1. This compound consists of one Li, one Al, and four H atoms. We set up the equation: \(+1 + (+3) + 4(-1) = 0\). Solving this equation, we find that the oxidation state of H in LiAlH4 is -1.

Determine the Oxidation State in OF2

Fluorine (F) always has an oxidation state of -1 due to its high electronegativity. In OF2, there is one O and two F atoms. The equation for the compound becomes \(x + 2(-1) = 0\), where \(x\) is the oxidation state of O. Solving this gives \(x = +2\), so the oxidation state of O in OF2 is +2.

Key concepts

Superoxides

Superoxides are unique compounds where oxygen exhibits a rare oxidation state of

• -1/2: This is much different than the common -2 oxidation state.
• Appears when oxygen bonds with highly electropositive elements like alkali metals.

Such as in potassium superoxide ( KO2 ) where potassium has a +1 oxidation state. Superoxides can effectively store oxygen, making them useful in many applications including rebreathers and closed-circuit oxygen systems. To further explore, consider how potassium interacts with two oxygen atoms, each with an oxidation state of -1/2, balancing out the +1 charge of potassium. This results in a total of zero, consistent with the neutral charge of the overall compound.

Peroxides

Peroxides are another class of compounds characterized by an unusual oxidation state of oxygen:

• The oxidation state of oxygen in peroxides is -1.
• Examples include compounds like hydrogen peroxide and sodium peroxide ( Na2O2 ).

In sodium peroxide, each sodium atom donates one electron because it typically has an oxidation state of +1. This allows two -1 charged oxygen atoms to pair, resulting in the peroxide bond. The high reactivity of peroxides can be attributed to the 2 > bond, which can release oxygen upon decomposition. This chemical property is widely utilized in applications such as bleaching, disinfection, and even in rocketry.

Metal Hydrides

Metal hydrides are compounds where hydrogen is bonded with metals and takes an unusual negative oxidation state of

• -1.

This opposes its usual +1 state in most compounds. This interesting trait can be observed in compounds like MgH2 and LiAlH4 . In these compounds:

• Magnesium and lithium act as electropositive elements, donating their electrons.
• Magnesium typically showcases a +2 oxidation state, requiring two hydrogen atoms to balance out its positive charge.
• Similarly, in LiAlH4 , lithium’s +1 and aluminum's +3 oxidation states are counterbalanced by hydrogen's -1 state to maintain neutrality.

These hydrides tend to be strong reducing agents, making them valuable in various chemical syntheses.

Fluorine Compounds

Fluorine is known for exhibiting a consistent

• -1 oxidation state due to its high electronegativity.

This property makes it a unique participant in compounds such as oxygen difluoride ( OF2 ), where oxygen interestingly has a

• +2 oxidation state to balance out the fluorine’s influence.

The highly reactive nature of fluorine arises from its ability to pull electrons from other elements, including oxygen, which generally holds onto its electrons tightly. This push-pull dynamic is the reason why oxygen adopts a positive oxidation state in such compounds, showcasing very unusual behavior. This makes fluorine compounds particularly powerful as oxidizing agents, reacting energetically with many other substances to form stable products.