Question

Hydrogen and oxygen usually have oxidation states of +1 and \(-2,\) respectively, in their compounds. The following cases serve to remind us that there are exceptions. What are the oxidation states of the atoms in each of the following compounds? (a) \(\mathrm{MgH}_{2}\) (b) \(\mathrm{CsO}_{3} ;\) (c) HOF; (d) \(\mathrm{NaAlH}_{4}\)

Short answer

The oxidation states of the atoms in the compounds are: (a) Mg (+2), H (-1); (b) Cs (+1), O (-2); (c) H (+1), O (-2), F (+1); (d) Na (+1), Al (+3), H (-1).

Step-by-step solution

Assign Oxidation States for \(\mathrm{MgH}_{2}\)

Magnesium (Mg) is in group 2 and thus has an oxidation state of +2. Hydrogen generally has an oxidation state of +1, but in this case there are two hydrogens attaching to magnesium, therefore hydrogen needs an oxidation state of -1 to balance the +2 charge. Thus, the oxidation states are Mg (+2) and H (-1).

Assign Oxidation States for \(\mathrm{CsO}_{3}\)

Cesium (Cs) is in group 1 and thus has an oxidation state of +1. Oxygen generally has an oxidation state of -2, and there are three of them in the compound. Therefore, the oxidation states are Cs (+1) and O (-2).

Assign Oxidation States for HOF

In this molecule, oxygen has its usual oxidation state of -2. Since the total charge of the molecule is 0, the charges of hydrogen and fluorine must balance out the -2 charge. Hydrogen usually has an oxidation state of +1. In order for the total charge to be zero, fluorine must have an oxidation state of +1. Thus, the oxidation states are H (+1), O (-2), and F (+1).

Assign Oxidation States for \(\mathrm{NaAlH}_{4}\)

Sodium (Na) is in group 1 and thus has an oxidation state of +1. Aluminum (Al) is in group 13 and thus has an oxidation state of +3. Multiple hydrogen atoms are present and they balance the charge from sodium and aluminum by taking an oxidation state of -1. Thus, the oxidation states are Na (+1), Al (+3), and H (-1).

Key concepts

Oxidation Number Rules

Understanding oxidation numbers is crucial in predicting and balancing chemical reactions. They indicate the number of electrons an atom gains, loses, or shares, and help in determining the electron distribution in chemical compounds. To assign oxidation states correctly, follow these fundamental rules:

• The oxidation state of a free element (uncombined element) is zero. For example, each atom in \( ext{N}_2\) or \( ext{O}_2\) has an oxidation state of 0.
• The oxidation state of a monatomic (single atom) ion is the same as its charge; for example, \( ext{Na}^+\) has an oxidation state of +1.
• In compounds, hydrogen usually has an oxidation state of +1, and oxygen usually has an oxidation state of -2.
• The sum of oxidation states in a neutral compound must be zero. For a polyatomic ion, it must equal the charge of the ion.

Using these rules, you can assign oxidation states to each element in a compound accurately.

Hydrogen Compounds

Hydrogen is one of the simplest and most common elements in compounds. Typically, it has an oxidation state of +1. This is because it often loses its single electron when forming bonds with more electronegative elements. However, there are exceptions to this rule.
In metal hydrides such as \(\text{MgH}_2\) or \(\text{NaAlH}_4\), hydrogen has an unusual oxidation state of -1. This occurs because the metals in these compounds, like magnesium and sodium, are less electronegative than hydrogen. Thus, hydrogen gains an electron from these metals, adopting a negative charge.
Understanding these variations is important for accurately determining the oxidation states in different types of compounds.

Oxygen Compounds

Oxygen is a highly electronegative element and generally forms compounds with other elements by gaining electrons. It typically has an oxidation state of -2, reflecting its tendency to attract two additional electrons to fill its outer shell.
In most compounds, this oxidation state is consistent, such as in water \(\text{H}_2\text{O}\) or carbon dioxide \(\text{CO}_2\). However, there are scenarios where oxygen exhibits different oxidation states. In the \(\text{CsO}_3\) molecule from the exercise, even though each oxygen still holds the -2 state, the structure of the molecule accommodates cesium's need to balance out its own +1 oxidation state.
Recognizing oxygen's typical behavior allows chemists to anticipate the outcome of reactions involving oxygen compounds effectively.

Exceptions in Oxidation States

Although there are general rules for assigning oxidation states, certain compounds display exceptions which are crucial to note. These exceptions arise due to the complex interactions between specific elements in a compound.
For example, in \(\text{HOF}\), fluorine, being more electronegative, takes the -1 oxidation state, despite usually being -2. This is an exception because hydrogen maintains its +1 state, while oxygen unusually ends up with a 0 oxidation state to balance the molecule. This peculiar distribution demonstrates how unique combinations lead to shifts in typical oxidation numbers.
Being aware of these exceptions is vital for correctly interpreting chemical formulas and ensuring accurate chemical balancing.