Question

Cr atoms can have a wide range of oxidation numbers. Assign oxidation numbers for the Cr atom in each compound, all of which are known compounds. a) \(\mathrm{Na}_{2} \mathrm{Cr} \mathrm{O}_{4}\) b) \(\mathrm{Na}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) c) \(\mathrm{CrF}_{5}\) d) \(\mathrm{CrCl}_{3}\) e) \(\mathrm{CrCl}_{2}\)

Short answer

a) +6, b) +6, c) +5, d) +3, e) +2.

Step-by-step solution

Determine the oxidation number of each element in Na2CrO4

In the compound \( \mathrm{Na}_{2} \mathrm{Cr} \mathrm{O}_{4} \), sodium (Na) has a known oxidation number of +1 and each oxygen (O) atom has an oxidation number of -2. Since there are two sodium atoms, the total oxidation number contribution of sodium is +2. There are four oxygen atoms, contributing a total of -8. The sum of the oxidation numbers in a neutral compound is 0, so we set up the equation: \( 2(+1) + x + 4(-2) = 0 \), where \( x \) is the oxidation number of Cr. Solving this gives: \( +2 + x - 8 = 0 \) which results in \( x = +6 \). Therefore, Cr in \( \mathrm{Na}_{2} \mathrm{Cr} \mathrm{O}_{4} \) has an oxidation number of +6.

Determine the oxidation number of each element in Na2Cr2O7

In the compound \( \mathrm{Na}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} \), sodium (Na) has an oxidation number of +1 and each oxygen (O) atom has an oxidation number of -2. There are two Na atoms contributing a total of +2 and seven O atoms contributing a total of -14. The sum of the oxidation numbers in the compound must be 0, so the equation is: \( 2(+1) + 2x + 7(-2) = 0 \), where \( x \) is the oxidation number of Cr. Solving this: \( +2 + 2x - 14 = 0 \) results in \( 2x = +12 \), so \( x = +6 \). Therefore, Cr in \( \mathrm{Na}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} \) has an oxidation number of +6.

Determine the oxidation number in CrF5

In \( \mathrm{CrF}_{5} \), the fluorine (F) atom typically has an oxidation number of -1. With five fluorine atoms, their total contribution is -5. This is not a neutral compound, but generally, chromium (Cr) must counterbalance the oxidation state's total to match the charge of the compound (which is neutral), so the sum of the oxidation states should equal 0. Therefore \( x + 5(-1) = 0 \), solving gives \( x = +5 \). Therefore, Cr in \( \mathrm{CrF}_{5} \) has an oxidation number of +5.

Determine the oxidation number in CrCl3

In \( \mathrm{CrCl}_{3} \), chlorine (Cl) has a typical oxidation number of -1. With three chlorine atoms, this results in a total of -3. Because the compound is neutral, the oxidation numbers must sum to 0: \( x + 3(-1) = 0 \) where \( x \) is the oxidation number of Cr. Solving this equation gives \( x = +3 \). Thus, Cr in \( \mathrm{CrCl}_{3} \) has an oxidation number of +3.

Determine the oxidation number in CrCl2

In the compound \( \mathrm{CrCl}_{2} \), chlorine (Cl) possesses a known oxidation number of -1. With two chlorine atoms, their contribution is -2. Being a neutral compound, their total with chromium's oxidation must equal 0, so \( x + 2(-1) = 0 \) where \( x \) is the oxidation number of Cr. Solving this gives \( x = +2 \). Therefore, Cr in \( \mathrm{CrCl}_{2} \) has an oxidation number of +2.

Key concepts

Chromium Oxidation States

Chromium is an interesting element due to its ability to exhibit multiple oxidation states. This is because of its electron configuration, which often allows electrons to be easily removed or paired, changing the oxidation state of the element. In the examples given, we encounter chromium in a range of oxidation states: +6 in both \( \mathrm{Na}_{2}\mathrm{CrO}_{4} \) and \( \mathrm{Na}_{2}\mathrm{Cr}_{2}\mathrm{O}_{7} \); +5 in \( \mathrm{CrF}_{5} \); +3 in \( \mathrm{CrCl}_{3} \); and +2 in \( \mathrm{CrCl}_{2} \).
Each oxidation state reflects chromium's interaction with different ligands, affecting its chemical properties and reactivity.
Chromium's adaptive nature makes it versatile in both industrial and chemical applications. By understanding its oxidation states, chemists can predict and control reactions involving chromium more effectively. Always pay close attention to the compound context when determining chromium's oxidation number.

Neutral Compounds

Neutral compounds are those in which the total charge of all atoms present is zero. This balance is crucial for many compounds to exist stably under normal conditions. When we calculate the oxidation number for each atom in a compound, it helps us ensure that the sum equals zero, reflecting the compound's neutrality. For instance, in \( \mathrm{Na}_{2}\mathrm{CrO}_{4} \) and \( \mathrm{CrCl}_{3} \), the positive and negative charges offset each other perfectly.
This concept is of paramount importance when working with several types of compounds.

• The sum of oxidation numbers in neutral compounds always equals zero.
• This helps chemists cross-verify if their calculations are correct.
• Understanding neutrality helps in predicting the behavior of a compound in chemical reactions.

By practicing the scenarios in which these rules apply, you will become more adept at quickly determining whether a compound is likely to be neutral or ionic.

Balancing Oxidation Equations

Balancing oxidation equations requires an understanding of electron transfer between atoms or groups. When one atom is oxidized, another is typically reduced, which is essential for maintaining charge balance in reactions. The skills for balancing oxidation numbers are foundational when writing chemical equations, especially in redox reactions. The key is to remember the following:

• Identify the atoms undergoing changes in oxidation state.
• Determine the increase or decrease in oxidation states as the reaction progresses.
• Use electron balancing to ensure both sides of the equation maintain equivalent charge and mass.

Practicing with various compounds will help you recall common oxidation states and predict possible changes.
By keeping these points in mind, you grow more confident in balancing complex equations involving electron transfers.

Oxidation Number Calculation

Calculating the oxidation number of an element within a compound is a critical skill in chemistry. It involves using known oxidation numbers of other elements to deduce the unknown one. Here are the basic steps:

• Identify all known oxidation numbers (like +1 for \( \mathrm{Na} \) or -2 for \( \mathrm{O} \)).
• Add these values based on the number of atoms present.
• Set up an equation where their sum, including the unknown, equals zero for a neutral compound.

For example, in \( \mathrm{Na}_{2}\mathrm{CrO}_{4} \), knowing the contributions of sodium and oxygen helps you find chromium's oxidation state as +6. This method is systematic and follows integer arithmetic to ensure precision.
Mastering it not only aids in understanding compound structure but also provides insights into chemical reactivity and bonding.