Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 20: Problem 41
Give the oxidation state (number) of chromium in each of these compounds. (a) \(\mathrm{FeCr}_{2} \mathrm{O}_{4}\) (b) \(\mathrm{Cr}_{2} \mathrm{O}_{5}\) (c) \(\mathrm{K}_{2}\left[\mathrm{CrF}_{6}\right]\) (d) \(\left[\mathrm{Cr}(\mathrm{en})_{3}\right] \mathrm{Cl}_{2}\)
Short Answer
Expert verified
(a) +3, (b) +5, (c) +4, (d) +2.
Step by step solution
01
Understanding the Compound
For compound (a) \(\mathrm{FeCr}_{2} \mathrm{O}_{4}\), recognize it as a spinel. The general formula is \( \mathrm{AB}_2\mathrm{X}_4 \) where \( \mathrm{A} \) and \( \mathrm{B} \) are metal cations and \( \mathrm{X} \) is an oxygen anion.
02
Assign Known Oxidation States
Iron (Fe) usually has an oxidation state of +2 or +3. Oxygen (O) has a known oxidation state of -2.
03
Calculate Oxidation State of Chromium
Since there are four oxygens, the total negative charge contributed by oxygen is \(4 \times (-2) = -8\). Assume iron is +2 for simplicity. Therefore, for the compound to be neutral, the oxidation states must sum to zero: \(+2 + 2x + (-8) = 0 \). Solving for \(x\), we find \(2x = +6\), so \( x = +3 \). Thus, each Cr atom has an oxidation state of +3 in \( \mathrm{FeCr}_{2} \mathrm{O}_{4} \).
04
Analyzing the Next Compound
For compound (b) \(\mathrm{Cr}_{2} \mathrm{O}_{5}\), recall that oxygen contributes -2 each. With five oxygens, the total is \(5 \times (-2) = -10\).
05
Determine Chromium's State
The compound needs to be neutral, so the total oxidation state of Cr must balance \(-10\). Let each Cr have an oxidation state \(x\). Thus, \(2x + (-10) = 0\). Solving for \(x\), we get \(2x = +10\), and \(x = +5\). Hence, chromium is in the +5 state in \(\mathrm{Cr}_{2} \mathrm{O}_{5}\).
06
Consideration of Complex Compound
For (c) \(\mathrm{K}_{2}\left[\mathrm{CrF}_{6}\right]\), potassium (K) has an oxidation state of +1. Since there are two potassium atoms, the total contribution from K is \(+2 \).
07
Analyzing Charges in the Complex Ion
The charge on the complex ion \([\mathrm{CrF}_{6}]^{2-}\) means that the sum of oxidation states in the bracket is -2. Each fluoride (F) has an oxidation state of -1, so \(-6\) total from the six fluorides. For the sum to be -2, \(x + (-6) = -2\), solving gives \(x = +4\). Therefore, Cr is +4 in \(\mathrm{K}_{2}\left[\mathrm{CrF}_{6}\right] \).
08
Evaluate the Coordination Compound
In (d) \(\left[\mathrm{Cr}({\rm en})_3\right]\mathrm{Cl}_2\), 'en' refers to ethylenediamine, a neutral ligand, and Cl, chloride, is -1.
09
Use Charge Neutrality
The compound overall must be neutral. There are two Cl- ions contributing -2, indicating the complex ion \([\mathrm{Cr}({\rm en})_3]^{2+}\) has a charge of +2. Since 'en' is neutral, Cr must solely account for this charge, so Cr's state is +2.
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91影视!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Chromium Compounds
Chromium is a versatile element found in various chemical compounds, each with unique properties and oxidation states. In its compounds, chromium can exhibit several oxidation states, commonly +2, +3, +4, and +6. This variability is due to chromium's ability to lose different numbers of electrons, making it useful in various applications including alloys, pigments, and catalysts.
- **Oxidation State +3:** Found in compounds like \( \mathrm{FeCr}_2\mathrm{O}_4 \), chromium stabilizes the spinel structure, making it a prevalent state for color in pigments.
- **Oxidation State +5:** Exhibited in \( \mathrm{Cr}_2\mathrm{O}_5 \), this state is rarer but reveals chromium鈥檚 flexibility in bonding with oxygen.
- **Oxidation State +4:** Appears in complex ions like \( \mathrm{K}_2[\mathrm{CrF}_6] \), where chromium forms bonds with fluorine.
Spinel Structure
The spinel structure is a crystalline form prominently seen in minerals where oxide anions pack into a cubic lattice, and metal cations fit into tetrahedral and octahedral sites. \( \mathrm{FeCr}_2\mathrm{O}_4 \) is an example of a spinel compound.
- **General Formula:** This structure follows the formula \( \mathrm{AB}_2\mathrm{X}_4 \). Here, \( \mathrm{A} \) and \( \mathrm{B} \) are different metal cations, typically divalent and trivalent, respectively.
- **Cation Replacement:** In \( \mathrm{FeCr}_2\mathrm{O}_4 \), iron (\( \mathrm{Fe} \)) usually has a +2 state while chromium (\( \mathrm{Cr} \)) is +3, nestled within oxygen (\( \mathrm{O} \)) anions.
Complex Ions
Complex ions consist of a central metal atom bonded to surrounding molecules or ions, called ligands. The arrangement and oxidation state of the metal determine the ion's properties, seen in compounds like \( \mathrm{K}_2[\mathrm{CrF}_6] \).
- **Central Metal:** Chromium acts as the central metal, influencing the ion's geometry and charge. In \( \mathrm{CrF}_6^{4-} \), chromium is in the +4 state.
- **Influence of Ligands:** Ligands like fluorine contribute to the complex's charge and stability. Each fluorine has an oxidation state of \(-1\), collectively influencing chromium鈥檚 behavior.
Ligand Charges
Ligands are atoms or molecules that donate electron pairs to a central metal, forming a complex. Their charges are crucial in determining the complex鈥檚 overall charge and characteristics.
- **Neutral Ligands:** Some ligands, like ethylenediamine (en) in \([\mathrm{Cr}({\rm en})_3]^{2+} \), do not affect the charge directly but influence the complex's geometry and stability.
- **Anionic Ligands:** Ligands like chloride (\( \mathrm{Cl}^- \)) contribute negative charges. In \( \left[\mathrm{Cr}({\rm en})_3\right]\mathrm{Cl}_2 \), chlorides each have a \(-1\) charge, balancing the complex鈥檚 positive charge.
