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.
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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’s 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’s 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’s 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’s positive charge.