Chapter 22: Problem 54
Which of the following high-spin complexes has the greatest number of unpaired electrons? (a) \(\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}\) \((c)\left[\operatorname{re}\left(H_{2} O\right)_{6}\right]^{2+}\) \((b)\left[\operatorname{Mn}\left(H_{2} O\right)_{6}\right]^{2+}\) \((d)\left[N i\left(H_{2} O\right)_{6}\right]^{2+}\)
Short Answer
Step by step solution
Identify the Oxidation State of the Metal
Determine Electron Configuration of Metal Ions
Identify High-Spin Configuration and Count Unpaired Electrons
Conclusion
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
High-Spin Complexes
This allows electrons to occupy higher energy orbitals rather than pairing up in lower ones, creating unpaired electrons. As a result, high-spin complexes tend to be paramagnetic.
These properties impact the magnetic behavior of the complexes and their reactivity. Common metal ions like \(\text{Fe}^{2+}\), \(\text{Mn}^{2+}\), and others surrounded by weak field ligands typically form high-spin complexes.
- High-spin complexes exhibit weak field ligands.
- They display smaller d-orbital splitting, resulting in higher energy orbital occupation.
- Leading to more unpaired electrons, they exhibit paramagnetism.
Unpaired Electrons
These unpaired electrons contribute to a complex's paramagnetic nature, as they are attracted to magnetic fields. The number of unpaired electrons directly correlates with the level of magnetism—more unpaired electrons mean stronger paramagnetism.
In the selection problem, counting the unpaired electrons helps identify which complex is the most paramagnetic. \(\text{Mn}^{2+}\) in its high-spin state confirms five unpaired electrons, representing maximum paramagnetism among the options.
- Unpaired electrons affect how compounds interact with magnetic fields.
- They thus play a vital role in applications involving magnetic materials and devices.
- More unpaired electrons equate to stronger paramagnetic properties.
Electron Configuration
Transitional metal ions like \(\text{Cr}^{3+}, \ ext{Fe}^{2+}, \text{Mn}^{2+},\) and \(\text{Ni}^{2+}\) each have specific configurations based on their oxidation states. These configurations tell us the number of unpaired electrons and whether a complex will be high-spin or low-spin.
For example, \(\text{Mn}^{2+}\) exhibits a \(3d^5\) configuration, leading to five unpaired electrons in a high-spin state. Understanding these configurations gives insight into an ion's chemical and physical properties.
- Electron configurations reflect an ion’s electronic structure in various chemical environments.
- They determine the magnetic and optical properties of metal complexes.
- Analyzing electron configurations guides the identification and prediction of high-spin and low-spin states.
Oxidation States
In our exercise, recognizing the correct oxidation states of \(\text{Cr}^{3+}\), \(\text{Fe}^{2+}\), \(\text{Mn}^{2+}\), and \(\text{Ni}^{2+}\) is the initial step. This permits accurate determination of each metal's electron configuration.
Identifying oxidation states also reveals how the metal interacts with ligands, influencing the energy levels and photochemical behaviors of the complex.
- An oxidation state is crucial for understanding coordination chemistry.
- Determines electron removal or addition count for defining orbital occupation.
- It affects the reactivity and stability of coordination complexes.