Which of these ions cannot form both high- and low-spin octahedral complexes: (a) \(\mathrm{Ti}^{3+}\) (b) \(\mathrm{Co}^{2+} ;\) (c) \(\mathrm{Fe}^{2+} ;\) (d) \(\mathrm{Cu}^{2+} ?\)

Short Answer

Expert verified
The ion that cannot form both high- and low-spin octahedral complexes is \(\text{Ti}^{3+}\).

Step by step solution

01

Understand high-spin and low-spin complexes

High-spin and low-spin complexes depend on the arrangement of electrons in the orbitals of the central metal ion in an octahedral field. High-spin complexes have more unpaired electrons, whereas low-spin complexes have fewer unpaired electrons due to pairing up in lower energy orbitals.
02

Examine each ion's electronic configuration

Determine the electronic configuration of each ion: (a) \(\text{Ti}^{3+}\) has configuration \[ \text{Ar} \, 3d^1 \](b) \(\text{Co}^{2+}\) has configuration \[ \text{Ar} \, 3d^7 \](c) \(\text{Fe}^{2+}\) has configuration \[ \text{Ar} \, 3d^6 \](d) \(\text{Cu}^{2+}\) has configuration \[ \text{Ar} \, 3d^9 \]
03

Analyze each ion for their possibility of forming high- and low-spin states

Compare electron configurations to high- and low-spin scenarios:(a) \(\text{Ti}^{3+}\) with \[3d^1\] has no possibility for different spin states since it has only one electron.(b) \(\text{Co}^{2+}\) with \[3d^7\] can form both high- and low-spin configurations.(c) \(\text{Fe}^{2+}\) with \[3d^6\] can form both high- and low-spin configurations.(d) \(\text{Cu}^{2+}\) with \[3d^9\] is highly unlikely to form both high- and low-spin states due to the near-filled d-orbital.
04

Determine the ion that cannot form both high- and low-spin complexes

Out of the given ions, \(\text{Ti}^{3+}\) has only one electron in the \(\text{d}\) orbital and cannot exhibit both high- and low-spin configurations.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

High-Spin Complexes
High-spin complexes occur when the electrons of a transition metal in an octahedral field occupy the higher energy orbitals to remain unpaired. This happens when the energy difference between the high and low orbitals (known as the crystal field splitting energy, \(\text{Δ} \text{_o}\)) is small.
When the crystal field splitting energy is less than the pairing energy (the energy required to pair two electrons in an orbital), electrons prefer to occupy different orbitals to stay unpaired.
Here are some key points to understand high-spin complexes:
  • High-spin complexes have a larger number of unpaired electrons.
  • They typically form in the presence of weak field ligands (like \( \text{Cl}^\text{-} \) or \( \text{F}^\text{-} \)).
  • High-spin complexes are usually paramagnetic due to the presence of unpaired electrons.
An example of a high-spin complex is \( \text{[Mn(H2O)6]}^{2+} \), where \( \text{Mn}^\text{2+} \) in an octahedral field of water ligands has a \(3d^5\) configuration with all electrons unpaired.
Low-Spin Complexes
Low-spin complexes occur when electrons of a transition metal in an octahedral field prefer to pair up in lower energy orbitals due to a large crystal field splitting energy, \(\text{Δ} \text{_o}\).
When the crystal field splitting energy is greater than the pairing energy, electrons tend to double up in lower energy orbitals rather than occupying higher ones.
Here are some key points to understand low-spin complexes:
  • Low-spin complexes have fewer unpaired electrons.
  • They typically form in the presence of strong field ligands (like \( \text{CN}^- \) or \( \text{CO} \)).
  • Low-spin complexes can be either paramagnetic (if there are unpaired electrons) or diamagnetic (if all electrons are paired).
An example of a low-spin complex is \( \text{[Fe(CN)_6]}^{4-} \), where \( \text{Fe}^\text{2+} \) has a \(3d^6\) configuration with electrons pairing up in the lower energy orbitals.
Electronic Configuration
Electronic configuration is the arrangement of electrons in the orbitals of an atom or ion. It determines the chemical and physical properties of the element.
For transition metals, the electronic configuration is especially important in understanding their behavior in complexes.
Some basic rules to determine electronic configurations include:
  • The Aufbau principle: Electrons fill orbitals starting from the lowest energy level.
  • Hund's rule: Electrons occupy degenerate orbitals singly first, and then start to pair up.
  • The Pauli Exclusion Principle: No two electrons in an atom can have the same set of four quantum numbers.
For example, in the problem, \( \text{Ti}^{3+} \) with configuration \( [\text{Ar}] 3d^1 \) has only one electron and cannot exhibit different spin states.
On the other hand, \( \text{Co}^{2+} \) with a configuration \( [\text{Ar}] 3d^7 \) can form both high- and low-spin configurations depending on the ligand field strength.

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Most popular questions from this chapter

Ionic liquids have many applications in engineering and materials science. The dissolution of the metavanadate ion in chloroaluminate ionic liquids has been studied: $$\mathrm{VO}_{3}^{-}+\mathrm{AlCl}_{4}^{-} \longrightarrow \mathrm{VO}_{2}\mathrm{Cl}_{2}^{-}+\mathrm{AlOCl}_{2}^{-}$$ (a) What is the oxidation number of \(\mathrm{V}\) and \(\mathrm{Al}\) in each ion? (b) In reactions of \(\mathrm{V}_{2} \mathrm{O}_{5}\) with \(\mathrm{HCl}\), acid concentration affects the product. At low acid concentration, \(\mathrm{VO}_{2} \mathrm{Cl}_{2}^{-}\) and \(\mathrm{VO}_{3}^{-}\) form: $$\mathrm{V}_{2} \mathrm{O}_{5}+\mathrm{HCl} \longrightarrow \mathrm{VO}_{2}\mathrm{Cl}_{2}^{-}+\mathrm{VO}_{3}^{-}+\mathrm{H}^{+}$$ At high acid concentration, \(\mathrm{VOCl}_{3}\) forms: $$\mathrm{V}_{2} \mathrm{O}_{5}+\mathrm{HCl} \longrightarrow \mathrm{VOCl}_{3}+\mathrm{H}_{2} \mathrm{O}$$ Balance each equation, and state which, if either, is a redox process. (c) What mass of \(\mathrm{VO}_{2} \mathrm{Cl}_{2}^{-}\) or \(\mathrm{VOCl}_{3}\) can form from \(12.5 \mathrm{~g}\) of \(\mathrm{V}_{2} \mathrm{O}_{5}\) and the appropriate concentration of acid?

\(\mathrm{VF}_{5}\) is a liquid that boils at \(48^{\circ} \mathrm{C},\) whereas \(\mathrm{VF}_{3}\) is a solid that melts above \(800^{\circ} \mathrm{C}\). Explain this difference in properties.

(a) What is the maximum number of unpaired electrons in a lanthanide ion? (b) How does this number relate to occupancy of the \(4 f\) subshell?

When neptunium (Np) and plutonium (Pu) were discovered, the periodic table did not include the actinides, so these elements were placed in Groups \(7 \mathrm{~B}(7)\) and \(8 \mathrm{~B}(8) .\) When americium (Am) and curium (Cm) were synthesized, they were placed in Groups \(8 \mathrm{~B}(9)\) and \(8 \mathrm{~B}(10) .\) However, during chemical isolation procedures, Glenn Seaborg and his colleagues, who had synthesized these elements, could not find their compounds among other compounds of members of the same groups, which led Seaborg to suggest they were part of a new inner transition series. (a) How do the electron configurations of these elements support Seaborg's suggestion? (b) The highest fluorides of \(\mathrm{Np}\) and \(\mathrm{Pu}\) are hexafluorides, and the highest fluoride of uranium is also the hexafluoride. How does this chemical evidence support the placement of \(\mathrm{Np}\) and \(\mathrm{Pu}\) as inner transition elements rather than transition elements?

Using the periodic table to locate each element, write the electron configuration of (a) Os; (b) \(\mathrm{Co} ;\) (c) Ag.

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