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From each pair of complexes, select the one that has the greater crystal field splitting. (a) \(\left[\mathrm{Rh}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}\) or \(\left[\mathrm{Rh}(\mathrm{CN})_{6}\right]^{3-}\) (b) \(\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) or \(\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}\) (c) \(\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}\) or \(\left[\mathrm{Rh}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}\) (d) \(\left[\mathrm{T} \mathrm{i} \mathrm{F}_{6}\right]^{3-}\) or \(\left[\mathrm{Ti}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}\)

Short Answer

Expert verified
(a) [Rh(CN)鈧哴鲁鈦, (b) [Fe(H鈧侽)鈧哴鲁鈦, (c) [Rh(H鈧侽)鈧哴鲁鈦, (d) [Ti(H鈧侽)鈧哴鲁鈦.

Step by step solution

01

Analyze Crystal Field Splitting

Crystal field splitting ( 螖) depends on the ligand's ability to split the d-orbitals. Stronger field ligands cause higher splitting. Order from strong to weak field ligands: CN鈦 > NH鈧 > H鈧侽 > F鈦.
02

Compare Rhodium Complexes (a)

For [Rh(NH鈧)鈧哴鲁鈦 and [Rh(CN)鈧哴鲁鈦, CN鈦 is a stronger field ligand than NH鈧. Thus, [Rh(CN)鈧哴鲁鈦 has the greater crystal field splitting.
03

Compare Iron Complexes (b)

For [Fe(H鈧侽)鈧哴虏鈦 and [Fe(H鈧侽)鈧哴鲁鈦, having a higher charge on the metal ion increases the field splitting since the central metal ion's ability to attract electrons increases. Therefore, [Fe(H鈧侽)鈧哴鲁鈦 has the greater crystal field splitting.
04

Compare Cobalt and Rhodium Complexes (c)

For [Co(H鈧侽)鈧哴鲁鈦 and [Rh(H鈧侽)鈧哴鲁鈦, Rh鲁鈦 is in the 4d series, whereas Co鲁鈦 is in the 3d series. 4d metals tend to have greater splitting than 3d for similar ligands. Therefore, [Rh(H鈧侽)鈧哴鲁鈦 has greater splitting.
05

Compare Titanium Complexes (d)

For [TiF鈧哴鲁鈦 and [Ti(H鈧侽)鈧哴鲁鈦, F鈦 is generally a weaker ligand than H鈧侽, leading to less splitting. However, the charge on Ti affects splitting: [Ti(H鈧侽)鈧哴鲁鈦 has higher field splitting due to larger charge (higher effective nuclear charge).
06

Conclusion

From each pair of complexes, the ones with greater crystal field splitting are: (a) [Rh(CN)鈧哴鲁鈦, (b) [Fe(H鈧侽)鈧哴鲁鈦, (c) [Rh(H鈧侽)鈧哴鲁鈦, and (d) [Ti(H鈧侽)鈧哴鲁鈦.

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Key Concepts

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

Ligand Field Strength
The ligand field strength is a critical factor in determining the extent of crystal field splitting in transition metal complexes. Ligands can be thought of as ions or molecules that donate a pair of electrons to a metal atom or ion, forming a complex.

The strength of these ligands can vary significantly, and this influences the energy difference between the different d-orbitals of the central metal atom. Strong-field ligands, like cyanide (CN鈦), lead to a larger splitting of the d-orbitals. This means that the electrons within these orbitals will experience a significant energy gap. On the other hand, weak-field ligands, such as fluoride (F鈦), result in a smaller gap and thus weaker splitting.

Here is a simple way to visualize it:
  • Strong-field ligand: Large d-orbital splitting, higher 螖 (difference in energy)
  • Weak-field ligand: Small d-orbital splitting, lower 螖
Understanding the ligand field strength helps us predict the behavior and properties of the metal complex.
Complex Ions
Complex ions are formed when a central metal ion is bonded to surrounding molecules or ions, called ligands. These formations result in a stable arrangement due to the donation of electron pairs from the ligands to the metal.

In transition metal complexes, the nature of the ligands and their strength directly affect the properties of the complex ion. Factors such as the oxidation state of the metal, the arrangement of the ligands around the metal, and the number of d-electrons influence the complexity. For instance, comparing different complexes with the same central metal but varying ligands, like \([\mathrm{Rh}(\mathrm{NH}_3)_6]^{3+} and [\mathrm{Rh}(\mathrm{CN})_6]^{3-}\), showcases the impact of ligand strength on the properties of the ion.

The configuration and energy of the d-electrons get altered based on these interactions, leading to phenomena like color changes or different magnetic properties in the complexes. Comprehending these effects allows chemists to design specific complexes for applications like catalysts or electronic materials.
d-Orbitals Splitting
The splitting of d-orbitals is a fascinating concept within crystal field theory, primarily influencing the stability and properties of the resulting complexes. In an isolated metal ion, d-orbitals are degenerate, implying they all have the same energy level. However, when ligands approach the metal ion, they interact with these orbitals differently.

In an octahedral crystal field, for example, the presence of six ligands splitting the orbitals into two groups鈥擻(t_{2g}\) (lower energy) and \(e_g\) (higher energy) sets鈥攊s observed. The energy difference between these sets is known as the crystal field splitting energy \(螖\).
  • The larger the value of \(螖\), the more stable the complex in low-spin configurations, where electrons pair up in lower-energy orbitals.
  • With weaker ligands, a smaller \(螖\) results, often leading to high-spin states where electrons remain unpaired.
This splitting is key not only for understanding the color and magnetic properties but also for explaining the thermodynamic and kinetic properties of complexes in various chemical reactions.

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Most popular questions from this chapter

Ethylenediamine (en) is a bidentate ligand, and the ligand diethylenetriamine (dien) is the tridentate analog of en, \(\mathrm{NH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{NHCH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}\) (the donor atoms are in color). Draw the Lewis structure of dien and all possible geometric isomers of \(\left[\mathrm{Co}(\mathrm{dien})_{2}\right]^{3+} .\)

For each \(d\) electron configuration, state the number of unpaired electrons expected in octahedral complexes. Give an example complex for each case. (Two answers are possible for some of these cases.) (a) \(d^{2}\) (b) \(d^{4}\) (c) \(d^{6}\) (d) \(d^{8}\)

A brown-yellow complex of vanadium is analyzed and found to contain \(49.75 \% \mathrm{~V}, 15.62 \% \mathrm{O},\) and \(34.63 \% \mathrm{Cl}\). The substance boils at \(127^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\) of pressure. What is the empirical formula of the compound? How many electrons are in the vanadium \(3 d\) orbitals? Predict whether the bonding is mainly covalent or ionic.

Only the Group \(1 \mathrm{~B}\) transition elements form simple compounds in which the oxidation state of the metal is \(+1 .\) For all of the other transition elements, the lowest positive oxidation state is +2 . What common feature in the electron configuration of the transition elements contributes to this fact?

The ratio of the density of tantalum to that of niobium is \(1.94,\) which is nearly identical to the 1.95 ratio of their atomic weights. Explain how this similarity is a result of the lanthanide contraction.

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