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Explain why \(\mathrm{Mn}^{3+}\) forms high-spin and low-spin octahedral complexes but \(\mathrm{Mn}^{4+}\) does not.

Short Answer

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Question: Explain the difference in behavior between Mn鲁鈦 and Mn鈦粹伜 in forming high-spin and low-spin octahedral complexes. Answer: Mn鲁鈦 can form both high-spin and low-spin octahedral complexes due to its 3d鈦 configuration, which allows for the accommodation of electrons in both the t2g and eg orbitals. In contrast, Mn鈦粹伜 cannot form distinct high-spin and low-spin octahedral complexes as its 3d鲁 configuration leads to the presence of electrons only in the lower energy t2g orbitals, resulting in no significant difference between the two configurations.

Step by step solution

01

Determine the electronic configurations of Mn鲁鈦 and Mn鈦粹伜

The electronic configuration of manganese (Mn) in its ground state is [Ar] 3d鈦 4s虏. In order to find the electronic configuration for Mn鲁鈦, we remove three electrons from the manganese atom. First, the two 4s electrons are removed, and then one more electron from the 3d orbitals. This results in the electronic configuration for Mn鲁鈦 as [Ar] 3d鈦. For Mn鈦粹伜, we remove four electrons, taking one more electron from the 3d orbitals, giving us an electronic configuration of [Ar] 3d鲁.
02

Analyze Mn鲁鈦 in forming high-spin and low-spin octahedral complexes

In octahedral complexes, the ligands approach the central metal ion and cause the energy levels of the d-orbitals to split into two sets: a lower energy level (t2g) containing three orbitals and a higher energy level (eg) containing two orbitals. For high-spin complexes, the ligand's effect on the d-orbitals is weak, and so electrons prefer to occupy all five d-orbitals singly before starting to pair up in the lower energy t2g orbitals. In the case of Mn鲁鈦 with its 3d鈦 configuration, there are enough unpaired electrons to distribute into the t2g orbitals and partially fill one of the eg orbitals, making it possible for the Mn鲁鈦 ion to form a high-spin complex. In low-spin complexes, the ligand has a stronger effect on the d-orbitals, causing the energy difference between the t2g and eg orbitals to be larger. Electrons will then pair up in the lower energy t2g orbitals before moving to the eg orbitals. Mn鲁鈦 can also form low-spin octahedral complexes because the four d-electrons can be accommodated by the lower energy t2g orbitals.
03

Analyze Mn鈦粹伜 in forming high-spin and low-spin octahedral complexes

When it comes to Mn鈦粹伜, the situation is different. Mn鈦粹伜 has an electronic configuration of [Ar] 3d鲁, which means there are fewer electrons to fill the d-orbitals. In both high-spin and low-spin octahedral complexes, the three unpaired electrons will first populate the t2g orbitals. In this case, there is no difference between the high-spin and low-spin configurations because all the electrons are already in the lower energy t2g orbitals. Consequently, Mn鈦粹伜 cannot form high-spin and low-spin octahedral complexes as there is no significant difference between the two. In conclusion, Mn鲁鈦 can form both high-spin and low-spin octahedral complexes due to its 3d鈦 configuration, which allows for the accommodation of electrons in both the t2g and eg orbitals. On the other hand, Mn鈦粹伜 cannot form distinct high-spin and low-spin octahedral complexes as its 3d鲁 configuration leads to the presence of electrons only in the lower energy t2g orbitals, making the two configurations essentially the same.

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Key Concepts

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

Electronic Configuration of Transition Metal Ions
Understanding the electronic configuration of transition metal ions is crucial when evaluating their chemical behavior. The element manganese (Mn) has a standard electronic configuration of [Ar] 3d鈦 4s虏. When manganese forms cations such as Mn鲁鈦 and Mn鈦粹伜, it does so by losing electrons.

When Mn loses three electrons to become Mn鲁鈦, the configuration becomes [Ar] 3d鈦. This is because the two 4s electrons are removed first, followed by one more from the 3d orbitals. Similarly, when it loses an additional electron to form Mn鈦粹伜, its configuration is [Ar] 3d鲁. The number and arrangement of these d electrons play a pivotal role in the magnetic properties and the types of complexes that the ion can form.
High-Spin and Low-Spin Complexes
The distinction between high-spin and low-spin complexes lies in the strength of the ligands' field. This field influences the energy separation between the two sets of d-orbitals in an octahedral complex, known as t2g and eg.

In high-spin complexes, weak field ligands cause a smaller energy gap between t2g and eg orbitals. This smaller gap allows electrons to occupy higher energy eg orbitals to minimize electron pairing, resulting in more unpaired electrons and hence a high-spin complex. In contrast, low-spin complexes have strong field ligands, inducing a greater energy separation, which enables electrons to pair up in the lower energy t2g orbitals before the higher eg orbitals are utilized.

Mn鲁鈦, with its four d-electrons, can form both types of complexes. Electrons can either fill the t2g and eg orbitals in a high-spin configuration or pair up within the t2g in a low-spin configuration. Mn鈦粹伜, however, remains in a low-spin-like state regardless of ligand strength due to having only three d-electrons to fill the lower t2g orbitals.
Ligand Field Theory
Ligand field theory (LFT) is a sophisticated theory that explains the electronic structures of transition metal complexes. It considers the interaction between the ligands and the d-orbitals of the central metal ion. In an octahedral field, the d-orbitals are split into two sets where three orbitals (t2g) have a lower energy level, and two orbitals (eg) have a higher energy level.

According to LFT, the split is influenced by the ligand's electric field, which can vary greatly depending on the nature of the liganship. For instance, ligands such as fluoride and water are considered weak field ligands and result in a smaller split. In contrast, strong field ligands like cyanide cause a larger split between t2g and eg.

This theory provides insight into why Mn鲁鈦 can exhibit both high-spin and low-spin configurations. The strength of the ligand's field can significantly alter the energy layout and thus the electron distribution in the d-orbitals. On the other hand, Mn鈦粹伜, with one less electron, cannot take advantage of this variable energy landscape due to its limited number of electrons, resulting in only one possible arrangement.

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Most popular questions from this chapter

Bipyridyl (Bipy) is a molecule commonly used as a bidentate ligand. When \(0.17 \mathrm{~mol}\) of bipyridyl is dissolved in \(2.4 \mathrm{~L}\) of a solution that contains \(0.052 \mathrm{M} \mathrm{Fe}^{2+},\left[\mathrm{Fe}(\mathrm{bipy})_{3}\right]^{2+}\left(K_{\mathrm{f}}=1.6 \times 10^{17}\right)\) is formed. What are the con- centrations of all species when equilibrium is established?

WEB Platinum(II) forms many complexes, among them those with the following ligands. Give the formula and charge of each complex. (a) two ammonia molecules and one oxalate ion \(\left(\mathrm{C}_{2} \mathrm{O}_{4}{ }^{2-}\right)\) (b) two ammonia molecules, one thiocyanate ion \(\left(\mathrm{SCN}^{-}\right)\), and one bromide ion (c) one ethylenediamine molecule and two nitrite ions

[ \(\left[\mathrm{Cr}(\mathrm{CN})_{6}\right]^{4-}\) is less paramagnetic than \(\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\). Explain.

Consider three complexes of \(\mathrm{Ag}^{+}\) and their formation constants, \(K_{\mathrm{f}}\) $$\begin{array}{ll}\hline \text { Complex lon } & K_{\mathrm{f}} \\\\\hline \mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}+ & 1.6 \times 10^{7} \\ \mathrm{Ag}(\mathrm{CN})_{2}^{-} & 5.6 \times 10^{18} \\\\\mathrm{AgBr}_{2}^{-} & 1.3 \times 10^{7} \\ \hline\end{array}$$ Which statements are true? (a) \(\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}{ }^{+}\) is more stable than \(\mathrm{Ag}(\mathrm{CN})_{2}^{-}\). (b) Adding a strong acid \(\left(\mathrm{HNO}_{3}\right)\) to a solution that is \(0.010 \mathrm{M}\) in \(\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}\) will tend to dissociate the complex ion into \(\mathrm{Ag}^{+}\) and \(\mathrm{NH}_{4}^{+} .\) (c) Adding a strong acid \(\left(\mathrm{HNO}_{3}\right)\) to a solution that is \(0.010 \mathrm{M}\) in \(\mathrm{AgBr}_{2}^{-}\) will tend to dissociate the complex ion into \(\mathrm{Ag}^{+}\) and \(\mathrm{Br}^{-} .\) (d) To dissolve AgI, one can add either \(\mathrm{NaCN}\) or \(\mathrm{HCN}\) as a source of the cyanide-complexing ligand. Fewer moles of NaCN would be required. (e) Solution \(A\) is \(0.10 M\) in \(B r^{-}\) and contains the complex ion \(\mathrm{AgBr}_{2}^{-}\). Solution B is \(0.10 M\) in \(\mathrm{CN}^{-}\) and contains the complex ion \(\mathrm{Ag}(\mathrm{CN})_{2}-\). Solution B will have more particles of complex ion per particle of \(\mathrm{Ag}^{+}\) than solution \(\mathrm{A}\).

Sketch the geometry of (a) \(\left[\mathrm{Zn}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}\right]\) (tetrahedral) (b) cis \(-\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4} \mathrm{Cl}_{2}\right]^{+}\) (c) trans- \(\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Br}_{2}\right]^{2+}\) (d) trans- \(\left[\mathrm{Ni}(\mathrm{ox})_{2}(\mathrm{OH})_{2}\right]^{3-}\) (e) \([\mathrm{Au}(\mathrm{CN}) \mathrm{Br}]^{+}\)

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