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Which of the following statement(s) is(are) true? a. The coordination number of a metal ion in an octahedral complex ion is 8. b. All tetrahedral complex ions are low-spin. c. The formula for triaquatriamminechromium(III) sulfate is \(\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{3}\left(\mathrm{NH}_{3}\right)_{3}\right]_{2}\left(\mathrm{SO}_{4}\right)_{3}\) d. The electron configuration of \(\mathrm{Hf}^{2+}\) is \([\mathrm{Xe}] 4 f^{12} 6 s^{2}\) e. Hemoglobin contains \(\mathrm{Fe}^{3+}\)

Short Answer

Expert verified
Statements b and c are true. All tetrahedral complex ions are low-spin, and the formula for triaquatriamminechromium(III) sulfate is $\left[\mathrm{Cr}\left(\mathrm{H}_{2}\mathrm{O}\right)_{3}\left(\mathrm{NH}_{3}\right)_{3}\right]_{2}\left(\mathrm{SO}_{4}\right)_{3}$.

Step by step solution

01

Statement a

The coordination number of a metal ion in an octahedral complex ion is the number of ligands surrounding the central metal atom/ion. In an octahedral complex, there are 6 ligands surrounding the central metal atom/ion. So, statement a is incorrect as the coordination number in an octahedral complex is not 8, but 6.
02

Statement b

Tetrahedral complex ions are composed of a central metal atom/ion surrounded by four ligands in a tetrahedral arrangement. These complexes typically have high-spin configurations when the central metal ion has d orbitals that can accommodate additional unpaired electrons. However, tetrahedral complexes do not exhibit the phenomenon of crystal-field splitting as significantly as octahedral complexes. Thus, they cannot undergo strong field ligand-induced stabilization, leading to low-spin configurations. So, statement b is correct, as all tetrahedral complex ions are low-spin.
03

Statement c

The formula for triaquatriamminechromium(III) sulfate is given as $\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{3}\left(\mathrm{NH}_{3}\right)_{3}\right]_{2}\left(\mathrm{SO}_{4}\right)_{3}\(. From the name, we know that the complex contains three water molecules and three ammonia molecules as ligands. The chromium(III) ion has a charge of +3, so the overall charge of the complex is +3. The formula of sulfate is \)\mathrm{SO}_{4}^{2-}$, so we need two of these ions to balance the charges for a total of -6. We need two of the complex ions (each with charge +3) to balance the overall charges. The given formula is correct, so statement c is true.
04

Statement d

The electron configuration of the neutral Hafnium atom is \(\left[\mathrm{Xe}\right] 4 f^{14} 5 d^{2} 6 s^{2}\). Upon losing two electrons to form \(\mathrm{Hf}^{2+}\), the electron configuration becomes \(\left[\mathrm{Xe}\right] 4 f^{14} 5 d^{0} 6 s^{2}\). The electron configuration provided in statement d is incorrect. So, statement d is false.
05

Statement e

Hemoglobin is a protein present in red blood cells responsible for carrying oxygen. It contains a central iron atom/ion as part of the heme group. The iron ion in hemoglobin is in the +2 oxidation state (Fe2+). When hemoglobin loses its oxygen-carrying ability, it gets oxidized to methemoglobin, which contains \(\mathrm{Fe}^{3+}\). So, statement e is false as hemoglobin contains \(\mathrm{Fe}^{2+}\), not \(\mathrm{Fe}^{3+}\).

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Key Concepts

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

Coordination Number
In the study of coordination chemistry, the coordination number is a fundamental concept. It refers to the total number of ligand atoms that are directly bonded to the central metal ion in a coordination complex. For instance, if a metal ion is surrounded by six ligands, its coordination number is 6. This is frequently observed in octahedral complexes, which are very common in transition metal chemistry.
Coordination numbers can vary, but 4 and 6 are among the most typical. This affects the geometric arrangement of the complex. Understanding the coordination number is crucial as it also plays a role in determining the complex's properties, such as color and magnetism.
  • Octahedral complexes: Coordination number = 6
  • Tetrahedral complexes: Coordination number = 4
It's essential to match the coordination number with the correct type of geometric configuration, as this can significantly impact the stability and reactivity of the coordination complex.
Octahedral and Tetrahedral Complexes
Coordination complexes can be categorized based on their geometry, and two common types are octahedral and tetrahedral complexes. An octahedral complex features a central metal surrounded by six ligands, forming an eight-sided geometrical shape (octahedron). This is a very stable and common type of configuration due to symmetric distribution of ligands. It is usually associated with transition metals and their d-electron interactions.
Tetrahedral complexes have a different configuration, with four ligands. They form a symmetrical three-dimensional shape, where each atom is equidistant from each other. This arrangement is less common than octahedral but is typical for metals with large ligands or those unable to accommodate six ligands due to steric hindrance.
  • Octahedral: Higher coordination number, more common in transition metal complexes.
  • Tetrahedral: Often leads to high-spin configurations due to smaller crystal field splitting.
Both of these geometries have distinct electronic characteristics and influence the overall reactivity and magnetic properties of the complexes.
Electron Configuration
The electron configuration of an atom or ion shows how its electrons are distributed in its atomic orbitals. This concept is important in predicting the chemical characteristics and reactivity of an element. Transition metals, because they can lose different numbers of electrons, often form ions with varying electronic configurations.
Taking the example of hafnium (\(\mathrm{Hf}\)), its neutrality begins with the electronic configuration \[\left[\mathrm{Xe}\right] 4f^{14} 5d^{2} 6s^{2}\]. Upon losing two electrons to form the \(\mathrm{Hf}^{2+}\) ion, the configuration changes to \[\left[\mathrm{Xe}\right] 4f^{14} 5d^{0} 6s^{2}\]. Understanding these configurations helps in foretelling the complexities in the behavior of different elements and their interactions.
  • Electron configuration helps in predicting the chemical properties of an ion.
  • It is crucial for understanding bonding and reactivity, especially in transition metals.
This concept is particularly essential in coordination chemistry, where the electron distribution around a metal ion determines the type and strength of bonds it can make.
Hemoglobin
Hemoglobin is a critical protein found in red blood cells, primarily responsible for oxygen transport throughout the body. Each hemoglobin molecule is composed of four subunits, and each subunit contains a heme group, where an iron ion is centrally located. This iron ion plays a vital role in binding oxygen molecules.
In its active form, hemoglobin features iron in the \(\mathrm{Fe}^{2+}\) oxidation state, which easily binds with oxygen. However, in situations where it becomes oxidized, it converts into methemoglobin, containing \(\mathrm{Fe}^{3+}\), which cannot bind oxygen effectively. Understanding the oxidation states of iron in hemoglobin is essential in medical chemistry as variations can indicate or result in medical conditions.
  • Hemoglobin primarily carries oxygen in the blood.
  • \(\mathrm{Fe}^{2+}\) is the active iron state in oxygen transport.
  • It provides insight into various physiological and pathological states.
The function and efficiency of hemoglobin in oxygen delivery are partially determined by the coordination chemistry involving iron's state and the surrounding heme structure.

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Most popular questions from this chapter

What is the maximum number of unpaired \(d\) electrons that an octahedral transition metal complex ion can have? Predict a compound that would have this number of unpaired electrons.

Which is more likely to be paramagnetic, \(\mathrm{Fe}(\mathrm{CN})_{6}^{4-}\) or \(\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}\) ? Explain.

Consider the complex ions \(\mathrm{Co}\left(\mathrm{NH}_{3}\right) 6^{3+}, \mathrm{Co}(\mathrm{CN})_{6}^{3-},\) and \(\mathrm{CoF}_{6}^{3-} .\) The wavelengths of absorbed electromagnetic radiation for these compounds (in no specific order) are \(770 \mathrm{nm},\) \(440 \mathrm{nm},\) and 290 \(\mathrm{nm} .\) Match the complex ion to the wave- length of absorbed electromagnetic radiation.

Use standard reduction potentials to calculate \(\mathscr{C}^{\circ}, \Delta G^{\circ},\) and \(K\) (at 298 K) for the reaction that is used in production of gold: $$2 \mathrm{Au}(\mathrm{CN})_{2}-(a q)+\mathrm{Zn}(s) \longrightarrow 2 \mathrm{Au}(s)+\mathrm{Zn}(\mathrm{CN})_{4}^{2-}(a q)$$ The relevant half-reactions are $$\begin{aligned} \operatorname{Au}(\mathrm{CN})_{2}^{-}+\mathrm{e}^{-} \longrightarrow \mathrm{Au}+2 \mathrm{CN}^{-} & \mathscr{C}^{\circ}=-0.60 \mathrm{V} \\ \mathrm{Zn}(\mathrm{CN})_{4}^{2-}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Zn}+4 \mathrm{CN}^{-} & \mathscr{C}^{\circ}=-1.26 \mathrm{V} \end{aligned}$$

The ferrate ion, \(\mathrm{FeO}_{4}^{2-}\) , is such a powerful oxidizing agent that in acidic solution, aqueous ammonia is reduced to elemental nitrogen along with the formation of the iron(III) ion. a. What is the oxidation state of iron in FeO \(_{4}^{2-},\) and what is the electron configuration of iron in this polyatomic ion? b. If 25.0 \(\mathrm{mL}\) of a 0.243 \(\mathrm{M} \mathrm{FeO}_{4}^{2-}\) solution is allowed to react with 55.0 \(\mathrm{mL}\) of 1.45 \(\mathrm{M}\) aqueous ammonia, what volume of nitrogen gas can form at \(25^{\circ} \mathrm{C}\) and 1.50 \(\mathrm{atm}\) ?

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