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Chapter 1: Problem 22

Which of the following ions would have the most unpaired electrons? (A) \(\mathrm{Mn}^{2+}\) (B) \(\mathrm{Ni}^{3+}\) (C) \(\mathrm{Ti}^{2+}\) (D) \(\mathrm{Cr}^{6+}\)

Short Answer

Expert verified
The Mn^2+ ion has the most unpaired electrons.

Step by step solution

01

Determine electron configurations

First, it's necessary to find out the electron configurations for all four ions. Remember that positive ion means that electrons are removed from the atom. For respective atom: Mn has electron configuration [Ar] 3d^5 4s^2, so Mn^2+ ion has configuration [Ar] 3d^3 4s^2. Ni has configuration [Ar] 3d^8 4s^2, so Ni^3+ has [Ar] 3d^7. Ti has configuration [Ar] 3d^2 4s^2, so Ti^2+ has [Ar] 3d^2. Cr has configuration [Ar] 3d^54s^1, so Cr^6+ has [Ar].
02

Analyze each ion

Now let's look at each ion individually. Mn^2+ has two unpaired electrons in the 3d orbital. Ni^3+ has one unpaired electron in the 3d orbital. Ti^2+ doesn't have any unpaired electrons, while all of Cr^6+'s electrons are paired.
03

Compare the number of unpaired electrons

Comparing the number of unpaired electrons in all four ions, it can be seen that Mn^2+ ion has the most unpaired electrons (two) among the four.

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Key Concepts

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

Electron Configuration
Understanding electron configuration is essential for predicting the properties of an atom or ion. Electron configuration details how electrons are distributed in an atom's orbitals, arranged by increasing energy levels. Electrons fill orbitals starting from the lowest available energy levels before moving to higher ones. The general order of filling is 1s, 2s, 2p, 3s, 3p, 4s, 3d, and so forth. This arrangement minimizes the energy of the atom or ion, achieving a stable electronic structure.
For transition metals, the electron configuration can become more complex due to the presence of d orbitals. In these metals, electrons fill the 4s orbital before the 3d orbital, though they might be removed from 4s first during ionization. This peculiar behavior is crucial when determining the electron configuration of transition metal ions.
Transition Metal Ions
Transition metals are known for their ability to form ions with various positive charges, or oxidation states. This happens because these metals have electrons available in both their s and d orbitals, which they can lose. As a result, transition metal ions often exhibit multiple oxidation states, allowing them to participate in a wide variety of chemical reactions.
  • Transition metals often form colored ions due to d-orbital electron transitions.
  • The multiple oxidation states make transition metals crucial in catalysis and industrial applications.
  • They can form complex salts with different structures and properties.
Understanding how the d-orbital plays a role in transition metal chemistry is essential, as it affects important characteristics such as magnetism, coordination chemistry, and reactivity.
Oxidation States
The oxidation state of an element in a compound or ion shows the degree of oxidation, often corresponding to the hypothetical charge if all bonds were ionic. It is important to identify the oxidation state as it sheds light on the electron loss or gain of the element.
Transition metals, because of their various electronic configurations, can exhibit a range of oxidation states; this is especially apparent when electrons in the s and d orbitals are involved. For example, Mn can exist in +2, +3, +4, +6, and even +7 states. In the specific exercise mentioned, Mn in \( ext{Mn}^{2+} \) shows a common +2 oxidation state by losing two electrons from the 4s orbital. Recognizing these states is vital for predicting chemical behavior and bonding characteristics.
Electron Removal
When forming ions, transition metals lose electrons, a process known as electron removal. Understanding which electrons are removed first is crucial. Generally, electrons are removed from the outermost shell; for most transition metals, the s-electrons are lost before the d-electrons.
For example, manganese (Mn) has the electron configuration [Ar] 3d鈦� 4s虏, and when it forms a \( ext{Mn}^{2+} \) ion, it loses two electrons from the 4s orbital, resulting in [Ar] 3d鈦�. This rule predominately accounts for transition metals, ensuring ion formation aligns with observed electronic structures and properties. This understanding aids in correctly predicting the magnetic properties of an ion, based on the number of unpaired d-electrons remaining.

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Most popular questions from this chapter

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A mixture of helium and neon gases has a total pressure of 1.2 atm. If the mixture contains twice as many moles of helium as neon, what is the partial pressure due to neon? (A) 0.2 atm (B) 0.3 atm (C) 0.4 atm (D) 0.8 atm

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Directions: Questions 1-3 are long free-response questions that require about 23 minutes each to answer and are worth 10 points each. Write your response in the space provided following each question. Examples and equations may be included in your responses where appropriate. For calculations, clearly show the method used and the steps involved in arriving at your answers. You must show your work to receive credit for your answer. Pay attention to significant figures. A student is tasked with determining the identity of an unknown carbonate compound with a mass of 1.89 g. The compound is first placed in water, where it dissolves completely. The \(K_{s p}\) value for several carbonate-containing compounds are given below. $$\begin{array}{|c|c|}\hline \text { Compound } & {K_{s p}} \\ \hline \text { Lithium carbonate } & {8.15 \times 10^{-4}} \\ \hline \text { Nickel (II) carbonate } & {1.42 \times 10^{-7}} \\ \hline \text { Strontium carbonate } & {5.60 \times 10^{-10}} \\ \hline\end{array}$$ (a) In order to precipitate the maximum amount of the carbonate ions from solution, which of the following should be added to the carbonate solution: \(\operatorname{LiNO}_{3}, \mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2},\) or \(\mathrm{Sr}\left(\mathrm{NO}_{3}\right)_{2} ?\) Justify your answer. (b) For the carbonate compound that contains the cation chosen in part (a), determine the concentration of each ion of that compound in solution at equilibrium. (c) When mixing the solution, should the student ensure the carbonate solution or the nitrate solution is in excess? Justify your answer. (d) After titrating sufficient solution to precipitate out all of the carbonate ions, the student filters the solution before placing it in a crucible and heating it to drive off the water. After several heatings, the final mass of the precipitate remains constant and is determined to be 2.02 g. (i) Determine the number of moles of precipitate. (ii) Determine the mass of carbonate present in the precipitate. (e) Determine the percent, by mass, of carbonate in the original sample. (f) Is the original compound most likely lithium carbonate, sodium carbonate, or potassium carbonate? Justify your answer.
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