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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

During a chemical reaction, NO(g) gets reduced and no nitrogen- containing compound is oxidized. Which of the following is a possible product of this reaction? (A) \(\mathrm{NO}_{2}(g)\) (B) \(\mathrm{N}_{2}(g)\) (C) \(\mathrm{NO}_{3}^{-}(a q)\) (D) \(\mathrm{NO}_{2}^{-}(a q)\)

A student added 1 liter of a 1.0\(M\) \(\mathrm{KCl}\) solution to 1 liter of a 1.0 \(M\) \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) solution. A lead chloride precipitate formed, and nearly all of the lead ions disappeared from the solution. Which of the following lists the ions remaining in the solution in order of decreasing concentration? (A) \(\left[\mathrm{NO}_{3}^{-}\right]>\left[\mathrm{K}^{+}\right]>\left[\mathrm{Pb}^{2+}\right]\) (B) \(\left[\mathrm{NO}_{3}^{-}\right]>\left[\mathrm{Pb}^{2+}\right]>\left[\mathrm{K}^{+}\right]\) (C) \(\left[\mathrm{K}^{+}\right]>\left[\mathrm{Pb}^{2+}\right]>\left[\mathrm{NO}_{3}^{-}\right]\) (D) \(\left[\mathrm{K}^{+}\right]>\left[\mathrm{NO}_{3}^{-}\right]>\left[\mathrm{Pb}^{2+}\right]\)

The value of \(K_{a}\) for \(\mathrm{HSO}_{4}^{-}\) is \(1 \times 10^{-2} .\) What is the value of \(K_{b}\) for \(\mathrm{SO}_{4}^{2-} ?\) (A) \(K_{b}=1 \times 10^{-12}\) (B) \(K_{b}=1 \times 10^{-8}\) (C) \(K_{b}=1 \times 10^{-2}\) (D) \(K_{b}=1 \times 10^{2}\)

The following mechanism is proposed for a reaction: \(\begin{array}{ll}{2 \mathrm{A} \rightarrow \mathrm{B}} & {\text { (fast equilibrium) }} \\ {\mathrm{C}+\mathrm{B} \rightarrow \mathrm{D}} & {\text { (slow) }} \\ {\mathrm{D}+\mathrm{A} \rightarrow \mathrm{E}} & {\text { (fast) }}\end{array}\) Which of the following is the correct rate law for compete reaction? (A) Rate \(=k[\mathrm{C}]^{2}[\mathrm{B}]\) (B) Rate \(=k\left[\mathrm{Cl}[\mathrm{A}]^{2}\right.\) (C) Rate \(=k[\mathrm{C}][\mathrm{A}]^{3}\) (D) Rate \(=k[\mathrm{D}][\mathrm{A}]\)

A sample of a hydrate of \(\mathrm{CuSO}_{4}\) with a mass of 250 grams was heated until all the water was removed. The sample was then weighed and found to have a mass of 160 grams. What is the formula for the hydrate? (A) \(\mathrm{CuSO}_{4} \cdot 10 \mathrm{H}_{2} \mathrm{O}\) (B) \(\mathrm{CuSO}_{4} \cdot \mathrm{TH}_{2} \mathrm{O}\) (C) \(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}\) (D) \(\mathrm{CuSO}_{4} \cdot 2 \mathrm{H}_{2} \mathrm{O}\)

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