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The +2 oxidation state is common for almost all the transition metals. Suggest an explanation.

Short Answer

Expert verified
The +2 oxidation state is common for almost all transition metals because they typically have 2 electrons in their outermost s-orbital (ns^2), which can be easily lost, resulting in a stable electronic configuration with a fully or half-filled d-orbital. Furthermore, low electronegativity and smaller atomic size of transition metals also contribute to the tendency of forming a +2 oxidation state.

Step by step solution

01

Understanding transition metals

Transition metals are the elements found in the d-block of the periodic table. These metals have similar properties, such as forming colored ions, being good conductors of heat and electricity, and having a high density and melting point. One of the key properties of transition metals is their ability to form different oxidation states.
02

Identifying electron configurations

In transition metals, the electron configurations involve the filling of the 3d, 4d, or 5d orbitals. In general, the electron configuration of a transition metal can be represented as [noble gas] ns^2 (n-1)d^x, where X can be any number from 0 to 10. For example, the electron configuration of Iron (Fe) is [Ar] 4s^2 3d^6.
03

Explaining the +2 oxidation state

The oxidation state of an element is the number of electrons lost or gained during a chemical reaction. Transition metals can lose electrons from both the s and d orbitals to form positive ions or cations. Most transition metals have 2 electrons in their outermost s-orbital (ns^2), so it is relatively easy for them to lose these 2 electrons and achieve a +2 oxidation state. Losing these 2 electrons results in a stable electronic configuration with a fully or half-filled d-orbital, which provides stability to the metal. For example, Manganese (Mn) with an electron configuration of [Ar] 4s^2 3d^5 can lose 2 electrons from the 4s orbital, resulting in a +2 oxidation state with the configuration of [Ar] 3d^5. This new arrangement leads to a half-filled 3d orbital, which is more stable.
04

Considering other factors

Besides the electron configurations, other factors like electronegativity and atomic size also contribute to the common +2 oxidation state of transition metals. These metals have relatively low electronegativity, making them more likely to lose electrons and form positive ions. The atomic size of transition metals is also smaller than the elements in the s-block, which makes it easier for them to lose outer electrons. In conclusion, the +2 oxidation state is common for almost all transition metals due to the ease of losing 2 electrons from the outermost s-orbital, resulting in a stable electronic configuration. Additionally, factors like low electronegativity and smaller atomic size contribute to this tendency as well.

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Key Concepts

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

Electron Configurations
Transition metals are unique in their ability to exhibit a variety of electron configurations. This versatility is largely due to the filling of the d-orbitals, which are energy levels that can house up to ten electrons, with configurations such as [noble gas] ns2 (n-1)dx, where 'x' can range from 0 to 10.

The significance of electron configurations lies in the way these configurations contribute to the metal's chemical properties. For example, iron (Fe), which has an electron configuration of [Ar] 4s2 3d6, can easily achieve a +2 oxidation state. It does this by losing the two 4s electrons, creating a more stable arrangement in the d-orbital.

Furthermore, when transition metals lose these electrons and form a +2 oxidation state, there is a preference for half-filled (e.g., d5) or completely filled (e.g., d10) d-orbitals, as these configurations are particularly stable due to electron pairing and exchange energy considerations.
Transition Metal Characteristics
Transition metals exhibit a set of shared characteristics that influence their chemical behavior. Not only are they known for high melting points, densities, and being good conductors, but they also have the distinctive propensity to form colorful compounds and complex ions.

One of the characteristic features highlighted in the exercise is the +2 oxidation state commonality among transition metals. The s-orbital electrons are often the first to be ionized due to energy considerations, leading to the prevalence of this oxidation state.

Variable Oxidation States

Unlike main-group elements, transition metals can access several oxidation states. This is due to the relatively low energy gap between their s and d orbitals, allowing electrons from both to be involved in bonding and reactions.

Magnetic Properties

Unfilled d-orbitals can lead to interesting magnetic properties. For instance, compounds having unpaired d electrons manifest magnetic characteristics, which render them important in various technological applications.
Chemical Stability
Chemical stability in transition metals is influenced by their ability to adopt a +2 oxidation state, but it's also reliant on the electronic structure. Full or half-full d-orbitals are energetically favorable, which leads to enhanced stability of the metal ion.

This stability is pivotal not just for the +2 oxidation state, but also for other oxidation states transition metals can assume. For instance, the +3 oxidation state in some metals is also stable, but it requires additional energy to remove the third electron, which is why +2 remains the most common.

It's also important to recognize that the stability of transition metals is affected by other factors, such as ligand interactions. Ligands can influence the oxidation state by stabilizing certain electron configurations, thereby playing a role in the color and reactivity of transition metal compounds. Understanding these interactions and their impact on stability is crucial in fields ranging from coordination chemistry to materials science.

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Most popular questions from this chapter

Write balanced chemical equations to represent the following observations. (In some instances the complex involved has been discussed previously in the text.) (a) Solid silver chloride dissolves in an excess of aqueous ammonia. (b) The green complex \(\left[\mathrm{Cr}(\mathrm{en})_{2} \mathrm{Cl}_{2}\right] \mathrm{Cl},\) on treatment with water over a long time, converts to a brown-orange complex. Reaction of \(\mathrm{AgNO}_{3}\) with a solution of the product precipitates \(3 \mathrm{~mol}\) of AgCl per mole of Cr present. (Write two chemical equations.) (c) When an NaOH solution is added to a solution of \(\mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2},\) a precipitate forms. Addition of excess \(\mathrm{NaOH}\) solution causes the precipitate to dissolve. (Write two chemical equations.) (d) A pink solution of \(\mathrm{Co}\left(\mathrm{NO}_{3}\right)_{2}\) turns deep blue on addition of concentrated hydrochloric acid.

Suppose that a transition-metal ion was in a lattice in which it was in contact with just two nearby anions, located on opposite sides of the metal. Diagram the splitting of the metal \(\bar{d}\) orbitals that would result from such a crystal field. Assuming a strong field, how many unpaired electrons would you expect for a metal ion with six \(d\) electrons? (Hint: Consider the linear axis to be the \(z\) -axis)

Consider the tetrahedral anions \(\mathrm{VO}_{4}^{3-}\) (orthovanadate ion), \(\mathrm{CrO}_{4}^{2-}(\) chromate ion \(),\) and \(\mathrm{MnO}_{4}^{-}\) (permanganate ion). (a) These anions are isoelectronic. What does this statement mean? (b) Would you expect these anions to exhibit \(d-d\) transitions? Explain. (c) As mentioned in "A Closer Look" on charge-transfer color, the violet color of \(\mathrm{MnO}_{4}^{-}\) is due to a ligand-to-metal charge transfer (LMCT) transition. What is meant by this term? (d) The LMCT transition in \(\mathrm{MnO}_{4}^{-}\) occurs at a wavelength of \(565 \mathrm{nm}\). The \(\mathrm{CrO}_{4}^{2-}\) ion is yellow. Is the wavelength of the LMCT transition for chromate larger or smaller than that for \(\mathrm{MnO}_{4}^{-}\) ? Explain. (e) The \(\mathrm{VO}_{4}{ }^{3-}\) ion is colorless. Do you expect the light absorbed by the LMCT to fall in the UV or the IR region of the electromagnetic spectrum? Explain your reasoning.

\(\begin{array}{lllll}\text { Sketch all the possible } & \text { stereoisomers } & \text { of }\end{array}\) (a) \(\left[\mathrm{Rh}(\text { bipy })(o \text { -phen })_{2}\right]^{3+}\), (b) \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{3}(\text { bipy }) \mathrm{Br}\right]^{2+},\) (c) square-planar \(\left[\mathrm{Pd}(\mathrm{en})(\mathrm{CN})_{2}\right]\).

Draw the crystal-field energy-level diagrams and show the placement of \(d\) electrons for each of the following: (a) \(\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) (four unpaired electrons), (b) \(\left[\mathrm{Mn}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) (high spin), (c) \(\left[\mathrm{Ru}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{H}_{2} \mathrm{O}\right]^{2+}\) (low spin), (d) \(\left[\operatorname{Ir} \mathrm{Cl}_{6}\right]^{2-}\) (low spin), (e) \(\left[\mathrm{Cr}(\mathrm{en})_{3}\right]^{3+},(\mathrm{f})\left[\mathrm{NiF}_{6}\right]^{4-}\)

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