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Describe how the transition elements compare with main-group metals (such as group 2 ) with respect to oxidation states, formation of complexes, colors of compounds, and magnetic properties.

Short Answer

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The main differences are that transition elements often have multiple oxidation states due to the incomplete filling of d-orbitals, form a large variety of complex compounds, have colored compounds due to the presence of unpaired d electrons, and are paramagnetic because of these unpaired electrons. On the other hand, group 2 metals usually display an oxidation state of +2, do not form a wide range of complexes, have colorless compounds, and are diamagnetic.

Step by step solution

01

Comparison of Oxidation states

Transition elements often have a larger variety of oxidation states. This is because they have incompletely filled d-orbitals; thus, the electrons in both the s and d orbitals can be involved in bond formation. The oxidation states can vary from +1 to the maximum oxidation state which is equal to the total number of unpaired electrons in the s and d orbitals of their atoms. Group 2 main group metals (alkaline earth metals) can only display an oxidation state of +2 due to the two valence electrons in the s orbital.
02

Comparison of Complex formation

Transition metals form a wide range of complex compounds due to their small sizes, high ionic charges, and the availability of vacant d orbital where they can accept a pair of electrons from the ligand. The formation of the complex can lead to the stabilization of the metal ion in various oxidation states. On the contrary, Group 2 metals do not form a wide range of complexes.
03

Comparison of Colors of the Compounds

Transition element compounds are often colored due to the presence of unpaired d electrons. The absorption of certain wavelengths of light by these unpaired d electrons, countries the complementary color. Group 2 metals typically form colorless compounds since they do not have any unpaired d electrons.
04

Comparison of Magnetic Properties

Transition metals and their compounds show paramagnetism due to the presence of unpaired d electrons. The magnetic moment depends on the number of unpaired electrons. In contrast, Group 2 elements and their compounds are usually diamagnetic, since their atoms have all paired electrons.

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Key Concepts

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

Oxidation States
In the world of chemistry, oxidation states play a crucial role in understanding the behavior of elements, particularly metals. Transition elements display a remarkable range of possible oxidation states that relate directly to their electron configurations. Unlike main-group metals, such as the alkaline earth metals in Group 2, transition metals have partially filled d-orbitals. This allows electrons from both the s- and d-orbitals to participate in bonding, resulting in various oxidation states.

Take, for instance, iron. It can exist in the +2 or +3 oxidation state, among others, which plays a vital part in processes like oxygen transport in hemoglobin. On the other side, a Group 2 metal like calcium reliably exhibits a +2 oxidation state, given its full s-orbital and lack of d-electrons for additional variability. This distinct difference defines the broad versatility seen in transition elements compared to their main-group counterparts.
Complex Formation
Complex formation is akin to a dance between metal ions and ligands, entities that can donate a pair of electrons. Transition metals, with their intricate electronic arrangements, are like versatile dancers capable of performing complex moves. Their small size, high charge, and especially the presence of vacant d-orbitals allow them to accept electrons from a variety of ligands, forming complexes.

Imagine a metal ion at the center, surrounded by a coordinated group of molecules or ions—these are its dance partners that define the structure and stability of the resultant complex. Conversely, main-group metals like those in Group 2 usually do not indulge in such extensive complex formation. Without accessible d-orbitals, they are the wallflowers of the dance, forming fewer and simpler complexes.
Compound Colors
The vibrant palette of transition element compounds can be attributed to the electronic transitions that occur within the d-orbitals. This phenomenon occurs when light is absorbed, causing an electron to jump to a higher energy state. Each specific jump corresponds to a certain color in the visible spectrum. Thus, when an element like copper forms a compound, it might exude a rich blue or green hue.

In stark contrast, Group 2 metals, such as magnesium, lead to colorless compounds. The absence of d-electrons equates to no exciting electron jumps within the visible spectrum, which means there's no distinct color to witness. This distinction in compound colors is a direct result of the differences in electronic configuration between transition elements and main-group metals.
Magnetic Properties
The presence or absence of unpaired electrons determines the magnetic nature of an element. Transition metals often have unpaired d electrons, which give rise to paramagnetism. This means that in the presence of a magnetic field, the compounds will be weakly attracted. Such magnetic sensitivity varies with the number of unpaired electrons – the more there are, the stronger the attraction.

For example, iron exhibits strong magnetic properties because of its unpaired electrons. Meanwhile, the alkaline earth metals, residing in Group 2, usually have paired electrons and exhibit diamagnetism. Their compounds, therefore, are typically not attracted to magnets. These contrasting magnetic properties highlight another fundamental difference between transition elements and main-group metals.

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Most popular questions from this chapter

In the metallurgical extraction of silver and gold, an alloy of the two metals is often obtained. The alloy can be separated into Ag and Au either with concentrated \(\mathrm{HNO}_{3}\) or boiling concentrated \(\mathrm{H}_{2} \mathrm{SO}_{4},\) in a process called parting. Write chemical equations to show how these separations work.

You are given these three reducing agents: \(\mathrm{Zn}(\mathrm{s})\) \(\mathrm{Sn}^{2+}(\mathrm{aq}),\) and \(\mathrm{I}^{-}(\mathrm{aq}) .\) Use data from Appendix \(\mathrm{D}\) to determine which of them can, under standard-state conditions in acidic solution, reduce (a) \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(\mathrm{aq})\) to \(\mathrm{Cr}^{3+}(\mathrm{aq})\) (b) \(\mathrm{Cr}^{3+}(\text { aq })\) to \(\mathrm{Cr}^{2+}(\mathrm{aq})\) (c) \(\mathrm{SO}_{4}^{2-}(\text { aq })\) to \(\mathrm{SO}_{2}(\mathrm{g})\)

Nitinol is a nickel-titanium alloy known as memory metal. The name nitinol is derived from the symbols for nickel (Ni), titanium (Ti), and the acronym for the Naval Ordinance Laboratory (NOL), where it was discovered. If an object made out of nitinol is heated to about \(500^{\circ} \mathrm{C}\) for about an hour and then allowed to cool, the original shape of the object is "remembered," even if the object is deformed into a different shape. The original shape can be restored by heating the metal. Because of this property, nitinol has found many uses, especially in medicine and orthodontics (for braces). Nitinol exists in a number of different solid phases. In the so- called austerite phase, the metal is relatively soft and elastic. The crystal structure for the austerite phase can be described as a simple cubic lattice of Ti atoms with Ni atoms occupying cubic holes in the lattice of Ti atoms. What is the empirical formula of nitinol and what is the percent by mass of titanium in the alloy?

By means of orbital diagrams, write electron configurations for the following transition element atom and ions: \((a) \mathrm{Ti} ;(\mathbf{b}) \mathrm{V}^{3+} ;(\mathrm{c}) \mathrm{Cr}^{2+} ;(\mathrm{d}) \mathrm{Mn}^{4+} ;(\mathrm{e}) \mathrm{Mn}^{2+} ;(\mathrm{f}) \mathrm{Fe}^{3+}\).

Nickel can be determined as nickel dimethylglyoximate, a brilliant scarlet precipitate that has the composition \(20.31 \% \mathrm{Ni}, 33.26 \% \mathrm{C}, 4.88 \% \mathrm{H}, 22.15 \% \mathrm{O},\) and \(19.39 \%\) N. \(\mathrm{A} 15.020 \mathrm{g}\) steel sample is dissolved in concentrated HCl(aq). The solution obtained is suitably treated to remove interfering ions, to establish the proper \(\mathrm{pH},\) and to obtain a final solution volume of \(250.0 \mathrm{mL} .\) A \(10.00 \mathrm{mL}\) sample of this solution is then treated with dimethylglyoxime. The mass of purified, dry nickel dimethylglyoximate obtained is \(0.104 \mathrm{g}\) (a) What is the empirical formula of nickel dimethylglyoximate? (b) What is the mass percent nickel in the steel sample?

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