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For which lanthanide are \(+2\) and \(+3\) oxidation states are common? (a) Eu (b) \(\mathrm{Na}\) (c) \(\mathrm{Ce}\) (d) \(\mathrm{La}\)

Short Answer

Expert verified
The lanthanide with common +2 and +3 oxidation states is Eu (Europium).

Step by step solution

01

Identify Lanthanides

First, identify which of the given options are lanthanides. Lanthanides are elements with atomic numbers from 57 to 71. In the list provided, Eu (Europium), Ce (Cerium), and La (Lanthanum) are lanthanides. Na (Sodium) is not a lanthanide.
02

Remember Common Oxidation States

Recall common oxidation states for the lanthanides. Most lanthanides typically exhibit a +3 oxidation state. However, a few also commonly exhibit a +2 oxidation state, such as Europium (Eu) and Ytterbium (Yb).
03

Analyze Option A

Examine option (a): Eu (Europium). Europium is known for having both +2 and +3 common oxidation states.
04

Analyze Other Options

Check options (c): Ce (Cerium) and (d): La (Lanthanum). While Cerium can have a +4 state, it usually shows +3; Lanthanum typically has a +3 oxidation state and does not commonly exhibit +2.
05

Select the Correct Answer

Based on the above analysis, the element that commonly exhibits both +2 and +3 oxidation states is Europium (Eu).

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Key Concepts

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

Europium oxidation states
Europium, a member of the lanthanide series, is well-known for its distinct oxidation states. Unlike most lanthanides that predominantly exhibit a +3 oxidation state, Europium is unique in that it commonly exhibits both +2 and +3 oxidation states. This characteristic is due to its electron configuration, which allows it to lose or gain electrons more flexibly compared to other lanthanides.
  • +2 Oxidation State: Europium's +2 state is relatively stable, making it more chemically reactive. This state arises when the europium atom loses two electrons, resulting in a stable electronic configuration. This stability is comparable to that of alkaline earth metals.

  • +3 Oxidation State: The +3 state is more typical among lanthanides. In this state, europium loses three electrons, contributing to its role in forming various compounds, such as europium oxide (Eu2O3).
Understanding these oxidation states of europium is crucial because they illustrate the element's versatility and its unique position among the lanthanide series.
lanthanide series
The lanthanide series, often referred to as the "rare earth elements," comprises 15 metallic chemical elements with atomic numbers from 57 to 71, starting from Lanthanum to Lutetium. These elements are grouped together because of their similar properties, such as their shiny appearance and high reactivity with oxygen and other non-metals.
  • Similar Properties: Lanthanides are known for their magnetic characteristics, high melting points, and the ability to conduct electricity. They are often used in manufacturing electronics and strong magnets.

  • Variable Oxidation States: Although the most stable oxidation state for these elements is +3, several lanthanides, such as Europium and Ytterbium, can also exhibit +2 oxidation states. This is due to the filling of f-orbitals, which allows for multiple stable configurations.
The lanthanide series is essential for industries focusing on advanced technological applications, making their study vital for understanding modern technology's foundation.
oxidation states in elements
Oxidation states, also known as oxidation numbers, represent the degree of oxidation of an atom in a chemical compound. They help chemists understand the electron distribution in compounds and predict the possible reactions that elements can undergo.
  • Definition: An oxidation state reflects the number of electrons an atom loses or gains compared to its neutral atom form. For instance, if an atom loses two electrons, it has a +2 oxidation state.

  • Application in Lanthanides: Typically, elements within the same group tend to have similar oxidation states. However, as seen in the lanthanide series, different elements can deviate. Europium's ability to show both +2 and +3 oxidation states is an example of this deviation and indicates its unique reactivity patterns.
Knowing how oxidation states work is instrumental in predicting chemical reactions, balancing equations, and understanding complex formation. It forms the backbone of inorganic chemistry, guiding the discovery and utility of new compounds.

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Most popular questions from this chapter

The IUPAC name of the coordination compound \(\mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]\) is (a) potassium hexacyanoferrate (III) (b) potassium hexacyanoferrate (II) (c) tripotassium hexacyaniron (II) (d) potassium hexacyanoiron (II)

Amongst the following, the lowest degree of paramagnetism per mole of the compound at \(298 \mathrm{~K}\) will be shown by (a) \(\mathrm{MnSO}_{4} \cdot 4 \mathrm{H}_{2} \mathrm{O}\) (b) \(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}\) (c) \(\mathrm{FeSO}_{4} \cdot 6 \mathrm{H}_{2} \mathrm{O}\) (d) \(\mathrm{NiSO}_{4} \cdot 6 \mathrm{H}_{2} \mathrm{O}\)

Iron is rendered passive by treatment with concentrated (a) \(\mathrm{HCl}\) (b) \(\mathrm{HNO}_{3}\) (c) \(\mathrm{H}_{3} \mathrm{PO}_{4}\) (d) \(\mathrm{H}_{2} \mathrm{SO}_{4}\)

The transition elements with some exceptions can show a large number of oxidation states. The various oxidation states are related to the electronic configuration of their atoms. The variable oxidation states of a transition metal is due to the involvement of \((\mathrm{n}-1) \mathrm{d}\) and outer \(\mathrm{ns}\)-electrons. For the first five elements of 3 d-transition series. The minimum oxidation state is equal to the number of electrons in 4s shell and the maximum oxidation state is equal to the sum of \(4 \mathrm{~s}\) and \(3 \mathrm{~d}\)-electrons. The relative stability of various oxidation state of a given element can be explained on the basis of stability of \(\mathrm{d}^{0}, \mathrm{~d}^{5}\) and \(\mathrm{d}^{10}\) configurations. Identify the correct statement (a) \(\mathrm{Ti}^{4+}, \mathrm{Mn}^{2+}\) are stable oxidation states (b) The most common oxidation state of 3 d-series is \(+2\) (c) The lowest oxidation state of \(\mathrm{Cr}\) and \(\mathrm{Cu}\) is \(+1\) while for others it is \(+2\). (d) All of these

Larger number of oxidation states are exhibited by the actinoids than those by the lanthanoids, the main reason being (a) 4 f-orbitals more diffused than the 5 f-orbitals (b) Lesser energy difference between \(5 \mathrm{f}\) and 6 d than between \(4 \mathrm{f}\) and \(5 \mathrm{~d}\) orbitals (c) More energy difference between \(5 \mathrm{f}\) and \(6 \mathrm{~d}\) than between \(4 \mathrm{f}\) and \(5 \mathrm{~d}\) orbitals (d) More reactive nature of the actinoids than the lanthanoids.

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