Problem 4 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

The electronic configurations of bivalent europium and trivalent cerium are: (atomic number : \(\mathrm{Xe}=54, \mathrm{Ce}=58, \mathrm{Eu}=63\) ) (a) \([\mathrm{Xe}] 4 f^{2}\) and \([\mathrm{Xe}] 4 f^{7}\) (b) \([\mathrm{Xe}] 4 f^{7}\) and \([\mathrm{Xe}] 4 f^{1}\) (c) \([\mathrm{Xe}] 4 f^{7} 6 \mathrm{~s}^{2}\) and \([\mathrm{Xe}] 4 f^{2} 6 \mathrm{~s}^{2}\) (d) \([\mathrm{Xe}] 4 f^{4}\) and \([\mathrm{Xe}] 4 f^{9}\)

Step-by-Step Solution

Answer

Option (b): \\([\mathrm{Xe}] 4f^7\\) and \\([\mathrm{Xe}] 4f^1\\).

Step 1: Identify Atomic Numbers

The atomic number of xenon (Xe) is 54, cerium (Ce) is 58, and europium (Eu) is 63.

Step 2: Understand Oxidation States

Europium commonly exists in a +2 oxidation state and cerium in a +3 oxidation state. This means europium loses 2 electrons and cerium loses 3 electrons from its neutral state.

Step 3: Determine Neutral Atom Configurations

The electronic configuration of cerium (Ce) in its neutral state is \([\mathrm{Xe}] 4f^1 5d^1 6s^2\), and the configuration of europium (Eu) is \([\mathrm{Xe}] 4f^7 6s^2\).

Step 4: Subtract Electrons for Ions

For europium, remove 2 electrons: the loss typically occurs from the 6s orbital first. The configuration becomes \([\mathrm{Xe}] 4f^7\). For cerium, remove 3 electrons: this typically involves the 6s and 5d orbitals, leading to the configuration \([\mathrm{Xe}] 4f^1\).

Step 5: Match to Options

Compare the modified configurations to the options given. For europium in the +2 state, it's \([\mathrm{Xe}] 4f^7\), and for cerium in the +3 state, it's \([\mathrm{Xe}] 4f^1\). These match option (b): \([\mathrm{Xe}] 4f^7\) and \([\mathrm{Xe}] 4f^1\).

Key Concepts

Oxidation States

Oxidation states are a crucial concept in chemistry, especially when discussing electron configurations. They refer to the charge of an atom when it has either gained, lost, or shared electrons in a chemical reaction. This change is what defines the atom's oxidation number, which can be positive, negative, or zero.

For example, europium typically has an oxidation state of +2, while cerium often displays a +3 oxidation state. This implies that europium loses two electrons, and cerium loses three when they form compounds. Understanding oxidation states helps predict how an element will react and what kind of compounds it will form.

Here's a simple way to remember the oxidation state changes:

• Positive oxidation states (like +2 or +3) mean electrons are lost.
• Negative oxidation states would mean electrons are gained, but this is less common for elements like europium and cerium.
• Each loss or gain of an electron alters the element's chemical behavior significantly.

Lanthanides

Lanthanides are a series of elements in the periodic table from atomic numbers 57 (lanthanum) to 71 (lutetium). These elements are also known as rare earth metals. They are f-block elements, meaning their electrons fill the 4f sub-level. This unique electron configuration gives lanthanides their characteristic properties.

Europium and cerium, both discussed in the exercise, are part of the lanthanide series. Lanthanides are known for their similar chemical properties, such as high magnetic susceptibility, which results from their unpaired 4f electrons. They are used widely in technology and industry for making powerful magnets, batteries, and phosphors for screens.

Key points about lanthanides include:

• They are known for their shiny and silvery appearance.
• Largely, they remain stable with high atomic and ionic sizes.
• Their chemistry is dominated by the stability of trivalent states (like Ce³⁺), with a few exceptions like europium showing a +2 state.

Understanding lanthanides' unique traits helps in studying their role in modern applications.

Electron Subtraction

Electron subtraction refers to the process of removing electrons from an atom, which is critical in determining the electronic configurations of ions. When electrons are removed, the atom forms a positive ion (cation), changing the neutral electron configuration to reflect the electrons lost during ionization.

In the exercise, europium and cerium undergo electron subtraction. For europium, we start with the neutral state \( [\mathrm{Xe}] 4f^7 6s^2 \). By losing two electrons, it adjusts to \( [\mathrm{Xe}] 4f^7 \). Cerium's electronic configuration \( [\mathrm{Xe}] 4f^1 5d^1 6s^2 \) changes to \( [\mathrm{Xe}] 4f^1 \) after losing three electrons.

Several factors influence electron subtraction:

• The electrons are usually removed from the highest energy level first (like 6s in these cases).
• Elements form ions by reaching a more stable configuration, often resembling the nearest noble gas.
• This process impacts physical and chemical properties, such as ionic size and reactivity.

Understanding this concept helps clarify why certain configurations are more stable and the nature of oxidation states.