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The elements \(\mathrm{Cu}, \mathrm{O}, \mathrm{La}, \mathrm{Y}, \mathrm{Ba}, \mathrm{Tl},\) and \(\mathrm{Bi}\) are all found in high-temperature ceramic superconductors. Write the expected electron configuration for these atoms.

Short Answer

Expert verified
The electron configurations for the given elements are: Cu: \([Ar] 3d^9 4s^2\) O: \([He] 2s^2 2p^4\) La: \([Xe] 4f^1 5d^1 6s^2\) Y: \([Kr] 4d^1 5s^2\) Ba: \([Xe] 6s^2\) Tl: \([Xe] 4f^14 5d^10 6s^2 6p^1\) Bi: \([Xe] 4f^14 5d^10 6s^2 6p^3\)

Step by step solution

01

Find the atomic numbers for each element

Use a periodic table to find the atomic number for each element: Cu (29), O (8), La (57), Y (39), Ba (56), Tl (81), and Bi (83).
02

Write electron configurations using the Aufbau Principle

Using the Aufbau Principle, fill the orbitals in the same order as they appear in the periodic table. The order of filling is: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p.
03

Write the electron configurations for each element

Cu (29): 1s虏, 2s虏, 2p鈦, 3s虏, 3p鈦, 4s虏, 3d鈦 \( \Rightarrow \) [Ar] 3d鈦4s虏 O (8): 1s虏, 2s虏, 2p鈦 \( \Rightarrow \) [He] 2s虏2p鈦 La (57): 1s虏, 2s虏, 2p鈦, 3s虏, 3p鈦, 4s虏, 3d鹿鈦, 4p鈦, 5s虏, 4d鹿鈦, 5p鈦, 6s虏, 4f鹿 \( \Rightarrow \) [Xe] 4f鹿5d鹿6s虏 Y (39): 1s虏, 2s虏, 2p鈦, 3s虏, 3p鈦, 4s虏, 3d鹿鈦, 4p鈦, 5s虏, 4d鹿 \( \Rightarrow \) [Kr] 4d鹿5s虏 Ba (56): 1s虏, 2s虏, 2p鈦, 3s虏, 3p鈦, 4s虏, 3d鹿鈦, 4p鈦, 5s虏, 4d鹿鈦, 5p鈦, 6s虏 \( \Rightarrow \) [Xe] 6s虏 Tl (81): 1s虏, 2s虏, 2p鈦, 3s虏, 3p鈦, 4s虏, 3d鹿鈦, 4p鈦, 5s虏, 4d鹿鈦, 5p鈦, 6s虏, 4f鹿鈦, 5d鹿鈦, 6p鹿 \( \Rightarrow \) [Xe] 4f鹿鈦5d鹿鈦6s虏6p鹿 Bi (83): 1s虏, 2s虏, 2p鈦, 3s虏, 3p鈦, 4s虏, 3d鹿鈦, 4p鈦, 5s虏, 4d鹿鈦, 5p鈦, 6s虏, 4f鹿鈦, 5d鹿鈦, 6p鲁 \( \Rightarrow \) [Xe] 4f鹿鈦5d鹿鈦6s虏6p鲁

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Key Concepts

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

Aufbau Principle
The Aufbau Principle is a fundamental guideline used in quantum chemistry and atomic physics to determine the electron configuration of an atom. The principle's name comes from the German word 'aufbauen', meaning 'to build up'. It suggests that as we move up in atomic number, electrons are added one by one to the lowest energy orbitals available before moving to higher energy levels.

According to the Aufbau Principle, electrons occupy orbitals starting from the lowest available (which is the 1s orbital) and progress to higher energy levels in a specific sequence. This sequence reflects the order in which energy levels and sublevels are filled, and it is a crucial aspect for students to grasp when writing electron configurations. A commonly used shorthand to remember this order is: 1s, 2s, 2p, 3s, and so on.
Periodic Table
The periodic table is much more than just a table of elements; it's a powerful tool that reflects the periodicity and trends in properties of the elements based on their atomic structure. It is organized by increasing atomic number (number of protons in the nucleus) and arranged in such a way that elements with similar chemical properties fall into the same columns, known as groups or families.

For students learning about electron configurations, the periodic table helps visualize the patterns in which electrons are distributed around the nucleus and provides a handy reference for determining the atomic number, which is crucial for writing out electron configurations. Elements are also divided into blocks corresponding to their outermost electron subshells, such as the 's-block', 'p-block', 'd-block', and 'f-block', which align with the typical orbital filling order.
Atomic Number
The atomic number plays a pivotal role in determining an element's position in the periodic table and its electron configuration. It is equal to the number of protons found in the nucleus of an atom and by extension鈥攖hanks to the charge-neutral condition of atoms鈥攊t also equals the number of electrons orbiting the nucleus in a neutral atom. This simple number defines the unique identity of each element.

Understanding the significance of the atomic number is crucial when it comes to determining the filling order of electrons in an element's orbitals. For example, hydrogen, with an atomic number of 1, has just one electron that will occupy the first available orbital (1s), while carbon, with an atomic number of 6, will have its electrons fill up to the 2p orbital.
Orbital Filling Order
Orbital filling order refers to the sequence in which orbitals are populated with electrons. The ordering is based on the principle of increasing energy; orbitals that are lower in energy are filled before those that are higher in energy. This hierarchy originates from the quantum mechanics governing electron behavior and the shapes of atom's probability distributions.

The filling order is generally: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, and so on. It's important to note the exception that occurs after the 4s subshell; the 3d comes into play before the 4p, even though 'd' would normally suggest a higher energy level. This order of filling is key in predicting and understanding the electron configuration of elements, especially in transition metals and inner transition metals where the 'd' and 'f' subshells are involved.

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Most popular questions from this chapter

Identify the following elements. a. An excited state of this element has the electron configuration \(1 s^{2} 2 s^{2} 2 p^{5} 3 s^{1}\). b. The ground-state electron configuration is \([\mathrm{Ne}] 3 s^{2} 3 p^{4}\). c. An excited state of this element has the electron configuration \([\mathrm{Kr}] 5 s^{2} 4 d^{6} 5 p^{2} 6 s^{1}\). d. The ground-state electron configuration contains three unpaired \(6 p\) electrons.

The Heisenberg uncertainty principle can be expressed in the form $$ \Delta E \cdot \Delta t \geqq \frac{h}{4 \pi} $$ where \(E\) represents energy and \(t\) represents time. Show that the units for this form are the same as the units for the form used in this chapter: $$ \Delta x \cdot \Delta(m v) \geq \frac{h}{4 \pi} $$

In the second row of the periodic table, \(\mathrm{Be}, \mathrm{N},\) and \(\mathrm{Ne}\) all have positive (unfavorable) electron affinities, whereas the other second-row elements have negative (favorable) electron affinities. Rationalize why Be, \(N,\) and Ne have unfavorable electron affinities.

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