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Identify the following three elements. a. The ground-state electron configuration is \([\mathrm{Kr}] 5 s^{2} 4 d^{10} 5 p^{4}\) b. The ground-state electron configuration is \([\mathrm{Ar}] 4 s^{2} 3 d^{10} 4 p^{2}\) c. An excited state of this element has the electron configuration \(1 s^{2} 2 s^{2} 2 p^{4} 3 s^{1}\)

Short Answer

Expert verified
a. Tellurium (Te) b. Germanium (Ge) c. Fluorine (F) in an excited state

Step by step solution

01

Determine the number of electrons for each configuration

: To determine the element, we first need to find the total number of electrons (which will be equal to the atomic number) using the electron configurations provided. For configuration a, [\(\mathrm{Kr}\) 5s虏 4d鹿鈦 5p鈦碷, count the superscript numbers in the configuration: 2 (from 5s虏) + 10 (from 4d鹿鈦) + 4 (from 5p鈦) = 16 electrons For configuration b, [\(\mathrm{Ar}\) 4s虏 3d鹿鈦 4p虏], count the superscript numbers in the configuration: 2 (from 4s虏) + 10 (from 3d鹿鈦) + 2 (from 4p虏) = 14 electrons For configuration c (excited state), count the superscript numbers in the configuration: 2 (from 1s虏) + 2 (from 2s虏) + 4 (from 2p鈦) + 1 (from 3s鹿) = 9 electrons
02

Determine the atomic numbers

Next, we will add the atomic numbers of the noble gases in the given electron configurations to the number of electrons: For configuration a: Kr has an atomic number of 36, so: 36 (Kr) + 16 (additional electrons) = 52 For configuration b: Ar has an atomic number of 18, so: 18 (Ar) + 14 (additional electrons) = 32 For configuration c (excited state): Since this configuration does not have a noble gas in it, the total number of electrons directly determines the atomic number: 9
03

Find the elements in the periodic table

Using the atomic numbers found in step 2, we can now identify the elements: For a: An element with an atomic number of 52 is Tellurium (Te). For b: An element with an atomic number of 32 is Germanium (Ge). For c (excited state): An element with an atomic number of 9 is Fluorine (F). So, the elements are: a. Tellurium (Te) b. Germanium (Ge) c. Fluorine (F) in an excited state

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Key Concepts

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

Ground-State Electron Configuration
Understanding the ground-state electron configuration is essential for grasping the basics of atomic structure and chemistry. This configuration is the arrangement of electrons around the nucleus of an atom in the lowest energy state, which is the most stable form. Electrons fill subshells in an order that minimizes the energy of the atom.

The Aufbau principle guides the order in which subshells are filled, governed by increasing energy levels. According to this principle, electrons occupy the orbitals starting from the lowest energy level, moving to higher levels only after the lower levels are full. The electron configuration for an element provides a roadmap showing which orbitals the electrons occupy for a given element in its ground state.

For example, the electron configuration \( [\mathrm{Kr}] 5s^{2} 4d^{10} 5p^{4} \) corresponds to the element in the first exercise example (tellurium), which follows the noble gas (\( \mathrm{Kr} \), krypton) notation.
  • \( 5s^{2} \) indicates there are two electrons in the 5s subshell,
  • \( 4d^{10} \) shows a completely filled 4d subshell with ten electrons, and
  • \( 5p^{4} \) has four electrons in the 5p subshell.
By knowing these configurations and following the rules, such as Hund's rule and the Pauli exclusion principle, you can determine the electronic structure of an atom.
Atomic Number
The atomic number is a fundamental property of an element, signifying the number of protons found in the nucleus of an atom. Not only does it define the identity of the element, but it also indicates the number of electrons in a neutral atom, as the number of protons and electrons are equal in a neutral atom to ensure electrical neutrality.

This property is so fundamental that the elements in the periodic table are arranged in order of increasing atomic number. In the example exercises, the atomic number is crucial in identifying the elements. It's the sum of the noble gas's atomic number and the number of additional electrons as indicated by the electron configuration. For instance, for krypton (Kr) with an atomic number of 36 plus 16 additional electrons, the atomic number for tellurium (Te) is 52.

Knowing the atomic number, one can reference the periodic table to determine the element, as done in the solution process. Each element in the periodic table has a unique atomic number, which also informs us about the chemical properties and the placement of elements in different groups and periods of the periodic table.
Periodic Table Elements
The periodic table is the tabular arrangement of all chemical elements, ordered by their atomic number, electron configurations, and recurring chemical properties. Elements are grouped into rows called periods and columns known as groups, with metals placed on the left and non-metals on the right.

The periodic table is an invaluable tool for chemists. It not only allows scientists and students to quickly identify elements like Tellurium (Te), Germanium (Ge), and Fluorine (F), as seen in the exercise, but it also enables predictions about the types of chemical reactions that elements are likely to undergo.
  • Periods correlate with the highest energy level of electrons in an atom,
  • While groups correlate with the number of valence electrons, which in turn affects the element鈥檚 reactivity.
Thus, elements within the same group typically exhibit similar properties such as valency and electronegativity. By referring to the periodic table and understanding the significance of the arrangement, students can effectively engage with exercises based on electron configuration and the properties associated with various elements.

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Most popular questions from this chapter

Give the electron configurations for the following atoms. Do not use the noble gas notation. Write out the complete electron configuration.

Compare the ionization energies of each pair of atoms. State the atom with the larger ionization energy for each pair.

Without referring to your textbook or a periodic table, write the full electron configuration, the orbital box diagram, and the noble gas shorthand configuration for the elements with the following atomic numbers. a. \(Z=19\) b. \(Z=22\) c. \(Z=14\) d. \(Z=26\) e. \(Z=30\)

Arrange the following atoms in order of increasing size (assuming all atoms are in their ground states). a. \([\mathrm{Kr}] 5 s^{2} 4 d^{10} 5 p^{6}\) b. \([\mathrm{Kr}] 5 s^{2} 4 d^{10} 5 p^{1}\) c. \([\mathrm{Kr}] 5 s^{2} 4 d^{10} 5 p^{3}\)

In the text (Section 11.6 ) it was mentioned that current theories of atomic structure suggest that all matter and all energy demonstrate both particle- like and wave-like properties under the appropriate conditions, although the wave-like nature of matter becomes apparent only in very small and very fast- moving particles. The relationship between wavelength \((\lambda)\) observed for a particle and the mass and velocity of that particle is called the de Broglie relationship. It is $$ \lambda=h / m v $$ in which \(h\) is Planck's constant \(\left(6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}\right), * m\) represents the mass of the particle in kilograms, and \(v\) represents the velocity of the particle in meters per second. Calculate the "de Broglie wavelength" for each of the following, and use your numerical answers to explain why macroscopic (large) objects are not ordinarily discussed in terms of their "wave-like" properties. a. an electron moving at 0.90 times the speed of light b. a \(150-\mathrm{g}\) ball moving at a speed of \(10 . \mathrm{m} / \mathrm{s}\) c. a 75 -kg person walking at a speed of \(2.0 \mathrm{~km} / \mathrm{h}\)

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