/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 29 A recently named element is darm... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A recently named element is darmstadtium (Ds), which has 110 electrons. Assume that you can put the 110 electrons into the atomic shells one by one and can neglect any electron-electron interaction. With the atom in ground state, what is the spectroscopic notation for the quantum number \(\ell\) for the last electron?

Short Answer

Expert verified
The quantum number \(\ell\) for the last electron in darmstadtium is \(\ell = 2\).

Step by step solution

01

Identify Electron Configuration

Darmstadtium (Ds) has an atomic number of 110, which means it has 110 electrons. The electrons fill up the atomic orbitals according to the Aufbau principle, starting from the lowest energy level and moving upwards.
02

Determine Orbital Order

The order in which orbitals are filled can be listed as follows: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p. We will fill these orbitals sequentially until all 110 electrons are placed.
03

Assign Electrons to Orbitals

Fill the orbitals using the order and maximum capacity: - 1s: 2 electrons - 2s: 2 electrons - 2p: 6 electrons - 3s: 2 electrons - 3p: 6 electrons - 4s: 2 electrons - 3d: 10 electrons - 4p: 6 electrons - 5s: 2 electrons - 4d: 10 electrons - 5p: 6 electrons - 6s: 2 electrons - 4f: 14 electrons - 5d: 10 electrons - 6p: 6 electrons - 7s: 2 electrons - 5f: 14 electrons - 6d: 10 electrons (only 8 needed to reach 110) The electron configuration for darmstadtium ends at 6d with 8 electrons.
04

Determine Quantum Number \\ell

The 6d orbital corresponds to the last electron. In quantum mechanics, the azimuthal quantum number, denoted by \ell, defines the shape of the orbital and is linked to the subshell:- s: \(\ell = 0\)- p: \(\ell = 1\)- d: \(\ell = 2\)- f: \(\ell = 3\)For the d subshell, \ell = 2.
05

Conclusion

The spectroscopic notation for the quantum number \ell for the last electron in darmstadtium, which is in the 6d subshell, is \(\ell = 2\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Quantum Numbers
Quantum numbers are like an address system for electrons in an atom. They tell us where an electron can be found around the nucleus. There are four quantum numbers, each with a special role:
  • **The Principal Quantum Number** (\( n \)): Tells us the energy level or shell of an electron. Higher values mean the electron is further from the nucleus.
  • **The Azimuthal Quantum Number** (\( \ell \)): Indicates the shape of the orbital. It is linked to the subshells: s, p, d, and f.
  • **The Magnetic Quantum Number** (\( m_\ell \)): Describes the orientation of the orbital in space.
  • **The Spin Quantum Number** (\( m_s \)): Tells the spin of the electron, which can be +1/2 or -1/2.
In the case of darmstadtium, the azimuthal quantum number is significant. For its last electron in the 6d subshell, \( \ell \) becomes 2, which confirms the orbital shape. Understanding these numbers helps you predict how electrons fill different orbitals depending on their energy levels, shapes, and orientations within the atom.
Aufbau Principle
The Aufbau principle is a guide that helps us figure out the arrangement of electrons in an atom's orbitals. In German, "Aufbau" means "building up" or "to construct." This principle suggests that electrons occupy orbitals starting from the lowest energy level before filling higher energy ones.
Consider this principle like a step-by-step manual:
  • **Start with the lowest energy orbitals**: Fill the 1s orbital first, followed by 2s, then 2p, and so on.
  • **Fill orbitals in a sequence**: Continue filling according to the increasing energy level order.
  • **Respect the maximum capacity**: Each orbital has a specific maximum number of electrons it can hold, like 2 for s, 6 for p, 10 for d, and 14 for f.
For darmstadtium, the Aufbau principle determines that the electrons up to the 6d subshell must be filled sequentially, ensuring the proper distribution of all 110 electrons across available energy levels. This process not only illustrates electron configuration but also emphasizes the energy efficiency maintained in atomic structure.
Darmstadtium
Darmstadtium is a chemically intriguing element, designated by the symbol Ds, and holds atomic number 110. It was first created in the lab in Darmstadt, Germany, which is where it gains its name. As a heavy, synthetic element, it doesn't occur naturally on Earth, but its production in particle accelerators has enabled us to study its properties. Some fascinating tidbits about darmstadtium include:
  • **Highly unstable**: It has a very short half-life, making it challenging to study.
  • **Electron configuration**: Due to its high atomic number, filling its electron orbitals follows the complex pathway up to 6d.
  • **Applications**: Has mostly scientific applications since its instability limits practical uses.
By learning about such elements, chemists continue to unravel the mysteries of atomic structure and behavior, expanding our knowledge of the material world, one electron at a time.
Atomic Orbitals
Atomic orbitals are regions around the nucleus of an atom where electrons are most likely to be found. Each orbital is defined by a set of quantum numbers and has a specific shape, based on its\( \ell \) value:
  • **s orbitals**: Spherical in shape, with\( \ell = 0 \)
  • **p orbitals**: Dumbbell-shaped, having\( \ell = 1 \)
  • **d orbitals**: Complex, having a cloverleaf shape or other unique structures, characterized by\( \ell = 2 \)
  • **f orbitals**: Even more complex shapes, with\( \ell = 3 \)
In darmstadtium, by the time you're adding electrons to the 6d subshell, you're dealing with d orbitals, which are distinct due to their multifaceted shapes. These orbitals illustrate the rich complexity of atomic structures at higher atomic numbers. Understanding how orbitals work highlights not only the beautiful geometry of an atom but also the principles that govern electron arrangement and chemical reactivity. This comprehension is key to mastering the chemistry of elements, especially those further along in the periodic table.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Ruby lases at a wavelength of \(694 \mathrm{nm}\). A certain ruby crystal has \(4.00 \times 10^{19} \mathrm{Cr}\) ions (which are the atoms that lase). The lasing transition is between the first excited state and the ground state, and the output is a light pulse lasting \(2.00 \mu \mathrm{s}\). As the pulse begins, \(60.0 \%\) of the \(\mathrm{Cr}\) ions are in the first excited state and the rest are in the ground state. What is the average power emitted during the pulse? (Hint: Don't just ignore the ground-state ions.)

If orbital angular momentum \(\vec{L}\) is measured along, say, a \(z\) axis to obtain a value for \(L_{z}\), show that $$\left(L_{x}^{2}+L_{y}^{2}\right)^{1 / 2}=\left[\ell(\ell+1)-m_{\ell}^{2}\right]^{1 / 2} h$$ is the most that can be said about the other two components of the orbital angular momentum.

Calculate the (a) smaller and (b) larger value of the semiclassical angle between the electron spin angular momentum vector and the magnetic field in a Stern-Gerlach experiment. Bear in mind that the orbital angular momentum of the valence electron in the silver atom is zero.

A molybdenum \((Z=42)\) target is bombarded with \(35.0 \mathrm{keV}\) electrons and the \(x\) -ray spectrum of Fig. \(40-13\) results. The \(K_{\beta}\) and \(K_{a}\) wavelengths are 63.0 and \(71.0 \mathrm{pm},\) respectively. What photon energy corresponds to the (a) \(K_{\beta}\) and (b) \(K_{a}\) radiation? The two radiations are to be filtered through one of the substances in the following table such that the substance absorbs the \(K_{\beta}\) line more strongly than the \(K_{a}\) line. A substance will absorb radiation \(x_{1}\) more strongly than it absorbs radiation \(x_{2}\) if a photon of \(x_{1}\) has enough energy to eject a \(K\) electron from an atom of the substance but a photon of \(x_{2}\) does not. The table gives the ionization energy of the \(K\) electron in molybdenum and four other substances. Which substance in the table will serve (c) best and (d) second best as the filter? $$\begin{array}{llllll}\hline & \mathrm{Zr} & \mathrm{Nb} & \mathrm{Mo} & \mathrm{Tc} & \mathrm{Ru} \\\\\hline Z & 40 & 40 & 42 & 43 & 44 \\ E_{K}(\mathrm{keV}) & 18.00 & 18.99 & 20.00 & 21.04 & 22.12 \\\\\hline\end{array}$$

In the subshell \(\ell=3,\) (a) what is the greatest (most positive) \(m_{f}\) value, (b) how many states are available with the greatest \(m_{\ell}\) value, and (c) what is the total number of states available in the subshell?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.