Write the expected electron configurations for each of the following atoms: Cl, Sb, Sr, W, Pb, Cf.

Short Answer

Expert verified
The electron configurations for each atom are: 1. Cl: \(1s^2 2s^2 2p^6 3s^2 3p^5\) 2. Sb: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^3\) 3. Sr: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10}\) 4. W: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6 6s^2 4f^{14} 5d^4\) 5. Pb: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6 6s^2 4f^{14} 5d^{10} 6p^2\) 6. Cf: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6 6s^2 4f^{14} 5d^{10} 6p^6 7s^2 5f^{10}\)

Step by step solution

01

Identify the atomic numbers for each element

Use the periodic table to find the atomic number of each element, which represents the number of electrons in each atom. - Cl: 17 - Sb: 51 - Sr: 38 - W: 74 - Pb: 82 - Cf: 98
02

Write the electron configurations

By following the aufbau principle, we will write the electron configuration for each atom: 1. Chlorine (Cl, atomic number 17) \[1s^2 2s^2 2p^6 3s^2 3p^5\] 2. Antimony (Sb, atomic number 51) \[1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^3\] 3. Strontium (Sr, atomic number 38) \[1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10}\] 4. Tungsten (W, atomic number 74) \[1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6 6s^2 4f^{14} 5d^4\] 5. Lead (Pb, atomic number 82) \[1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6 6s^2 4f^{14} 5d^{10} 6p^2\] 6. Californium (Cf, atomic number 98) \[1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6 6s^2 4f^{14} 5d^{10} 6p^6 7s^2 5f^{10}\]

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 Vaia!

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

In each of the following sets, which atom or ion has the smallest ionization energy? a. \(\mathrm{Ca}, \mathrm{Sr}, \mathrm{Ba}\) b. \(\mathrm{K}, \mathrm{Mn}, \mathrm{Ga}\) c. \(\mathrm{N}, \mathrm{O}, \mathrm{F}\) d. \(\mathrm{S}^{2-}, \mathrm{S}, \mathrm{S}^{2+}\) e. \(\mathrm{Cs}, \mathrm{Ge}, \mathrm{Ar}\)

An ionic compound of potassium and oxygen has the empirical formula KO. Would you expect this compound to be potassium(II) oxide or potassium peroxide? Explain.

It takes 208.4 kJ of energy to remove 1 mole of electrons from an atom on the surface of rubidium metal. How much energy does it take to remove a single electron from an atom on the surface of solid rubidium? What is the maximum wavelength of light capable of doing this?

One bit of evidence that the quantum mechanical model is “correct” lies in the magnetic properties of matter. Atoms with unpaired electrons are attracted by magnetic fields and thus are said to exhibit paramagnetism. The degree to which this effect is observed is directly related to the number of unpaired electrons present in the atom. Consider the ground-state electron configurations for Li, N, Ni, Te, Ba, and Hg. Which of these atoms would be expected to be paramagnetic, and how many unpaired electrons are present in each paramagnetic atom?

In the ground state of element 115, Uup, a. how many electrons have \(n=5\) as one of their quantum numbers? b. how many electrons have \(\ell=3\) as one of their quantum numbers? c. how many electrons have \(m_{\ell}=1\) as one of their quantum numbers? d. how many electrons have \(m_{s}=-\frac{1}{2}\) as one of their quantum numbers?

See all solutions

Recommended explanations on Chemistry 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.

Sign-up for free