Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Identify the group of elements that corresponds to each of the following generalized electron configurations and indicate the number of unpaired electrons for each: (a) \([\) noble gas \(] n s^{2} n p^{5}\) (b) \([\) noble gas \(] n s^{2}(n-1) d^{2}\) (c) \([\) noble gas \(] n s^{2}(n-1) d^{10} n p^{1}\) (d) \([\) noble gas \(] n s^{2}(n-2) f^{6}\)

Short Answer

Expert verified
The elements corresponding to the given generalized electron configurations belong to the following groups in the periodic table, and the number of unpaired electrons is: (a) Group 17 (Halogens) with 1 unpaired electron. (b) Group 4 (Transition metals) with 2 unpaired electrons. (c) Group 13 with 1 unpaired electron. (d) Group 2 with 6 unpaired electrons.

Step by step solution

01

Understanding generalized electron configurations notation

In generalized electron configurations, the notation \([\)noble gas\(]\) indicates the electron configuration of the preceding noble gas element in the periodic table. \(n\) represents the principal quantum number of the current energy level and the superscripts represent the number of electrons in the given sub-levels (e.g., s, p, d, or f).
02

Identify the group and unpaired electrons for (a) #[noble gas] ns^{2} np^{5}

For the configuration \([\)noble gas\(] ns^{2} np^{5}\), the element will have 2 electrons in the s sub-level and 5 electrons in the p sub-level of the valence shell (with principal quantum number n). To belong to a particular group in the periodic table, the total number of electrons in the valence shell is important. In this case, the element will have 2 + 5 = 7 valence electrons, which means that it belongs to Group 17 (also known as Halogens). In the p sub-level (which has 3 orbitals), there will be two completely filled orbitals and one orbital with a single electron. Therefore, there is 1 unpaired electron.
03

Identify the group and unpaired electrons for (b) #[noble gas] ns^{2}(n-1) d^{2}

For the configuration \([\)noble gas\(] ns^{2}(n-1) d^{2}\), the element will have 2 electrons in the s sub-level and 2 electrons in the d sub-level of the previous energy level (with principal quantum number n-1). The element will have 2 valence electrons in the s sub-level (principal quantum number n) and 2 electrons in the d sub-level of the previous principal quantum number (n-1) energy level. This means that the element belongs to Group 4 (transition metals group). In the d sub-level (which has 5 orbitals), there will be two singly occupied orbitals, and the other orbitals will be empty. Therefore, there are 2 unpaired electrons.
04

Identify the group and unpaired electrons for (c) #[noble gas] ns^{2}(n-1) d^{10} np^{1}

For the configuration \([\)noble gas\(] ns^{2}(n-1) d^{10} np^{1}\), the element will have 2 electrons in the s sub-level, 10 electrons in the d sub-level of the previous energy level (with principal quantum number n-1), and 1 electron in the p sub-level of the valence shell (principal quantum number n). The total valence electrons for this element are 2 (from the s sub-level) and 1 (from the p sub-level), which gives a total of 3 valence electrons. Therefore, the element belongs to Group 13. As all the orbitals in the d sub-level are completely filled, there are no unpaired electrons in that sub-level. The p sub-level has one orbital with a single electron, giving 1 unpaired electron.
05

Identify the group and unpaired electrons for (d) #[noble gas] ns^{2}(n-2) f^{6}

For the configuration \([\)noble gas\(] ns^{2}(n-2) f^{6}\), the element will have 2 electrons in the s sub-level of the valence shell (principal quantum number n) and 6 electrons in the f sub-level of the energy level with principal quantum number (n-2). As the electrons in the f sub-level are not considered valence electrons, the element will have only 2 valence electrons (from the s sub-level). Therefore, the element belongs to Group 2. In the f sub-level (which has 7 orbitals), there will be 6 singly occupied orbitals, and the remaining one will be empty. Therefore, there are 6 unpaired electrons.

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!

Key Concepts

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

Periodic Table Groups
Elements in the periodic table are organized into groups. A group, also known as a column, is composed of elements that share the same number of valence electrons and exhibit similar chemical properties.
Based on these shared characteristics, elements within the same group often behave similarly in chemical reactions. Understanding these groups is key for predicting the chemical behavior of an element.
  • Group 17: Known as the Halogens, these elements have 7 electrons in their valence shell, characterized by the electron configuration ewline \([ ext{noble gas}] ns^{2} np^{5}\).
  • Group 4: Part of the transition metals, these elements typically have 4 electrons in the combined s and d sub-levels, such as \([ ext{noble gas}] ns^{2}(n-1)d^2\). This influences their ability to form different oxidation states.
  • Group 13: Elements here often have 3 valence electrons, as shown by configurations like \([ ext{noble gas}] ns^{2}(n-1)d^{10} np^1\).
  • Group 2: Representatives of alkaline earth metals, members of this group have 2 valence electrons with configurations like \([ ext{noble gas}] ns^{2}(n-2)f^6\).
Unpaired Electrons
Unpaired electrons are those electrons which do not have a partner in the same atomic orbital. They greatly influence an element's magnetic properties and chemical reactivity, as unpaired electrons can participate in bonding.
Understanding how to identify unpaired electrons can give invaluable insight into the magnetic nature of the element, which is crucial in areas like material science and chemistry.
  • Example Configuration: For an element with the electron configuration \([ ext{noble gas}] ns^{2} np^{5}\), there is 1 unpaired electron in the np sub-level.
  • d-Orbital Unpaired Electrons: Configurations such as \([ ext{noble gas}] ns^{2}(n-1)d^{2}\) show 2 unpaired electrons in the \(d\)-orbital due to scatter over 5 orbitals.
  • f-Orbital Complexity: The f sub-level can accommodate 14 electrons; when it contains only 6 electrons, like in \([ ext{noble gas}] ns^{2}(n-2)f^6\), expect 6 unpaired electrons.
Valence Shell
The valence shell refers to the outermost shell of an atom that contains electrons. It plays a significant role in determining chemical bonding and reactivity. The electrons in this shell are involved in forming bonds with other atoms.
Knowledge of the valence shell is essential in predicting how atoms will interact, form compounds, and determine the element's physical properties.
  • Occupation of the Orbitals: For Halogens, the general configuration \([ ext{noble gas}] ns^{2} np^{5}\) shows 7 valence electrons, reflecting strong reactive behavior as they seek one more electron to complete their octet.
  • Transition Metals: Although transition metals like those in Group 4 have inner d orbital electrons involved in reactions, the valence electrons are primarily in their s sub-level, as seen in \([ ext{noble gas}] ns^{2}(n-1)d^{2}\).
  • P-Block and Group 13: With configurations like \([ ext{noble gas}] ns^{2}(n-1)d^{10}np^{1}\), the valence shell contains 3 electrons, which aids in understanding elements like aluminum’s behavior.

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

The energy from radiation can be used to cause the rupture of chemical bonds. A minimum energy of \(941 \mathrm{~kJ} / \mathrm{mol}\) is required to break the nitrogen-nitrogen bond in \(\mathrm{N}_{2}\). What is the longest wavelength of radiation that possesses the necessary energy to break the bond? What type of electromagnetic radiation is this?

Molybdenum metal must absorb radiation with a minimum frequency of \(1.09 \times 10^{15} \mathrm{~s}^{-1}\) before it can eject an electron from its surface via the photoelectric effect. (a) What is the minimum energy needed to eject an electron? (b) What wavelength of radiation will provide a photon of this energy? (c) If molybdenum is irradiated with light of wavelength of \(120 \mathrm{nm}\), what is the maximum possible kinetic energy of the emitted electrons?

Using the periodic table as a guide, write the condensed electron configuration and determine the number of unpaired electrons for the ground state of (a) \(\mathrm{Si},\) (b) \(\mathrm{Zn}\), (c) \(\mathrm{Zr},(\mathrm{d}) \mathrm{Sn}\) (e) \(\mathrm{Ba},(\mathrm{f}) \mathrm{Tl}\)

Scientists have speculated that element 126 might have a moderate stability, allowing it to be synthesized and characterized. Predict what the condensed electron configuration of this element might be.

The discovery of hafnium, element number \(72,\) provided a controversial episode in chemistry. G. Urbain, a French chemist, claimed in 1911 to have isolated an element number 72 from a sample of rare earth (elements \(58-71\) ) compounds. However, Niels Bohr believed that hafnium was more likely to be found along with zirconium than with the rare earths. D. Coster and G. von Hevesy, working in Bohr's laboratory in Copenhagen, showed in 1922 that element 72 was present in a sample of Norwegian zircon, an ore of zirconium. (The name hafnium comes from the Latin name for Copenhagen, Hafnia). (a) How would you use electron configuration arguments to justify Bohr's prediction? (b) Zirconium, hafnium's neighbor in group \(4 \mathrm{~B}\), can be produced as a metal by reduction of solid \(\mathrm{ZrCl}_{4}\) with molten sodium metal. Write a balanced chemical equation for the reaction. Is this an oxidation- reduction reaction? If yes, what is reduced and what is oxidized? (c) Solid zirconium dioxide, \(\mathrm{ZrO}_{2}\), is reacted with chlorine gas in the presence of carbon. The products of the reaction are \(\mathrm{ZrCl}_{4}\) and two gases, \(\mathrm{CO}_{2}\) and CO in the ratio 1: 2 . Write a balanced chemical equation for the reaction. Starting with a 55.4-g sample of \(\mathrm{ZrO}_{2}\), calculate the mass of \(\mathrm{ZrCl}_{4}\) formed, assuming that \(\mathrm{ZrO}_{2}\) is the limiting reagent and assuming \(100 \%\) yield. (d) Using their electron configurations, account for the fact that \(\mathrm{Zr}\) and \(\mathrm{Hf}\) form chlorides \(\mathrm{MCl}_{4}\) and oxides \(\mathrm{MO}_{2}\)

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