/*! 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 76 Write the condensed electron con... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Write the condensed electron configurations for the following atoms and indicate how many unpaired electrons each has: (a) \(\mathrm{Mg},(\mathbf{b}) \mathrm{Ge},(\mathbf{c}) \mathrm{Br},(\mathbf{d}) \mathrm{V},(\mathbf{e}) \mathrm{Y},(\mathbf{f}) \mathrm{Lu}\)

Short Answer

Expert verified
The condensed electron configurations and unpaired electrons for the given elements are as follows: (a) Mg: \([\mathrm{Ne}]\ 3s^2\], 0 unpaired electrons (b) Ge: \([\mathrm{Ar}]\ 4s^2 3d^10 4p^2\], 2 unpaired electrons (c) Br: \([\mathrm{Ar}]\ 4s^2 3d^10 4p^5\], 1 unpaired electron (d) V: \([\mathrm{Ar}]\ 4s^2 3d^3\], 3 unpaired electrons (e) Y: \([\mathrm{Kr}]\ 5s^2 4d^1\], 1 unpaired electron (f) Lu: \([\mathrm{Xe}]\ 6s^2 4f^14 5d^1\], 1 unpaired electron

Step by step solution

01

Determine the atomic numbers

Look up the atomic numbers of the given elements using the periodic table: (a) Mg (magnesium) has an atomic number of 12. (b) Ge (germanium) has an atomic number of 32. (c) Br (bromine) has an atomic number of 35. (d) V (vanadium) has an atomic number of 23. (e) Y (yttrium) has an atomic number of 39. (f) Lu (lutetium) has an atomic number of 71.
02

Write the condensed electron configurations

Using the periodic table, determine the condensed electron configurations for the given elements: (a) Mg: \[1s^2 2s^2 2p^6 3s^2\] or \[[\mathrm{Ne}]\ 3s^2\] (b) Ge: \[1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^2\] or \[[\mathrm{Ar}]\ 4s^2 3d^10 4p^2\] (c) Br: \[1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^5\] or \[[\mathrm{Ar}]\ 4s^2 3d^10 4p^5\] (d) V: \[1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^3\] or \[[\mathrm{Ar}]\ 4s^2 3d^3\] (e) Y: \[1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6 5s^2 4d^1\] or \[[\mathrm{Kr}]\ 5s^2 4d^1\] (f) Lu: \[1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6 5s^2 4d^10 5p^6 6s^2 4f^14 5d^1\] or \[[\mathrm{Xe}]\ 6s^2 4f^14 5d^1\]
03

Determine the number of unpaired electrons

Count the number of unpaired electrons in the electron configurations: (a) Mg: 0 unpaired electrons (b) Ge: 2 unpaired electrons (c) Br: 1 unpaired electron (d) V: 3 unpaired electrons (e) Y: 1 unpaired electron (f) Lu: 1 unpaired electron In summary, we have the following condensed electron configurations and unpaired electrons: (a) Mg: \[[\mathrm{Ne}]\ 3s^2\], 0 unpaired electrons (b) Ge: \[[\mathrm{Ar}]\ 4s^2 3d^10 4p^2\], 2 unpaired electrons (c) Br: \[[\mathrm{Ar}]\ 4s^2 3d^10 4p^5\], 1 unpaired electron (d) V: \[[\mathrm{Ar}]\ 4s^2 3d^3\], 3 unpaired electrons (e) Y: \[[\mathrm{Kr}]\ 5s^2 4d^1\], 1 unpaired electron (f) Lu: \[[\mathrm{Xe}]\ 6s^2 4f^14 5d^1\], 1 unpaired electron

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.

Atomic Number
The atomic number is truly a cornerstone in chemistry and physics, representing the unique identity of an element. It is defined as the number of protons in the nucleus of an atom. This number not only identifies an element but also determines its position in the periodic table.

Each element has a distinct atomic number. For example, magnesium (Mg) has an atomic number of 12, meaning it has 12 protons in its nucleus. Similarly, germanium (Ge) has an atomic number of 32. This concept is crucial because it tells us how many electrons an atom has in a neutral state, since the number of protons equals the number of electrons.

The atomic number not only determines the element's identity but also its chemical properties and placement on the periodic table. As you move across the table from left to right, each consecutive element has an atomic number one greater than the previous one.
Unpaired Electrons
Unpaired electrons have a significant role in the chemistry of an element. They are electrons that remain single in an atom's electron configuration. These unpaired electrons reside in orbitals that are not completely filled.

Unpaired electrons often indicate the magnetic properties of an element, contributing to paramagnetism, where atoms are attracted to a magnetic field. For example, in the bromine (Br) atom, there is 1 unpaired electron, revealing its magnetic nature.

To determine the number of unpaired electrons, examine the electron configuration of an atom, and identify if any orbitals contain unpaired electrons. For instance, vanadium (V) has three unpaired electrons in the 3d subshell according to its configuration \[\mathrm{[Ar] 4s^2 3d^3}\], making vanadium exhibit paramagnetic properties.

Remember, the presence of unpaired electrons is key in understanding the chemical reactivity and bonding behavior of elements.
Periodic Table
The periodic table is an essential tool for organizing elements that shows physical and chemical properties in a systematic way. Composed of rows called periods and columns known as groups or families, the periodic table allows us to predict the behavior and properties of elements.

Elements are arranged in increasing order of their atomic numbers. For instance, magnesium (Mg) is in the second period and group 2, indicating its place in the table with an atomic number 12.

The periodic table reveals trends such as atomic radii, ionization energy, and electronegativity. It also highlights categories like metals, nonmetals, and metalloids. This organization helps us understand not only individual elements but also their interactions.

The table is our guide to writing electron configurations and determining an element's valence electrons, which affect its chemical behavior. For instance, looking at yttrium (Y) in group 3, we see it typically forms 3+ ions, consistent with losing its three outermost electrons in the 5s and 4d subshells.

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

Give the values for \(n, l,\) and \(m_{l}\) for (a) each orbital in the 2\(p\) subshell, (b) each orbital in the 5\(d\) subshell.

(a) Calculate the energy of a photon of electromagnetic radiation whose frequency is \(2.94 \times 10^{14} \mathrm{s}^{-1} .\) (b) Calculate the energy of a photon of radiation whose wavelength is 413 nm. (c) What wavelength of radiation has photons of energy \(6.06 \times 10^{-19} \mathrm{J} ?\)

Determine which of the following statements are false and correct them. (a) The frequency of radiation increases as the wavelength increases. (b) Electromagnetic radiation travels through a vacuum at a constant speed, regardless of wavelength. (c) Infrared light has higher frequencies than visible light. (d) The glow from a fireplace, the energy within a microwave oven, and a foghorn blast are all forms of electromagnetic radiation.

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 hafnum 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, \(\mathrm{ZrO}_{2},\) reacts with chlorine gas in the presence of carbon. The products of the reaction are \(Z r \mathrm{Cl}_{4}\) and two gases, \(\mathrm{CO}_{2}\) and \(\mathrm{CO}\) in the ratio \(1 : 2 .\) Write a balanced chemical equation for the reaction. Starting with a \(55.4-\mathrm{g}\) sample of \(\mathrm{ZrO}_{2},\) calculate the mass of \(\mathrm{ZrCl}_{4}\) formed, assuming that \(Z r O_{2}\) 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 \(\mathrm{MO}_{2}\)

Among the elementary subatomic particles of physics is the muon, which decays within a few microseconds after formation. The muon has a rest mass 206.8 times that of an electron. Calculate the de Broglie wavelength associated with a muon traveling at \(8.85 \times 10^{5} \mathrm{cm} / \mathrm{s}\) .

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.