Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 9: Problem 88

For the atom \(^{119} \mathrm{Sn}\), indicate the number of (a) protons in the nucleus; (b) neutrons in the nucleus; (c) \(4 d\) electrons; (d) 3s electrons; (e) 5 \(p\) electrons; (f) electrons in the valence shell.

Short Answer

Expert verified
The atom \(^{119} Sn\) has (a) 50 protons, (b) 69 neutrons, (c) 10 4d electrons, (d) 2 3s electrons, (e) 2 5p electrons, and (f) 4 electrons in the valence shell.

Step by step solution

01

Protons from Atomic Number

Search for Sn on the periodic table, it has atomic number 50. Hence, it has 50 protons.
02

Neutrons from Mass Number

Subtract atomic number from mass number, i.e., \(neutrons = 119 - 50 = 69\).
03

Electrons in the 4d Orbital

According to the electron distribution in atomic orbitals, 4d orbital fills after the 5s orbital. Tin corresponds to the electron configuration [Kr]4d105s25p2. Since the 4d orbital is filled before the 5s and 5p orbitals, it's full with 10 electrons.
04

Electrons in the 3s Orbital

The 3s orbital fills before the 3p and 4s orbitals. It always holds only 2 electrons when full. Therefore, there are 2 electrons in the 3s orbital of Tin.
05

Electrons in the 5p Orbital

In configuration [Kr]4d105s25p2, it's evident that 5p orbital has 2 electrons.
06

Electrons in the Valence Shell

Valence shell is the outermost shell, which is the 5th shell for Tin. This shell includes 5s and 5p orbitals, hence holds \(2+2=4\) 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.

Start your free trial

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

All the isoelectronic species illustrated in the text had the electron configurations of noble gases. Can two ions be isoelectronic without having noble-gas electron configurations? Explain.

In estimating the boiling point and melting point of bromine in Example \(9-5,\) could we have used Celsius or Fahrenheit instead of Kelvin temperature? Explain.

Write electron configurations to show the first two ionizations for Cs. Explain why the second ionization energy is much greater than the first.

Refer only to the periodic table on the inside front cover, and arrange the following ionization energies in order of increasing value: \(I_{1}\) for \(\mathrm{F} ; I_{2}\) for \(\mathrm{Ba}\) \(I_{3}\) for \(\mathrm{Sc} ; I_{2}\) for \(\mathrm{Na} ; I_{3}\) for \(\mathrm{Mg}\). Explain the basis of any uncertainties.

Which of the following ions are unlikely to be found in chemical compounds: \(\mathrm{K}^{+}, \mathrm{Ga}^{4+}, \mathrm{Fe}^{6+} \mathrm{S}^{2-}, \mathrm{Ge}^{5+},\) or \(\mathrm{Br}^{-} ?\) Explain briefly.
See all solutions

Recommended explanations on Chemistry Textbooks

Kinetics

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Inorganic Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Making Measurements

Read Explanation

Chemical Reactions

Read Explanation
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