Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 2: Problem 105

The two species that have the same number of electrons as \(^{3}\) th S are (a) \(^{32} \mathrm{Cl} ;\) (b) \(^{34} \mathrm{S}^{+} ;\) (c) \(^{33} \mathrm{P}^{+} ;\) (d) \(^{28} \mathrm{Si}^{2-}\) (e) \(^{35} S^{2-} ;(f)^{40} A r^{2+} ;(g)^{40} C a^{2+}\)

Short Answer

Expert verified
The exercise presented doesn't have species that hold exactly the same number of electrons as \(^{3}\) th S, as all the options presented resulted in a different number of electrons.

Step by step solution

01

Identifying sulfur's electron count

The elemental symbol S stands for sulfur. An atom of sulfur, which is neutral, holds 16 protons and therefore 16 electrons in its shells. In our exercise, the species we are trying to match to is \(^{3}\) th S, which has 3 more protons in its nucleus, but it's neutral, so it holds 19 electrons in total.
02

Analyzing option (a): \(^{32} \mathrm{Cl} ;\)

The elemental symbol Cl stands for chloride. A standard chloride atom has 17 protons and therefore 17 electrons. Since \(^{32} \mathrm{Cl}\) has no charge, it does not gain or lose electrons and still has 17 electrons, which doesn't equal to the 19 electrons in \(^{3}\) th S.
03

Analyzing option (b): \(^{34} \mathrm{S}^{+} ;\)

This species is a sulfur ion. A standard sulfur ion has 16 protons, so a neutral one would also have 16 electrons. However, the superscript '+' means that this ion has lost one electron. Therefore, it now has 15 electrons, which doesn't match the 19 electrons of \(^{3}\) th S.
04

Analyzing option (c): \(^{33} \mathrm{P}^{+} ;\)

The elemental symbol P stands for phosphorus. A neutral atom of phosphorus would have 15 protons and therefore 15 electrons. This ion is positively charged, meaning it lost an electron and now has 14 electrons in total, which doesn't match with the 19 electrons of \(^{3}\) th S.
05

Analyzing option (d): \(^{28} \mathrm{Si}^{2-}\)

The elemental symbol Si stands for silicon. A neutral silicon atom has 14 protons and therefore 14 electrons. This ion is negatively charged, which means it gained two electrons and now has 16 electrons in total. Still, this doesn't match with the 19 electrons of \(^{3}\) th S.
06

Analyzing option (e): \(^{35} S^{2-} ;\)

This represents a sulfur ion. A normal sulfur atom with 16 protons would also have 16 electrons. But this ion is negatively charged, meaning it gained two electrons for a total of 18, which doesn't match the 19 electrons of \(^{3}\) th S.
07

Analyzing option (f): \(^{40} A r^{2+} ;\)

The elemental symbol Ar stands for argon. An argon atom has 18 protons and therefore 18 electrons. This ion has a 2+ charge, meaning it lost two electrons and now has 16, which does not match the 19 electrons of \(^{3}\) th S.
08

Analyzing option (g): \(^{40} C a^{2+}\)

The elemental symbol Ca stands for calcium. A calcium atom has 20 protons and therefore 20 electrons. This ion is positively charged, so it lost two electrons and now has 18, which doesn't match the 19 electrons of \(^{3}\) th S.

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

Phosphorus forms two compounds with chlorine. In the first compound, \(1.000 \mathrm{g}\) of phosphorus is combined with \(3.433 \mathrm{g}\) chlorine, and in the second, \(2.500 \mathrm{g}\) phosphorus is combined with \(14.308 \mathrm{g}\) chlorine. Show that these results are consistent with Dalton's law of multiple proportions.

In an experiment, \(125 \mathrm{cm}^{3}\) of zinc and \(125 \mathrm{cm}^{3}\) of iodine are mixed together and the iodine is completely converted to \(164 \mathrm{cm}^{3}\) of zinc iodide. What volume of zinc remains unreacted? The densities of zinc, iodine, and zinc iodide are \(7.13 \mathrm{g} / \mathrm{cm}^{3}, 4.93 \mathrm{g} / \mathrm{cm}^{3}\) and \(4.74 \mathrm{g} / \mathrm{cm}^{3}\), respectively.

In each case, identify the element in question. (a) The mass number of an atom is 234 and the atom has \(60.0 \%\) more neutrons than protons. (b) An ion with a \(2+\) charge has \(10.0 \%\) more protons than electrons. (c) An ion with a mass number of 110 and a \(2+\) charge has \(25.0 \%\) more neutrons than electrons.

William Prout (1815) proposed that all other atoms are built up of hydrogen atoms, suggesting that all elements should have integral atomic masses based on an atomic mass of one for hydrogen. This hypothesis appeared discredited by the discovery of atomic masses, such as 24.3 u for magnesium and 35.5 u for chlorine. In terms of modern knowledge, explain why Prout's hypothesis is actually quite reasonable.

Appendix E describes a useful study aid known as concept mapping. Using the method presented in Appendix \(\mathrm{E}\), construct a concept map illustrating the different concepts in Sections \(2-7\) and \(2-8\).
See all solutions

Recommended explanations on Chemistry Textbooks

Chemistry Branches

Read Explanation

Inorganic Chemistry

Read Explanation

Kinetics

Read Explanation

Making Measurements

Read Explanation

Organic Chemistry

Read Explanation

Nuclear Chemistry

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