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Explain the contributions of Johann Dobereiner and John

Chemistry: A Molecular Approach | 3rd Edition | ISBN: 9780321809247 | Authors: Nivaldo J. Tro ISBN: 9780321809247 1

Solution for problem 3E Chapter 8

Chemistry: A Molecular Approach | 3rd Edition

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Chemistry: A Molecular Approach | 3rd Edition | ISBN: 9780321809247 | Authors: Nivaldo J. Tro

Chemistry: A Molecular Approach | 3rd Edition

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Problem 3E

Problem 3E

Explain the contributions of Johann Dobereiner and John Newlands to the organization of elements according to their properties.

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Solution 3E:The formation of NaClSodium (electronic configuration : 2,8,1 ) has 1 electron more than a stable noble gas structure +(2,8). So it gave away that electron it would become more stable. Na has smaller in size due tothe removal of the electron from the last shell. Similarly Chlorine (electronic configuration :2,8,7) has 1 electron short of a stable noble gas structure (2,8,8). If it could gain an electron from -somewhere it too would become more stable.Whereas the size of Cl remain same and which is +bigger than Na . + (a) The smaller balls 2, 4,6 and 8 are Na ions. This is because the due to the removal of electrons from last shell. It reduces the size of the atom. Therefore, smaller red balls are + Na ion. -(b) The bigger balls 1, 3,5 and 7 are Cl ions. This is because the due to the presence of anadditional p subshell . Therefore the green balls of 1, 3, 5 and 7 are Cl ions. -\n(C) -From figure, it has been shown that, ion 5 which is Cl ion (anion) is surrounded by fourpositively charged cations which are Na ions of 2, 4, 6 and 8.Therefore, in ion 5, four electrostatic attractive interactions are shown.(D) -From figure, it has been shown that, ion 5 which is Cl ion (anion) is surrounded by fournegatively charged anions which are 1, 3, 7 and 9.Therefore, in ion 5, four electrostatic repulsive interactions are shown.(e)From figure it is shown that, ion 5 is surrounded by other ions in equal distances. Thus, the sumof four electrostatic attraction is same as the four electrostatic repulsive interactions.Therefore, the electrostatic attractive forces are cancel out to each other and net force will bezero at ion 5. Similarly, the electrostatic repulsive forces are also cancel out to each other and thenet force around ion 5 ion will be zero.(f)The magnitude of the lattice energy of an ionic solid depends (a) on the charges of the ions, (b) their sizes, (c) and their arrangement in the solid.Therefore,\nQ1 and Q2 are the charges of the ions.d is the distance between the ions. k is the constant.So if the lattice structure is extended in 2-dimension, then new bonds will be formed. Therefore,large amount of energy will be released.Thus, the lattice energy will be negative.

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Chapter 8, Problem 3E is Solved
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Textbook: Chemistry: A Molecular Approach
Edition: 3
Author: Nivaldo J. Tro
ISBN: 9780321809247

Chemistry: A Molecular Approach was written by and is associated to the ISBN: 9780321809247. This textbook survival guide was created for the textbook: Chemistry: A Molecular Approach, edition: 3. The full step-by-step solution to problem: 3E from chapter: 8 was answered by , our top Chemistry solution expert on 02/22/17, 04:35PM. This full solution covers the following key subjects: according, contributions, dobereiner, elements, explain. This expansive textbook survival guide covers 82 chapters, and 9454 solutions. The answer to “Explain the contributions of Johann Dobereiner and John Newlands to the organization of elements according to their properties.” is broken down into a number of easy to follow steps, and 18 words. Since the solution to 3E from 8 chapter was answered, more than 370 students have viewed the full step-by-step answer.

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