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Write Lewis structures for each molecule or ion. Include

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

Solution for problem 76E Chapter 9

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 76E

Problem 76E

Write Lewis structures for each molecule or ion. Include resonance structures if necessary and assign formal charges to all atoms. If necessary, expand the octet on the central atom to lower formal charge.

a. SO42-

b. HSO4-

c. SO3

d. BrO2-

Step-by-Step Solution:
Step 1 of 3

Solution:Step-1(a)Peroxide ion (O ) has longer bond length than the superoxide ion (O ) - 2 2It is known that the length of the bond is determined by the number of bonded electrons (thebond order). The higher the bond order, the stronger the pull between the two atoms and theshorter the bond length.The reason behind difference between the bond length of this two ions is due the differencein bond order . 2-The bond order of peroxide ion (O ) is, 2Bond order in Oxygen molecule (O ) 2- 2=½ [(Number of bonding electrons) – (number of anti-bonding electrons)]= ½ [10 – 8]= 1With the increase in number of electrons in anti-bonding orbital the stability and bond orderof the molecule decreases. 2-O2 -Superoxide ion (O ) 2Bond order in Oxygen molecule (O2-)=½ [(Number of bonding electrons) – (number of anti-bonding electrons)]= ½ [10 – 7] 3= 2= 1.5\nO2- :Higher the bond order, the stronger the pull between the two atoms and the shorter the bond 2- -length. In between Peroxide ion (O ) and superox2e ion (O ), the bond order of pero2de 2- -ion (O )2s 1 and superoxide ion (O ) is 1.5. Thus 2roxide has longer bond length due tolow bond order as compared to superoxide ion.Step-2(b)The MO diagram of B 2The MO diagram of B shows that 2 has two unpaired electrons 2p and 2p (which x zmakes it paramagnetic) and these electrons are in bonding molecular orbitals. 42Hence the bond order of B = 2 2 =1As B i2paramagnetic (2 unpaired electrons) it indicates that the 2p has lower energyotherwise if 2p orbital has lower energy , than the molecule does not have any unpairedelectrons.\nStep-3 2+(c)The O-O in O 2 is stronger than O 2It is known that bond order is inversely proportional to the bond length i.e higher the bondorder shorter will be the bond length. 2+ 104Bond order of O 2 = 2 =3 84Bond order of O = 2 2 =2 2+The bond order of O 2 is 3 which is higher than O (2). S2higher the bond order shorter will 2+be the bond length and the bond will be more stronger.Thus the O-O in O 2 is stronger thanO . 2

Step 2 of 3

Chapter 9, Problem 76E is Solved
Step 3 of 3

Textbook: Chemistry: A Molecular Approach
Edition: 3
Author: Nivaldo J. Tro
ISBN: 9780321809247

Since the solution to 76E from 9 chapter was answered, more than 354 students have viewed the full step-by-step answer. The answer to “Write Lewis structures for each molecule or ion. Include resonance structures if necessary and assign formal charges to all atoms. If necessary, expand the octet on the central atom to lower formal charge.a. SO42-b. HSO4-c. SO3d. BrO2-” is broken down into a number of easy to follow steps, and 37 words. 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: 76E from chapter: 9 was answered by , our top Chemistry solution expert on 02/22/17, 04:35PM. This full solution covers the following key subjects: formal, structures, necessary, include, central. This expansive textbook survival guide covers 82 chapters, and 9454 solutions.

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Write Lewis structures for each molecule or ion. Include