Explanation Lewis acid-base reaction of PF 5 with F − : Lewis structure of PF 5 ; Given species is PF 5 . The total number of valence electron present in PF 5 is calculated as shown below; Total number of valence electron = 1 P ( 5 electron ) + 5 F ( 7 electron ) = 5 + 35 = 40 Skeletal structure for PF 5 is drawn as shown below; Ten electrons are used up in the skeletal structure. Thirty electrons are left out. Six electrons are placed over fluorine atom each. Therefore, the Lewis structure of PF 5 is drawn as; Lewis structure of F − ; Given species is F − . Atomic number of fluorine is 9 . The given species F − is formed by adding one electron. The total number of valence electron present in F − is eight. Lewis structure is same as F − is drawn as by placing all the eight electrons as dots around the symbol of fluorine with a negative charge. Reaction between PF 5 and F − results in formation of PF 6 − . This can be represented as shown below; PF 5 + F − → PF 6 − Lewis structure of PF 6 − ; Given species is PF 6 − . The total number of valence electron present in PF 6 − is calculated as shown below; Total number of valence electron = 1 P ( 5 electron ) + 6 F ( 7 electron ) + 1 e l e c t r o n = 5 + 42 + 1 = 48 Skeletal structure for PF 6 − is drawn as shown below; Twelve electrons are used up in the skeletal structure. Thirty six electrons are left out. Six electrons are placed over fluorine atom each. Therefore, the Lewis structure of PF 6 − is drawn as; Lewis acid-base reaction of Cl − with SO 2 : Lewis structure of Cl − ; Given species is Cl −

ACHIEVE/CHEMICAL PRINCIPLES ACCESS 2TERM

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ISBN: 9781319403959

Author: ATKINS

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Chapter 6, Problem 6A.15E

Interpretation Introduction

Interpretation:

Lewis structure for the reactant and the product has to be drawn.

Concept Introduction:

Electron pair acceptor is known as Lewis acid while electron pair donor is known as Lewis base. Lewis structure is a simple representation considering the valence electrons present in a molecule. Electrons are shown using dots around the atoms in a molecule.

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Chapter 6, Problem 6A.14E

Chapter 6, Problem 6A.16E

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ACHIEVE/CHEMICAL PRINCIPLES ACCESS 2TERM

Ch. 6 - Prob. 6A.1AST Ch. 6 - Prob. 6A.1BST Ch. 6 - Prob. 6A.2AST Ch. 6 - Prob. 6A.2BST Ch. 6 - Prob. 6A.3AST Ch. 6 - Prob. 6A.3BST Ch. 6 - Prob. 6A.1E Ch. 6 - Prob. 6A.2E Ch. 6 - Prob. 6A.3E Ch. 6 - Prob. 6A.4E

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Solution Summary: The author explains the Lewis structure for the reactant and the product has to be drawn.

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