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Draw the Lewis structures for TeCl 4 , ICl 5 , PCl 5 , KrCl 4 , and XeCl 2 . Which of the compounds exhibit at least one bond angle that is approximately 120 degrees? Which of the compounds exhibit d 2 sp 3 hybridization? Which of the compounds have a square planar molecular structure? Which of the compounds are polar?

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Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243

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BuyFindarrow_forward

Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 4, Problem 104CWP
Textbook Problem
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Draw the Lewis structures for TeCl4, ICl5, PCl5, KrCl4, and XeCl2. Which of the compounds exhibit at least one bond angle that is approximately 120 degrees? Which of the compounds exhibit d2sp3 hybridization? Which of the compounds have a square planar molecular structure? Which of the compounds are polar?

Interpretation Introduction

Interpretation: The Lewis structures for the given compounds are to be drawn. The compounds which are polar and exhibit at least one bond angle of 120ο are to be stated. Also, the compounds which exhibit d2sp3 hybridization and those have a square planar molecular structure are to be stated.

Concept introduction: The Lewis structure shows the connectivity by identifying the lone pairs of electrons in a compound. Hybridization is the process of mixing atomic orbitals into hybrid orbitals to identify the structure of the compound. Bond angle is defined as the angle between three or at least two atoms. When two atoms shared electrons pair with each other in an equal extent then the bond is non polar in nature but when a pair of electrons is unequally shared between two atoms then a polar chemical bond is formed.

To determine: The Lewis structures for the given compounds. The polar compounds and the compounds those exhibit at least one bond angle of 120ο . The compounds which exhibit d2sp3 hybridization and those have a square planar molecular structure.

Explanation of Solution

Explanation

The Lewis structure of TeCl4 is,

Figure 1

The given compound is TeCl4 containing one tellurium atom as a central atom and four chlorine atoms. The valence electrons in tellurium are six and in chlorine are seven. The total valence electrons are calculated as,

1Te+4Cl=1×6+4×7=34

First, the skeleton structure is drawn in which the chlorine atoms are single bonded to the central atom then the electron pairs around every atom is drawn so that each gets an octet. The Lewis structure of TeCl4 is shown in Figure 1.

The Lewis structure of ICl5 is,

Figure 2

The given compound is ICl5 containing one iodine atom and five chlorine atoms. Iodine and chlorine have 7 valence electrons. The total valence electrons in ICl5 is calculated as,

1I+5Cl=1×7+5×7=42

Iodine and chlorine in order to complete its octet needs one more electron. Hence, they form five covalent bonds by sharing their valence electrons in order to complete their octet. The Lewis structure of ICl5 is shown in Figure 2.

The Lewis structure of PCl5 is,

Figure 3

The given compound is PCl5 having phosphorus as a central metal atom surrounded by five chlorine atoms. Phosphorus contains 5 valence electrons and chlorine has seven valence electrons. The total valence electrons are calculated as,

1P+5Cl=1×5+5×7=40

The skeleton structure is drawn in which the phosphorus and chlorine atoms are single bonded then the electron pairs around every atom is drawn so that each gets an octet. The Lewis structure of PCl5 is shown in Figure 3.

The Lewis structure of KrCl4 is,

Figure 4

The given compound is KrCl4 having krypton as a central metal atom surrounded by chlorine atoms. The number of valence electrons in krypton is eight and chlorine contains seven valence electrons respectively. The structure is drawn in which the atoms are single bonded to each other then the electron pairs around every atom is drawn so that each gets an octet

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Chemistry: An Atoms First Approach
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