Chapter 8, Problem 74A

(a)

Interpretation Introduction

Interpretation: The geometry of ${\text{SiCl}}_{\text{4}}$ needs to be predicted using VSEPR theory.

Concept Introduction: VSEPR theory is used to predict the molecular shape with the help of electron pairs around the central atoms of the molecule. It is based on the assumption that the shape of the molecule is such that there is minimum electronic repulsion between the valence shell and atom.

Explanation of Solution

The given compound is ${\text{SiCl}}_{\text{4}}$ . Here, Si has 4 valence electrons. It shares its 4 electrons with 4 chlorine atoms to form Si-Cl single bonds. The number of valence electrons in Cl is 7 thus, each chlorine atom has 3 lone pairs of electrons. The Si atom is a central atom with no lone pair of electrons. The hybridization of central atom Si will be $s{p}^3$ . Thus, it has tetrahedral geometry. According to VSEPR theory since, the central atom has $s{p}^3$ hybridization with no lone pair of electrons, the shape will also be tetrahedral.

(b)

Interpretation Introduction

Interpretation: The geometry of ${\text{CO}}_3^{2-}$ needs to be predicted using VSEPR theory.

Concept Introduction: VSEPR theory is used to predict the molecular shape with the help of electron pairs around the central atoms of the molecule. It is based on the assumption that the shape of the molecule is such that there is minimum electronic repulsion between the valence shell and the atom.

are represented as dots in pairs around the symbol of atoms of the molecule.

Explanation of Solution

The given compound is ${\text{CO}}_3^{2-}$ . Here, C has 4 valence electrons. There are 3 oxygen atoms with 6 valence electrons. The atoms are bonded together in such a way that one C-O bond is a double bond and two C-O bonds are single bonds with a negative charge on the O atom.

Here, central atom C has no lone pair of electrons with $s{p}^2$ hybridization. Thus, the geometry of the compound will be trigonal planar.

According to VSEPR theory, since the central atom has $s{p}^2$ hybridization with no lone pair of electrons on the C atom, the shape of the compound will also be trigonal planar.

(c)

Interpretation Introduction

Interpretation: The geometry of ${\text{CCl}}_{\text{4}}$ needs to be predicted using VSEPR theory.

Concept Introduction: VSEPR theory is used to predict the molecular shape with the help of electron pairs around the central atoms of the molecule. It is based on the assumption that the shape of the molecule is such that there is minimum electronic repulsion between the valence shell and the atom.

Explanation of Solution

The given compound is ${\text{CCl}}_{\text{4}}$ . Here, C is the central atom with 4 Cl atoms bonded with single covalent bonds. The C atom here is $s{p}^3$ hybridized; thus, the geometry will be tetrahedral.

According to VSEPR theory since, the central atom has $s{p}^3$ hybridization with no lone pair of electrons, the shape will also be tetrahedral.

(d)

Interpretation Introduction

Interpretation: The geometry of ${\text{SCl}}_2$ needs to be predicted using VSEPR theory.

Concept Introduction: VSEPR theory is used to predict the molecular shape with the help of electron pairs around the central atoms of the molecule. It is based on the assumption that the shape of the molecule is such that there is minimum electronic repulsion between the valence shell and the atom.

Explanation of Solution

The given compound is ${\text{SCl}}_2$ . Here, S is the central atom with 2 Cl atoms bonded with single covalent bonds. There are 6 valence electrons in the S atom. Here, 2 atoms are shared with Cl atoms thus, 4 valence electrons are represented as 2 lone pairs of electrons on the central S atom. Thus, S will be $s{p}^3$ hybridized. The geometry will be tetrahedral. According to VSEPR theory since, the central atom has $s{p}^3$ hybridization with two lone pairs of electrons, the shape will be bent.