Chapter 3, Problem 1P

Give the structures of the following molecules: (a) ${\text{BCl}}_{\text{3}}$ (b) ${\text{SO}}_{\text{2}}$ (c) ${\text{PBr}}_{\text{3}}$ (d) ${\text{CS}}_{\text{2}}$ (e) ${\text{CHF}}_{\text{3}}$. Which molecules are polar?

Interpretation Introduction

Interpretation:

The structures of the following molecules: (a) ${\text{BCl}}_{\text{3}}$ (b) ${\text{SO}}_{\text{2}}$ (c) ${\text{PBr}}_{\text{3}}$ (d) ${\text{CS}}_{\text{2}}$ (e) ${\text{CHF}}_{\text{3}}$ are to be given andmolecules are polar or not, are to be determined.

Concept Introduction:

The VSEPR provides the simple model for predicting the shapes of molecular species.

Electron-electron repulsions decrease in the order:

Lone pair-lone pair > lone pair-bond pair > bond pair-bond pair

Where the central atom is involved in multiple bond formation to other atoms, electron-electron repulsion decrease in the order:

Triple bond-single bond > double bond-single bond > single bond single bond

The atoms share or give electrons to form molecules. If the electrons are shared equally by the atoms or the atoms which share the electrons are of equal electronegativity, then there is no resulting charge, and the molecule is nonpolar. Polar molecules are the opposite and have a positive or negative charge.

Answer to Problem 1P

Solution:

1. ${\text{BCl}}_{\text{3}}$ molecule has trigonal planar geometry and it is non polar.

2. ${\text{SO}}_{\text{2}}$ molecule has bent shape and it is polar.

3. ${\text{PBr}}_{\text{3}}$ molecule has trigonal pyramidal geometry and it is polar.

4. ${\text{CS}}_{\text{2}}$ molecule has linear shape and it is non polar.

5. ${\text{CHF}}_{\text{3}}$ molecule has tetrahedral geometry and it is polar.

Explanation of Solution

The valence shell electron pair repulsion (VSEPR) model is used to rationalize the shapes of molecular species and the shapes of such species depend on the number of electrons in the valence shell of the central atom.

With the information about valence electrons and electron dot formula, the geometry of a molecule can be determined.

Refer to the Fig. 2.16 and table 2.3 for the shapes and structure of molecules.

1. ${\text{BCl}}_{\text{3}}$ molecule

   Atomic number of boron = 5

   The electronic configuration of B = ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^1$

   Number of valence electron present in B = ${\text{2s}}^{\text{2}}{\text{2p}}^1$ = 3

   Atomic number of chlorine = 17

   The electronic configuration of Cl = ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^6{\text{3s}}^{\text{2}}{\text{3p}}^5$

   Number of valence electron present in Cl = ${\text{3s}}^{\text{2}}{\text{3p}}^5$ = 7

   The electron dot formula of ${\text{BCl}}_{\text{3}}$ is

   

   There is three single covalent bonds between one boron atom and three chlorine atoms and no lone pair electron present around the central atom.

   Thus, molecule has trigonal planar geometry and itis non polar because dipoles in the molecule cancel out.

2. ${\text{SO}}_{\text{2}}$ molecule

   Atomic number of Sulphur = 16

   The electronic configuration of S = ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^6{\text{3s}}^{\text{2}}{\text{3p}}^4$

   Number of valence electron present in S = ${\text{3s}}^{\text{2}}{\text{3p}}^4$ = 6

   Atomic number of oxygen = 17

   The electronic configuration of O = ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^4$

   Number of valence electron present in O = ${\text{2s}}^{\text{2}}{\text{2p}}^4$ = 6

   The electron dot formula of ${\text{SO}}_{\text{2}}$ is

   $:\underset{··}{\overset{··}{\text{O}}}:\underset{··}{\text{S}}::\underset{··}{\overset{··}{\text{O}}}$

   There is two doublecovalent bonds between one sulphuratom and two oxygen atoms andone lone pair of electron present around the central atom.

   Thus, molecule has bent shape and it is polar because dipoles in the molecule do not cancel out.

3. ${\text{PBr}}_{\text{3}}$ molecule

   Atomic number of phosphorous = 15

   The electronic configuration of P = ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^6{\text{3s}}^{\text{2}}{\text{3p}}^3$

   Number of valence electron present in P = ${\text{3s}}^{\text{2}}{\text{3p}}^3$ = 5

   Atomic number of bromine = 35

   The electronic configuration of Br = ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^6{\text{3s}}^{\text{2}}{\text{3p}}^6{\text{3d}}^{10}{\text{4s}}^2{\text{4p}}^5$

   Number of valence electron present in Br = ${\text{4s}}^2{\text{4p}}^5$ = 7

   The electron dot formula of ${\text{PBr}}_{\text{3}}$ is

   Locate bonding electrons around P

   

   There is three single covalent bonds between one boron atom and three chlorine atoms and one lone pair electron present around the central atom.

   Thus, molecule has trigonal pyramidal geometry and itis polar because dipoles in the molecule do not cancel out.

4. ${\text{CS}}_{\text{2}}$ molecule

   Atomic number of carbon= 6

   The electronic configuration of C = ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^2$

   Number of valence electron present in C = ${\text{2s}}^{\text{2}}{\text{2p}}^2$ = 4

   Atomic number of Sulphur = 16

   The electronic configuration of S = ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^6{\text{3s}}^{\text{2}}{\text{3p}}^4$

   Number of valence electron present in S = ${\text{3s}}^{\text{2}}{\text{3p}}^4$ = 6

   The electron dot formula of ${\text{CS}}_{\text{2}}$ is

   $\underset{··}{\overset{··}{\text{S}}}::\text{C}::\underset{··}{\overset{··}{\text{S}}}$

   There is two double covalent bonds between one carbon atom and two sulphur atoms and no lone pair of electron present around the central atom.

   Thus, molecule has linear shape and it is non polar because dipoles in the molecule cancel out.

5. ${\text{CHF}}_{\text{3}}$ molecule

   Atomic number of carbon = 6

   The electronic configuration of C = ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^2$

   Number of valence electron present in C = ${\text{2s}}^{\text{2}}{\text{2p}}^2$ = 4

   Atomic number of hydrogen = 1

   The electronic configuration of H = ${\text{1s}}^1$

   Number of valence electron present in H = ${\text{1s}}^1$ = 1

   Atomic number of fluorine = 9

   The electronic configuration of F = ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^5$

   Number of valence electron present in F = ${\text{2s}}^{\text{2}}{\text{2p}}^5$ = 7

   The electron dot formula of ${\text{CHF}}_{\text{3}}$ is

   

   There is four single covalent bonds between one carbon atom and three fluorine atoms and no lone pair electron present around the central atom.

   Thus, all the bonds are not identical, molecule has tetrahedral geometry and it is polar because dipoles in the molecule do not cancel out.