Chapter 8.4, Problem 67SSC

Interpretation Introduction

Interpretation:

The Lewis structure, molecular shape, bond angle and hybrid orbitals in ${\text{CS}}_{\text{2}}{\text{, CH}}_{\text{2}}{\text{O, H}}_{\text{2}}{\text{O, CCl}}_{\text{2}}{\text{F}}_{\text{2}}{\text{, NCl}}_{\text{3}}$ molecules needs to be determined.

Concept introduction:

VSEPR theory stands for Valence Shell Electron Pair Repulsion Theory. It helps to predict the molecular shape or geometry of the molecule with the help of the number of bond pairs or lone pairs present in it.

According to the VSEPR theory, the presence of lone pair on the central atom of a molecule causes deviation from standard molecular geometry. Thus, valence electrons provide a Lewis structure, which gives an idea about electron pair geometry and hybridization.

Answer to Problem 67SSC

	Lewis structure	molecular shape,	bond angle	hybrid orbitals
${\text{CS}}_{\text{2}}$		Linear	180°	sp hybrid orbitals
${\text{CH}}_{\text{2}}\text{O}$		Trigonal planer	120°	sp2 hybrid orbitals
${\text{H}}_{\text{2}}\text{O}$		Bent	104°	sp3 hybrid orbitals
${\text{CCl}}_{\text{2}}{\text{F}}_{\text{2}}$		Tetrahedral	109°	sp3 hybrid orbitals
${\text{NCl}}_{\text{3}}$		Pyramid	107°	sp3 hybrid orbitals

Explanation of Solution

Given information:

${\text{CS}}_{\text{2}}{\text{, CH}}_{\text{2}}{\text{O, H}}_{\text{2}}{\text{O, CCl}}_{\text{2}}{\text{F}}_{\text{2}}{\text{, NCl}}_{\text{3}}$

In the ${\text{CS}}_{\text{2}}$ molecule, C atom contains 4 valence electrons, and S has 6 valence electrons. The total number of valence electron can be determined in it as follows:

Number of valence electron = Number of atom C (valence electron in C) + Number of atom S (valence electron in S) = $\text{(1}\times \text{4) + (2}\times \text{6) = 18}$

Hence the Lewis structure must be:

$\begin{array}{l}\text{Hybridization = Number of sigma bonds + Number of lone pairs on bonded atoms}\\ \text{Hybridization = 2 + 0 = sp}\end{array}$

Geometry = Linear, bond angle = 180°

In the ${\text{CH}}_{\text{2}}\text{O}$ molecule, C atom contains 4 valence electron, and O has 6 valence electrons. The total number of valence electron can be determined in it as follows:

Number of valence electron = Number of atom C (valence electron in C) + Number of atom O (valence electron in O)+ Number of atom H (valence electron in H) = $\text{(1}\times \text{4) + (1}\times \text{6)+ (2}\times \text{1) = 12}$

Hence the Lewis structure must be:

$\begin{array}{l}\text{Hybridization = Number of sigma bonds + Number of lone pairs on bonded atoms}\\ {\text{Hybridization = 3 + 0 = sp}}^{\text{2}}\end{array}$

Geometry = Trigonal planer, bond angle = 120°

In the ${\text{H}}_{\text{2}}\text{O}$ molecule, H atom contains 1 valence electron, and O has 6 valence electrons. The total number of valence electron can be determined in it as follows:

Number of valence electron = Number of atom O (valence electron in O) + Number of atom H (valence electron in H) = $\text{(1}\times \text{6)+ (2}\times \text{1) = 8}$

Hence the Lewis structure must be:

$\begin{array}{l}\text{Hybridization = Number of sigma bonds + Number of lone pairs on bonded atoms}\\ {\text{Hybridization = 2 + 2 = sp}}^3\end{array}$

Geometry = bent, bond angle = 104°

In ${\text{CCl}}_{\text{2}}{\text{F}}_{\text{2}}$ molecule, the central C atom involves in ${\mathrm{sp}}^3$ hybridization with one 2s and three 2p orbitals to form four ${\mathrm{sp}}^3$ hybridized orbitals. The orbitals arrange in tetrahedral manner ton overlap with 2p orbital of Cl and F atoms to form ${\text{CCl}}_{\text{2}}{\text{F}}_{\text{2}}$ molecule.

In the ${\text{PF}}_{\text{3 }}$ molecule, P atom contains 5 valence electrons, and Cl has 7 valence electrons. The total number of valence electron can be determined in it as follows:

Number of valence electron = Number of atom P (valence electron in P) + Number of atom Cl (valence electron in Cl) = $\text{(1}\times \text{5) + (3}\times \text{7) = 26}$

$\begin{array}{l}\text{Hybridization = Number of sigma bonds + Number of lone pairs on bonded atoms}\\ {\text{Hybridization = 3 + 1 = sp}}^{\text{3 }}\end{array}$

With sp³ hybridization, the geometry must be tetrahedral but the presence of one lone pair on central P atom alters the standard geometry of molecule to pyramid shape due to repulsion between lone pair and bond pair of the molecule.

Conclusion

Thus,

	Lewis structure	molecular shape,	bond angle	hybrid orbitals
${\text{CS}}_{\text{2}}$		Linear	180°	sp hybrid orbitals
${\text{CH}}_{\text{2}}\text{O}$		Trigonal planer	120°	sp2 hybrid orbitals
${\text{H}}_{\text{2}}\text{O}$		Bent	104°	sp3 hybrid orbitals
${\text{CCl}}_{\text{2}}{\text{F}}_{\text{2}}$		Tetrahedral	109°	sp3 hybrid orbitals
${\text{NCl}}_{\text{3}}$		Pyramid	107°	sp3 hybrid orbitals