Chapter 9, Problem 8PS

Specify the electron-pair and molecular geometry for each underlined atom in the following list. Describe the hybrid orbital set used by this atom in each molecule or ion.

(a) CSe₂

(b) SO₂

(c) CH₂O

(d) NH₄^ss

Interpretation Introduction

Interpretation:

The electron pair and molecular geometry for the atom underlined in the given molecule has to be specified and the hybrid orbital set used by this atom in the molecule should be described.

• $\underset{\_}{C}S{e}_2$

Concept Introduction:

• Hybridization is the mixing of valence atomic orbitals to get equivalent hybridized orbitals that having similar characteristics and energy.
• Geometry of a molecule can be predicted by knowing its hybridization.
• ${\text{sp}}^{\text{3}}$ hybrid orbitals is produced by hybridization of single s-orbital and three p-orbital.
• ${\text{sp}}^2$ hybrid orbitals is produced by hybridization between one s-orbital and two p-orbitals.
• $\text{sp }$ hybrid orbitals is produced by hybridization of single s-orbital and single p-orbital.

Electron geometry is the shape of a molecule predicted by considering both bond electron pairs and lone pair of electrons.

Molecular geometry is the shape of a molecule predicted by considering only bond pair of electrons.

Geometry of different types of molecule with respect to the hybridizations are mentioned are mentioned below,

$\begin{array}{l}\begin{array}{cccc}Typeofmolecule& Hybridaization& \begin{array}{l}Atomicorbitals\\ usedforhybridaization\end{array}& Geometry\\ A{X}_2& sp& 1s+1p& Linear\\ A{X}_3,A{X}_{2}B& s{p}^2& 1s+2p& Trigonalplanar\\ A{X}_4,A{X}_{3}B,A{X}_2{B}_2& s{p}^3& 1s+3p& Tetrahedral\\ A{X}_5,A{X}_{4}B,A{X}_3{B}_2,A{X}_2{B}_3& s{p}^{3}d& 1s+3p+1d& Trigonalbipyramidal\\ A{X}_6,A{X}_{5}B,A{X}_4{B}_2& s{p}^3{d}^2& 1s+3p+2d& Octahedral\end{array}\\ A\to Centralatom\\ X\to AtomsbondedtoA\\ B\to NonbondingelectronpairsonA\end{array}$

Geometry of different type of molecules with respect to the number of electron pairs are mentioned below,

$\begin{array}{ccccc}\begin{array}{l}Typeof\\ Molecule\end{array}& \begin{array}{l}No.ofatoms\\ bondedtocentralatoms\end{array}& \begin{array}{l}No.oflonepairs\\ oncentralatom\end{array}& \begin{array}{l}Arrangementof\\ electronpairs\end{array}& \begin{array}{l}Molecular\\ Geometry\end{array}\\ A{B}_2& 2& 0& Linear& Linear\\ A{B}_3& 3& 0& Trigonalplanar& Trigonalplanar\\ A{B}_4& 4& 0& Tetrahedral& Tetrahedral\\ A{B}_5& 5& 0& Trigonalbipyramidal& Trigonalbipyramidal\\ A{B}_6& 6& 0& Octahedral& Octahedral\end{array}$

Explanation of Solution

In the given molecule, $\underset{\_}{C}S{e}_2$ (According to the above mentioned table it is an example for $A{X}_2$ type molecule) two Se atoms are bonded to the carbon atom. Since there are two atoms bonded to the central atom without any lone pair of electrons on central atom, both the electron pair and molecular geometries are linear.

A molecule with $A{X}_2$ type ($\underset{\_}{C}S{e}_2$) possesses $sp$ hybridization

Interpretation Introduction

Interpretation:

The electron pair and molecular geometry for the atom underlined in the given molecule has to be specified and the hybrid orbital set used by this atom in the molecule should be described.

• $\underset{\_}{S}{O}_2$

Concept Introduction:

• Hybridization is the mixing of valence atomic orbitals to get equivalent hybridized orbitals that having similar characteristics and energy.
• Geometry of a molecule can be predicted by knowing its hybridization.
• ${\text{sp}}^{\text{3}}$ hybrid orbitals is produced by hybridization of single s-orbital and three p-orbital.
• ${\text{sp}}^2$ hybrid orbitals is produced by hybridization between one s-orbital and two p-orbitals.
• $\text{sp }$ hybrid orbitals is produced by hybridization of single s-orbital and single p-orbital.

Electron geometry is the shape of a molecule predicted by considering both bond electron pairs and lone pair of electrons.

Molecular geometry is the shape of a molecule predicted by considering only bond pair of electrons.

Geometry of different types of molecule with respect to the hybridizations are mentioned are mentioned below,

$\begin{array}{l}\begin{array}{cccc}Typeofmolecule& Hybridaization& \begin{array}{l}Atomicorbitals\\ usedforhybridaization\end{array}& Geometry\\ A{X}_2& sp& 1s+1p& Linear\\ A{X}_3,A{X}_{2}B& s{p}^2& 1s+2p& Trigonalplanar\\ A{X}_4,A{X}_{3}B,A{X}_2{B}_2& s{p}^3& 1s+3p& Tetrahedral\\ A{X}_5,A{X}_{4}B,A{X}_3{B}_2,A{X}_2{B}_3& s{p}^{3}d& 1s+3p+1d& Trigonalbipyramidal\\ A{X}_6,A{X}_{5}B,A{X}_4{B}_2& s{p}^3{d}^2& 1s+3p+2d& Octahedral\end{array}\\ A\to Centralatom\\ X\to AtomsbondedtoA\\ B\to NonbondingelectronpairsonA\end{array}$

Geometry of different type of molecules with respect to the number of electron pairs are mentioned below,

$\begin{array}{ccccc}\begin{array}{l}Typeof\\ Molecule\end{array}& \begin{array}{l}No.ofatoms\\ bondedtocentralatoms\end{array}& \begin{array}{l}No.oflonepairs\\ oncentralatom\end{array}& \begin{array}{l}Arrangementof\\ electronpairs\end{array}& \begin{array}{l}Molecular\\ Geometry\end{array}\\ A{B}_2& 2& 0& Linear& Linear\\ A{B}_3& 3& 0& Trigonalplanar& Trigonalplanar\\ A{B}_4& 4& 0& Tetrahedral& Tetrahedral\\ A{B}_5& 5& 0& Trigonalbipyramidal& Trigonalbipyramidal\\ A{B}_6& 6& 0& Octahedral& Octahedral\end{array}$

Explanation of Solution

In the given molecule, $\underset{\_}{S}{O}_2$ (According to the above mentioned table it is an example for $A{X}_2$ type molecule) two oxygen atoms are bonded to the sulfur atom. Since there are two atoms bonded to the central atom without any lone pair of electrons on central atom, both the electron pair and molecular geometries are linear.

A molecule with $A{X}_2$ type ($\underset{\_}{S}{O}_2$) possesses $sp$ hybridization.

Interpretation Introduction

Interpretation:

The electron pair and molecular geometry for the atom underlined in the given molecule has to be specified and the hybrid orbital set used by this atom in the molecule should be described.

• $\underset{\_}{C}{H}_{2}O$

Concept Introduction:

• Hybridization is the mixing of valence atomic orbitals to get equivalent hybridized orbitals that having similar characteristics and energy.
• Geometry of a molecule can be predicted by knowing its hybridization.
• ${\text{sp}}^{\text{3}}$ hybrid orbitals is produced by hybridization of single s-orbital and three p-orbital.
• ${\text{sp}}^2$ hybrid orbitals is produced by hybridization between one s-orbital and two p-orbitals.
• $\text{sp }$ hybrid orbitals is produced by hybridization of single s-orbital and single p-orbital.

Electron geometry is the shape of a molecule predicted by considering both bond electron pairs and lone pair of electrons.

Molecular geometry is the shape of a molecule predicted by considering only bond pair of electrons.

Geometry of different types of molecule with respect to the hybridizations are mentioned are mentioned below,

$\begin{array}{l}\begin{array}{cccc}Typeofmolecule& Hybridaization& \begin{array}{l}Atomicorbitals\\ usedforhybridaization\end{array}& Geometry\\ A{X}_2& sp& 1s+1p& Linear\\ A{X}_3,A{X}_{2}B& s{p}^2& 1s+2p& Trigonalplanar\\ A{X}_4,A{X}_{3}B,A{X}_2{B}_2& s{p}^3& 1s+3p& Tetrahedral\\ A{X}_5,A{X}_{4}B,A{X}_3{B}_2,A{X}_2{B}_3& s{p}^{3}d& 1s+3p+1d& Trigonalbipyramidal\\ A{X}_6,A{X}_{5}B,A{X}_4{B}_2& s{p}^3{d}^2& 1s+3p+2d& Octahedral\end{array}\\ A\to Centralatom\\ X\to AtomsbondedtoA\\ B\to NonbondingelectronpairsonA\end{array}$

Geometry of different type of molecules with respect to the number of electron pairs are mentioned below,

$\begin{array}{ccccc}\begin{array}{l}Typeof\\ Molecule\end{array}& \begin{array}{l}No.ofatoms\\ bondedtocentralatoms\end{array}& \begin{array}{l}No.oflonepairs\\ oncentralatom\end{array}& \begin{array}{l}Arrangementof\\ electronpairs\end{array}& \begin{array}{l}Molecular\\ Geometry\end{array}\\ A{B}_2& 2& 0& Linear& Linear\\ A{B}_3& 3& 0& Trigonalplanar& Trigonalplanar\\ A{B}_4& 4& 0& Tetrahedral& Tetrahedral\\ A{B}_5& 5& 0& Trigonalbipyramidal& Trigonalbipyramidal\\ A{B}_6& 6& 0& Octahedral& Octahedral\end{array}$

Explanation of Solution

In the given molecule, $\underset{\_}{C}{H}_{2}O$ (According to the above mentioned table it is an example for $A{X}_{2}B$ type molecule) two hydrogen atoms and an oxygen atom are bonded to the carbon atom. Since there are three atoms bonded to the central atom without any lone pair of electrons on central atom, both the electron pair and molecular geometries are trigonal planar.

A molecule with $A{X}_{2}B$ type ($\underset{\_}{C}{H}_{2}O$) possesses $s{p}^2$ hybridization.

Interpretation Introduction

Interpretation:

The electron pair and molecular geometry for the atom underlined in the given molecule has to be specified and the hybrid orbital set used by this atom in the molecule should be described.

• $\underset{\_}{N}{H}_4^{+}$

Concept Introduction:

• Hybridization is the mixing of valence atomic orbitals to get equivalent hybridized orbitals that having similar characteristics and energy.
• Geometry of a molecule can be predicted by knowing its hybridization.
• ${\text{sp}}^{\text{3}}$ hybrid orbitals is produced by hybridization of single s-orbital and three p-orbital.
• ${\text{sp}}^2$ hybrid orbitals is produced by hybridization between one s-orbital and two p-orbitals.
• $\text{sp }$ hybrid orbitals is produced by hybridization of single s-orbital and single p-orbital.

Electron geometry is the shape of a molecule predicted by considering both bond electron pairs and lone pair of electrons.

Molecular geometry is the shape of a molecule predicted by considering only bond pair of electrons.

Geometry of different types of molecule with respect to the hybridizations are mentioned are mentioned below,

$\begin{array}{l}\begin{array}{cccc}Typeofmolecule& Hybridaization& \begin{array}{l}Atomicorbitals\\ usedforhybridaization\end{array}& Geometry\\ A{X}_2& sp& 1s+1p& Linear\\ A{X}_3,A{X}_{2}B& s{p}^2& 1s+2p& Trigonalplanar\\ A{X}_4,A{X}_{3}B,A{X}_2{B}_2& s{p}^3& 1s+3p& Tetrahedral\\ A{X}_5,A{X}_{4}B,A{X}_3{B}_2,A{X}_2{B}_3& s{p}^{3}d& 1s+3p+1d& Trigonalbipyramidal\\ A{X}_6,A{X}_{5}B,A{X}_4{B}_2& s{p}^3{d}^2& 1s+3p+2d& Octahedral\end{array}\\ A\to Centralatom\\ X\to AtomsbondedtoA\\ B\to NonbondingelectronpairsonA\end{array}$

Geometry of different type of molecules with respect to the number of electron pairs are mentioned below,

$\begin{array}{ccccc}\begin{array}{l}Typeof\\ Molecule\end{array}& \begin{array}{l}No.ofatoms\\ bondedtocentralatoms\end{array}& \begin{array}{l}No.oflonepairs\\ oncentralatom\end{array}& \begin{array}{l}Arrangementof\\ electronpairs\end{array}& \begin{array}{l}Molecular\\ Geometry\end{array}\\ A{B}_2& 2& 0& Linear& Linear\\ A{B}_3& 3& 0& Trigonalplanar& Trigonalplanar\\ A{B}_4& 4& 0& Tetrahedral& Tetrahedral\\ A{B}_5& 5& 0& Trigonalbipyramidal& Trigonalbipyramidal\\ A{B}_6& 6& 0& Octahedral& Octahedral\end{array}$

Explanation of Solution

In the given molecule, $\underset{\_}{N}{H}_4^{+}$ (According to the above mentioned table it is an example for $A{X}_4$ type molecule) four hydrogen atoms are bonded to the nitrogen atom. Since there are four atoms bonded to the central atom without any lone pair of electrons on central atom, both the electron pair and molecular geometries are tetrahedral.

A molecule with $A{X}_4$ type ($\underset{\_}{N}{H}_4^{+}$) possesses $s{p}^3$ hybridization.