Chapter 23, Problem 2PS

(a)

Interpretation Introduction

Interpretation:

The hybridization and bond angles of carbon is bonded via four single bonds to adjacent atoms should be determined.

Concept introduction:

Carbon atomic number 6. Electronic configuration -${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^2$.

Carbon has tetra valency. It is bonded with four bonds to adjacent atoms or molecules.

Single covalent bond  - one pair of each electrons are shared.

Double covalent bond – two pair of electrons are shared.

Triple covalent bond – Three pairs of electron are shared.

Hybridization: The phenomenon of formation new orbitals by the mixing of atomic orbital’s with equal energy.

Sp hybridization: Mixing of one –‘s’ orbital and one ‘p’ orbital. And form new hybrid orbital. Angle is ${\text{180}}^{\text{0}}$.

Example - ${\text{BeCl}}_{\text{2}}$

${\text{Sp}}^{\text{2}}$ hybridization: Mixing of one –‘s’ orbital and two ‘p’ orbital. And form new hybrid orbitals. Angle is ${\text{120}}^{\text{0}}$.

Example –  Ethylene.

${\text{Sp}}^3$ hybridization: Mixing of one –‘s’ orbital and three ‘p’ orbital. And form new hybrid orbitals. Angle is $\text{109}{\text{.28}}^0$.

Example –  Ethane.

${\text{Sp}}^{\text{3}}\text{d}$ hybridization: Mixing of one –‘s’ orbital and three ‘p’ orbital and one ‘d’ orbital. And form new hybrid orbitals.

Example –  ${\text{PCl}}_{\text{5}}$

Bond angle is the angle between two bonds of a molecule and it is determined based on the electron-domain geometry.

[Bond angles: tetrahedral = $\text{109}{\text{.5}}^{\text{o}}$, trigonal planar = ${\text{120}}^{\text{o}}$, T-shape = ${\text{90}}^{\text{o}}$]

Answer to Problem 2PS

The carbon is ${\text{sp}}^{\text{3}}$ hybridized and the bond angle is $\text{109}{\text{.28}}^0$

Explanation of Solution

Carbon is bonded via four single bonds to adjacent atoms. Carbon is bonded with four single four hydrogen atoms.

The Lewis structure as shown below.

Let’s write the carbon electronic configuration:

$\begin{array}{l}\text{Electronic}\text{configuration}\text{of}1s2s2p\\ \text{C}\text{in}\text{ground}\text{state}\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow }\boxed{\uparrow }\boxed{{}^{}}\\ \\ \text{Electronic}\text{configuration}\text{of}1s2s2p\\ \text{C}\text{in}\text{excited}\text{state}\boxed{\uparrow \downarrow }\boxed{\boxed{\uparrow }\boxed{\uparrow }\boxed{\uparrow }\boxed{\uparrow }}\\ \\ {\text{sp}}^{\text{3}}h\text{ybridization}\text{in}\text{C}\boxed{\uparrow \downarrow }\boxed{\uparrow }\boxed{\uparrow }\boxed{\uparrow }\boxed{\uparrow }\\ \text{four}{\text{sp}}^{\text{3}}\\ \text{hybridized}\text{orbitals}\end{array}$

Here, the carbon is ${\text{sp}}^{\text{3}}$ hybridized and thus the bond angle is $\text{109}{\text{.28}}^0$

(b)

Interpretation Introduction

Interpretation:

The hybridization and bond angle of carbon which is bonded via two single bonds and one pi bond should be determined.

Concept introduction:

Carbon atomic number 6. Electronic configuration -${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^2$.

Carbon has tetra valency. It is bonded with four bonds to adjacent atoms or molecules.

Single covalent bond  - one pair of each electrons are shared.

Double covalent bond – two pair of electrons are shared.

Triple covalent bond – Three pairs of electron are shared.

Hybridization: The phenomenon of formation new orbitals by the mixing of atomic orbital’s with equal energy.

Sp hybridization: Mixing of one –‘s’ orbital and one ‘p’ orbital. And form new hybrid orbital. Angle is ${\text{180}}^{\text{0}}$.

Example - ${\text{BeCl}}_{\text{2}}$

${\text{Sp}}^{\text{2}}$ hybridization: Mixing of one –‘s’ orbital and two ‘p’ orbital. And form new hybrid orbitals. Angle is ${\text{120}}^{\text{0}}$.

Example –  Ethylene.

${\text{Sp}}^3$ hybridization: Mixing of one –‘s’ orbital and three ‘p’ orbital. And form new hybrid orbitals. Angle is $\text{109}{\text{.28}}^0$.

Example –  Ethane.

Bond angle is the angle between two bonds of a molecule and it is determined based on the electron-domain geometry.

[Bond angles: tetrahedral = $\text{109}{\text{.5}}^{\text{o}}$, trigonal planar = ${\text{120}}^{\text{o}}$, T-shape = ${\text{90}}^{\text{o}}$]

Answer to Problem 2PS

The carbon is ${\text{sp}}^2$ hybridized and the bond angle is ${\text{120}}^{\text{0}}$.

Explanation of Solution

Carbon is bonded via two single bonds and one pi bond.

Side overlapping of the two adjacent carbon atoms of orbitals form pi-bond.

The Lewis structure as shown below.

Let’s find the hybridization:

$\begin{array}{l}\text{Electronic}\text{configuration}\text{of}1s2s2p\\ \text{C}\text{in}\text{ground}\text{state}\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow }\boxed{\uparrow }\boxed{{}^{}}\\ \\ \text{Electronic}\text{configuration}\text{of}1s2s2p\\ \text{C}\text{in}\text{excited}\text{state}\boxed{\uparrow \downarrow }\boxed{\boxed{\uparrow }\boxed{\uparrow }\boxed{\uparrow }}\boxed{\uparrow }\\ \\ {\text{sp}}^{2}h\text{ybridization}\text{in}\text{C}\boxed{\uparrow \downarrow }\boxed{\uparrow }\boxed{\uparrow }\boxed{\uparrow }\boxed{\uparrow }\\ \text{Three}{\text{sp}}^{2}un\text{hybridized}\\ \text{hybridized}\text{orbitals}\text{orbital}\end{array}$

Here, the carbon is ${\text{sp}}^2$ hybridized and thus the bond angle is ${\text{120}}^{\text{0}}$

(c)

Interpretation Introduction

Interpretation:

The hybridization and bond angles of carbon is bonded via one single bond and one triple bond should be determined.

Concept introduction:

Carbon atomic number 6. Electronic configuration -${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^2$.

Carbon has tetra valency. It is bonded with four bonds to adjacent atoms or molecules.

Single covalent bond  - one pair of each electrons are shared.

Double covalent bond – two pair of electrons are shared.

Triple covalent bond – Three pairs of electron are shared.

Hybridization: The phenomenon of formation new orbitals by the mixing of atomic orbital’s with equal energy.

Sp hybridization: Mixing of one –‘s’ orbital and one ‘p’ orbital. And form new hybrid orbital. Angle is ${\text{180}}^{\text{0}}$.

Example - ${\text{BeCl}}_{\text{2}}$

${\text{Sp}}^{\text{2}}$ hybridization: Mixing of one –‘s’ orbital and two ‘p’ orbital. And form new hybrid orbitals. Angle is ${\text{120}}^{\text{0}}$.

Example –  Ethylene.

Bond angle is the angle between two bonds of a molecule and it is determined based on the electron-domain geometry.

[Bond angles: tetrahedral = $\text{109}{\text{.5}}^{\text{o}}$, trigonal planar = ${\text{120}}^{\text{o}}$, T-shape = ${\text{90}}^{\text{o}}$]

Answer to Problem 2PS

The carbon is $\text{sp}$ hybridized and the bond angle is ${\text{180}}^{\text{0}}$

Explanation of Solution

Carbon is bonded via one single bond and one triple bond. Three pairs of electron are shared by two adjacent carbon atoms.

The Lewis structure as shown below.

Let’s find the Hybridization:

$\begin{array}{l}\text{Electronic}\text{configuration}\text{of}1s2s2p\\ \text{C}\text{in}\text{ground}\text{state}\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow }\boxed{\uparrow }\boxed{{}^{}}\\ \\ \text{Electronic}\text{configuration}\text{of}1s2s2p\\ \text{C}\text{in}\text{excited}\text{state}\boxed{\uparrow \downarrow }\boxed{\boxed{\uparrow }\boxed{\uparrow }}\boxed{\uparrow }\boxed{\uparrow }\\ \\ \text{sp}h\text{ybridization}\text{in}\text{C}\boxed{\uparrow \downarrow }\boxed{\uparrow }\boxed{\uparrow }\boxed{\uparrow }\boxed{\uparrow }\\ \text{Two}\text{sp}un\text{hybridized}\\ \text{hybridized}\text{orbitals}\text{orbital}\end{array}$

Here, the carbon is $\text{sp}$ hybridized and thus the bond angle is ${\text{180}}^{\text{0}}$

(d)

Interpretation Introduction

Interpretation:

The hybridization and bond angles of carbon which are bonded via two double bonds should be determined.

Concept introduction:

Carbon atomic number 6. Electronic configuration -${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^2$.

Carbon has tetra valency. It is bonded with four bonds to adjacent atoms or molecules.

Single covalent bond  - one pair of each electrons are shared.

Double covalent bond – two pair of electrons are shared.

Triple covalent bond – Three pairs of electron are shared.

Hybridization: The phenomenon of formation new orbitals by the mixing of atomic orbital’s with equal energy.

Sp hybridization: Mixing of one –‘s’ orbital and one ‘p’ orbital. And form new hybrid orbital. Angle is ${\text{180}}^{\text{0}}$.

Example - ${\text{BeCl}}_{\text{2}}$

${\text{Sp}}^{\text{2}}$ hybridization: Mixing of one –‘s’ orbital and two ‘p’ orbital. And form new hybrid orbitals. Angle is ${\text{120}}^{\text{0}}$.

Example –  Ethylene.

${\text{Sp}}^3$ hybridization: Mixing of one –‘s’ orbital and three ‘p’ orbital. And form new hybrid orbitals. Angle is $\text{109}{\text{.28}}^0$.

Example –  Ethane.

Bond angle is the angle between two bonds of a molecule and it is determined based on the electron-domain geometry.

[Bond angles: tetrahedral = $\text{109}{\text{.5}}^{\text{o}}$, trigonal planar = ${\text{120}}^{\text{o}}$, T-shape = ${\text{90}}^{\text{o}}$]

Answer to Problem 2PS

Explanation of Solution

Carbon is bonded via two double bonds. The three adjacent carbon atoms orbitals overlap to form two pi bonds.

The Lewis structure as shown below.

Let’s find the hybridization:

1. (a) Hybridization of the terminal carbon atoms

$\begin{array}{l}\text{Electronic}\text{configuration}\text{of}1s2s2p\\ \text{C}\text{in}\text{ground}\text{state}\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow }\boxed{\uparrow }\boxed{{}^{}}\\ \\ \text{Electronic}\text{configuration}\text{of}1s2s2p\\ \text{C}\text{in}\text{excited}\text{state}\boxed{\uparrow \downarrow }\boxed{\boxed{\uparrow }\boxed{\uparrow }\boxed{\uparrow }}\boxed{\uparrow }\\ \\ {\text{sp}}^{2}h\text{ybridization}\text{in}\text{C}\boxed{\uparrow \downarrow }\boxed{\uparrow }\boxed{\uparrow }\boxed{\uparrow }\boxed{\uparrow }\\ \text{Three}{\text{sp}}^{2}un\text{hybridized}\\ \text{hybridized}\text{orbitals}\text{orbital}\end{array}$

Here, the carbon is ${\text{sp}}^2$ hybridized and thus the bond angle is ${\text{120}}^{\text{0}}$

1. (b) Hybridization of the central carbon atom

$\begin{array}{l}\text{Electronic}\text{configuration}\text{of}1s2s2p\\ \text{C}\text{in}\text{ground}\text{state}\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow }\boxed{\uparrow }\boxed{{}^{}}\\ \\ \text{Electronic}\text{configuration}\text{of}1s2s2p\\ \text{C}\text{in}\text{excited}\text{state}\boxed{\uparrow \downarrow }\boxed{\boxed{\uparrow }\boxed{\uparrow }}\boxed{\uparrow }\boxed{\uparrow }\\ \\ \text{sp}h\text{ybridization}\text{in}\text{C}\boxed{\uparrow \downarrow }\boxed{\uparrow }\boxed{\uparrow }\boxed{\uparrow }\boxed{\uparrow }\\ \text{Two}\text{sp}un\text{hybridized}\\ \text{hybridized}\text{orbitals}\text{orbital}\end{array}$

Here, the carbon is $\text{sp}$ hybridized and thus the bond angle is ${\text{180}}^{\text{0}}$

Therefore,

The hybridization and bond angle is