Question

What is the type of hybridisation of each carbon in the following compound?
(a) $C{H}_{3}Cl$
(b) ${\left(C{H}_3\right)}_{2}CO$
(c) $C{H}_{3}CN$
(d) $HCON{H}_2$
(e) $C{H}_{3}CH=CHCN$

Answer 1

Step 1: Definition of hybridization:
Hybridization is intermixing of atomic orbitals of different shapes and nearly the same energy to give the same number of orbitals of the same shape, equal energy and orientation such that there is minimum repulsion between these hybridized orbitals. The new orbitals thus formed are known as hybrid orbitals.

a) $C{H}_{3}Cl$

Step 1: Definition of hybridization:
Hybridization is intermixing of atomic orbitals of different shapes and nearly the same energy to give the same number of orbitals of the same shape, equal energy and orientation such that there is minimum repulsion between these hybridized orbitals. The new orbitals thus formed are known as hybrid orbitals.

Step 2: Lewis’s structure of the compound:
Draw the Lewis structure of given compound to visualize the number and types of bonds of each carbon atoms. After that number each carbon atoms to analyze the hybridization. Lewis’s structure of given molecular formula can be drawn as.


Step 3: Hybridization of carbon:
Carbon can make maximum $4$ bonds

If all are sigma $\sigma$ bonds $\Rightarrow s{p}^3$ hybridisation
If $3$ sigma and one $\pi$ bonds $\Rightarrow s{p}^2$ hybridisation
If $2$ sigma and $2\pi$ bonds $\Rightarrow sp$ hybridization.

$\because$ We know,
$\text{Single bond}=\text{Always sigma bond}$
$\text{Double bond}=1\sigma \text{bond}+1\pi \text{bond}$
$\text{Triple bond}=1\sigma \text{bond}+2\pi \text{bond}$


There is only one carbon atom present in given molecular formula and according to structure, it is making $4$ sigma bonds (single bond) with the help of one $s$ hybrid orbital $\&3p$ hybrid orbital. Hence, given carbon is $s{p}^3$ hybridized.

Final answer: Hybridisation of carbon is $s{p}^3$

(b) ${\left(C{H}_3\right)}_{2}CO$

Step 1: Definition of hybridization:
Hybridization is intermixing of atomic orbitals of different shapes and nearly the same energy to give the same number of orbitals of the same shape, equal energy and orientation such that there is minimum repulsion between these hybridized orbitals. The new orbitals thus formed are known as hybrid orbitals.

Step 2: Lewis’s structure of the compound:
Draw the Lewis structure of given compound to visualize the number and types of bonds of each carbon atoms. After that number each carbon atoms to analyze the hybridization.

Lewis’s structure of given molecular formula can be drawn as.


Step 3: Hybridization of carbon: Carbon can make maximum $4$ bonds.

If all are sigma $\sigma$ bonds $\Rightarrow s{p}^3$ hybridisation
If $3$ sigma and one $\pi$ bonds $\Rightarrow s{p}^2$ hybridisation
If $2$ sigma and $2\pi$ bonds $\Rightarrow sp$ hybridization.

$\because$ We know,
$\text{Single bond}=\text{Always sigma bond}$
$\text{Double bond}=1\sigma \text{bond}+1\pi \text{bond}$
$\text{Triple bond}=1\sigma \text{bond}+2\pi \text{bond}$


There are three carbon atoms present in given molecular formula and according to structure, $C_1$ and $C_3$ are making $4$ sigma bonds (single bond) with the help of one $s$ hybrid orbital $\&3p$ hybrid orbital and $C_2$ making $3\sigma$ bonds $(3\sigma \text{bond}+1\pi \text{bond})$ with the help of one $s$ hybrid orbital $\&2p$ hybrid orbital. Hence, $C_1$ and $C_3$ carbons are $s{p}^3$ and $C_2$ carbon is $s{p}^2$ hybridized.

Final answer: Hybridisation of carbons: $C_1{C}_2{C}_3$
$s{p}^{3}s{p}^{2}s{p}^3$

(c) $C{H}_{3}CN$

Step 1: Definition of hybridization:
Hybridization is intermixing of atomic orbitals of different shapes and nearly the same energy to give the same number of orbitals of the same shape, equal energy and orientation such that there is minimum repulsion between these hybridized orbitals. The new orbitals thus formed are known as hybrid orbitals.

Step 2: Lewis’s structure of the compound:
Draw the Lewis structure of given compound to visualize the number and types of bonds of each carbon atoms. After that number each carbon atoms to analyze the hybridization.
Lewis’s structure of given molecular formula can be drawn as.


Step 3: Hybridization of carbon: Carbon can make maximum $4$ bonds
If all are sigma $\sigma$ bonds $\Rightarrow s{p}^3$ hybridisation
If $3$ sigma and one $\pi$ bonds $\Rightarrow s{p}^2$ hybridisation
If $2$ sigma and $2\pi$ bonds $\Rightarrow sp$ hybridization.

$\because$ We know,
$\text{Single bond}=\text{Always sigma bond}$
$\text{Double bond}=1\sigma \text{bond}+1\pi \text{bond}$
$\text{Triple bond}=1\sigma \text{bond}+2\pi \text{bond}$


There are two carbon atoms present in given molecular formula and according to structure, $C_1$ is making $4$ sigma bonds (single bond) with the help of one $s$ hybrid orbital $\&3p$ hybrid orbital and $C_2$ making $2\sigma$ bonds $(2\sigma \text{bonds}+2\pi \text{bonds})$ with the help of one $s$ hybrid orbital $\&$ one $p$ hybrid orbital.
Hence, $C_1$ is $s{p}^3$ and $C_2$ is $sp$ hybridized.

Final answer: Hybridisation of carbons: $C_1{C}_2$
$s{p}^{3}sp$


(d) $HCON{H}_2$

Step 1: Definition of hybridization:
Hybridization is intermixing of atomic orbitals of different shapes and nearly the same energy to give the same number of orbitals of the same shape, equal energy and orientation such that there is minimum repulsion between these hybridized orbitals. The new orbitals thus formed are known as hybrid orbitals.

Step 2: Lewis’s structure of the compound:
Draw the Lewis structure of given compound to visualize the number and types of bonds of each carbon atoms. After that number each carbon atoms to analyze the hybridization.

Lewis’s structure of given molecular formula can be drawn as.


Step 3: Hybridization of carbon: Carbon can make maximum $4$ bonds
If all are sigma $\sigma$ bonds $\Rightarrow s{p}^3$ hybridisation
If $3$ sigma and one $\pi$ bonds $\Rightarrow s{p}^2$ hybridisation
If $2$ sigma and $2\pi$ bonds $\Rightarrow sp$ hybridization.

$\because$ We know,
$\text{Single bond}=\text{Always sigma bond}$
$\text{Double bond}=1\sigma \text{bond}+1\pi \text{bond}$
$\text{Triple bond}=1\sigma \text{bond}+2\pi \text{bond}$


There is only one carbon atom present in given molecular formula and according to structure, it is making $3$ sigma bonds $(3\sigma \text{bond}+1\pi \text{bond})$ with the help of $1s$ hybrid orbital $\&2p$ hybrid orbital. Hence, given carbon is $s{p}^2$ hybridized..

Final answer: Hybridisation of carbons is $s{p}^2$

(e) $C{H}_{3}CH=CHCN$

Step 1: Definition of hybridization:
Hybridization is intermixing of atomic orbitals of different shapes and nearly the same energy to give the same number of orbitals of the same shape, equal energy and orientation such that there is minimum repulsion between these hybridized orbitals. The new orbitals thus formed are known as hybrid orbitals.

Step 2: Lewis’s structure of the compound:
Draw the Lewis structure of given compound to visualize the number and types of bonds of each carbon atoms. After that number each carbon atoms to analyze the hybridization.

Lewis’s structure of given molecular formula can be drawn as.


Step 3: Hybridization of carbon: Carbon can make maximum $4$ bonds.

If all are sigma $\sigma$ bonds $\Rightarrow s{p}^3$ hybridisation
If $3$ sigma and one $\pi$ bonds $\Rightarrow s{p}^2$ hybridisation
If $2$ sigma and $2\pi$ bonds $\Rightarrow sp$ hybridization.

$\because$ We know,
$\text{Single bond}=\text{Always sigma bond}$
$\text{Double bond}=1\sigma \text{bond}+1\pi \text{bond}$
$\text{Triple bond}=1\sigma \text{bond}+2\pi \text{bond}$

There are four carbon atoms present in given molecular formula and according to structure, $C_1$ making $4$ sigma bonds (single bond) with the help of one $s$ hybrid orbital $\&3p$ hybrid orbital and $C_2$ and $C_3$ making $3\sigma$ bonds $(3\sigma \text{bond}+1\pi \text{bond})$ with the help of one $s$ hybrid orbital $\&2p$ hybrid orbital while $C_4$ making $2\sigma$ bonds $(2\sigma \text{bond}+2\pi \text{bond})$ with the help of one $s$ hybrid orbital and one $p$ hybrid orbital.
Hence, $C_1$ is $s{p}^3$, $C_2$, $C_3$ are $s{p}^2$ and $C_4$ is $sp$ hybridized.

Final answer: Hybridisation of carbons:
$C_1{C}_2{C}_3{C}_4$
$s{p}^{3}s{p}^{2}s{p}^{2}sp$