Question

Write the state of hybridisation of carbon in the following compounds and shapes of each of the molecules.
(a) $H_{2}C=0$
(b) $C{H}_{3}F$
(c) $HC\equiv N$

Answer 1

(a) 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

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

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

According to structure, it is making $3$ sigma bonds $\left(3\sigma \text{bond + 1}\pi \text{bond}\right)$ with the help of one $s$ hybrid orbital & $2p$ hybrid orbital. Hence, given carbon is $s{p}^2$ hybridized.

Step-4: shape of the molecule
Hybridization of molecules and corresponding shape according to the VSEPR theory shown below.

S.No	Hybridization	Shape of the molecule
$1$	$s{p}^3$	tetrahedral
$2$	$s{p}^2$	trigonal planar
$3$	$sp$	linear

Since here carbon $s{p}^2$ hybridized. Hence, the shape of the molecule is trigonal planar.
Final answer: $\text{Hybridisation of carbon}\Rightarrow s{p}^2$
$\text{Shape of the molecule}\Rightarrow \text{trigonal planar}$


(b) 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
$\text{If all are sigma}\sigma \text{bonds}\Rightarrow s{p}^3\text{hybridisation}$
$\text{If 3 sigma and one}\pi \text{bonds}\Rightarrow s{p}^2\text{hybridisation}$
$\text{If 2 sigma and one 2}\pi \text{bonds}\Rightarrow sp\text{hybridisation}$

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

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.

Step-4: shape of the molecule
Hybridization of molecules and corresponding shape according to the VSEPR theory shown below.

S.No	Hybridization	Shape of the molecule
$1$	$s{p}^3$	tetrahedral
$2$	$s{p}^2$	trigonal planar
$3$	$sp$	linear

Since here carbon $s{p}^3$ hybridized. Hence, the shape of the molecule is trigonal planar.
Final answer: $\text{Hybridisation of carbon}\Rightarrow s{p}^3$
$\text{Shape of the molecule}\Rightarrow \text{tetrahedral}$

(c) 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
$\text{If all are sigma}\sigma \text{bonds}\Rightarrow s{p}^3\text{hybridisation}$
$\text{If 3 sigma and one}\pi \text{bonds}\Rightarrow s{p}^2\text{hybridisation}$
$\text{If 2 sigma and one 2}\pi \text{bonds}\Rightarrow sp\text{hybridisation}$

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

According to structure, it is making $2$ sigma bonds $(2\sigma \text{bond}+2\pi \text{bond})$ with the help of one s hybrid orbital & one p hybrid orbital. Hence, given carbon is sp hybridized.

Step-4: shape of the molecule
Hybridization of molecules and corresponding shape according to the VSEPR theory shown below.

S.No	Hybridization	Shape of the molecule
$1$	$s{p}^3$	tetrahedral
$2$	$s{p}^2$	trigonal planar
$3$	$sp$	linear

Since here carbon $sp$ hybridized. Hence, the shape of the molecule is trigonal planar.
Final answer: $\text{Hybridisation of carbon}\Rightarrow sp$
$\text{Shape of the molecule}\Rightarrow \text{linear}$