Question
Explain geometry of methane molecule on the basis of Hybridisation.
✓
Answer
Formation of methane $(CH_4)$ molecule on the basis of $sp^3$ hybridization:
i. Methane molecule $(CH_4)$ has one carbon atom and four hydrogen atoms.
ii. The ground state electronic configuration of $C (Z = 6)$ is $1s^2$ $2 p_{\frac{1}{x}}^1 2 p _{ y }^1 2 p _{ z }^1$
Electronic configuration of carbon:
iii. In order to form four equivalent bonds with hydrogen, the $2\ s$ and $2\ p$ orbitals of $C$ -atom undergo $sp ^3$ hybridization.
iv. One electron from the $2\ s$ orbital of carbon atom is excited to the $2\ pz$ orbital. Then the four orbitals $2 s, px , py$ and pz mix and recast to form four new sp ${ }^3$ hybrid orbitals having same shape and equal energy. They are maximum apart and have tetrahedral geometry with $H-C-H$ bond angle of $109^{\circ} 28^{\prime}$. Each hybrid orbital contains one unpaired electron.
v. Each of these $sp ^3$ hybrid orbitals with one electron overlap axially with the $1\ s$ orbital of hydrogen atom to form one $C - H$ sigma bond. Thus, in $CH _4$ molecule, there are four $C - H$ bonds formed by the $sp ^3$-s overlap.
Diagram:
i. Methane molecule $(CH_4)$ has one carbon atom and four hydrogen atoms.
ii. The ground state electronic configuration of $C (Z = 6)$ is $1s^2$ $2 p_{\frac{1}{x}}^1 2 p _{ y }^1 2 p _{ z }^1$
Electronic configuration of carbon:
iii. In order to form four equivalent bonds with hydrogen, the $2\ s$ and $2\ p$ orbitals of $C$ -atom undergo $sp ^3$ hybridization.
iv. One electron from the $2\ s$ orbital of carbon atom is excited to the $2\ pz$ orbital. Then the four orbitals $2 s, px , py$ and pz mix and recast to form four new sp ${ }^3$ hybrid orbitals having same shape and equal energy. They are maximum apart and have tetrahedral geometry with $H-C-H$ bond angle of $109^{\circ} 28^{\prime}$. Each hybrid orbital contains one unpaired electron.
v. Each of these $sp ^3$ hybrid orbitals with one electron overlap axially with the $1\ s$ orbital of hydrogen atom to form one $C - H$ sigma bond. Thus, in $CH _4$ molecule, there are four $C - H$ bonds formed by the $sp ^3$-s overlap.
Diagram:
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