Question

The shape of $\text{CH}_{\text{3}}^{\text{+}}$ is ……………….. (planar/bent).

Answer 1

In $s{{p}^{2}}$ molecule one pi-bond is required to link the double bond between the carbons. Also, only three sigma bonds are formed per carbon atom. The positive charge makes $3$ available electrons. It forms $3$ single bonds with hydrogen atoms.

Complete step by step answer:
In VSEPR theory, hybridization is the process of adding atomic orbital’s with the new hybrid orbitals, which is appropriate for the pairing of electrons to form chemical bonds. This theory is used to explain molecular geometry and atomic bonding properties.
In $s{{p}^{2}}$ molecule, carbon is $s{{p}^{2}}$hybridized. Here, one pi-bond $(\pi -\text{bond})$ is required to link the double bond between the carbons. Also, only three sigma bonds $(\sigma -\text{bond})$are formed per carbon atom. So, simply it involves the promotion of one electron in the $s$- orbital to one $2p$ orbital and thereby, the one $s$ and two $p$ orbitals of the same shell and of equal energies get mixed to form $3$ equivalent orbital’s. So, in $s{{p}^{2}}$ hybridization, the $2s$ orbital is mixed with the two available $2p$ orbitals (known as $2px$ and $2py$ orbitals) out of three $2p$ orbitals. So, the last or $2pz$orbital remains unhybridized. So, we get the formation of the total of the three $s{{p}^{2}}$ orbital’s having one unhybridized $p$ orbital.
Now, $\text{CH}_{3}^{+}$ has $s{{p}^{2}}$ carbon atom. So, that means in its carbonation, carbon is $s{{p}^{2}}$ hybridized. There are $4$ valence electrons in the $\text{CH}_{3}^{+}$molecule. The positive charge makes $3$ available electrons. It forms $3$ single bonds with hydrogen $(\text{H})$ atoms and thereby, forms the planar structure with $s{{p}^{2}}$ hybridization. So, the shape of $\text{CH}_{3}^{+}$ is planer.

Additional information:
Redistribution of the energy of orbitals of individual atoms to give orbital’s of equivalent energy happening when two atomic orbitals combine to form hybrid orbital in a molecule. This process is called hybridization and new orbital’s is hybrid orbitals.
It is not necessary that all the half-filled orbitals must participate in hybridization, even the completely filled.
Hybridization happens only during the bond formation and not in an isolated gaseous atom.

Note: Things should be noted that this mixture of $s$ and $p$ orbital forms the trifocal symmetry and maintains $120{}^\circ$ between each bond. Also, it should be remembered that all the three orbitals are placed in one plane having $120{}^\circ$ between each bond. So, also, we can say that the shape of $\text{CH}_{3}^{+}$ is triangular planar.