Question

Why is a carbocation $s{{p}^{2}}$ hybridized around the central carbon?

Answer 1

In valence bond theory, orbital hybridisation is the notion of combining atomic orbitals into new hybrid orbitals (with different energies, shapes, and other properties than the component atomic orbitals) appropriate for electron pairing to create chemical bonds. The VSEPR theory is a chemical model that uses the number of electron pairs surrounding a core atom to predict the geometry of individual molecules.

Complete answer:
A carbocation, according to the IUPAC, is any cation with an even number of electrons and a major amount of the positive charge on a carbon atom. Carbocation and carbonium ion were used interchangeably before Olah and associates discovered five-coordinate carbocations. Olah recommended that the carbonium ion be redefined as a carbocation with any form of three-center two-electron bonding, while the carbenium ion be defined as a carbocation with only two-center two-electron bonds and a three-coordinate positive carbon.

One s and two p atomic orbitals are mixed in the $s{{p}^{2}}$ hybridization, which entails the promotion of one electron from the s orbital to one of the 2p atomic orbitals. When these atomic orbitals are combined, three new hybrid orbitals with the same energy level are created. The hybrid orbitals have a greater energy than the s orbital but a lower energy than the p orbitals, but they are closer to the p orbitals in terms of energy. The newly generated hybrid orbitals produce trigonal structures with a 120-degree molecular geometry.
The valence shell electron pair repulsion hypothesis posits that electron pairs surrounding a core atom form Platonic solids, which reduces electrostatic repulsion between bonded and lone pairs. Two electron pairs are organised linearly, three electron pairs in a trigonal plane, and four electron pairs in a tetrahedron. Because there are three bonding electron pairs around the carbon in $^{+}C{{R}_{3}}$ , the bond angle $\angle R-C-R$ is ${{120}^{o}}$ . Because boron halides and alkyls are isoelectronic with $^{+}C{{R}_{3}}$, $B{{R}_{3}}$ is neutral in this situation.

Note:
Carbon will be $s{{p}^{2}}$ hybridised in carbocation, and its form will be trigonal planar. There's also an empty p orbital, indicating that it's electron-poor. The valence shell of carbon has 6 electrons. It is an electron-deficient species, commonly known as an electrophile, as a result of this.