Question

Question #e4180

Answer 1

You do the basic steps as you would for any molecule.

Start by drawing the Lewis structure of the `CH_3^(+)` ion, also known as a carbocation. The total number of valence electrons for the carbocation is 6 - each hydrogen atom brings 1, and the carbon atom brings 3 instead of the usual 4, hence the plus charge.

To determine hybridization, count all the electron-dense regions, i.e covalent bonds or lone pairs, that are located around the central atom - this is know as the steric number.

Notice that the carbon atom bonded to three hydrogen atoms, which implies that its steric number is equal to 3.

This means that the central atom is `"sp"^(2)` hybridized. Carbon will have two of its three p-orbitals hybridized and bonded to the s-orbital of the hydrogen atoms, and one p-orbital, where another pair of bonding electrons would have been, unoccupied (empty).

The geometry of the molecule will be trigonal planar with `120^@` bond angles.

Answer 2

Methylium cation, `CH_3^+`, has 6 valence electrons, all of which are used to make three equivalent covalent C-H bonds, so the orbitals on the central carbon atom are `sp^2` hybridized.

The leftover `2p_z` orbital on carbon is unchanged, and is unoccupied due to the net +1 charge on the ion.