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.

![http://en.wikipedia.org/wiki/Reactive_intermediate]()

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.

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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).

![http://quizlet.com/62637858/molecular-geometry-and-hybridization-flash-cards/]()

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.