How many electrons are in the p-orbital of the methyl radical (CH3^.)
6C is 1s2 2s2 2p2
Unpair one of the two electrons in the 2s orbital and promote it to the 2p. It now looks like this:
1s2 2s1 2p3
Now hybridize the 2s and 2p orbitals to give sp3 hybrid orbitals (that's four places to bond with hydrogen so the H adds one electron to each of the s and 3p hybrid orbitals. The compound now looks like this:
1s2 [2s2 2p6] with the brackets denoting sp3 hybridization. The extra four electrons have come from 4 H atoms; i.e., 6 from C and 4 from 4 hydrogen atoms = 10 electrons total AND there are now 8 electrons in the sp3 hybrid orbitals (or 2 in the s and 6 in the p if you are separating them). So if we remove one of the H atoms from the p orbitals, that will leave a +1 charge on the CH3^+ ion and I still count 8 electrons total or 2 in the s and 6 in the p if you are separating them. We had 1 e in the s orbital and 3 in the p before the C combined with anything; we now have 2 in the s and 6 in the p. Note: I don't think this agrees with an earlier answer I gave because I counted only electrons from C in that earlier response. Check my thinking.
To determine the number of electrons in the p-orbital of the methyl radical (CH3^.), we need to know the electronic configuration of carbon in the radical.
Carbon (C) has an atomic number of 6, meaning it has 6 electrons. In its ground state, the electronic configuration of carbon is 1s^2 2s^2 2p^2.
The methyl radical (CH3^.) is formed when one electron is removed from a carbon atom, resulting in the loss of one electron from the 2s orbital or one of the three 2p orbitals. Let's assume that one electron is removed from a 2p orbital.
Each p-orbital can hold a maximum of two electrons, and there are three p-orbitals (2px, 2py, and 2pz) in the 2p subshell. Thus, a maximum of six electrons can occupy the p-orbitals.
Since we removed one electron from a 2p orbital in the methyl radical, the number of electrons in the p-orbital will be five (6 - 1 = 5).
Therefore, there are five electrons in the p-orbital of the methyl radical (CH3^.).
To determine the number of electrons in the p-orbital of the methyl radical (CH3^.), we need to consider the electron configuration of carbon (C) and the presence of an additional unpaired electron due to the radical (•) symbol.
Carbon has an atomic number of 6, meaning it has 6 electrons. The electron configuration of carbon is 1s^2 2s^2 2p^2.
In the formation of the methyl radical (CH3^.), one electron is removed, resulting in the loss of one electron from the p-orbital. Since each p-orbital can hold a maximum of 2 electrons, with the loss of one electron, the p-orbital will now have only 1 electron.
Therefore, there is 1 electron in the p-orbital of the methyl radical (CH3^.).