To determine the number of valence electrons in the (CH3)3C+ cation, we should first count the total number of valence electrons for the neutral molecule (CH3)3C and then subtract one electron because of the positive charge.
In the neutral molecule, carbon (C) has 4 valence electrons, and hydrogen (H) has 1 valence electron.
The structure of (CH3)3C, also known as tert-butyl group, is such that the central carbon atom is bonded to three methyl groups (CH3) and typically has one more bond, possibly to another atom or group in a larger molecule. For the sake of counting valence electrons, we'll just consider the (CH3)3C part.
Each CH3 group:
- 1 carbon with 4 valence electrons
- 3 hydrogens with 1 valence electron each
So for each CH3 group, that's 4 + (3 * 1) = 7 valence electrons.
Since there are three CH3 groups, the total valence electrons for these groups are 3 * 7 = 21 valence electrons.
Now add the valence electrons for the central carbon atom, which has 4 more valence electrons.
The total number of valence electrons in the neutral (CH3)3C molecule is:
21 (from three CH3 groups) + 4 (from the central C) = 25 valence electrons.
However, we are dealing with the (CH3)3C+ cation, which means we must remove one electron due to the positive charge.
25 valence electrons (neutral) - 1 electron (positive charge) = 24 valence electrons.
So, the (CH3)3C+ cation has 24 valence electrons.
How many valence electrons in (CH3)3C+
1 answer