According to VSEPR theory, in which species do all the atoms lie in the same plane?
$\ce{CH3+}$
$\ce{CH3-}$
(A) 1 only (B) 2 only (C) both 1 and 2 (D) neither 1 nor 2
My try: Because $\ce{CH3}$ in both forms only are bonded with three hydrogens, they both have one lone pair, which contributes to their tetrahedral structure. So I got D. However, the answer key says that B is the correct answer. How do I get this answer?
The VSEPR model is a rather simple model that actually helps in this case. It assumes that anything ‘on’ a central atom — atoms that it is bound to and lone pairs — requests almost equal amounts of space, and therefore the compound’s final geometry is dictated by the number (and type; lone pairs are deemed ‘invisible’) of substituents only.
It is important to recognise that only lone pairs (and radicals) count towards the VSEPR model; empty orbitals do not. The reasoning is that something cannot take up space if it is empty.
Applying the theory to the methyl cation should give you three substituents: the three hydrogens. These three will, according to the theory, align themselves in a trigonal planar fashion around the central carbon, giving a perfect bond angle of $120^\circ$ between them. Thus, the methyl cation is planar.
Applying it to the methyl anion should give you four substituents: the three hydrogens we had previously plus the lone pair. The lone pair takes up space, so including it into our visualisation would give us a tetrahedral geometry. However, it is also invisible (we can only ‘see’ full atoms) hence the final structure is predicted to be trigonal pyramidal with the four atoms not occupying the same plane.
Therefore, assuming you copied accurately, the answer key should have given A as the correct answer.
Unless some more advanced concept is at play, the answers are incorrect and the correct answer should be A.
This is because with structure I, the carbon has no lone pairs making it trigonal planar, while in structure II, the carbon has a lone pair, making it trigonal pyramidal
