What is the geometry of an alkyl radical? I encountered this question which asks how many possible radicals are formed when $\ce{CH3CH2C(CH3)3}$ is monosubstituted by $\ce{Br2}$. The answer given is 3 while I think it  should be 4, reasoning that there is a "chiral" carbon if the radical is formed on carbon 3. While arguing with my tutor two questions came to my mind:


What is the geometry of the $\ce{CH3\dot{\ce{C}}HC(CH3)3}$ radical?
If the radical is trigonal pyramidal, does it transition quickly between the two possible states? If so are they considered two radicals?




My findings so far:

This answer on SE hints that the exact shape depends on the substituents on the alkyl radical.

Modern Physical Organic Chemistry states that all other (non-methyl) localized radicals are not planar.

Organic Reaction Mechanisms states that the geometry of free radicals is still controversial.
First off, it was very astute of you to recognize that if the radical formed at carbon 3 (the methylene carbon in the starting compound) is pyramidal, then the radical would be chiral ( the lone radical electron serving as the 4th different substituent on that carbon) and 4 different radicals would be possible.

In a typical carbocation, the carbon bearing the positive charge is $\ce{sp^2}$ hybridized, meaning that the p-orbital at the cationic center is empty and the carbocation is planar.  On the other hand, in a typical carbanion, the negatively charged carbon is pyramidal and approaches $\ce{sp^3}$ hybridization. This $\ce{sp^3}$-like orbital contains 2 electrons.  Therefore, it would be reasonable to expect that a typical hydrocarbon radical, where the orbital holds 1 electron would be somewhere in between these two extremes and be slightly pyramidal.  



This is indeed the case and the degree of non-planarity is very sensitive to the substituents attached to the radical center.  In the case of hydrocarbon radicals, such as those created from your starting material, the barrier to inversion from one pyramidal form to the other is extremely low, on the order of a few kcal/mole or so.  Therefore, interconversion between the two enantiomeric forms of the radical would be extremely rapid at room temperature.  

So yes, the $\ce{CH_3\dot{C}HC(CH3)3}$ radical is slightly pyramidal and it does transition rapidly between the two pyramidal enantiomers.  Your answer of 4 radicals is correct, as long as you understand that 2 of them rapidly interconvert at room temperature.