How can you order a free group? A left order on a (discrete) group $G$ is a total order on $G$ satisfying $\forall g,h,k \in G: g < h \implies kg < kh$. A right order is defined symmetrically, and a biorder is an order that is both at once. An order is scattered if it has no densely ordered subset of cardinality at least two, where a set $S$ is densely ordered if $a, c \in S \wedge a \neq c \implies \exists b \in S: a < b < c$.
I'm interested in order types of left orders (and biorders, why not). It's easy to show that left orders on the integers $\mathbb{Z}$ all have order type $\mathbb{Z}$, and on $\mathbb{Z}^2$ there are two constructions of left orders, and as far as I can tell one gives order type $\mathbb{Z}^2$ in lex order (so scattered), and the other orders are dense (so not scattered).
Let $F_2$ be the free nonabelian group on two generators.

Does $F_2$ admit a scattered left order?
Does $F_2$ admit a scattered biorder?

I had an awesome application for this, but I broke it already. Now I'm just curious. I don't actually know what you get from the Magnus embedding $a \mapsto 1+a$, I admit I was too lazy (or scared?) to even give it real thought, and I did not notice a statement in the literature.
More generally, one may ask:

What are the order-types of left orders on $F_2$?
What are the order-types of biorders on $F_2$?

More generally, I'm interested in information on order types of orders on any torsion-free groups, there's plenty of literature on orders but I haven't seen much about order types.
 A: Let us first clarify the relationship between scattered, discrete, and dense orders. The last two notions are standard in the theory of ordered groups.
An order (left or two-sided) is discrete if there exists a smallest (necessarily unique) positive element. If such an element does not exist, the order is called dense.
We claim that no dense order (left or two-sided) is scattered. Moreover, we claim that, if the order is dense, then the entire group is a densely ordered set. Indeed, if $a < b$ then $e < a^{-1}b$ and, since the order is dense, there exists $c$ such that $e < c < a^{-1}b$ and we obtain $a < ac < b$.
On the other hand, discrete orders are not necessarily scattered. For instance the lexicographic bi-order on $\mathbb{Z} \times F_2$, with any bi-order on $F_2$, is discrete with smallest element $(1,e)$, but there are densely ordered subsets, namely the copy of $F_2$.

Going back to the posed question, since free groups of rank > 1 do not admit discrete bi-orders (in fact, no centerless group admits a discrete bi-order) they do not admit scattered bi-orders.
On the other hand, there are discrete left-orders on free groups of rank > 1, and, therefore, the question is more interesting there.
The claim that centerless groups do not admit discrete bi-orders follows from Theorem 2.1 in the paper cited below, while the claim that free groups admit discrete orders (that are, moreover Conradian) is Corollary 3.6 in the same paper.
Linnell, Peter A.; Rhemtulla, Akbar; Rolfsen, Dale P. O., Discretely ordered groups, Algebra Number Theory 3, No. 7, 797-807 (2009). ZBL1229.06008.
