The coordination number of Th in $$K_4[Th(C_2O_4)_4(H_2O)_2]$$ is: ($$C_2O_4^{2-}$$ = oxalato)

We begin by recalling that the coordination number of the central metal ion is defined as the total number of donor atoms of all the ligands that are directly bonded (co-ordinated) to that metal.

The complex given in the question is $$K_4[Th(C_2O_4)_4(H_2O)_2]$$. Inside the square brackets we have the complex ion $$[Th(C_2O_4)_4(H_2O)_2]^{4-}$$ with thorium, $$Th$$, as the central metal.

Now we examine each type of ligand present and count how many donor atoms each contributes.

First, consider the oxalato ligand, $$C_2O_4^{2-}$$. Oxalate contains two oxygen atoms, each possessing a lone pair that can form a coordinate bond with the metal. Hence:

$$C_2O_4^{2-}$$ is a bidentate ligand (denticity $$= 2).$$

There are four such oxalate ligands in the complex. Therefore, the total number of donor atoms supplied by all oxalate ligands is

$$\text{Donor atoms from oxalate} = 4 \times 2 = 8.$$

Next, look at the water molecules, $$H_2O$$. A water molecule donates one lone pair from the oxygen atom; thus it behaves as a monodentate ligand:

$$H_2O$$ is a monodentate ligand (denticity $$= 1).$$

There are two water molecules present, so the donor atoms contributed by water are

$$\text{Donor atoms from water} = 2 \times 1 = 2.$$

We now add the contributions from both kinds of ligands to obtain the total number of donor atoms (and hence the coordination number) around thorium:

$$\begin{aligned} \text{Coordination number of }Th &= 8 \;(\text{from }C_2O_4^{2-}) + 2 \;(\text{from }H_2O) \\ &= 10. \end{aligned}$$

Therefore, thorium is surrounded by ten donor atoms in this complex.

Hence, the correct answer is Option D.