Spin only magnetic moment in BM of FeCO$$_4$$C$$_2$$O$$_4^+$$ is:

We begin by finding the oxidation state of iron in the complex $$\text{Fe(CO)}_4\text{(C}_2\text{O}_4)^+.$$

CO is a neutral ligand, so each of the four CO groups contributes zero charge. The oxalato ligand $$\text{C}_2\text{O}_4^{2-}$$ carries a charge of $$-2.$$ Let the oxidation state of iron be $$x.$$ Writing the charge-balance equation, we have

$$x \;+\; 4(0) \;+\; (-2) \;=\; +1.$$

Simplifying,

$$x - 2 = +1 \;\;\Longrightarrow\;\; x = +3.$$

So the metal centre is $$\text{Fe}^{3+}.$$

Next we count the d-electrons. The electronic configuration of neutral iron (atomic number 26) is $$[ \text{Ar} ]\,3d^6\,4s^2.$$ Removing three electrons to obtain $$\text{Fe}^{3+}$$ first removes the two 4s electrons and one 3d electron, leaving

$$\text{Fe}^{3+}: [ \text{Ar} ]\,3d^5.$$

Hence there are five d-electrons (a $$d^5$$ configuration).

Now we decide whether the complex is low-spin or high-spin. The ligand set consists of four CO groups and one oxalato ligand. CO is a very strong-field (strong π-acceptor) ligand, while oxalato is a weaker, purely σ-donor ligand. Because four of the five donor sites are occupied by the strong-field ligand CO, the overall ligand field is strong enough to favour electron pairing. We therefore assume a low-spin configuration for the $$d^5$$ ion.

In an octahedral low-spin $$d^5$$ set-up the five electrons all occupy the lower-energy $$t_{2g}$$ orbitals:

$$t_{2g}^5\,e_g^0.$$

According to Hund’s rule, the first three electrons fill the three $$t_{2g}$$ orbitals singly, and the next two electrons pair up in two of those orbitals. Consequently there is exactly one unpaired electron.

We now calculate the spin-only magnetic moment. The formula for the spin-only magnetic moment (in Bohr magnetons, BM) is

$$\mu_s = \sqrt{n(n+2)} \;\text{BM},$$

where $$n$$ is the number of unpaired electrons.

Here $$n = 1,$$ so

$$\mu_s = \sqrt{1(1 + 2)} = \sqrt{3} \approx 1.73\;\text{BM}.$$

Hence, the correct answer is Option D.