1. Find oxidation state of Fe in $[Fe(CO)_4(C_2O_4)]^+$

   • $CO$ is a neutral ligand.
   • Oxalate, $C_2O_4^{2-}$, has charge $-2$.
   • Let oxidation state of Fe be $x$.

   $x + 4(0) + (-2) = +1$ $x = +3$

   So, iron is $Fe^{3+}$.

2. Find d-electron count

   Iron has atomic number $26$.

   Neutral Fe: $[Ar]3d^64s^2$

   For $Fe^{3+}$, remove 2 electrons from $4s$ and 1 from $3d$:

   $Fe^{3+} : 3d^5$

3. Determine the nature of ligands and pairing

   • $CO$ is a very strong field ligand.
   • Oxalate is weaker than $CO$, but the presence of four $CO$ ligands makes the complex strongly field-stabilized.
   • Coordination number is $6$ here: four $CO$ donors + bidentate oxalate.

   Thus, the complex is expected to be low spin octahedral.

4. Electronic configuration in octahedral low-spin field

   For $d^5$ low spin:

   $t_{2g}^5 e_g^0$

   This gives one unpaired electron.

5. Calculate spin-only magnetic moment

   Formula:

   $\mu = \sqrt{n(n+2)}\ \text{BM}$

   where $n =$ number of unpaired electrons.

   Here, $n=1$:

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

6. Check options

   • A: $5.92$ BM $\rightarrow$ corresponds to 5 unpaired electrons, incorrect
   • B: $0$ BM $\rightarrow$ no unpaired electrons, incorrect
   • C: $1$ BM $\rightarrow$ not the spin-only value
   • D: $1.73$ BM $\rightarrow$ correct

Final Answer: $\boxed{1.73\ \text{BM}}$ (Option D)