1. Find oxidation state of Fe in both complexes

For $[\mathrm{Fe}(\mathrm{CN})_6]^{3-}:$ Let oxidation state of Fe be $x$. Since each $\mathrm{CN}^-$ has charge $-1$, $x+6(-1)=-3$ $x-6=-3$ $x=+3$

For $[\mathrm{Fe}(\mathrm{H_2O})_6]^{3+}:$ Since $\mathrm{H_2O}$ is neutral, $x=+3$

So, in both complexes Fe is indeed in the $+3$ oxidation state. Hence, Reason R is true.

2. Electronic configuration of $\mathrm{Fe^{3+}}$

Atomic number of Fe = 26

Neutral Fe: $[\mathrm{Ar}]\,3d^6 4s^2$

For $\mathrm{Fe^{3+}}$: remove 2 electrons from $4s$ and 1 from $3d$: $[\mathrm{Ar}]\,3d^5$

Thus both complexes contain $d^5$ metal ion.

3. Nature of ligands and spin state

• $\mathrm{CN}^-$ is a strong field ligand.
• $\mathrm{H_2O}$ is a weak field ligand.

So for octahedral $d^5$ complexes:

• $[\mathrm{Fe}(\mathrm{CN})_6]^{3-}$ is low spin: $t_{2g}^5 e_g^0$ Number of unpaired electrons, $n=1$

• $[\mathrm{Fe}(\mathrm{H_2O})_6]^{3+}$ is high spin: $t_{2g}^3 e_g^2$ Number of unpaired electrons, $n=5$

4. Calculate spin-only magnetic moment

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

For $[\mathrm{Fe}(\mathrm{CN})_6]^{3-}:$ $\mu=\sqrt{1(1+2)}=\sqrt{3}\approx 1.73\,\text{BM}$ which is given as $1.74\,\text{BM}$, so correct.

For $[\mathrm{Fe}(\mathrm{H_2O})_6]^{3+}:$ $\mu=\sqrt{5(5+2)}=\sqrt{35}\approx 5.92\,\text{BM}$ So this is also correct.

Hence, Assertion A is true.

5. Check whether R explains A

Reason R says only that Fe is in $+3$ oxidation state in both complexes. That is true, but this alone does not explain the different magnetic moments.

The actual reason for different magnetic moments is:

• $\mathrm{CN}^-$ is strong field $\Rightarrow$ low spin $\Rightarrow$ 1 unpaired electron
• $\mathrm{H_2O}$ is weak field $\Rightarrow$ high spin $\Rightarrow$ 5 unpaired electrons

So R is not the correct explanation of A.

6. Conclusion

• Assertion A: True
• Reason R: True
• R is not the correct explanation of A

Therefore, the correct option is: $\boxed{\text{D}}$