1. Identify the complexes and determine which are inner-orbital / outer-orbital

We are given:

• $FeCl_3 \cdot 3H_2O$
• $K_3[Fe(CN)_6]$
• $[Co(NH_3)_6]Cl_3$

We need the inner-orbital complex among these that absorbs light at shortest wavelength.

Recall:

• Shorter absorbed wavelength $\Rightarrow$ larger crystal field splitting $\Delta$ because $\Delta = \frac{hc}{\lambda}$ so smaller $\lambda$ means larger $\Delta$.
• Inner-orbital complexes are typically formed with strong-field ligands, causing pairing and use of $(n-1)d$ orbitals.

2. Examine each complex

(a) $FeCl_3 \cdot 3H_2O$

This corresponds to a hydrated iron(III) chloride species with weak ligands like $Cl^-$ and $H_2O$.

• $Fe^{3+}$ is $3d^5$.
• With weak-field ligands, it is high spin.
• Hence it is outer-orbital, not inner-orbital.

So this is not the required complex.

(b) $K_3[Fe(CN)_6]$

Complex ion: $[Fe(CN)_6]^{3-}$

• Oxidation state of Fe: $x + 6(-1) = -3 \Rightarrow x = +3$ so $Fe^{3+} = 3d^5$.
• $CN^-$ is a strong-field ligand, so pairing occurs.
• Octahedral strong-field $d^5$ gives $t_{2g}^5 e_g^0$ This is a low-spin inner-orbital complex.

(c) $[Co(NH_3)_6]Cl_3$

Complex ion: $[Co(NH_3)_6]^{3+}$

• Oxidation state of Co is $+3$.
• $Co^{3+} = 3d^6$.
• In octahedral field with $NH_3$ and high charge on $Co^{3+}$, it becomes low spin: $t_{2g}^6 e_g^0$ This is also an inner-orbital complex.

3. Find which inner-orbital complex absorbs shortest wavelength

Among the inner-orbital complexes:

• $[Fe(CN)_6]^{3-}$
• $[Co(NH_3)_6]^{3+}$

We compare crystal field splitting.

Using spectrochemical series: $CN^- > NH_3$ So $CN^-$ produces a larger $\Delta$ than $NH_3$.

Therefore, $[Fe(CN)_6]^{3-}$ has larger splitting and hence absorbs shorter wavelength light.

So the required complex is: $K_3[Fe(CN)_6]$

4. Calculate its spin-only magnetic moment

For $[Fe(CN)_6]^{3-}$:

• $Fe^{3+} : 3d^5$
• Low-spin octahedral configuration: $t_{2g}^5 e_g^0$ This has 1 unpaired electron.

Spin-only magnetic moment: $\mu = \sqrt{n(n+2)}$ where $n =$ number of unpaired electrons.

So, $\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \text{ B.M.}$ Nearest integer: $2$

5. Final answer

The spin-only magnetic moment is $\boxed{2 \text{ B.M.}}$ (nearest integer)

6. Comparison with stored correct answer

Stored correct answer = $2$

Our derived answer also = $2$. So, they agree.