1. Find oxidation state and electronic configuration of chromium in both complexes

In both complexes, chromium is in the $+3$ oxidation state:

• In $[\mathrm{Cr}(\mathrm{CN})_6]^{3-}$: $x+6(-1)=-3 \Rightarrow x=+3$
• In $[\mathrm{Cr}(\mathrm{H_2O})_6]^{3+}$: Since $\mathrm{H_2O}$ is neutral, $x=+3$

So in both cases, the metal ion is $\mathrm{Cr}^{3+}$.

Atomic number of Cr $=24$. Neutral Cr: $[\mathrm{Ar}]\,3d^5 4s^1$ Therefore, $\mathrm{Cr}^{3+}=[\mathrm{Ar}]\,3d^3$

2. Distribute electrons in octahedral crystal field

For an octahedral complex, the $d$-orbitals split into $t_{2g}$ and $e_g$ levels.

Since $\mathrm{Cr}^{3+}$ is $d^3$, the three electrons occupy: $t_{2g}^3 e_g^0$

This happens regardless of whether the ligand is strong field ($\mathrm{CN^-}$) or weak field ($\mathrm{H_2O}$), because for $d^3$ there is no pairing ambiguity in octahedral field.

Thus both complexes have:

• Number of unpaired electrons, $n=3$

3. Calculate spin-only magnetic moment

Spin-only magnetic moment is: $\mu_{\text{eff}}=\sqrt{n(n+2)}\,\text{BM}$

With $n=3$: $\mu_{\text{eff}}=\sqrt{3(3+2)}=\sqrt{15}\,\text{BM}$

So, $\mu_{\text{eff}}\big([\mathrm{Cr}(\mathrm{CN})_6]^{3-}\big)=\sqrt{15}$ $\mu_{\text{eff}}\big([\mathrm{Cr}(\mathrm{H_2O})_6]^{3+}\big)=\sqrt{15}$

4. Take the ratio

$\frac{\mu_{\text{eff}}\left([\mathrm{Cr}(\mathrm{CN})_6]^{3-}\right)}{\mu_{\text{eff}}\left([\mathrm{Cr}(\mathrm{H_2O})_6]^{3+}\right)}=\frac{\sqrt{15}}{\sqrt{15}}=1$

5. Comparison with stored answer

Derived answer = $1$.

This matches the stored correct answer.