Step 1: Spin-only formula।
\[ \mu_s = \sqrt{n(n+2)} \text{ BM} \]
Step 2: $ [Co(NH_3)_6]^{3+} $।
$ Co^{3+} $: $ 3d^6 $। $ NH_3 $ strong field ligand, low spin: $ t_{2g}^6 e_g^0 $, $ n = 0 $, $ \mu = 0 $।
Step 3: $ [Ti(H_2O)_6]^{3+} $।
$ Ti^{3+} $: $ 3d^1 $, $ n = 1 $, $ \mu = \sqrt{3} \approx 1.73 $ BM।
Step 4: $ [Mn(CN)_6]^{3-} $।
$ Mn^{3+} $: $ 3d^4 $। $ CN^- $ strong field, low spin: $ t_{2g}^4 e_g^0 $, 3 paired + 1 unpaired: $ n = 2 $, $ \mu = \sqrt{8} \approx 2.83 $ BM।
Step 5: $ [Fe(CN)_6]^{3-} $।
$ Fe^{3+} $: $ 3d^5 $। $ CN^- $ strong field, low spin: $ t_{2g}^5 $, $ n = 1 $, $ \mu = \sqrt{3} $ BM।
Step 6: उत्तर।
\[ \boxed{[Mn(CN)_6]^{3-} \text{ (} n = 2 \text{, } \mu = \sqrt{8} \text{ BM)}} \]