Which one of the following complexes will have $$\Delta_0 = 0$$ and $$\mu = 5.96$$ B.M.?

The crystal-field stabilization energy in an octahedral field is $$CFSE = (-0.4x + 0.6y)\,\Delta_0$$, where $$x$$ electrons occupy the $$t_{2g}$$ level and $$y$$ electrons occupy the $$e_g$$ level.

For a high-spin $$d^5$$ ion the electronic distribution is $$t_{2g}^3e_g^2$$, giving
$$CFSE = [(-0.4)(3) + (0.6)(2)]\Delta_0 = (-1.2 + 1.2)\Delta_0 = 0$$.
Hence a high-spin $$d^5$$ complex has $$\Delta_0 = 0$$.

The spin-only magnetic moment is $$\mu = \sqrt{n(n+2)} \text{ B.M.}$$, where $$n$$ is the number of unpaired electrons.

For high-spin $$d^5$$, $$n = 5$$, so
$$\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \text{ B.M.}$$ (≈ 5.96 B.M.).

Therefore we must locate the complex containing a high-spin $$d^5$$ metal ion.

Option A: $$[Fe(CN)_6]^{4-}$$  ⟹  $$Fe^{2+}$$ (since $$x-6=-4$$) is $$d^6$$. CN⁻ is a strong-field ligand (low-spin), not $$d^5$$.

Option B: $$[Co(NH_3)_6]^{3+}$$  ⟹  $$Co^{3+}$$ is $$d^6$$. Not $$d^5$$.

Option C: $$[FeF_6]^{4-}$$  ⟹  $$Fe^{2+}$$, $$d^6$$ (weak-field, high-spin). Still $$d^6$$, not $$d^5$$.

Option D: $$[Mn(SCN)_6]^{4-}$$  ⟹  $$Mn^{2+}$$ (since $$x-6=-4$$) is $$d^5$$. SCN⁻ is a weak/medium-field ligand, so the complex is high-spin $$d^5$$, giving $$\Delta_0 = 0$$ and $$\mu \approx 5.92$$ B.M.

Hence the correct complex is $$[Mn(SCN)_6]^{4-}$$.

Correct answer: Option D.