The observed magnetic moment of the complex $$[Mn(NCS)_6]^{x-}$$ is 6.06 BM. The numerical value of x is _____.

We are given that the magnetic moment of $$[Mn(NCS)_6]^{x-}$$ is 6.06 BM.

Using the spin-only magnetic moment formula $$\mu = \sqrt{n(n+2)}$$ BM, where $$n$$ is the number of unpaired electrons:

$$6.06 = \sqrt{n(n+2)}$$, so $$36.72 = n(n+2)$$.

For $$n = 5$$: $$5(5+2) = 35$$, giving $$\mu = \sqrt{35} = 5.92$$ BM. The observed value of 6.06 BM is slightly higher than the spin-only value due to orbital contribution, which is characteristic of high-spin $$d^5$$ systems. So we have $$n = 5$$ unpaired electrons.

Now, $$Mn^{2+}$$ has the electronic configuration $$[Ar] 3d^5$$, which gives 5 unpaired electrons. Since $$NCS^-$$ is a weak field ligand, no pairing occurs in the high-spin octahedral complex, giving configuration $$t_{2g}^3 e_g^2$$ with 5 unpaired electrons. This matches.

Each $$NCS^-$$ ligand has a charge of $$-1$$. With 6 ligands, total ligand charge = $$-6$$. The overall charge on the complex is $$x-$$:

$$+2 + 6(-1) = -x$$, so $$-4 = -x$$, giving $$x = 4$$.

Hence, the answer is $$x = 4$$.