The spin-only magnetic moment value for the complex [Co(CN)$$_6$$]$$^{4-}$$ is ___ BM. (nearest integer value)
[At. no. of Co = 27]

For the complex $$[\text{Co(CN)}_6]^{4-}$$, we need to determine the oxidation state of cobalt. CN⁻ has a charge of −1, so with 6 CN⁻ ligands the total ligand charge is −6. The overall complex charge is −4:

$$x + (-6) = -4 \implies x = +2$$

So cobalt is in the +2 oxidation state (Co²⁺). The electronic configuration of Co is [Ar] 3d⁷ 4s², and Co²⁺ has the configuration [Ar] 3d⁷ (7 d-electrons).

CN⁻ is a very strong-field ligand. In an octahedral complex with strong-field ligands, the crystal field splitting $$\Delta_o$$ is large, causing maximum pairing of electrons (low-spin configuration). For a d⁷ ion in a low-spin octahedral field, the electron configuration is $$t_{2g}^6 \, e_g^1$$, giving 1 unpaired electron.

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

$$\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.732 \text{ BM}$$

Rounding to the nearest integer, $$\mu \approx 2 \text{ BM}$$.

Therefore, the spin-only magnetic moment is $$\boxed{2}$$ BM.