The magnetic moment of the complex anion $$[Cr(NO)(NH_3)(CN)_4]^{2-}$$ is :

Let $$x$$ be the oxidation state of Chromium:

$$x + (0) + (0) + 4(-1) = -2$$ ($$NO$$, $$NH_3$$ are neutral ligands and $$CN$$ has an oxidation state of -1)

$$x = +2$$

Thus, Chromium is in the $$+2$$ oxidation state ($$Cr^{2+}$$).

$$Cr^{2+} = [Ar] 3d^4$$

Cyanide ($$CN^-$$) is a strong field ligand, which causes a large crystal field splitting ($$\Delta_o$$). This forces the electrons to pair up in the lower energy orbitals, resulting in a low spin complex.

For a $$d^4$$ low spin configuration:

The first three electrons fill the $$t_{2g}$$ orbitals singly. The fourth electron pairs up in one of the $$t_{2g}$$ orbitals.

Unpaired electrons ($$n$$): $$2$$

$$\mu_{eff} = \sqrt{n(n + 2)} \text{ BM}$$

$$\mu_{eff} = \sqrt{2(2 + 2)}$$

$$\mu_{eff} \approx 2.828 \text{ BM}$$