\((1)\) \(C r^{3+} \rightarrow 3 d^1 4 s^2\) no. of unpaired \(e^{-}=1\)
\((2)\) \(M n^{3+} \rightarrow 3 d^3 4 s^2\) no. of unpaired \(e^{-}=1\)
\((3)\) \(F e^{3+} \rightarrow 3 d^3 4 s^2 \quad\) no. of unpaired \(e^{-}=1\)
\((4)\) \(Co ^{3+} \rightarrow 3 d^6 4 s^2\) no. of unpaired \(e^{-}=0\)
So \(C o\) in \(\left[\operatorname{Co}(C N)_6\right]^{3-}\) is having less number of unpaired electrons
comparatively i.e 0 . So it is a diamagnetic compound means less paramagnetic comparatively.